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- this is a rectangle . and yes , this is a rectangle . let 's measure the green line below using both rectangles and a centimeter ruler . so , we have this green line and we 're going to measure it in terms of rectangles and we 're going to measure it on this ruler . and you immediately see the green line is equal to one rectangle long , but if we measure it in centimeters it 's equal to six centimeters long . so we can say the line is one rectangle long , but it 's six centimeters long and that makes sense because the rectangle is much more than a centimeter . actually , the length of a rectangle is six centimeters . so , the length of a rectangle is longer than a centimeter . you see that . this is the length of the rectangle , it 's the same as the line , while centimeter is this length . the rectangle is actually six centimeters long . so it takes blank rectangles , than centimeters , to measure the line . well , since the rectangle is longer , it takes fewer rectangles than centimeters to measure the line . so it 's fewer rectangles than centimeters and you see that . fewer rectangles , one is fewer than six . it took one , it only took one rectangle to measure the line because a rectangle 's so long and it took six centimeters to measure the line . so it took fewer rectangles than centimeters to measure the line . let 's keep going . alright , so we have . this is a rectangle . yes , that is a rectangle . let 's measure the blue line below using both a rectangle and a centimeter ruler . so it 's a similar idea . so , when we look at this blue line , the line is now two of these green rectangles long . so it is two rectangles long . now , in terms of centimeters , it is four , four centimeters long . four centimeters long . the length of each rectangle is blank than a centimeter . well , each rