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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule .
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what is a steriogenic centre ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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it has an atomic number 6 . hydrogen is 1 . you probably know that already .
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is it possible to simply draw an arrow from 1 to 2 without redrawing it so 4 is at the back ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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it has an atomic number 6 . hydrogen is 1 . you probably know that already .
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how do i know which way to turn the molecule/ which molecules move to make the hydrogen go in back ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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it has an atomic number 6 . hydrogen is 1 . you probably know that already .
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i understand imagining looking at it as if i was behind the hydrogen , but what if the hydrogen was in the plane of the page ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is .
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0 when sal begins rotating the groups to get # 4 in the back , does it matter what axis i choose to rotate around ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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but to decide whether we branch off , it does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl group , which is not as simple as maybe it could be .
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when determining the ranking of substituent chains after there is a tie on the first molecule of those chains , can you only use molecules attached to the first molecule in the chain , or can you use molecules farther down the chain ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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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 .
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so like , if you had a bromine attached to the second carbon in a substituent chain versus an oxygen attached to the first carbon in a different substituent chain , which would be ranked higher ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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we 've named it using the r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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if sal did not draw for us how 1-bromo-3-fluoro-2,3-dimethylbutane 's atoms are oriented in 3d space , and only gave us a bond-line structure for instance , how would we determine the 3d orientation of the functional groups of the chiral center ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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it has an atomic number 6 . hydrogen is 1 . you probably know that already .
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why does the # 1 group stay in place and the rest of the groups rotate ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it has three hydrogens , so definitely not attached to four different groups .
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how do you rank them if the constituents are just methyl , ethyl , propyl , isopropyl , butyl , etc ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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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 .
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how can it be methyl if only two hydrogens are attached to the carbon ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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it has an atomic number 6 . hydrogen is 1 . you probably know that already .
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are we always rotating while keeping the # 1 atom stationary ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system .
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would the first molecule also be include with cis in its name ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules .
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hey assume we have two chiral carbons.. is there any possibility that u know the configuration is r with respect to one c-atom and s with respect to another.. ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule .
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can you do rs on cyclic systems please ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule .
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how do you make the distinction ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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it has an atomic number 6 . hydrogen is 1 . you probably know that already .
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do you always rotate the molecule about the # 1 group ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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it has an atomic number 6 . hydrogen is 1 . you probably know that already .
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how do you know when to make a certain atom go behind or forward like the hydrogen and methyl in this question ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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so hydrogen is definitely the number four group . 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 three 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 .
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how do we know which carbon to start naming from ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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so number three 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 one , two to three . and that 's pretty straightforward .
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why is the turn going counter clockwise ( s ) if the groups should go in the direction of the 1st ( or most ) priority to the fourth-priority ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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we 've named it using the r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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so when dealing with the groups of hydrogen ( # 4 ) , of c-f ( # 2 ) and the methyl ( # 3 ) , should n't the arrow start at # 2 to # 3 to # 4 ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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we 've named it using the r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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at the end ( ) how are you able to tell that # 1 stays the same but # 2 and # 3 switch ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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so we 're done . we 've named it using the r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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what is the difference between r-s and l-d ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page .
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is it possible to convert directly from the one-dimensional version of this molecule to a fischer projection ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group .
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how would the molecule shown be depicted ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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so we 're done . we 've named it using the r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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i had a doubt.. what if double bonds were present in the compound ... how do we name them using r or s nomenclature ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system .
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how to name if the molecule contains a double or a triple bond ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four group is right now . so if we were to redraw that , let 's redraw our chiral carbon .
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or is it really a must for the group to rotate around the molecule with the largest atomic number ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that .
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why ch3 on left side below the plane was not taken during r-s nomenclature ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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it has an atomic number 6 . hydrogen is 1 . you probably know that already .
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in assigning chirality r and s , what do you do if the hydrogen is not in a backward or forward position ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 .
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at 6.13 , does 'highest to highest ' mean that it does n't matter if there is 10 flourines ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system .
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why do you have to put the lowest molecule at the back first ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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and if that 's a tie , then you keep going on and on and on . so on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it .
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also , what would be a higher number in the case of a chiral carbon being bonded to : bromine and ch2br ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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and that 's pretty straightforward . to go from one to two to three , we have to go counterclockwise . or another way to think of it , we 're going to the left , counterclockwise .
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our only criteria for rotating is that # 4 should go below the plane right ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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this carbon is just a methyl group . so this right here does look like a chiral center . it does look like a chiral carbon , and the other ones do n't .
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can we have enantiomer of a molecule that has no chiral center ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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we 've named it using the r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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for instance would the molecule be named ( 2s ) -1-bromo-3-fluoro-2,3-dimethylbutane ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this .
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is n't the more reactive halogen group should get a lower number ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed .
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can the molecule be rotated about any axis ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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if we went from here , we have one , two , three . we can 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 but- , or it will be a butane , because they 're all single bonds here , so it is a butane .
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i was just wondering , why the carbon on the left , connected to the f and two ch3 groups was not chosen as the chiral center , both carbons seem to be connected to four groups ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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this carbon is just a methyl group . so this right here does look like a chiral center . it does look like a chiral carbon , and the other ones do n't .
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is this a multi-chiral center molecule ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule .
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well , as far as i know , atoms can not be rotated around double bonds .so if i 'm trying to use the cahn ingold prelog system , rotating the molecule such that the lowest ranked atom be into the page would n't be possible .how can i proceed in such a case ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can just use our standard nomenclature rules and we 'd be done .
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at 9.10 ... why does n't this atom structure indicates whether it is s / r ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is .
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how do you know which will be the axis around which the other atoms or groups rotate ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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so hydrogen is definitely the number four group . 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 three 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 .
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so the fluoro group on the third carbon should n't it be the first point of difference and not the bromine on the first carbon ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page .
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or would you still count it as one flourine constituent as your highest attachment ( not add the flourines , only count it as one ) which would still be less than bromine ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane .
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at 07 , why is it that at the 3rd carbon it is not 3-dimethyl since you have two methyl groups coming off ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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same thing about this carbon right here . this carbon right here is attached to a fluorine , but then it 's attached to two methyl groups . so it 's the same group , so this is also not a chiral carbon or an asymmetric carbon .
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so even if the carbon next to the chiral center has a f plus 2 ch3 groups attached to it , we still consider the c attached to the br as the higher one ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it has three hydrogens , so definitely not attached to four different groups .
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when naming the molecule , why is the bottom methyl group chosen instead of the leftmost methyl group ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four group is right now . so if we were to redraw that , let 's redraw our chiral carbon .
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2 when you do a rotation , why do n't you also rotate the br group ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four group is right now . so if we were to redraw that , let 's redraw our chiral carbon .
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0 , does the number 1 group have to be the `` stem '' of the umbrella and not rotate ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules .
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okay so anytime there is a chiral center you will have to differentiate the difference between the original and its mirror image using r or s even if there is onlt one chiral center ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed .
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at < < 8.34 > > how did you rotated the molecule ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 .
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what is the criteria to decide which group has to stick into the page , or pop out , or be at the top while drawing a 3d image of the given compound ?
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let 's see if we can name this molecule using the -- sometimes called the r-s system , or the cahn-ingold-prelog system . and the first thing to do is just to see if there are any chiral centers in this molecule . if there are n't , then we do n't even have to use the r-s system . we can 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 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 . 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 . it does look like a chiral carbon , and the other ones do n't . this is just a methyl group . it 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 r-s naming system will apply . but a good starting point will just be naming it using our standard nomenclature rules . and to do that we look for the longest carbon chain here . let 's see , if we start over here , and i do n'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 can 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 but- , 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 does n't matter whether we use this ch3 or this ch3 , they 're the same group . but to decide whether we use this 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 a bromomethyl 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 is closest to something being attached to it , and that bromine is right there . so there 's 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 just a hydrogen . then three we have a fluoro , so on a carbon three , we have a fluoro , and then on carbon three , we also have a methyl group right here , so we also have a 3-methyl . so when we name it , we put in alphabetical order . bromo comes first , so this thing right here is 1-bromo . then alphabetically , fluoro comes next , 1-bromo-3-fluoro . we have two methyls , so it 's going to be 2 comma 3-dimethyl . and remember , the d does n't count in alphabetical order . dimethylbutane , because we have the longest chain is four carbons . dimethylbutane . so that 's just the standard nomenclature rules . we still have n't used the r-s system . now we can do that . now to think about that , 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 cahn-ingold-prelog rules , and we 'll do all that in this example . so let 's look at the different groups attached to it . so when you look at it , this guy has three carbons and a hydrogen . carbon is definitely higher in atomic number on the periodic table . it has an atomic number 6 . hydrogen is 1 . you probably know that already . so hydrogen is definitely going to be number four . so let me put number four there next to the hydrogen . and let me find a nice color . i 'll do it in white . so hydrogen is definitely the number four group . 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 three 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 on this carbon right here , we have a bromine . bromine has an atomic number of 35 , which is higher than carbon . so this guy has a bromine attached to it . this guy only has hydrogen 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 three . the one has the bromine attached to it is going to be number one , and the one that has the fluorine attached to it is number two . and just a reminder , we were tied with the carbon , so we have to look at the next highest constituent , and even if this had three 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 . and i 'm just doing this for visualization purposes . and right here we have our number one group . i 'll literally just call that our number one group . so right there that is our number one group . it 's in the plane of the screen . so i 'll just call that our number one group . over here , also in the plane of the screen , i have our number two group . so let me do it like that . so then you have your number two group , just like that . and then you have your number three group behind the molecule right now the way it 's drawn . i 'll do that in magenta . so then you have your number three group . it 's behind the molecule , so i 'll draw it like this . this is our number three group . and then we have our number four group , which is the hydrogen pointing out right now . and i 'll just do that in a yellow . we have our number four group pointing out in front right now . so that is number four , just like that . 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 three group . the number three group should look like -- so this is our number three group . let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen 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 four group , i 'll just label it number four . it really is just a hydrogen , so i really did n't have to simplify it much there . now by the r-s system , or by the cahn-ingold-prelog system , we want our number four group to be the one furthest back . so we really want it where the number three position is . and so the easiest way i can think of doing that is you can imagine this is a tripod that 's leaning upside down . or another way to view it is you can view it as an umbrella , where this is the handle of the umbrella and that 's the top of the umbrella that would block the rain , i guess . but the easiest way to get the number four group that 's actually a hydrogen in the number three position would be to rotate it . you could imagine , rotate it around the axis defined by the number one group . so the number one group is just going to stay where it is . the number four is going to rotate to the number three group . number three is going to rotate around to the number two group , and then the number two group is going to rotate to where the number four 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 the little asterisk there to say that that 's our chiral carbon . the number four group is now behind . i 'll do it with the circles . it makes it look a little bit more like atoms . so the number four group is now behind where the number three group used to be , so number four is now there . number one has n't changed . that 's kind of the axis that we rotated around . so the number one group has not changed . number one is still there . number two is now where number four used to be , so number two is now jutting out of the page . and then we have number three is now where number two was . so number three 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 one , two to three . and that 's pretty straightforward . to go from one to two to three , 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 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 r-s system . this molecule is ( s ) -- sinister -- 1-bromo-3-fluoro-2,3-di --
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let me draw it a little bit more like this . the number three group is behind us . and then finally , you have your number four group in yellow , which is just a hydrogen that 's coming straight out .
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when it comes to rotating the molecule to position the smallest group behind the chiral centre , do you always have to hold the largest group in the same position when the others rotate ?
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now that you 've had a little bit of exposure to what a convolution is , i can introduce you to the convolution theorem , or at least in the context of -- there may be other convolution theorems -- but we 're talking about differential equations and laplace transforms . so this is the convolution theorem as applies to laplace transforms . and it tells us that if i have a function f of t -- and i can define its laplace transform as , let 's see , the laplace transform of f of t is capital f of s. we 've done that before . and if i have another function , g of t , and i take its laplace transform , that of course is capital g of s. then if we were to convolute these two functions , so if i were to take f and i were to convolute it with g , which is going to be another function of t -- and we already saw this . we saw that in the last video . i convoluted sine and cosine . so this is going to be a function of t. that the laplace transform of this thing , and this the crux of the theorem , the laplace transform of the convolution of these two functions is equal to the products of their laplace transforms . it equals f of s , big capital f of s , times big capital g of s. now , this might seem very abstract and very , you know , hard to kind of handle for you right now . so let 's do an actual example . and actually , even better , let 's do an inverse laplace transform with an example . and actually , let me write one more thing . if this is true , then we could also do it the other way . we could also say that f -- and i 'll just do it all in yellow ; it takes me too much time to keep switching colors -- that the convolution of f and g , this is just a function of t , i can just say it 's the inverse laplace transform . it 's just the inverse laplace transform of f of s times g of s. although i could n't resist it . let me switch colors . there you go . now , what good does all of this do ? well , we can take inverse laplace transforms . let 's just say that i had -- let me write it down here -- let 's say i told you that the following expression or function , let 's say h of s -- let me write it this way -- h of s is equal to 2s over s squared plus 1 . now , we did this long differential equations at the end , we end up with this thing and we have to take the inverse laplace transform of it . so we want to figure out the inverse laplace transform of h of s , or the inverse laplace transform of this thing right there . so we want to figure out the inverse laplace transform of this expression right here , 2s over s squared plus 1 squared . i do n't want to lose that . right there . now , can we write this as the product of two laplace transforms that we do know ? let 's try to do it . so we can rewrite this . and so this is the inverse laplace transform . so let me rewrite this expression down here . so i can rewrite 2s over s squared plus 1 squared . this is the same thing as -- let me write it this way -- 2 times 1 over s squared plus 1 , times s over s squared plus 1 . i just kind of broke it up . if you multiply the numerators here , you get 2 times 1 , times s , or 2s . if you multiply the denominators here , s squared plus 1 , times s squared plus 1 , well , that 's just s squared plus 1 squared . so this is the same thing . so if we want to take the inverse laplace transform of this , it 's the same thing as taking the inverse laplace transform of this right here . now , something should hopefully start popping out at you . if these were separate transforms , if they were on their own , we know what this is . if we call this f of s , if we said this is the laplace transform of some function , we know what that function is . this is this piece right here . i 'm just doing a little dotted line around it . this is the laplace transform of sine of t. and then if we draw a little box around this one right here , this is the laplace transform of cosine of t , g of s. so this is the laplace transform of sine of t , or we could write that this implies that f of t is equal to sine of t. you should recognize that one by now . and this implies that g of t , if we define this as the laplace transform of g , this means that g of t is equal to cosine of t. and , of course , when you take the inverse laplace transforms , you could take the 2 's out . so now what can we say ? we can now say that the -- let me write it this way -- the inverse -- so actually , let me write it this way . or , actually , a better thing to do , instead of taking the 2 out , so i can leave it nice and clean , we could , if we were to draw a box around this whole thing , and define this whole thing as f of s , then f of s is the laplace transform of 2 sine of t. i just wanted to include that 2 . i did n't want to leave that out and confuse the issue . i wanted a very pure f of s times g of s. so this expression right here is the product of the laplace transform of 2 sine of t , and the laplace transform of cosine of t. now , our convolution theorem told us this right here . that if we want to take the inverse laplace transform of the laplace transforms of two functions -- i know that sounds very confusing -- but you just kind of pattern match . you say , ok look , this thing that i had here , i could rewrite it as a product of two laplace transforms i can recognize . this right here is the laplace transform of 2 sine of t. this is the laplace transform of cosine of t. and we just wrote that as g of s , and f of s. so if i have an expression written like this , i can take the inverse laplace transform and it 'll be equal to the convolution of the original functions . it 'll be equal to the convolution of the inverse of g or the inverse of f. let me write it this way . i could write it like this . we know that f of t is equal to the inverse laplace transform of f of s. and we know that g -- i should have done it in a different color , but i 'll do g in green -- we know that g of t is equal to the inverse laplace transform of g of s. so we can rewrite the convolution theorem as the inverse -- and this might maybe confuse you more than help , but i 'll give my best shot . the inverse laplace transform of -- and i 'll try to stay true to the colors -- of f of s times g of s is equal to -- i 'm just restating this convolution theorem right here . this is equal to the convolution of the inverse laplace transform of f of s. so it 's equal to the convolution of the inverse laplace transform of f of s with the inverse laplace transform of g of s. with the inverse laplace transform of capital g , of g of s. i 'm not sure if that helps you or not , but if you go back to this example it might . this is f of s , this is f of s right here . 2 times -- i 'll do it in the light blue -- this is 2 over s squared plus 1 . that 's f of s in our example . and the g of s was s over s squared plus 1 . and all i got that from is i just broke this up into two things that i recognize . if i multiply this together , i get back to my original thing that i was trying to take the inverse laplace transform of . and so the convolution theorem just says that , ok , well , the inverse laplace transform of this is equal to the inverse laplace transform of 2 over s squared plus 1 , convoluted with the inverse laplace transform of our g of s , of s over s squared plus 1 . and we know what these things are . i already told them to you , but they should be somewhat second nature now . this is 2 times sine of t. you take the laplace transform of sine of t , you get 1 over s squared plus 1 , and then you multiply it by 2 , you get the 2 up there . and you 're going to have to convolute that with the inverse laplace transform of this thing here . and we already went over this . this is cosine of t. so our result so far -- let me be very clear . it 's always good to take a step back and just think about what we 're doing , much less why we 're doing it . but let 's see , the inverse laplace transform of this thing up in this top left corner , 2s over s squared plus 1 squared , which before we did what we 're doing now was very hard to figure out -- actually , this would be a curly bracket right here , but you get the idea -- is equal to this . it 's equal to 2 sine of t , convoluted with cosine of t. and you 're like , sal , throughout this whole process i 've already forgotten what it means to convolute two functions , so let 's convolute them . and i 'll just write the definition , or the definition we 're using of the convolution . that f convoluted with g -- it 's going to be a function of g. i 'll just write this short-hand -- is equal to the integral from 0 to t , of f of t minus tau , times g of tau , dtau . so 2 sine of t convoluted with cosine of t is equal to -- let me do a neutral color -- the integral from 0 to t , of 2 sine of t , minus tau , times the cosine of tau , dtau . now if you watched the very last video i made , i actually solved this , or i solved a very similar thing to this . if we take the 2 out we get 2 , times the integral from 0 to t , of sine of t minus tau , times the cosine of tau . i actually solved this in the previous video . this right here , this is the convolution of sine of t and cosine of t. it 's sine of t convoluted with cosine of t. and i show you in the previous video , just watch that video , where i introduce a convolution , that this thing right here is equal to 1/2t sine of t. now , if this thing is equal to 1/2 t sine of t , and i have to multiply it by 2 , then we get , our big result , that the inverse laplace transform of 2s over s squared plus 1 squared is equal to the convolution of 2 sine of t with cosine of t. which is just 2 times this thing here , which is 2 times 1/2 -- those cancel out -- so it equals t sine of t. and once you get the hang of it , you wo n't have to go through all of these steps . but the key is to recognize that this could be broken down as the products of two laplace transforms that you recognize . this could be broken down as the product of two laplace transforms we recognized . this is the laplace transform of 2 sine of t. this was the laplace transform of cosine of t. so the inverse laplace transform of our original thing , or original expression , is just the convolution of that with that . and if you watched the previous video , you 'd realize that actually calculating that convolution was no simple task , but it can be done . so you actually can get an integral form . even if it ca n't be done , you can get your answer , at least , in terms of some integral . so i have n't proven the convolution theorem to you just yet . i 'll do that in a future video . but hopefully , this gave you a little bit of a sense of how you can use it to actually take inverse laplace transforms . and remember , the reason why we 're learning to take inverse laplace transforms , and we have all of these tools to do it , is because that 's always that last step when you 're solving these differential equations , using your laplace transforms .
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so we can rewrite this . and so this is the inverse laplace transform . so let me rewrite this expression down here .
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why do the convolution to get the inverse laplace transform ?
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now that you 've had a little bit of exposure to what a convolution is , i can introduce you to the convolution theorem , or at least in the context of -- there may be other convolution theorems -- but we 're talking about differential equations and laplace transforms . so this is the convolution theorem as applies to laplace transforms . and it tells us that if i have a function f of t -- and i can define its laplace transform as , let 's see , the laplace transform of f of t is capital f of s. we 've done that before . and if i have another function , g of t , and i take its laplace transform , that of course is capital g of s. then if we were to convolute these two functions , so if i were to take f and i were to convolute it with g , which is going to be another function of t -- and we already saw this . we saw that in the last video . i convoluted sine and cosine . so this is going to be a function of t. that the laplace transform of this thing , and this the crux of the theorem , the laplace transform of the convolution of these two functions is equal to the products of their laplace transforms . it equals f of s , big capital f of s , times big capital g of s. now , this might seem very abstract and very , you know , hard to kind of handle for you right now . so let 's do an actual example . and actually , even better , let 's do an inverse laplace transform with an example . and actually , let me write one more thing . if this is true , then we could also do it the other way . we could also say that f -- and i 'll just do it all in yellow ; it takes me too much time to keep switching colors -- that the convolution of f and g , this is just a function of t , i can just say it 's the inverse laplace transform . it 's just the inverse laplace transform of f of s times g of s. although i could n't resist it . let me switch colors . there you go . now , what good does all of this do ? well , we can take inverse laplace transforms . let 's just say that i had -- let me write it down here -- let 's say i told you that the following expression or function , let 's say h of s -- let me write it this way -- h of s is equal to 2s over s squared plus 1 . now , we did this long differential equations at the end , we end up with this thing and we have to take the inverse laplace transform of it . so we want to figure out the inverse laplace transform of h of s , or the inverse laplace transform of this thing right there . so we want to figure out the inverse laplace transform of this expression right here , 2s over s squared plus 1 squared . i do n't want to lose that . right there . now , can we write this as the product of two laplace transforms that we do know ? let 's try to do it . so we can rewrite this . and so this is the inverse laplace transform . so let me rewrite this expression down here . so i can rewrite 2s over s squared plus 1 squared . this is the same thing as -- let me write it this way -- 2 times 1 over s squared plus 1 , times s over s squared plus 1 . i just kind of broke it up . if you multiply the numerators here , you get 2 times 1 , times s , or 2s . if you multiply the denominators here , s squared plus 1 , times s squared plus 1 , well , that 's just s squared plus 1 squared . so this is the same thing . so if we want to take the inverse laplace transform of this , it 's the same thing as taking the inverse laplace transform of this right here . now , something should hopefully start popping out at you . if these were separate transforms , if they were on their own , we know what this is . if we call this f of s , if we said this is the laplace transform of some function , we know what that function is . this is this piece right here . i 'm just doing a little dotted line around it . this is the laplace transform of sine of t. and then if we draw a little box around this one right here , this is the laplace transform of cosine of t , g of s. so this is the laplace transform of sine of t , or we could write that this implies that f of t is equal to sine of t. you should recognize that one by now . and this implies that g of t , if we define this as the laplace transform of g , this means that g of t is equal to cosine of t. and , of course , when you take the inverse laplace transforms , you could take the 2 's out . so now what can we say ? we can now say that the -- let me write it this way -- the inverse -- so actually , let me write it this way . or , actually , a better thing to do , instead of taking the 2 out , so i can leave it nice and clean , we could , if we were to draw a box around this whole thing , and define this whole thing as f of s , then f of s is the laplace transform of 2 sine of t. i just wanted to include that 2 . i did n't want to leave that out and confuse the issue . i wanted a very pure f of s times g of s. so this expression right here is the product of the laplace transform of 2 sine of t , and the laplace transform of cosine of t. now , our convolution theorem told us this right here . that if we want to take the inverse laplace transform of the laplace transforms of two functions -- i know that sounds very confusing -- but you just kind of pattern match . you say , ok look , this thing that i had here , i could rewrite it as a product of two laplace transforms i can recognize . this right here is the laplace transform of 2 sine of t. this is the laplace transform of cosine of t. and we just wrote that as g of s , and f of s. so if i have an expression written like this , i can take the inverse laplace transform and it 'll be equal to the convolution of the original functions . it 'll be equal to the convolution of the inverse of g or the inverse of f. let me write it this way . i could write it like this . we know that f of t is equal to the inverse laplace transform of f of s. and we know that g -- i should have done it in a different color , but i 'll do g in green -- we know that g of t is equal to the inverse laplace transform of g of s. so we can rewrite the convolution theorem as the inverse -- and this might maybe confuse you more than help , but i 'll give my best shot . the inverse laplace transform of -- and i 'll try to stay true to the colors -- of f of s times g of s is equal to -- i 'm just restating this convolution theorem right here . this is equal to the convolution of the inverse laplace transform of f of s. so it 's equal to the convolution of the inverse laplace transform of f of s with the inverse laplace transform of g of s. with the inverse laplace transform of capital g , of g of s. i 'm not sure if that helps you or not , but if you go back to this example it might . this is f of s , this is f of s right here . 2 times -- i 'll do it in the light blue -- this is 2 over s squared plus 1 . that 's f of s in our example . and the g of s was s over s squared plus 1 . and all i got that from is i just broke this up into two things that i recognize . if i multiply this together , i get back to my original thing that i was trying to take the inverse laplace transform of . and so the convolution theorem just says that , ok , well , the inverse laplace transform of this is equal to the inverse laplace transform of 2 over s squared plus 1 , convoluted with the inverse laplace transform of our g of s , of s over s squared plus 1 . and we know what these things are . i already told them to you , but they should be somewhat second nature now . this is 2 times sine of t. you take the laplace transform of sine of t , you get 1 over s squared plus 1 , and then you multiply it by 2 , you get the 2 up there . and you 're going to have to convolute that with the inverse laplace transform of this thing here . and we already went over this . this is cosine of t. so our result so far -- let me be very clear . it 's always good to take a step back and just think about what we 're doing , much less why we 're doing it . but let 's see , the inverse laplace transform of this thing up in this top left corner , 2s over s squared plus 1 squared , which before we did what we 're doing now was very hard to figure out -- actually , this would be a curly bracket right here , but you get the idea -- is equal to this . it 's equal to 2 sine of t , convoluted with cosine of t. and you 're like , sal , throughout this whole process i 've already forgotten what it means to convolute two functions , so let 's convolute them . and i 'll just write the definition , or the definition we 're using of the convolution . that f convoluted with g -- it 's going to be a function of g. i 'll just write this short-hand -- is equal to the integral from 0 to t , of f of t minus tau , times g of tau , dtau . so 2 sine of t convoluted with cosine of t is equal to -- let me do a neutral color -- the integral from 0 to t , of 2 sine of t , minus tau , times the cosine of tau , dtau . now if you watched the very last video i made , i actually solved this , or i solved a very similar thing to this . if we take the 2 out we get 2 , times the integral from 0 to t , of sine of t minus tau , times the cosine of tau . i actually solved this in the previous video . this right here , this is the convolution of sine of t and cosine of t. it 's sine of t convoluted with cosine of t. and i show you in the previous video , just watch that video , where i introduce a convolution , that this thing right here is equal to 1/2t sine of t. now , if this thing is equal to 1/2 t sine of t , and i have to multiply it by 2 , then we get , our big result , that the inverse laplace transform of 2s over s squared plus 1 squared is equal to the convolution of 2 sine of t with cosine of t. which is just 2 times this thing here , which is 2 times 1/2 -- those cancel out -- so it equals t sine of t. and once you get the hang of it , you wo n't have to go through all of these steps . but the key is to recognize that this could be broken down as the products of two laplace transforms that you recognize . this could be broken down as the product of two laplace transforms we recognized . this is the laplace transform of 2 sine of t. this was the laplace transform of cosine of t. so the inverse laplace transform of our original thing , or original expression , is just the convolution of that with that . and if you watched the previous video , you 'd realize that actually calculating that convolution was no simple task , but it can be done . so you actually can get an integral form . even if it ca n't be done , you can get your answer , at least , in terms of some integral . so i have n't proven the convolution theorem to you just yet . i 'll do that in a future video . but hopefully , this gave you a little bit of a sense of how you can use it to actually take inverse laplace transforms . and remember , the reason why we 're learning to take inverse laplace transforms , and we have all of these tools to do it , is because that 's always that last step when you 're solving these differential equations , using your laplace transforms .
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so we can rewrite this . and so this is the inverse laplace transform . so let me rewrite this expression down here .
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why not apply laplace transform to do the the convolution ?
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now that you 've had a little bit of exposure to what a convolution is , i can introduce you to the convolution theorem , or at least in the context of -- there may be other convolution theorems -- but we 're talking about differential equations and laplace transforms . so this is the convolution theorem as applies to laplace transforms . and it tells us that if i have a function f of t -- and i can define its laplace transform as , let 's see , the laplace transform of f of t is capital f of s. we 've done that before . and if i have another function , g of t , and i take its laplace transform , that of course is capital g of s. then if we were to convolute these two functions , so if i were to take f and i were to convolute it with g , which is going to be another function of t -- and we already saw this . we saw that in the last video . i convoluted sine and cosine . so this is going to be a function of t. that the laplace transform of this thing , and this the crux of the theorem , the laplace transform of the convolution of these two functions is equal to the products of their laplace transforms . it equals f of s , big capital f of s , times big capital g of s. now , this might seem very abstract and very , you know , hard to kind of handle for you right now . so let 's do an actual example . and actually , even better , let 's do an inverse laplace transform with an example . and actually , let me write one more thing . if this is true , then we could also do it the other way . we could also say that f -- and i 'll just do it all in yellow ; it takes me too much time to keep switching colors -- that the convolution of f and g , this is just a function of t , i can just say it 's the inverse laplace transform . it 's just the inverse laplace transform of f of s times g of s. although i could n't resist it . let me switch colors . there you go . now , what good does all of this do ? well , we can take inverse laplace transforms . let 's just say that i had -- let me write it down here -- let 's say i told you that the following expression or function , let 's say h of s -- let me write it this way -- h of s is equal to 2s over s squared plus 1 . now , we did this long differential equations at the end , we end up with this thing and we have to take the inverse laplace transform of it . so we want to figure out the inverse laplace transform of h of s , or the inverse laplace transform of this thing right there . so we want to figure out the inverse laplace transform of this expression right here , 2s over s squared plus 1 squared . i do n't want to lose that . right there . now , can we write this as the product of two laplace transforms that we do know ? let 's try to do it . so we can rewrite this . and so this is the inverse laplace transform . so let me rewrite this expression down here . so i can rewrite 2s over s squared plus 1 squared . this is the same thing as -- let me write it this way -- 2 times 1 over s squared plus 1 , times s over s squared plus 1 . i just kind of broke it up . if you multiply the numerators here , you get 2 times 1 , times s , or 2s . if you multiply the denominators here , s squared plus 1 , times s squared plus 1 , well , that 's just s squared plus 1 squared . so this is the same thing . so if we want to take the inverse laplace transform of this , it 's the same thing as taking the inverse laplace transform of this right here . now , something should hopefully start popping out at you . if these were separate transforms , if they were on their own , we know what this is . if we call this f of s , if we said this is the laplace transform of some function , we know what that function is . this is this piece right here . i 'm just doing a little dotted line around it . this is the laplace transform of sine of t. and then if we draw a little box around this one right here , this is the laplace transform of cosine of t , g of s. so this is the laplace transform of sine of t , or we could write that this implies that f of t is equal to sine of t. you should recognize that one by now . and this implies that g of t , if we define this as the laplace transform of g , this means that g of t is equal to cosine of t. and , of course , when you take the inverse laplace transforms , you could take the 2 's out . so now what can we say ? we can now say that the -- let me write it this way -- the inverse -- so actually , let me write it this way . or , actually , a better thing to do , instead of taking the 2 out , so i can leave it nice and clean , we could , if we were to draw a box around this whole thing , and define this whole thing as f of s , then f of s is the laplace transform of 2 sine of t. i just wanted to include that 2 . i did n't want to leave that out and confuse the issue . i wanted a very pure f of s times g of s. so this expression right here is the product of the laplace transform of 2 sine of t , and the laplace transform of cosine of t. now , our convolution theorem told us this right here . that if we want to take the inverse laplace transform of the laplace transforms of two functions -- i know that sounds very confusing -- but you just kind of pattern match . you say , ok look , this thing that i had here , i could rewrite it as a product of two laplace transforms i can recognize . this right here is the laplace transform of 2 sine of t. this is the laplace transform of cosine of t. and we just wrote that as g of s , and f of s. so if i have an expression written like this , i can take the inverse laplace transform and it 'll be equal to the convolution of the original functions . it 'll be equal to the convolution of the inverse of g or the inverse of f. let me write it this way . i could write it like this . we know that f of t is equal to the inverse laplace transform of f of s. and we know that g -- i should have done it in a different color , but i 'll do g in green -- we know that g of t is equal to the inverse laplace transform of g of s. so we can rewrite the convolution theorem as the inverse -- and this might maybe confuse you more than help , but i 'll give my best shot . the inverse laplace transform of -- and i 'll try to stay true to the colors -- of f of s times g of s is equal to -- i 'm just restating this convolution theorem right here . this is equal to the convolution of the inverse laplace transform of f of s. so it 's equal to the convolution of the inverse laplace transform of f of s with the inverse laplace transform of g of s. with the inverse laplace transform of capital g , of g of s. i 'm not sure if that helps you or not , but if you go back to this example it might . this is f of s , this is f of s right here . 2 times -- i 'll do it in the light blue -- this is 2 over s squared plus 1 . that 's f of s in our example . and the g of s was s over s squared plus 1 . and all i got that from is i just broke this up into two things that i recognize . if i multiply this together , i get back to my original thing that i was trying to take the inverse laplace transform of . and so the convolution theorem just says that , ok , well , the inverse laplace transform of this is equal to the inverse laplace transform of 2 over s squared plus 1 , convoluted with the inverse laplace transform of our g of s , of s over s squared plus 1 . and we know what these things are . i already told them to you , but they should be somewhat second nature now . this is 2 times sine of t. you take the laplace transform of sine of t , you get 1 over s squared plus 1 , and then you multiply it by 2 , you get the 2 up there . and you 're going to have to convolute that with the inverse laplace transform of this thing here . and we already went over this . this is cosine of t. so our result so far -- let me be very clear . it 's always good to take a step back and just think about what we 're doing , much less why we 're doing it . but let 's see , the inverse laplace transform of this thing up in this top left corner , 2s over s squared plus 1 squared , which before we did what we 're doing now was very hard to figure out -- actually , this would be a curly bracket right here , but you get the idea -- is equal to this . it 's equal to 2 sine of t , convoluted with cosine of t. and you 're like , sal , throughout this whole process i 've already forgotten what it means to convolute two functions , so let 's convolute them . and i 'll just write the definition , or the definition we 're using of the convolution . that f convoluted with g -- it 's going to be a function of g. i 'll just write this short-hand -- is equal to the integral from 0 to t , of f of t minus tau , times g of tau , dtau . so 2 sine of t convoluted with cosine of t is equal to -- let me do a neutral color -- the integral from 0 to t , of 2 sine of t , minus tau , times the cosine of tau , dtau . now if you watched the very last video i made , i actually solved this , or i solved a very similar thing to this . if we take the 2 out we get 2 , times the integral from 0 to t , of sine of t minus tau , times the cosine of tau . i actually solved this in the previous video . this right here , this is the convolution of sine of t and cosine of t. it 's sine of t convoluted with cosine of t. and i show you in the previous video , just watch that video , where i introduce a convolution , that this thing right here is equal to 1/2t sine of t. now , if this thing is equal to 1/2 t sine of t , and i have to multiply it by 2 , then we get , our big result , that the inverse laplace transform of 2s over s squared plus 1 squared is equal to the convolution of 2 sine of t with cosine of t. which is just 2 times this thing here , which is 2 times 1/2 -- those cancel out -- so it equals t sine of t. and once you get the hang of it , you wo n't have to go through all of these steps . but the key is to recognize that this could be broken down as the products of two laplace transforms that you recognize . this could be broken down as the product of two laplace transforms we recognized . this is the laplace transform of 2 sine of t. this was the laplace transform of cosine of t. so the inverse laplace transform of our original thing , or original expression , is just the convolution of that with that . and if you watched the previous video , you 'd realize that actually calculating that convolution was no simple task , but it can be done . so you actually can get an integral form . even if it ca n't be done , you can get your answer , at least , in terms of some integral . so i have n't proven the convolution theorem to you just yet . i 'll do that in a future video . but hopefully , this gave you a little bit of a sense of how you can use it to actually take inverse laplace transforms . and remember , the reason why we 're learning to take inverse laplace transforms , and we have all of these tools to do it , is because that 's always that last step when you 're solving these differential equations , using your laplace transforms .
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now that you 've had a little bit of exposure to what a convolution is , i can introduce you to the convolution theorem , or at least in the context of -- there may be other convolution theorems -- but we 're talking about differential equations and laplace transforms . so this is the convolution theorem as applies to laplace transforms . and it tells us that if i have a function f of t -- and i can define its laplace transform as , let 's see , the laplace transform of f of t is capital f of s. we 've done that before .
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can we have the proof of convolution theorem ?
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now that you 've had a little bit of exposure to what a convolution is , i can introduce you to the convolution theorem , or at least in the context of -- there may be other convolution theorems -- but we 're talking about differential equations and laplace transforms . so this is the convolution theorem as applies to laplace transforms . and it tells us that if i have a function f of t -- and i can define its laplace transform as , let 's see , the laplace transform of f of t is capital f of s. we 've done that before . and if i have another function , g of t , and i take its laplace transform , that of course is capital g of s. then if we were to convolute these two functions , so if i were to take f and i were to convolute it with g , which is going to be another function of t -- and we already saw this . we saw that in the last video . i convoluted sine and cosine . so this is going to be a function of t. that the laplace transform of this thing , and this the crux of the theorem , the laplace transform of the convolution of these two functions is equal to the products of their laplace transforms . it equals f of s , big capital f of s , times big capital g of s. now , this might seem very abstract and very , you know , hard to kind of handle for you right now . so let 's do an actual example . and actually , even better , let 's do an inverse laplace transform with an example . and actually , let me write one more thing . if this is true , then we could also do it the other way . we could also say that f -- and i 'll just do it all in yellow ; it takes me too much time to keep switching colors -- that the convolution of f and g , this is just a function of t , i can just say it 's the inverse laplace transform . it 's just the inverse laplace transform of f of s times g of s. although i could n't resist it . let me switch colors . there you go . now , what good does all of this do ? well , we can take inverse laplace transforms . let 's just say that i had -- let me write it down here -- let 's say i told you that the following expression or function , let 's say h of s -- let me write it this way -- h of s is equal to 2s over s squared plus 1 . now , we did this long differential equations at the end , we end up with this thing and we have to take the inverse laplace transform of it . so we want to figure out the inverse laplace transform of h of s , or the inverse laplace transform of this thing right there . so we want to figure out the inverse laplace transform of this expression right here , 2s over s squared plus 1 squared . i do n't want to lose that . right there . now , can we write this as the product of two laplace transforms that we do know ? let 's try to do it . so we can rewrite this . and so this is the inverse laplace transform . so let me rewrite this expression down here . so i can rewrite 2s over s squared plus 1 squared . this is the same thing as -- let me write it this way -- 2 times 1 over s squared plus 1 , times s over s squared plus 1 . i just kind of broke it up . if you multiply the numerators here , you get 2 times 1 , times s , or 2s . if you multiply the denominators here , s squared plus 1 , times s squared plus 1 , well , that 's just s squared plus 1 squared . so this is the same thing . so if we want to take the inverse laplace transform of this , it 's the same thing as taking the inverse laplace transform of this right here . now , something should hopefully start popping out at you . if these were separate transforms , if they were on their own , we know what this is . if we call this f of s , if we said this is the laplace transform of some function , we know what that function is . this is this piece right here . i 'm just doing a little dotted line around it . this is the laplace transform of sine of t. and then if we draw a little box around this one right here , this is the laplace transform of cosine of t , g of s. so this is the laplace transform of sine of t , or we could write that this implies that f of t is equal to sine of t. you should recognize that one by now . and this implies that g of t , if we define this as the laplace transform of g , this means that g of t is equal to cosine of t. and , of course , when you take the inverse laplace transforms , you could take the 2 's out . so now what can we say ? we can now say that the -- let me write it this way -- the inverse -- so actually , let me write it this way . or , actually , a better thing to do , instead of taking the 2 out , so i can leave it nice and clean , we could , if we were to draw a box around this whole thing , and define this whole thing as f of s , then f of s is the laplace transform of 2 sine of t. i just wanted to include that 2 . i did n't want to leave that out and confuse the issue . i wanted a very pure f of s times g of s. so this expression right here is the product of the laplace transform of 2 sine of t , and the laplace transform of cosine of t. now , our convolution theorem told us this right here . that if we want to take the inverse laplace transform of the laplace transforms of two functions -- i know that sounds very confusing -- but you just kind of pattern match . you say , ok look , this thing that i had here , i could rewrite it as a product of two laplace transforms i can recognize . this right here is the laplace transform of 2 sine of t. this is the laplace transform of cosine of t. and we just wrote that as g of s , and f of s. so if i have an expression written like this , i can take the inverse laplace transform and it 'll be equal to the convolution of the original functions . it 'll be equal to the convolution of the inverse of g or the inverse of f. let me write it this way . i could write it like this . we know that f of t is equal to the inverse laplace transform of f of s. and we know that g -- i should have done it in a different color , but i 'll do g in green -- we know that g of t is equal to the inverse laplace transform of g of s. so we can rewrite the convolution theorem as the inverse -- and this might maybe confuse you more than help , but i 'll give my best shot . the inverse laplace transform of -- and i 'll try to stay true to the colors -- of f of s times g of s is equal to -- i 'm just restating this convolution theorem right here . this is equal to the convolution of the inverse laplace transform of f of s. so it 's equal to the convolution of the inverse laplace transform of f of s with the inverse laplace transform of g of s. with the inverse laplace transform of capital g , of g of s. i 'm not sure if that helps you or not , but if you go back to this example it might . this is f of s , this is f of s right here . 2 times -- i 'll do it in the light blue -- this is 2 over s squared plus 1 . that 's f of s in our example . and the g of s was s over s squared plus 1 . and all i got that from is i just broke this up into two things that i recognize . if i multiply this together , i get back to my original thing that i was trying to take the inverse laplace transform of . and so the convolution theorem just says that , ok , well , the inverse laplace transform of this is equal to the inverse laplace transform of 2 over s squared plus 1 , convoluted with the inverse laplace transform of our g of s , of s over s squared plus 1 . and we know what these things are . i already told them to you , but they should be somewhat second nature now . this is 2 times sine of t. you take the laplace transform of sine of t , you get 1 over s squared plus 1 , and then you multiply it by 2 , you get the 2 up there . and you 're going to have to convolute that with the inverse laplace transform of this thing here . and we already went over this . this is cosine of t. so our result so far -- let me be very clear . it 's always good to take a step back and just think about what we 're doing , much less why we 're doing it . but let 's see , the inverse laplace transform of this thing up in this top left corner , 2s over s squared plus 1 squared , which before we did what we 're doing now was very hard to figure out -- actually , this would be a curly bracket right here , but you get the idea -- is equal to this . it 's equal to 2 sine of t , convoluted with cosine of t. and you 're like , sal , throughout this whole process i 've already forgotten what it means to convolute two functions , so let 's convolute them . and i 'll just write the definition , or the definition we 're using of the convolution . that f convoluted with g -- it 's going to be a function of g. i 'll just write this short-hand -- is equal to the integral from 0 to t , of f of t minus tau , times g of tau , dtau . so 2 sine of t convoluted with cosine of t is equal to -- let me do a neutral color -- the integral from 0 to t , of 2 sine of t , minus tau , times the cosine of tau , dtau . now if you watched the very last video i made , i actually solved this , or i solved a very similar thing to this . if we take the 2 out we get 2 , times the integral from 0 to t , of sine of t minus tau , times the cosine of tau . i actually solved this in the previous video . this right here , this is the convolution of sine of t and cosine of t. it 's sine of t convoluted with cosine of t. and i show you in the previous video , just watch that video , where i introduce a convolution , that this thing right here is equal to 1/2t sine of t. now , if this thing is equal to 1/2 t sine of t , and i have to multiply it by 2 , then we get , our big result , that the inverse laplace transform of 2s over s squared plus 1 squared is equal to the convolution of 2 sine of t with cosine of t. which is just 2 times this thing here , which is 2 times 1/2 -- those cancel out -- so it equals t sine of t. and once you get the hang of it , you wo n't have to go through all of these steps . but the key is to recognize that this could be broken down as the products of two laplace transforms that you recognize . this could be broken down as the product of two laplace transforms we recognized . this is the laplace transform of 2 sine of t. this was the laplace transform of cosine of t. so the inverse laplace transform of our original thing , or original expression , is just the convolution of that with that . and if you watched the previous video , you 'd realize that actually calculating that convolution was no simple task , but it can be done . so you actually can get an integral form . even if it ca n't be done , you can get your answer , at least , in terms of some integral . so i have n't proven the convolution theorem to you just yet . i 'll do that in a future video . but hopefully , this gave you a little bit of a sense of how you can use it to actually take inverse laplace transforms . and remember , the reason why we 're learning to take inverse laplace transforms , and we have all of these tools to do it , is because that 's always that last step when you 're solving these differential equations , using your laplace transforms .
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i could write it like this . we know that f of t is equal to the inverse laplace transform of f of s. and we know that g -- i should have done it in a different color , but i 'll do g in green -- we know that g of t is equal to the inverse laplace transform of g of s. so we can rewrite the convolution theorem as the inverse -- and this might maybe confuse you more than help , but i 'll give my best shot . the inverse laplace transform of -- and i 'll try to stay true to the colors -- of f of s times g of s is equal to -- i 'm just restating this convolution theorem right here .
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is there some form of `` inverse '' convolution theorem giving l { f ( t ) g ( t ) } as some convolution of f ( s ) and g ( s ) ?
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now that you 've had a little bit of exposure to what a convolution is , i can introduce you to the convolution theorem , or at least in the context of -- there may be other convolution theorems -- but we 're talking about differential equations and laplace transforms . so this is the convolution theorem as applies to laplace transforms . and it tells us that if i have a function f of t -- and i can define its laplace transform as , let 's see , the laplace transform of f of t is capital f of s. we 've done that before . and if i have another function , g of t , and i take its laplace transform , that of course is capital g of s. then if we were to convolute these two functions , so if i were to take f and i were to convolute it with g , which is going to be another function of t -- and we already saw this . we saw that in the last video . i convoluted sine and cosine . so this is going to be a function of t. that the laplace transform of this thing , and this the crux of the theorem , the laplace transform of the convolution of these two functions is equal to the products of their laplace transforms . it equals f of s , big capital f of s , times big capital g of s. now , this might seem very abstract and very , you know , hard to kind of handle for you right now . so let 's do an actual example . and actually , even better , let 's do an inverse laplace transform with an example . and actually , let me write one more thing . if this is true , then we could also do it the other way . we could also say that f -- and i 'll just do it all in yellow ; it takes me too much time to keep switching colors -- that the convolution of f and g , this is just a function of t , i can just say it 's the inverse laplace transform . it 's just the inverse laplace transform of f of s times g of s. although i could n't resist it . let me switch colors . there you go . now , what good does all of this do ? well , we can take inverse laplace transforms . let 's just say that i had -- let me write it down here -- let 's say i told you that the following expression or function , let 's say h of s -- let me write it this way -- h of s is equal to 2s over s squared plus 1 . now , we did this long differential equations at the end , we end up with this thing and we have to take the inverse laplace transform of it . so we want to figure out the inverse laplace transform of h of s , or the inverse laplace transform of this thing right there . so we want to figure out the inverse laplace transform of this expression right here , 2s over s squared plus 1 squared . i do n't want to lose that . right there . now , can we write this as the product of two laplace transforms that we do know ? let 's try to do it . so we can rewrite this . and so this is the inverse laplace transform . so let me rewrite this expression down here . so i can rewrite 2s over s squared plus 1 squared . this is the same thing as -- let me write it this way -- 2 times 1 over s squared plus 1 , times s over s squared plus 1 . i just kind of broke it up . if you multiply the numerators here , you get 2 times 1 , times s , or 2s . if you multiply the denominators here , s squared plus 1 , times s squared plus 1 , well , that 's just s squared plus 1 squared . so this is the same thing . so if we want to take the inverse laplace transform of this , it 's the same thing as taking the inverse laplace transform of this right here . now , something should hopefully start popping out at you . if these were separate transforms , if they were on their own , we know what this is . if we call this f of s , if we said this is the laplace transform of some function , we know what that function is . this is this piece right here . i 'm just doing a little dotted line around it . this is the laplace transform of sine of t. and then if we draw a little box around this one right here , this is the laplace transform of cosine of t , g of s. so this is the laplace transform of sine of t , or we could write that this implies that f of t is equal to sine of t. you should recognize that one by now . and this implies that g of t , if we define this as the laplace transform of g , this means that g of t is equal to cosine of t. and , of course , when you take the inverse laplace transforms , you could take the 2 's out . so now what can we say ? we can now say that the -- let me write it this way -- the inverse -- so actually , let me write it this way . or , actually , a better thing to do , instead of taking the 2 out , so i can leave it nice and clean , we could , if we were to draw a box around this whole thing , and define this whole thing as f of s , then f of s is the laplace transform of 2 sine of t. i just wanted to include that 2 . i did n't want to leave that out and confuse the issue . i wanted a very pure f of s times g of s. so this expression right here is the product of the laplace transform of 2 sine of t , and the laplace transform of cosine of t. now , our convolution theorem told us this right here . that if we want to take the inverse laplace transform of the laplace transforms of two functions -- i know that sounds very confusing -- but you just kind of pattern match . you say , ok look , this thing that i had here , i could rewrite it as a product of two laplace transforms i can recognize . this right here is the laplace transform of 2 sine of t. this is the laplace transform of cosine of t. and we just wrote that as g of s , and f of s. so if i have an expression written like this , i can take the inverse laplace transform and it 'll be equal to the convolution of the original functions . it 'll be equal to the convolution of the inverse of g or the inverse of f. let me write it this way . i could write it like this . we know that f of t is equal to the inverse laplace transform of f of s. and we know that g -- i should have done it in a different color , but i 'll do g in green -- we know that g of t is equal to the inverse laplace transform of g of s. so we can rewrite the convolution theorem as the inverse -- and this might maybe confuse you more than help , but i 'll give my best shot . the inverse laplace transform of -- and i 'll try to stay true to the colors -- of f of s times g of s is equal to -- i 'm just restating this convolution theorem right here . this is equal to the convolution of the inverse laplace transform of f of s. so it 's equal to the convolution of the inverse laplace transform of f of s with the inverse laplace transform of g of s. with the inverse laplace transform of capital g , of g of s. i 'm not sure if that helps you or not , but if you go back to this example it might . this is f of s , this is f of s right here . 2 times -- i 'll do it in the light blue -- this is 2 over s squared plus 1 . that 's f of s in our example . and the g of s was s over s squared plus 1 . and all i got that from is i just broke this up into two things that i recognize . if i multiply this together , i get back to my original thing that i was trying to take the inverse laplace transform of . and so the convolution theorem just says that , ok , well , the inverse laplace transform of this is equal to the inverse laplace transform of 2 over s squared plus 1 , convoluted with the inverse laplace transform of our g of s , of s over s squared plus 1 . and we know what these things are . i already told them to you , but they should be somewhat second nature now . this is 2 times sine of t. you take the laplace transform of sine of t , you get 1 over s squared plus 1 , and then you multiply it by 2 , you get the 2 up there . and you 're going to have to convolute that with the inverse laplace transform of this thing here . and we already went over this . this is cosine of t. so our result so far -- let me be very clear . it 's always good to take a step back and just think about what we 're doing , much less why we 're doing it . but let 's see , the inverse laplace transform of this thing up in this top left corner , 2s over s squared plus 1 squared , which before we did what we 're doing now was very hard to figure out -- actually , this would be a curly bracket right here , but you get the idea -- is equal to this . it 's equal to 2 sine of t , convoluted with cosine of t. and you 're like , sal , throughout this whole process i 've already forgotten what it means to convolute two functions , so let 's convolute them . and i 'll just write the definition , or the definition we 're using of the convolution . that f convoluted with g -- it 's going to be a function of g. i 'll just write this short-hand -- is equal to the integral from 0 to t , of f of t minus tau , times g of tau , dtau . so 2 sine of t convoluted with cosine of t is equal to -- let me do a neutral color -- the integral from 0 to t , of 2 sine of t , minus tau , times the cosine of tau , dtau . now if you watched the very last video i made , i actually solved this , or i solved a very similar thing to this . if we take the 2 out we get 2 , times the integral from 0 to t , of sine of t minus tau , times the cosine of tau . i actually solved this in the previous video . this right here , this is the convolution of sine of t and cosine of t. it 's sine of t convoluted with cosine of t. and i show you in the previous video , just watch that video , where i introduce a convolution , that this thing right here is equal to 1/2t sine of t. now , if this thing is equal to 1/2 t sine of t , and i have to multiply it by 2 , then we get , our big result , that the inverse laplace transform of 2s over s squared plus 1 squared is equal to the convolution of 2 sine of t with cosine of t. which is just 2 times this thing here , which is 2 times 1/2 -- those cancel out -- so it equals t sine of t. and once you get the hang of it , you wo n't have to go through all of these steps . but the key is to recognize that this could be broken down as the products of two laplace transforms that you recognize . this could be broken down as the product of two laplace transforms we recognized . this is the laplace transform of 2 sine of t. this was the laplace transform of cosine of t. so the inverse laplace transform of our original thing , or original expression , is just the convolution of that with that . and if you watched the previous video , you 'd realize that actually calculating that convolution was no simple task , but it can be done . so you actually can get an integral form . even if it ca n't be done , you can get your answer , at least , in terms of some integral . so i have n't proven the convolution theorem to you just yet . i 'll do that in a future video . but hopefully , this gave you a little bit of a sense of how you can use it to actually take inverse laplace transforms . and remember , the reason why we 're learning to take inverse laplace transforms , and we have all of these tools to do it , is because that 's always that last step when you 're solving these differential equations , using your laplace transforms .
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i already told them to you , but they should be somewhat second nature now . this is 2 times sine of t. you take the laplace transform of sine of t , you get 1 over s squared plus 1 , and then you multiply it by 2 , you get the 2 up there . and you 're going to have to convolute that with the inverse laplace transform of this thing here .
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i wanted to ask if we take the laplace of ( t*sint ( t ) ) we are supposed to get 2s/ ( ( s^2 ) +1 ) ^2 again right ?
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now that you 've had a little bit of exposure to what a convolution is , i can introduce you to the convolution theorem , or at least in the context of -- there may be other convolution theorems -- but we 're talking about differential equations and laplace transforms . so this is the convolution theorem as applies to laplace transforms . and it tells us that if i have a function f of t -- and i can define its laplace transform as , let 's see , the laplace transform of f of t is capital f of s. we 've done that before . and if i have another function , g of t , and i take its laplace transform , that of course is capital g of s. then if we were to convolute these two functions , so if i were to take f and i were to convolute it with g , which is going to be another function of t -- and we already saw this . we saw that in the last video . i convoluted sine and cosine . so this is going to be a function of t. that the laplace transform of this thing , and this the crux of the theorem , the laplace transform of the convolution of these two functions is equal to the products of their laplace transforms . it equals f of s , big capital f of s , times big capital g of s. now , this might seem very abstract and very , you know , hard to kind of handle for you right now . so let 's do an actual example . and actually , even better , let 's do an inverse laplace transform with an example . and actually , let me write one more thing . if this is true , then we could also do it the other way . we could also say that f -- and i 'll just do it all in yellow ; it takes me too much time to keep switching colors -- that the convolution of f and g , this is just a function of t , i can just say it 's the inverse laplace transform . it 's just the inverse laplace transform of f of s times g of s. although i could n't resist it . let me switch colors . there you go . now , what good does all of this do ? well , we can take inverse laplace transforms . let 's just say that i had -- let me write it down here -- let 's say i told you that the following expression or function , let 's say h of s -- let me write it this way -- h of s is equal to 2s over s squared plus 1 . now , we did this long differential equations at the end , we end up with this thing and we have to take the inverse laplace transform of it . so we want to figure out the inverse laplace transform of h of s , or the inverse laplace transform of this thing right there . so we want to figure out the inverse laplace transform of this expression right here , 2s over s squared plus 1 squared . i do n't want to lose that . right there . now , can we write this as the product of two laplace transforms that we do know ? let 's try to do it . so we can rewrite this . and so this is the inverse laplace transform . so let me rewrite this expression down here . so i can rewrite 2s over s squared plus 1 squared . this is the same thing as -- let me write it this way -- 2 times 1 over s squared plus 1 , times s over s squared plus 1 . i just kind of broke it up . if you multiply the numerators here , you get 2 times 1 , times s , or 2s . if you multiply the denominators here , s squared plus 1 , times s squared plus 1 , well , that 's just s squared plus 1 squared . so this is the same thing . so if we want to take the inverse laplace transform of this , it 's the same thing as taking the inverse laplace transform of this right here . now , something should hopefully start popping out at you . if these were separate transforms , if they were on their own , we know what this is . if we call this f of s , if we said this is the laplace transform of some function , we know what that function is . this is this piece right here . i 'm just doing a little dotted line around it . this is the laplace transform of sine of t. and then if we draw a little box around this one right here , this is the laplace transform of cosine of t , g of s. so this is the laplace transform of sine of t , or we could write that this implies that f of t is equal to sine of t. you should recognize that one by now . and this implies that g of t , if we define this as the laplace transform of g , this means that g of t is equal to cosine of t. and , of course , when you take the inverse laplace transforms , you could take the 2 's out . so now what can we say ? we can now say that the -- let me write it this way -- the inverse -- so actually , let me write it this way . or , actually , a better thing to do , instead of taking the 2 out , so i can leave it nice and clean , we could , if we were to draw a box around this whole thing , and define this whole thing as f of s , then f of s is the laplace transform of 2 sine of t. i just wanted to include that 2 . i did n't want to leave that out and confuse the issue . i wanted a very pure f of s times g of s. so this expression right here is the product of the laplace transform of 2 sine of t , and the laplace transform of cosine of t. now , our convolution theorem told us this right here . that if we want to take the inverse laplace transform of the laplace transforms of two functions -- i know that sounds very confusing -- but you just kind of pattern match . you say , ok look , this thing that i had here , i could rewrite it as a product of two laplace transforms i can recognize . this right here is the laplace transform of 2 sine of t. this is the laplace transform of cosine of t. and we just wrote that as g of s , and f of s. so if i have an expression written like this , i can take the inverse laplace transform and it 'll be equal to the convolution of the original functions . it 'll be equal to the convolution of the inverse of g or the inverse of f. let me write it this way . i could write it like this . we know that f of t is equal to the inverse laplace transform of f of s. and we know that g -- i should have done it in a different color , but i 'll do g in green -- we know that g of t is equal to the inverse laplace transform of g of s. so we can rewrite the convolution theorem as the inverse -- and this might maybe confuse you more than help , but i 'll give my best shot . the inverse laplace transform of -- and i 'll try to stay true to the colors -- of f of s times g of s is equal to -- i 'm just restating this convolution theorem right here . this is equal to the convolution of the inverse laplace transform of f of s. so it 's equal to the convolution of the inverse laplace transform of f of s with the inverse laplace transform of g of s. with the inverse laplace transform of capital g , of g of s. i 'm not sure if that helps you or not , but if you go back to this example it might . this is f of s , this is f of s right here . 2 times -- i 'll do it in the light blue -- this is 2 over s squared plus 1 . that 's f of s in our example . and the g of s was s over s squared plus 1 . and all i got that from is i just broke this up into two things that i recognize . if i multiply this together , i get back to my original thing that i was trying to take the inverse laplace transform of . and so the convolution theorem just says that , ok , well , the inverse laplace transform of this is equal to the inverse laplace transform of 2 over s squared plus 1 , convoluted with the inverse laplace transform of our g of s , of s over s squared plus 1 . and we know what these things are . i already told them to you , but they should be somewhat second nature now . this is 2 times sine of t. you take the laplace transform of sine of t , you get 1 over s squared plus 1 , and then you multiply it by 2 , you get the 2 up there . and you 're going to have to convolute that with the inverse laplace transform of this thing here . and we already went over this . this is cosine of t. so our result so far -- let me be very clear . it 's always good to take a step back and just think about what we 're doing , much less why we 're doing it . but let 's see , the inverse laplace transform of this thing up in this top left corner , 2s over s squared plus 1 squared , which before we did what we 're doing now was very hard to figure out -- actually , this would be a curly bracket right here , but you get the idea -- is equal to this . it 's equal to 2 sine of t , convoluted with cosine of t. and you 're like , sal , throughout this whole process i 've already forgotten what it means to convolute two functions , so let 's convolute them . and i 'll just write the definition , or the definition we 're using of the convolution . that f convoluted with g -- it 's going to be a function of g. i 'll just write this short-hand -- is equal to the integral from 0 to t , of f of t minus tau , times g of tau , dtau . so 2 sine of t convoluted with cosine of t is equal to -- let me do a neutral color -- the integral from 0 to t , of 2 sine of t , minus tau , times the cosine of tau , dtau . now if you watched the very last video i made , i actually solved this , or i solved a very similar thing to this . if we take the 2 out we get 2 , times the integral from 0 to t , of sine of t minus tau , times the cosine of tau . i actually solved this in the previous video . this right here , this is the convolution of sine of t and cosine of t. it 's sine of t convoluted with cosine of t. and i show you in the previous video , just watch that video , where i introduce a convolution , that this thing right here is equal to 1/2t sine of t. now , if this thing is equal to 1/2 t sine of t , and i have to multiply it by 2 , then we get , our big result , that the inverse laplace transform of 2s over s squared plus 1 squared is equal to the convolution of 2 sine of t with cosine of t. which is just 2 times this thing here , which is 2 times 1/2 -- those cancel out -- so it equals t sine of t. and once you get the hang of it , you wo n't have to go through all of these steps . but the key is to recognize that this could be broken down as the products of two laplace transforms that you recognize . this could be broken down as the product of two laplace transforms we recognized . this is the laplace transform of 2 sine of t. this was the laplace transform of cosine of t. so the inverse laplace transform of our original thing , or original expression , is just the convolution of that with that . and if you watched the previous video , you 'd realize that actually calculating that convolution was no simple task , but it can be done . so you actually can get an integral form . even if it ca n't be done , you can get your answer , at least , in terms of some integral . so i have n't proven the convolution theorem to you just yet . i 'll do that in a future video . but hopefully , this gave you a little bit of a sense of how you can use it to actually take inverse laplace transforms . and remember , the reason why we 're learning to take inverse laplace transforms , and we have all of these tools to do it , is because that 's always that last step when you 're solving these differential equations , using your laplace transforms .
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and i 'll just write the definition , or the definition we 're using of the convolution . that f convoluted with g -- it 's going to be a function of g. i 'll just write this short-hand -- is equal to the integral from 0 to t , of f of t minus tau , times g of tau , dtau . so 2 sine of t convoluted with cosine of t is equal to -- let me do a neutral color -- the integral from 0 to t , of 2 sine of t , minus tau , times the cosine of tau , dtau .
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would it be right to say that : l { f * g } = f ( s ) x g ( s ) and ( f * g ) = l^-1 ( f x g ) where l is laplace transform , l^-1 is inverse laplace and f and g are l ( f ) and l ( g ) .. ?
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now that you 've had a little bit of exposure to what a convolution is , i can introduce you to the convolution theorem , or at least in the context of -- there may be other convolution theorems -- but we 're talking about differential equations and laplace transforms . so this is the convolution theorem as applies to laplace transforms . and it tells us that if i have a function f of t -- and i can define its laplace transform as , let 's see , the laplace transform of f of t is capital f of s. we 've done that before . and if i have another function , g of t , and i take its laplace transform , that of course is capital g of s. then if we were to convolute these two functions , so if i were to take f and i were to convolute it with g , which is going to be another function of t -- and we already saw this . we saw that in the last video . i convoluted sine and cosine . so this is going to be a function of t. that the laplace transform of this thing , and this the crux of the theorem , the laplace transform of the convolution of these two functions is equal to the products of their laplace transforms . it equals f of s , big capital f of s , times big capital g of s. now , this might seem very abstract and very , you know , hard to kind of handle for you right now . so let 's do an actual example . and actually , even better , let 's do an inverse laplace transform with an example . and actually , let me write one more thing . if this is true , then we could also do it the other way . we could also say that f -- and i 'll just do it all in yellow ; it takes me too much time to keep switching colors -- that the convolution of f and g , this is just a function of t , i can just say it 's the inverse laplace transform . it 's just the inverse laplace transform of f of s times g of s. although i could n't resist it . let me switch colors . there you go . now , what good does all of this do ? well , we can take inverse laplace transforms . let 's just say that i had -- let me write it down here -- let 's say i told you that the following expression or function , let 's say h of s -- let me write it this way -- h of s is equal to 2s over s squared plus 1 . now , we did this long differential equations at the end , we end up with this thing and we have to take the inverse laplace transform of it . so we want to figure out the inverse laplace transform of h of s , or the inverse laplace transform of this thing right there . so we want to figure out the inverse laplace transform of this expression right here , 2s over s squared plus 1 squared . i do n't want to lose that . right there . now , can we write this as the product of two laplace transforms that we do know ? let 's try to do it . so we can rewrite this . and so this is the inverse laplace transform . so let me rewrite this expression down here . so i can rewrite 2s over s squared plus 1 squared . this is the same thing as -- let me write it this way -- 2 times 1 over s squared plus 1 , times s over s squared plus 1 . i just kind of broke it up . if you multiply the numerators here , you get 2 times 1 , times s , or 2s . if you multiply the denominators here , s squared plus 1 , times s squared plus 1 , well , that 's just s squared plus 1 squared . so this is the same thing . so if we want to take the inverse laplace transform of this , it 's the same thing as taking the inverse laplace transform of this right here . now , something should hopefully start popping out at you . if these were separate transforms , if they were on their own , we know what this is . if we call this f of s , if we said this is the laplace transform of some function , we know what that function is . this is this piece right here . i 'm just doing a little dotted line around it . this is the laplace transform of sine of t. and then if we draw a little box around this one right here , this is the laplace transform of cosine of t , g of s. so this is the laplace transform of sine of t , or we could write that this implies that f of t is equal to sine of t. you should recognize that one by now . and this implies that g of t , if we define this as the laplace transform of g , this means that g of t is equal to cosine of t. and , of course , when you take the inverse laplace transforms , you could take the 2 's out . so now what can we say ? we can now say that the -- let me write it this way -- the inverse -- so actually , let me write it this way . or , actually , a better thing to do , instead of taking the 2 out , so i can leave it nice and clean , we could , if we were to draw a box around this whole thing , and define this whole thing as f of s , then f of s is the laplace transform of 2 sine of t. i just wanted to include that 2 . i did n't want to leave that out and confuse the issue . i wanted a very pure f of s times g of s. so this expression right here is the product of the laplace transform of 2 sine of t , and the laplace transform of cosine of t. now , our convolution theorem told us this right here . that if we want to take the inverse laplace transform of the laplace transforms of two functions -- i know that sounds very confusing -- but you just kind of pattern match . you say , ok look , this thing that i had here , i could rewrite it as a product of two laplace transforms i can recognize . this right here is the laplace transform of 2 sine of t. this is the laplace transform of cosine of t. and we just wrote that as g of s , and f of s. so if i have an expression written like this , i can take the inverse laplace transform and it 'll be equal to the convolution of the original functions . it 'll be equal to the convolution of the inverse of g or the inverse of f. let me write it this way . i could write it like this . we know that f of t is equal to the inverse laplace transform of f of s. and we know that g -- i should have done it in a different color , but i 'll do g in green -- we know that g of t is equal to the inverse laplace transform of g of s. so we can rewrite the convolution theorem as the inverse -- and this might maybe confuse you more than help , but i 'll give my best shot . the inverse laplace transform of -- and i 'll try to stay true to the colors -- of f of s times g of s is equal to -- i 'm just restating this convolution theorem right here . this is equal to the convolution of the inverse laplace transform of f of s. so it 's equal to the convolution of the inverse laplace transform of f of s with the inverse laplace transform of g of s. with the inverse laplace transform of capital g , of g of s. i 'm not sure if that helps you or not , but if you go back to this example it might . this is f of s , this is f of s right here . 2 times -- i 'll do it in the light blue -- this is 2 over s squared plus 1 . that 's f of s in our example . and the g of s was s over s squared plus 1 . and all i got that from is i just broke this up into two things that i recognize . if i multiply this together , i get back to my original thing that i was trying to take the inverse laplace transform of . and so the convolution theorem just says that , ok , well , the inverse laplace transform of this is equal to the inverse laplace transform of 2 over s squared plus 1 , convoluted with the inverse laplace transform of our g of s , of s over s squared plus 1 . and we know what these things are . i already told them to you , but they should be somewhat second nature now . this is 2 times sine of t. you take the laplace transform of sine of t , you get 1 over s squared plus 1 , and then you multiply it by 2 , you get the 2 up there . and you 're going to have to convolute that with the inverse laplace transform of this thing here . and we already went over this . this is cosine of t. so our result so far -- let me be very clear . it 's always good to take a step back and just think about what we 're doing , much less why we 're doing it . but let 's see , the inverse laplace transform of this thing up in this top left corner , 2s over s squared plus 1 squared , which before we did what we 're doing now was very hard to figure out -- actually , this would be a curly bracket right here , but you get the idea -- is equal to this . it 's equal to 2 sine of t , convoluted with cosine of t. and you 're like , sal , throughout this whole process i 've already forgotten what it means to convolute two functions , so let 's convolute them . and i 'll just write the definition , or the definition we 're using of the convolution . that f convoluted with g -- it 's going to be a function of g. i 'll just write this short-hand -- is equal to the integral from 0 to t , of f of t minus tau , times g of tau , dtau . so 2 sine of t convoluted with cosine of t is equal to -- let me do a neutral color -- the integral from 0 to t , of 2 sine of t , minus tau , times the cosine of tau , dtau . now if you watched the very last video i made , i actually solved this , or i solved a very similar thing to this . if we take the 2 out we get 2 , times the integral from 0 to t , of sine of t minus tau , times the cosine of tau . i actually solved this in the previous video . this right here , this is the convolution of sine of t and cosine of t. it 's sine of t convoluted with cosine of t. and i show you in the previous video , just watch that video , where i introduce a convolution , that this thing right here is equal to 1/2t sine of t. now , if this thing is equal to 1/2 t sine of t , and i have to multiply it by 2 , then we get , our big result , that the inverse laplace transform of 2s over s squared plus 1 squared is equal to the convolution of 2 sine of t with cosine of t. which is just 2 times this thing here , which is 2 times 1/2 -- those cancel out -- so it equals t sine of t. and once you get the hang of it , you wo n't have to go through all of these steps . but the key is to recognize that this could be broken down as the products of two laplace transforms that you recognize . this could be broken down as the product of two laplace transforms we recognized . this is the laplace transform of 2 sine of t. this was the laplace transform of cosine of t. so the inverse laplace transform of our original thing , or original expression , is just the convolution of that with that . and if you watched the previous video , you 'd realize that actually calculating that convolution was no simple task , but it can be done . so you actually can get an integral form . even if it ca n't be done , you can get your answer , at least , in terms of some integral . so i have n't proven the convolution theorem to you just yet . i 'll do that in a future video . but hopefully , this gave you a little bit of a sense of how you can use it to actually take inverse laplace transforms . and remember , the reason why we 're learning to take inverse laplace transforms , and we have all of these tools to do it , is because that 's always that last step when you 're solving these differential equations , using your laplace transforms .
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and i 'll just write the definition , or the definition we 're using of the convolution . that f convoluted with g -- it 's going to be a function of g. i 'll just write this short-hand -- is equal to the integral from 0 to t , of f of t minus tau , times g of tau , dtau . so 2 sine of t convoluted with cosine of t is equal to -- let me do a neutral color -- the integral from 0 to t , of 2 sine of t , minus tau , times the cosine of tau , dtau .
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is f * g = df/dt * integral ( g ( x ) dx ) a property of convolution ?
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now that you 've had a little bit of exposure to what a convolution is , i can introduce you to the convolution theorem , or at least in the context of -- there may be other convolution theorems -- but we 're talking about differential equations and laplace transforms . so this is the convolution theorem as applies to laplace transforms . and it tells us that if i have a function f of t -- and i can define its laplace transform as , let 's see , the laplace transform of f of t is capital f of s. we 've done that before . and if i have another function , g of t , and i take its laplace transform , that of course is capital g of s. then if we were to convolute these two functions , so if i were to take f and i were to convolute it with g , which is going to be another function of t -- and we already saw this . we saw that in the last video . i convoluted sine and cosine . so this is going to be a function of t. that the laplace transform of this thing , and this the crux of the theorem , the laplace transform of the convolution of these two functions is equal to the products of their laplace transforms . it equals f of s , big capital f of s , times big capital g of s. now , this might seem very abstract and very , you know , hard to kind of handle for you right now . so let 's do an actual example . and actually , even better , let 's do an inverse laplace transform with an example . and actually , let me write one more thing . if this is true , then we could also do it the other way . we could also say that f -- and i 'll just do it all in yellow ; it takes me too much time to keep switching colors -- that the convolution of f and g , this is just a function of t , i can just say it 's the inverse laplace transform . it 's just the inverse laplace transform of f of s times g of s. although i could n't resist it . let me switch colors . there you go . now , what good does all of this do ? well , we can take inverse laplace transforms . let 's just say that i had -- let me write it down here -- let 's say i told you that the following expression or function , let 's say h of s -- let me write it this way -- h of s is equal to 2s over s squared plus 1 . now , we did this long differential equations at the end , we end up with this thing and we have to take the inverse laplace transform of it . so we want to figure out the inverse laplace transform of h of s , or the inverse laplace transform of this thing right there . so we want to figure out the inverse laplace transform of this expression right here , 2s over s squared plus 1 squared . i do n't want to lose that . right there . now , can we write this as the product of two laplace transforms that we do know ? let 's try to do it . so we can rewrite this . and so this is the inverse laplace transform . so let me rewrite this expression down here . so i can rewrite 2s over s squared plus 1 squared . this is the same thing as -- let me write it this way -- 2 times 1 over s squared plus 1 , times s over s squared plus 1 . i just kind of broke it up . if you multiply the numerators here , you get 2 times 1 , times s , or 2s . if you multiply the denominators here , s squared plus 1 , times s squared plus 1 , well , that 's just s squared plus 1 squared . so this is the same thing . so if we want to take the inverse laplace transform of this , it 's the same thing as taking the inverse laplace transform of this right here . now , something should hopefully start popping out at you . if these were separate transforms , if they were on their own , we know what this is . if we call this f of s , if we said this is the laplace transform of some function , we know what that function is . this is this piece right here . i 'm just doing a little dotted line around it . this is the laplace transform of sine of t. and then if we draw a little box around this one right here , this is the laplace transform of cosine of t , g of s. so this is the laplace transform of sine of t , or we could write that this implies that f of t is equal to sine of t. you should recognize that one by now . and this implies that g of t , if we define this as the laplace transform of g , this means that g of t is equal to cosine of t. and , of course , when you take the inverse laplace transforms , you could take the 2 's out . so now what can we say ? we can now say that the -- let me write it this way -- the inverse -- so actually , let me write it this way . or , actually , a better thing to do , instead of taking the 2 out , so i can leave it nice and clean , we could , if we were to draw a box around this whole thing , and define this whole thing as f of s , then f of s is the laplace transform of 2 sine of t. i just wanted to include that 2 . i did n't want to leave that out and confuse the issue . i wanted a very pure f of s times g of s. so this expression right here is the product of the laplace transform of 2 sine of t , and the laplace transform of cosine of t. now , our convolution theorem told us this right here . that if we want to take the inverse laplace transform of the laplace transforms of two functions -- i know that sounds very confusing -- but you just kind of pattern match . you say , ok look , this thing that i had here , i could rewrite it as a product of two laplace transforms i can recognize . this right here is the laplace transform of 2 sine of t. this is the laplace transform of cosine of t. and we just wrote that as g of s , and f of s. so if i have an expression written like this , i can take the inverse laplace transform and it 'll be equal to the convolution of the original functions . it 'll be equal to the convolution of the inverse of g or the inverse of f. let me write it this way . i could write it like this . we know that f of t is equal to the inverse laplace transform of f of s. and we know that g -- i should have done it in a different color , but i 'll do g in green -- we know that g of t is equal to the inverse laplace transform of g of s. so we can rewrite the convolution theorem as the inverse -- and this might maybe confuse you more than help , but i 'll give my best shot . the inverse laplace transform of -- and i 'll try to stay true to the colors -- of f of s times g of s is equal to -- i 'm just restating this convolution theorem right here . this is equal to the convolution of the inverse laplace transform of f of s. so it 's equal to the convolution of the inverse laplace transform of f of s with the inverse laplace transform of g of s. with the inverse laplace transform of capital g , of g of s. i 'm not sure if that helps you or not , but if you go back to this example it might . this is f of s , this is f of s right here . 2 times -- i 'll do it in the light blue -- this is 2 over s squared plus 1 . that 's f of s in our example . and the g of s was s over s squared plus 1 . and all i got that from is i just broke this up into two things that i recognize . if i multiply this together , i get back to my original thing that i was trying to take the inverse laplace transform of . and so the convolution theorem just says that , ok , well , the inverse laplace transform of this is equal to the inverse laplace transform of 2 over s squared plus 1 , convoluted with the inverse laplace transform of our g of s , of s over s squared plus 1 . and we know what these things are . i already told them to you , but they should be somewhat second nature now . this is 2 times sine of t. you take the laplace transform of sine of t , you get 1 over s squared plus 1 , and then you multiply it by 2 , you get the 2 up there . and you 're going to have to convolute that with the inverse laplace transform of this thing here . and we already went over this . this is cosine of t. so our result so far -- let me be very clear . it 's always good to take a step back and just think about what we 're doing , much less why we 're doing it . but let 's see , the inverse laplace transform of this thing up in this top left corner , 2s over s squared plus 1 squared , which before we did what we 're doing now was very hard to figure out -- actually , this would be a curly bracket right here , but you get the idea -- is equal to this . it 's equal to 2 sine of t , convoluted with cosine of t. and you 're like , sal , throughout this whole process i 've already forgotten what it means to convolute two functions , so let 's convolute them . and i 'll just write the definition , or the definition we 're using of the convolution . that f convoluted with g -- it 's going to be a function of g. i 'll just write this short-hand -- is equal to the integral from 0 to t , of f of t minus tau , times g of tau , dtau . so 2 sine of t convoluted with cosine of t is equal to -- let me do a neutral color -- the integral from 0 to t , of 2 sine of t , minus tau , times the cosine of tau , dtau . now if you watched the very last video i made , i actually solved this , or i solved a very similar thing to this . if we take the 2 out we get 2 , times the integral from 0 to t , of sine of t minus tau , times the cosine of tau . i actually solved this in the previous video . this right here , this is the convolution of sine of t and cosine of t. it 's sine of t convoluted with cosine of t. and i show you in the previous video , just watch that video , where i introduce a convolution , that this thing right here is equal to 1/2t sine of t. now , if this thing is equal to 1/2 t sine of t , and i have to multiply it by 2 , then we get , our big result , that the inverse laplace transform of 2s over s squared plus 1 squared is equal to the convolution of 2 sine of t with cosine of t. which is just 2 times this thing here , which is 2 times 1/2 -- those cancel out -- so it equals t sine of t. and once you get the hang of it , you wo n't have to go through all of these steps . but the key is to recognize that this could be broken down as the products of two laplace transforms that you recognize . this could be broken down as the product of two laplace transforms we recognized . this is the laplace transform of 2 sine of t. this was the laplace transform of cosine of t. so the inverse laplace transform of our original thing , or original expression , is just the convolution of that with that . and if you watched the previous video , you 'd realize that actually calculating that convolution was no simple task , but it can be done . so you actually can get an integral form . even if it ca n't be done , you can get your answer , at least , in terms of some integral . so i have n't proven the convolution theorem to you just yet . i 'll do that in a future video . but hopefully , this gave you a little bit of a sense of how you can use it to actually take inverse laplace transforms . and remember , the reason why we 're learning to take inverse laplace transforms , and we have all of these tools to do it , is because that 's always that last step when you 're solving these differential equations , using your laplace transforms .
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it 's equal to 2 sine of t , convoluted with cosine of t. and you 're like , sal , throughout this whole process i 've already forgotten what it means to convolute two functions , so let 's convolute them . and i 'll just write the definition , or the definition we 're using of the convolution . that f convoluted with g -- it 's going to be a function of g. i 'll just write this short-hand -- is equal to the integral from 0 to t , of f of t minus tau , times g of tau , dtau .
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how does the definition of the laplace transform relate to the definition of the convolution integral ?
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now that you 've had a little bit of exposure to what a convolution is , i can introduce you to the convolution theorem , or at least in the context of -- there may be other convolution theorems -- but we 're talking about differential equations and laplace transforms . so this is the convolution theorem as applies to laplace transforms . and it tells us that if i have a function f of t -- and i can define its laplace transform as , let 's see , the laplace transform of f of t is capital f of s. we 've done that before . and if i have another function , g of t , and i take its laplace transform , that of course is capital g of s. then if we were to convolute these two functions , so if i were to take f and i were to convolute it with g , which is going to be another function of t -- and we already saw this . we saw that in the last video . i convoluted sine and cosine . so this is going to be a function of t. that the laplace transform of this thing , and this the crux of the theorem , the laplace transform of the convolution of these two functions is equal to the products of their laplace transforms . it equals f of s , big capital f of s , times big capital g of s. now , this might seem very abstract and very , you know , hard to kind of handle for you right now . so let 's do an actual example . and actually , even better , let 's do an inverse laplace transform with an example . and actually , let me write one more thing . if this is true , then we could also do it the other way . we could also say that f -- and i 'll just do it all in yellow ; it takes me too much time to keep switching colors -- that the convolution of f and g , this is just a function of t , i can just say it 's the inverse laplace transform . it 's just the inverse laplace transform of f of s times g of s. although i could n't resist it . let me switch colors . there you go . now , what good does all of this do ? well , we can take inverse laplace transforms . let 's just say that i had -- let me write it down here -- let 's say i told you that the following expression or function , let 's say h of s -- let me write it this way -- h of s is equal to 2s over s squared plus 1 . now , we did this long differential equations at the end , we end up with this thing and we have to take the inverse laplace transform of it . so we want to figure out the inverse laplace transform of h of s , or the inverse laplace transform of this thing right there . so we want to figure out the inverse laplace transform of this expression right here , 2s over s squared plus 1 squared . i do n't want to lose that . right there . now , can we write this as the product of two laplace transforms that we do know ? let 's try to do it . so we can rewrite this . and so this is the inverse laplace transform . so let me rewrite this expression down here . so i can rewrite 2s over s squared plus 1 squared . this is the same thing as -- let me write it this way -- 2 times 1 over s squared plus 1 , times s over s squared plus 1 . i just kind of broke it up . if you multiply the numerators here , you get 2 times 1 , times s , or 2s . if you multiply the denominators here , s squared plus 1 , times s squared plus 1 , well , that 's just s squared plus 1 squared . so this is the same thing . so if we want to take the inverse laplace transform of this , it 's the same thing as taking the inverse laplace transform of this right here . now , something should hopefully start popping out at you . if these were separate transforms , if they were on their own , we know what this is . if we call this f of s , if we said this is the laplace transform of some function , we know what that function is . this is this piece right here . i 'm just doing a little dotted line around it . this is the laplace transform of sine of t. and then if we draw a little box around this one right here , this is the laplace transform of cosine of t , g of s. so this is the laplace transform of sine of t , or we could write that this implies that f of t is equal to sine of t. you should recognize that one by now . and this implies that g of t , if we define this as the laplace transform of g , this means that g of t is equal to cosine of t. and , of course , when you take the inverse laplace transforms , you could take the 2 's out . so now what can we say ? we can now say that the -- let me write it this way -- the inverse -- so actually , let me write it this way . or , actually , a better thing to do , instead of taking the 2 out , so i can leave it nice and clean , we could , if we were to draw a box around this whole thing , and define this whole thing as f of s , then f of s is the laplace transform of 2 sine of t. i just wanted to include that 2 . i did n't want to leave that out and confuse the issue . i wanted a very pure f of s times g of s. so this expression right here is the product of the laplace transform of 2 sine of t , and the laplace transform of cosine of t. now , our convolution theorem told us this right here . that if we want to take the inverse laplace transform of the laplace transforms of two functions -- i know that sounds very confusing -- but you just kind of pattern match . you say , ok look , this thing that i had here , i could rewrite it as a product of two laplace transforms i can recognize . this right here is the laplace transform of 2 sine of t. this is the laplace transform of cosine of t. and we just wrote that as g of s , and f of s. so if i have an expression written like this , i can take the inverse laplace transform and it 'll be equal to the convolution of the original functions . it 'll be equal to the convolution of the inverse of g or the inverse of f. let me write it this way . i could write it like this . we know that f of t is equal to the inverse laplace transform of f of s. and we know that g -- i should have done it in a different color , but i 'll do g in green -- we know that g of t is equal to the inverse laplace transform of g of s. so we can rewrite the convolution theorem as the inverse -- and this might maybe confuse you more than help , but i 'll give my best shot . the inverse laplace transform of -- and i 'll try to stay true to the colors -- of f of s times g of s is equal to -- i 'm just restating this convolution theorem right here . this is equal to the convolution of the inverse laplace transform of f of s. so it 's equal to the convolution of the inverse laplace transform of f of s with the inverse laplace transform of g of s. with the inverse laplace transform of capital g , of g of s. i 'm not sure if that helps you or not , but if you go back to this example it might . this is f of s , this is f of s right here . 2 times -- i 'll do it in the light blue -- this is 2 over s squared plus 1 . that 's f of s in our example . and the g of s was s over s squared plus 1 . and all i got that from is i just broke this up into two things that i recognize . if i multiply this together , i get back to my original thing that i was trying to take the inverse laplace transform of . and so the convolution theorem just says that , ok , well , the inverse laplace transform of this is equal to the inverse laplace transform of 2 over s squared plus 1 , convoluted with the inverse laplace transform of our g of s , of s over s squared plus 1 . and we know what these things are . i already told them to you , but they should be somewhat second nature now . this is 2 times sine of t. you take the laplace transform of sine of t , you get 1 over s squared plus 1 , and then you multiply it by 2 , you get the 2 up there . and you 're going to have to convolute that with the inverse laplace transform of this thing here . and we already went over this . this is cosine of t. so our result so far -- let me be very clear . it 's always good to take a step back and just think about what we 're doing , much less why we 're doing it . but let 's see , the inverse laplace transform of this thing up in this top left corner , 2s over s squared plus 1 squared , which before we did what we 're doing now was very hard to figure out -- actually , this would be a curly bracket right here , but you get the idea -- is equal to this . it 's equal to 2 sine of t , convoluted with cosine of t. and you 're like , sal , throughout this whole process i 've already forgotten what it means to convolute two functions , so let 's convolute them . and i 'll just write the definition , or the definition we 're using of the convolution . that f convoluted with g -- it 's going to be a function of g. i 'll just write this short-hand -- is equal to the integral from 0 to t , of f of t minus tau , times g of tau , dtau . so 2 sine of t convoluted with cosine of t is equal to -- let me do a neutral color -- the integral from 0 to t , of 2 sine of t , minus tau , times the cosine of tau , dtau . now if you watched the very last video i made , i actually solved this , or i solved a very similar thing to this . if we take the 2 out we get 2 , times the integral from 0 to t , of sine of t minus tau , times the cosine of tau . i actually solved this in the previous video . this right here , this is the convolution of sine of t and cosine of t. it 's sine of t convoluted with cosine of t. and i show you in the previous video , just watch that video , where i introduce a convolution , that this thing right here is equal to 1/2t sine of t. now , if this thing is equal to 1/2 t sine of t , and i have to multiply it by 2 , then we get , our big result , that the inverse laplace transform of 2s over s squared plus 1 squared is equal to the convolution of 2 sine of t with cosine of t. which is just 2 times this thing here , which is 2 times 1/2 -- those cancel out -- so it equals t sine of t. and once you get the hang of it , you wo n't have to go through all of these steps . but the key is to recognize that this could be broken down as the products of two laplace transforms that you recognize . this could be broken down as the product of two laplace transforms we recognized . this is the laplace transform of 2 sine of t. this was the laplace transform of cosine of t. so the inverse laplace transform of our original thing , or original expression , is just the convolution of that with that . and if you watched the previous video , you 'd realize that actually calculating that convolution was no simple task , but it can be done . so you actually can get an integral form . even if it ca n't be done , you can get your answer , at least , in terms of some integral . so i have n't proven the convolution theorem to you just yet . i 'll do that in a future video . but hopefully , this gave you a little bit of a sense of how you can use it to actually take inverse laplace transforms . and remember , the reason why we 're learning to take inverse laplace transforms , and we have all of these tools to do it , is because that 's always that last step when you 're solving these differential equations , using your laplace transforms .
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now , what good does all of this do ? well , we can take inverse laplace transforms . let 's just say that i had -- let me write it down here -- let 's say i told you that the following expression or function , let 's say h of s -- let me write it this way -- h of s is equal to 2s over s squared plus 1 .
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why do i see fourier transforms in signal conditioning but laplace transforms in control systems ?
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now that you 've had a little bit of exposure to what a convolution is , i can introduce you to the convolution theorem , or at least in the context of -- there may be other convolution theorems -- but we 're talking about differential equations and laplace transforms . so this is the convolution theorem as applies to laplace transforms . and it tells us that if i have a function f of t -- and i can define its laplace transform as , let 's see , the laplace transform of f of t is capital f of s. we 've done that before . and if i have another function , g of t , and i take its laplace transform , that of course is capital g of s. then if we were to convolute these two functions , so if i were to take f and i were to convolute it with g , which is going to be another function of t -- and we already saw this . we saw that in the last video . i convoluted sine and cosine . so this is going to be a function of t. that the laplace transform of this thing , and this the crux of the theorem , the laplace transform of the convolution of these two functions is equal to the products of their laplace transforms . it equals f of s , big capital f of s , times big capital g of s. now , this might seem very abstract and very , you know , hard to kind of handle for you right now . so let 's do an actual example . and actually , even better , let 's do an inverse laplace transform with an example . and actually , let me write one more thing . if this is true , then we could also do it the other way . we could also say that f -- and i 'll just do it all in yellow ; it takes me too much time to keep switching colors -- that the convolution of f and g , this is just a function of t , i can just say it 's the inverse laplace transform . it 's just the inverse laplace transform of f of s times g of s. although i could n't resist it . let me switch colors . there you go . now , what good does all of this do ? well , we can take inverse laplace transforms . let 's just say that i had -- let me write it down here -- let 's say i told you that the following expression or function , let 's say h of s -- let me write it this way -- h of s is equal to 2s over s squared plus 1 . now , we did this long differential equations at the end , we end up with this thing and we have to take the inverse laplace transform of it . so we want to figure out the inverse laplace transform of h of s , or the inverse laplace transform of this thing right there . so we want to figure out the inverse laplace transform of this expression right here , 2s over s squared plus 1 squared . i do n't want to lose that . right there . now , can we write this as the product of two laplace transforms that we do know ? let 's try to do it . so we can rewrite this . and so this is the inverse laplace transform . so let me rewrite this expression down here . so i can rewrite 2s over s squared plus 1 squared . this is the same thing as -- let me write it this way -- 2 times 1 over s squared plus 1 , times s over s squared plus 1 . i just kind of broke it up . if you multiply the numerators here , you get 2 times 1 , times s , or 2s . if you multiply the denominators here , s squared plus 1 , times s squared plus 1 , well , that 's just s squared plus 1 squared . so this is the same thing . so if we want to take the inverse laplace transform of this , it 's the same thing as taking the inverse laplace transform of this right here . now , something should hopefully start popping out at you . if these were separate transforms , if they were on their own , we know what this is . if we call this f of s , if we said this is the laplace transform of some function , we know what that function is . this is this piece right here . i 'm just doing a little dotted line around it . this is the laplace transform of sine of t. and then if we draw a little box around this one right here , this is the laplace transform of cosine of t , g of s. so this is the laplace transform of sine of t , or we could write that this implies that f of t is equal to sine of t. you should recognize that one by now . and this implies that g of t , if we define this as the laplace transform of g , this means that g of t is equal to cosine of t. and , of course , when you take the inverse laplace transforms , you could take the 2 's out . so now what can we say ? we can now say that the -- let me write it this way -- the inverse -- so actually , let me write it this way . or , actually , a better thing to do , instead of taking the 2 out , so i can leave it nice and clean , we could , if we were to draw a box around this whole thing , and define this whole thing as f of s , then f of s is the laplace transform of 2 sine of t. i just wanted to include that 2 . i did n't want to leave that out and confuse the issue . i wanted a very pure f of s times g of s. so this expression right here is the product of the laplace transform of 2 sine of t , and the laplace transform of cosine of t. now , our convolution theorem told us this right here . that if we want to take the inverse laplace transform of the laplace transforms of two functions -- i know that sounds very confusing -- but you just kind of pattern match . you say , ok look , this thing that i had here , i could rewrite it as a product of two laplace transforms i can recognize . this right here is the laplace transform of 2 sine of t. this is the laplace transform of cosine of t. and we just wrote that as g of s , and f of s. so if i have an expression written like this , i can take the inverse laplace transform and it 'll be equal to the convolution of the original functions . it 'll be equal to the convolution of the inverse of g or the inverse of f. let me write it this way . i could write it like this . we know that f of t is equal to the inverse laplace transform of f of s. and we know that g -- i should have done it in a different color , but i 'll do g in green -- we know that g of t is equal to the inverse laplace transform of g of s. so we can rewrite the convolution theorem as the inverse -- and this might maybe confuse you more than help , but i 'll give my best shot . the inverse laplace transform of -- and i 'll try to stay true to the colors -- of f of s times g of s is equal to -- i 'm just restating this convolution theorem right here . this is equal to the convolution of the inverse laplace transform of f of s. so it 's equal to the convolution of the inverse laplace transform of f of s with the inverse laplace transform of g of s. with the inverse laplace transform of capital g , of g of s. i 'm not sure if that helps you or not , but if you go back to this example it might . this is f of s , this is f of s right here . 2 times -- i 'll do it in the light blue -- this is 2 over s squared plus 1 . that 's f of s in our example . and the g of s was s over s squared plus 1 . and all i got that from is i just broke this up into two things that i recognize . if i multiply this together , i get back to my original thing that i was trying to take the inverse laplace transform of . and so the convolution theorem just says that , ok , well , the inverse laplace transform of this is equal to the inverse laplace transform of 2 over s squared plus 1 , convoluted with the inverse laplace transform of our g of s , of s over s squared plus 1 . and we know what these things are . i already told them to you , but they should be somewhat second nature now . this is 2 times sine of t. you take the laplace transform of sine of t , you get 1 over s squared plus 1 , and then you multiply it by 2 , you get the 2 up there . and you 're going to have to convolute that with the inverse laplace transform of this thing here . and we already went over this . this is cosine of t. so our result so far -- let me be very clear . it 's always good to take a step back and just think about what we 're doing , much less why we 're doing it . but let 's see , the inverse laplace transform of this thing up in this top left corner , 2s over s squared plus 1 squared , which before we did what we 're doing now was very hard to figure out -- actually , this would be a curly bracket right here , but you get the idea -- is equal to this . it 's equal to 2 sine of t , convoluted with cosine of t. and you 're like , sal , throughout this whole process i 've already forgotten what it means to convolute two functions , so let 's convolute them . and i 'll just write the definition , or the definition we 're using of the convolution . that f convoluted with g -- it 's going to be a function of g. i 'll just write this short-hand -- is equal to the integral from 0 to t , of f of t minus tau , times g of tau , dtau . so 2 sine of t convoluted with cosine of t is equal to -- let me do a neutral color -- the integral from 0 to t , of 2 sine of t , minus tau , times the cosine of tau , dtau . now if you watched the very last video i made , i actually solved this , or i solved a very similar thing to this . if we take the 2 out we get 2 , times the integral from 0 to t , of sine of t minus tau , times the cosine of tau . i actually solved this in the previous video . this right here , this is the convolution of sine of t and cosine of t. it 's sine of t convoluted with cosine of t. and i show you in the previous video , just watch that video , where i introduce a convolution , that this thing right here is equal to 1/2t sine of t. now , if this thing is equal to 1/2 t sine of t , and i have to multiply it by 2 , then we get , our big result , that the inverse laplace transform of 2s over s squared plus 1 squared is equal to the convolution of 2 sine of t with cosine of t. which is just 2 times this thing here , which is 2 times 1/2 -- those cancel out -- so it equals t sine of t. and once you get the hang of it , you wo n't have to go through all of these steps . but the key is to recognize that this could be broken down as the products of two laplace transforms that you recognize . this could be broken down as the product of two laplace transforms we recognized . this is the laplace transform of 2 sine of t. this was the laplace transform of cosine of t. so the inverse laplace transform of our original thing , or original expression , is just the convolution of that with that . and if you watched the previous video , you 'd realize that actually calculating that convolution was no simple task , but it can be done . so you actually can get an integral form . even if it ca n't be done , you can get your answer , at least , in terms of some integral . so i have n't proven the convolution theorem to you just yet . i 'll do that in a future video . but hopefully , this gave you a little bit of a sense of how you can use it to actually take inverse laplace transforms . and remember , the reason why we 're learning to take inverse laplace transforms , and we have all of these tools to do it , is because that 's always that last step when you 're solving these differential equations , using your laplace transforms .
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now that you 've had a little bit of exposure to what a convolution is , i can introduce you to the convolution theorem , or at least in the context of -- there may be other convolution theorems -- but we 're talking about differential equations and laplace transforms . so this is the convolution theorem as applies to laplace transforms .
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what are the advantages/disadvantages of both ?
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now that you 've had a little bit of exposure to what a convolution is , i can introduce you to the convolution theorem , or at least in the context of -- there may be other convolution theorems -- but we 're talking about differential equations and laplace transforms . so this is the convolution theorem as applies to laplace transforms . and it tells us that if i have a function f of t -- and i can define its laplace transform as , let 's see , the laplace transform of f of t is capital f of s. we 've done that before . and if i have another function , g of t , and i take its laplace transform , that of course is capital g of s. then if we were to convolute these two functions , so if i were to take f and i were to convolute it with g , which is going to be another function of t -- and we already saw this . we saw that in the last video . i convoluted sine and cosine . so this is going to be a function of t. that the laplace transform of this thing , and this the crux of the theorem , the laplace transform of the convolution of these two functions is equal to the products of their laplace transforms . it equals f of s , big capital f of s , times big capital g of s. now , this might seem very abstract and very , you know , hard to kind of handle for you right now . so let 's do an actual example . and actually , even better , let 's do an inverse laplace transform with an example . and actually , let me write one more thing . if this is true , then we could also do it the other way . we could also say that f -- and i 'll just do it all in yellow ; it takes me too much time to keep switching colors -- that the convolution of f and g , this is just a function of t , i can just say it 's the inverse laplace transform . it 's just the inverse laplace transform of f of s times g of s. although i could n't resist it . let me switch colors . there you go . now , what good does all of this do ? well , we can take inverse laplace transforms . let 's just say that i had -- let me write it down here -- let 's say i told you that the following expression or function , let 's say h of s -- let me write it this way -- h of s is equal to 2s over s squared plus 1 . now , we did this long differential equations at the end , we end up with this thing and we have to take the inverse laplace transform of it . so we want to figure out the inverse laplace transform of h of s , or the inverse laplace transform of this thing right there . so we want to figure out the inverse laplace transform of this expression right here , 2s over s squared plus 1 squared . i do n't want to lose that . right there . now , can we write this as the product of two laplace transforms that we do know ? let 's try to do it . so we can rewrite this . and so this is the inverse laplace transform . so let me rewrite this expression down here . so i can rewrite 2s over s squared plus 1 squared . this is the same thing as -- let me write it this way -- 2 times 1 over s squared plus 1 , times s over s squared plus 1 . i just kind of broke it up . if you multiply the numerators here , you get 2 times 1 , times s , or 2s . if you multiply the denominators here , s squared plus 1 , times s squared plus 1 , well , that 's just s squared plus 1 squared . so this is the same thing . so if we want to take the inverse laplace transform of this , it 's the same thing as taking the inverse laplace transform of this right here . now , something should hopefully start popping out at you . if these were separate transforms , if they were on their own , we know what this is . if we call this f of s , if we said this is the laplace transform of some function , we know what that function is . this is this piece right here . i 'm just doing a little dotted line around it . this is the laplace transform of sine of t. and then if we draw a little box around this one right here , this is the laplace transform of cosine of t , g of s. so this is the laplace transform of sine of t , or we could write that this implies that f of t is equal to sine of t. you should recognize that one by now . and this implies that g of t , if we define this as the laplace transform of g , this means that g of t is equal to cosine of t. and , of course , when you take the inverse laplace transforms , you could take the 2 's out . so now what can we say ? we can now say that the -- let me write it this way -- the inverse -- so actually , let me write it this way . or , actually , a better thing to do , instead of taking the 2 out , so i can leave it nice and clean , we could , if we were to draw a box around this whole thing , and define this whole thing as f of s , then f of s is the laplace transform of 2 sine of t. i just wanted to include that 2 . i did n't want to leave that out and confuse the issue . i wanted a very pure f of s times g of s. so this expression right here is the product of the laplace transform of 2 sine of t , and the laplace transform of cosine of t. now , our convolution theorem told us this right here . that if we want to take the inverse laplace transform of the laplace transforms of two functions -- i know that sounds very confusing -- but you just kind of pattern match . you say , ok look , this thing that i had here , i could rewrite it as a product of two laplace transforms i can recognize . this right here is the laplace transform of 2 sine of t. this is the laplace transform of cosine of t. and we just wrote that as g of s , and f of s. so if i have an expression written like this , i can take the inverse laplace transform and it 'll be equal to the convolution of the original functions . it 'll be equal to the convolution of the inverse of g or the inverse of f. let me write it this way . i could write it like this . we know that f of t is equal to the inverse laplace transform of f of s. and we know that g -- i should have done it in a different color , but i 'll do g in green -- we know that g of t is equal to the inverse laplace transform of g of s. so we can rewrite the convolution theorem as the inverse -- and this might maybe confuse you more than help , but i 'll give my best shot . the inverse laplace transform of -- and i 'll try to stay true to the colors -- of f of s times g of s is equal to -- i 'm just restating this convolution theorem right here . this is equal to the convolution of the inverse laplace transform of f of s. so it 's equal to the convolution of the inverse laplace transform of f of s with the inverse laplace transform of g of s. with the inverse laplace transform of capital g , of g of s. i 'm not sure if that helps you or not , but if you go back to this example it might . this is f of s , this is f of s right here . 2 times -- i 'll do it in the light blue -- this is 2 over s squared plus 1 . that 's f of s in our example . and the g of s was s over s squared plus 1 . and all i got that from is i just broke this up into two things that i recognize . if i multiply this together , i get back to my original thing that i was trying to take the inverse laplace transform of . and so the convolution theorem just says that , ok , well , the inverse laplace transform of this is equal to the inverse laplace transform of 2 over s squared plus 1 , convoluted with the inverse laplace transform of our g of s , of s over s squared plus 1 . and we know what these things are . i already told them to you , but they should be somewhat second nature now . this is 2 times sine of t. you take the laplace transform of sine of t , you get 1 over s squared plus 1 , and then you multiply it by 2 , you get the 2 up there . and you 're going to have to convolute that with the inverse laplace transform of this thing here . and we already went over this . this is cosine of t. so our result so far -- let me be very clear . it 's always good to take a step back and just think about what we 're doing , much less why we 're doing it . but let 's see , the inverse laplace transform of this thing up in this top left corner , 2s over s squared plus 1 squared , which before we did what we 're doing now was very hard to figure out -- actually , this would be a curly bracket right here , but you get the idea -- is equal to this . it 's equal to 2 sine of t , convoluted with cosine of t. and you 're like , sal , throughout this whole process i 've already forgotten what it means to convolute two functions , so let 's convolute them . and i 'll just write the definition , or the definition we 're using of the convolution . that f convoluted with g -- it 's going to be a function of g. i 'll just write this short-hand -- is equal to the integral from 0 to t , of f of t minus tau , times g of tau , dtau . so 2 sine of t convoluted with cosine of t is equal to -- let me do a neutral color -- the integral from 0 to t , of 2 sine of t , minus tau , times the cosine of tau , dtau . now if you watched the very last video i made , i actually solved this , or i solved a very similar thing to this . if we take the 2 out we get 2 , times the integral from 0 to t , of sine of t minus tau , times the cosine of tau . i actually solved this in the previous video . this right here , this is the convolution of sine of t and cosine of t. it 's sine of t convoluted with cosine of t. and i show you in the previous video , just watch that video , where i introduce a convolution , that this thing right here is equal to 1/2t sine of t. now , if this thing is equal to 1/2 t sine of t , and i have to multiply it by 2 , then we get , our big result , that the inverse laplace transform of 2s over s squared plus 1 squared is equal to the convolution of 2 sine of t with cosine of t. which is just 2 times this thing here , which is 2 times 1/2 -- those cancel out -- so it equals t sine of t. and once you get the hang of it , you wo n't have to go through all of these steps . but the key is to recognize that this could be broken down as the products of two laplace transforms that you recognize . this could be broken down as the product of two laplace transforms we recognized . this is the laplace transform of 2 sine of t. this was the laplace transform of cosine of t. so the inverse laplace transform of our original thing , or original expression , is just the convolution of that with that . and if you watched the previous video , you 'd realize that actually calculating that convolution was no simple task , but it can be done . so you actually can get an integral form . even if it ca n't be done , you can get your answer , at least , in terms of some integral . so i have n't proven the convolution theorem to you just yet . i 'll do that in a future video . but hopefully , this gave you a little bit of a sense of how you can use it to actually take inverse laplace transforms . and remember , the reason why we 're learning to take inverse laplace transforms , and we have all of these tools to do it , is because that 's always that last step when you 're solving these differential equations , using your laplace transforms .
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so you actually can get an integral form . even if it ca n't be done , you can get your answer , at least , in terms of some integral . so i have n't proven the convolution theorem to you just yet . i 'll do that in a future video .
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so there 's a convolution theorem in case there is n't such a method that ca n't be computed by the former ?
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hey guys , so in the last video i was talking about how you can define a function whose graph is a plane , and moreover a plane that passes through a specified point and whose orientation you can somehow specify . and we ended up seeing how specifying that orientation comes down to certain partial derivative information . and first let me just kind of repeat what the conclusion was but i 'll put it in more abstract terms since i did it in a very specific example last time . so basically if you want some kind of function which gives you a plane that passes through a certain point . well first let 's see what that point is right . let 's say the point was x nought , y nought , and z nought . so these are constant values and this is my way of abstractly describing a single point in space using x nought to represent a constant x value , y nought to represent constant y value , that kind of thing . so what it is , is it 's gon na be some sort of other constant a multiplied by x minus x nought . so this white x here is the variable and then x nought is just a constant . now let me go ahead and make that parentheses there . then we add to that b multiplied by and then b is just some other constant just like a is some other constant multiplied by y minus y nought and then all of that you add z nought . and now if you were just presented this as it is it 's kind of a lot right , there 's five different constants going on . but really what this is saying is you want something where the partial derivative with respect to x is just some constant and you want to be able to specify what that constant is . and similarly , the partial derivative with respect to y is another constant . and you just want to ensure that this passes through this point x nought , y nought , z nought . and if you imagine plugging in x the variable equals x nought the constant this part goes to zero . similarly plugging in y nought the constant makes this part go to zero . so this is a way of specifying that when you evaluate the function x nought y nought equals z nought , and that 's what makes sure that the graph passes through that point . so with that said let 's start thinking about how you can find the tangent plane to a graph . and first of all let 's think about what that point is , how you specify such a point . instead of specifying any three numbers in space , because you have to make sure the point is somewhere on the graph , you instead only specify two . you 're basically gon na say what 's the x coordinate and in this case let 's say the x coordinate was one , and then the y coordinate which looks about like negative two . to make it easier i 'm just gon na say it is negative two . then the z coordinate is specified because this is a graph . z coordinate is forced to be whatever the output of the function is at one negative two . so this is gon na be whatever the output of our function is at one negative two . and f here , f is going to be whatever function gives us this graph . so maybe i should write down the actual function that i 'm using for this graph . in this case f which is a function of x and y is equal to three minus one third of x squared minus y squared . so this is the function that we 're using and you evaluate it at that point and this will give you your point in three dimensional space that our linear function , that our tangent plane has to pass through . so we can start writing out our linear function . we can say okay so our linear function has a function of x and y . it 's got ta make sure it goes through that one and that negative two , so this is gon na be some constant a that we 'll fill in in a moment , multiplied by x minus that one , plus and then b also a constant we 'll specify in a moment , times y minus that negative two , so it 's minus a negative two and then the thing that we add to it is f of one negative two . now let 's just go ahead and evaluate that . let 's say we plug in one to negative two . so if we go up here and we plug in three minus one third of if x equals one , one third of one squared , so that 's one third one squared , and then y is negative two . so that would be minus negative two squared . so that 's three minus a third minus four so the whole thing is equal to three minus four is negative one minus another third is negative four thirds . so that 's what we add to this entire thing . we add negative four thirds or maybe i should just kind of make clear the separation here . so this is our function but we do n't know what a and b are . those are things that we need to plug in . now the whole idea of the tangent plane is that the partial derivative with respect to x should match that of the original function . so if we go over to the graph here and start thinking about partial derivative information . if we want the partial derivative with respect to x then you imagine moving purely in the x direction here . this intersects the graph along some kind of curve and what the partial derivative with respect to x at this point tells you , is the slope of the tangent line , in that direction of that point . so that 's what the partial derivative with respect to x is telling you and what you want when you look at the tangent plane is that the tangent plane also has that same slope . if i lined things up here , you 'd want it also to have that same slope . so you can specify over here and say a . you want a to be equal to the partial derivative of the function with respect to x evaluated at this one negative two . evaluated at that special point , one negative two . and similarly b for pretty much the same reasons and i 'll draw it out here so let 's erase this line . so instead of intersecting it with that slice let 's see what movement in the y direction looks like . so in this case it looks like a very steep slope right because in this case the tangent line in that direction is a pretty steep slope and now when we bring in the tangent plane it should intersect with that constant x value plane along that same slope . made it kind of messy there but you can see the line formed by intersecting these two planes should be that desired tangent , and what that corresponds to in formulas is that this b which represents the partial derivative of l , l is the tangent plane function , that should be the same as if we took the partial derivative of f with respect to y at that point , at this point one negative two . and this is stuff that we can compute and that we can figure out . so let 's start plugging that in . first let me just copy this function because we 're gon na need it . now let 's go on down here . i 'm just gon na , let 's paste it down here in the bottom because that 's what we 'll need . so let 's compute the partial derivative of f with respect to x . so we look down here , the only place where x shows up is in this negative one third of x squared context so the partial derivative of f with respect to x is gon na be just the derivative of this little guy which is negative , we bring down the two , negative two thirds of x . so when we go ahead and plug in x equals one to see what it looks like when we evaluate at this point that 's just gon na be equal to negative two thirds . so that tells us that a is gon na have to be negative two thirds . now for similar reasons , let 's go ahead and compute the partial derivative with respect to y . we look down here , the only place that y shows up in the entire expression is this negative y squared . so the partial derivative of f with respect to y is equal to negative two y , negative two y . and now when we plug in y equals negative two what we get is negative two multiplied coincidentally by negative two , did n't have to be the case that those were the same , and that whole thing equals four . so the partial derivative of f with respect to y evaluated at this point one negative two is equal to four . so if we were to plug this information back up into our formula we would replace a with negative two thirds . it would say negative two thirds . and we would replace b with four , replace b with four . and that would give us the full formula , the full formula for the tangent plane . and this can be kind of a lot to look at at first because we have to specify the input point one negative two . and then we had to figure out where the function evaluates at that point . and then we had to figure out both of the partial derivatives with respect to x and with respect to y . but all in all , there 's not actually a lot to remember from how you go about computing this . looking at the graph actually makes things seem a lot more reasonable because each of those terms has an actual meaning . if we look at the one and negative two , that 's just telling us the input , the kind of x and y coordindates of the input and of course we have to evaluate that because that tells us the z coordinate that will put us on the graph corresponding to that point and then to get a tangent plane you just need to specify the two bits of partial differential information and that will tell you how this graph needs to be oriented , and once you start thinking of things in that way , geometrically , even though there 's a lot going on here , there 's five different numbers you have to put in , each one of them feels like of course you need that number otherwise you could n't specify a tangent plane . there 's kind of a lot of information required to put it on the appropriate spot . so with that i will see you next video .
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it would say negative two thirds . and we would replace b with four , replace b with four . and that would give us the full formula , the full formula for the tangent plane .
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do the constants a and b have anything to do with the vector normal to the plane ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that .
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as we know that the photon is massless then how the momentum is being conserved ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers .
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what if the frequency of the incident ray of light is exactly equal to the threshold frequency of the metal ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron .
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is work function same as ionization energy ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron .
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so what happens to the energy of the photon ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced .
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does the electron accept this energy ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron .
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and if it is massless then how is light energy affected by gravity ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced .
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would n't the combined energy be enough to create that photoelectron ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle .
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in a photoelectric cell where there is no voltage , what causes the ejected electrons to travel to the anode ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is .
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if so , then is there a possible velocity that a caesium photoelectron must achieve so that it can escape from the attractive force exerted by the piece of caesium ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron .
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therefore should n't the energy required to free an electron , be different for different shells ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again .
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then , when the work function of metallic caesium is given as 3.43 x 10^-19 j , is this the energy required to free an electron from the outermost shell or the innermost , or is it some kind of average value ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that .
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if no , where does the mass of the photon go once it hits an electron in the metal sheet ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 .
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how did you get the value of plancks constant ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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is it essential to use monochromatic light for photoelectric effect ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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or any light can be used for photoelectric effect ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced .
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we learned that a photon with a high enough energy is capable of exciting electrons of an atom to a higher level of energy , why is the electron completely knocked loose in this case ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons .
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does photoelectric effect also happens in the case of gases ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that .
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what is the charge of photon ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved .
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in particle sense of proton do it moves like wave as you can see above ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved .
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mostly particle move in straight line ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that .
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how can a photon be massless ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron .
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what if the energy of the photon is less than the work function ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that .
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when the photon hits the electron , and when the electron is free- where does it go ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about .
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are those waves actually photoelectrons being ejected or something entirely different ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons .
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the difference between spectroscopy and photoelectric effect is that in the former electrons are only transitioned between energy levels and emit photon whereas in photoelectric effect the entire electron is emitted ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron .
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if i keep giving the minimum frequency needed to eject electron , will all the electron from the atom or the element fall out ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 .
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after you convert the 525bnm to 525e-9 what is the unit for the 525e-9 at that point ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that .
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why is a photon massless ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein .
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but if you were to take a ridged piece of metal , broken up and busted , yet still shiny enough to bounce off light , would it make the light ridged physically ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein .
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what exactly is the intensity of light ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced .
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what happens to the proton involved in causing the emission of the electron ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron .
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so what exactly is the difference between the ionisation energy of an element and the work function of an element ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved .
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if light is in particle form , how does it have a wavelength ?
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- sometimes light seems to act as a wave , and sometimes light seems to act as a particle . and , an example of this , would be the photoelectric effect , as described by einstein . so let 's say you had a piece of metal , and we know the metal has electrons . i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle . so i 'm gon na draw in a particle of light which we call a photon , so this is massless , and the photon is going to hit this electron , and if the photon has enough energy , it can free the electron , right ? so we can knock it loose , and so let me go ahead and show that . so here , we 're showing the electron being knocked loose and so the electron 's moving in , let 's just say , this direction , with some velocity , v , and if the electron has mass , m , we know that there 's a kinetic energy . the kinetic energy of the electron would be equal to one half mv squared . this freed electron is usually referred to now as a photoelectron . so one photon creates one photoelectron . so one particle hits another particle . and , if you think about this in terms of classical physics , you could think about energy being conserved . so the energy of the photon , the energy that went in , so let me go ahead and write this here , so the energy of the photon , the energy that went in , what happened to that energy ? some of that energy was needed to free the electron . so the electron was bound , and some of the energy freed the electron . i 'm gon na call that e naught , the energy that freed the electron , and then the rest of that energy must have gone into the kinetic energy of the electron , and so we can write here kinetic energy of the photoelectron that was produced . so , kinetic energy of the photoelectron . so let 's say you wanted to solve for the kinetic energy of that photoelectron . so that would be very simple , it would just be kinetic energy would be equal to the energy of the photon , energy of the photon , minus the energy that was necessary to free the electron from the metallic surface . and this e naught , here i 'm calling it e naught , you might see it written differently , a different symbol , but this is the work function . let me go ahead and write work function here , and the work function is different for every kind of metal . so , it 's the minimum amount of energy that 's necessary to free the electron , and so obviously that 's going to be different depending on what metal you 're talking about . all right , let 's do a problem . now that we understand the general idea of the photoelectric effect , let 's look at what this problem asks us . so the problem says , `` if a photon of wavelength `` 525 nm hits metallic cesium ... '' and so here 's the work function for metallic cesium . `` what is the velocity of the photoelectron produced ? '' so they want to know the velocity of the photoelectron produced , which we know is hiding in the kinetic energy right here , and we also know what the work function is . so we know what e naught is here . what we do n't know is the energy of the photon so that 's what we need to calculate first . and so the energy of the photon , energy of the photon , is equal to h , which is planck 's constant , times the frequency , which is usually symbolized by nu . so , we got the frequency , but they gave us the wavelength in the problem here . they gave us wavelength , so we need to relate frequency to wavelength , and that 's related by c , which is the speed of light , is equal to lambda times nu . so , c is the speed of light , and that 's equal to the frequency times the wavelength . so we can substitute n for the frequency , all right , 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength . the frequency is equal to speed of light over lambda , so we can plug that into here , and so now we have the energy of the photon is equal to hc over lambda , and we can plug in those numbers . h is planck 's constant , which is 6.626 times 10 to the negative 34 . so , times 10 to the negative 34 here . c is the speed of light , which is 2.998 times 10 to the 8th meters over seconds , and all over lambda . lambda is the wavelength . that 's 525 nanometers . so 525 times 10 to the negative 9th meters . all right , so let 's get out our calculator and calculate the energy of the photon here . so , let 's go ahead and do that math , so we have 6.626 times 10 to the negative 34 , and we 're going to multiply that number by the speed of light , 2.998 times 10 to the 8th , and we get that number . we 're gon na divide it by the wavelength , 525 times 10 to the negative 9 , and we get 3.78 times 10 to the negative 19 . so , let me go ahead and write that down here . 3.78 times 10 to the negative 19 , and if you did you units up here , you would get joules , and so let 's think about this number for a second , 3.78 times 10 to the negative 19 is the energy of the photon . and that energy of the photon is greater than the work function , which means that that 's a high-energy photon . it 's able to knock the electron free , 'cause remember , this number right here , is the minimum amount of energy needed to free the electron and so we 've exceeded that minimum amount of energy , and so we will produce a photoelectron . so , this photon is high-energy enough to produce a photoelectron . so let 's go ahead and find the kinetic energy of the photoelectron that 's produced . so we 're gon na use this equation right up here . so let me just go and get some more room , and i will rewrite that equation . so we have the kinetic energy of the photoelectron , kinetic energy of the photoelectron , is equal to the energy of the photon , energy of the photon , minus the work function . so let 's plug in our numbers . the energy of the photon was 3.78 times 10 the negative 19 joules , and then the work function is right up here again , it 's 3.43 , so minus 3.43 times 10 to the negative 19 joules . so let 's get out the calculator again . so , from that we 're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20 . so let 's go ahead and write that . this is equal to 3.5 times 10 to the negative 20 joules . this is equal to the kinetic energy of the photoelectron , and we know that kinetic energy is equal to one half mv squared . the problem asked us to solve for the velocity of the photoelectron . so all we have to do is plug in the mass of an electron , which is 9.11 times 10 to the negative 31st kilograms , times v squared . this is equal to 3.5 times 10 to the negative 20 . so , let 's do that math . so we take 3.5 times 10 to the negative 20 , we multiply that by 2 , and then we divide by the mass of an electron , 9.11 times 10 to the negative 31st , and this gives us that number , which we need to take the square root of . so , square root of our answer gives us the velocity of the electron , 2.8 times 10 to the 5th . so if you look at your decimal place here , this 'll be one , two , three , four , five , so 2.8 times 10 to the 5th meters per second . so here 's the velocity of the photoelectron produced , 2.8 times 10 to the 5th meters per second , and if you increased the intensity of this light , so you had more photons , they would produce more photoelectrons . so one photon knocks out one photoelectron if it has enough energy to do so . so let 's think about this same problem , but let 's change the wavelength . so , what if your wavelength changed to 625 nanometers . so what would happen now ? well , to save time , i wo n't do the calculation , but all we would have to do is plug in 625 up here . so instead of 525 , just plug in 625 to calculate your energy , and if you did that , so if you used 625 times 10 to the negative 9 here , i 'll go ahead and give you the answer just to save some time , you would get 3.2 times 10 to the negative 19 joules . and that is lower than the work function . so let me go ahead and highlight that here . so this number is not as high as the work function . the work function was how much energy we needed to free that electron , and since this is lower than the work function that means we do not get a photoelectron . so , you have to have a high enough energy photon in order to produce a photoelectron . it would n't even matter if we increased the intensity . so if we had more and more and more of these photons at this wavelength , we still would n't produce any photoelectrons . and so , this is the idea of the photoelectric effect , which is best explained by thinking about light as a particle .
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i 'm gon na go ahead and draw one electron in here , and this electron is bound to the metal because it 's attracted to the positive charges in the nucleus . if you shine a light on the metal , so the right kind of light with the right kind of frequency , you can actually knock some of those electrons loose , which causes a current of electrons to flow . so this is kind of like a collision between two particles , if we think about light as being a particle .
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( about ) : if shining a light of the right frequency and wavelength on a substance can knock electrons loose , therefore creating ions , would n't that change the properties of the substance ?
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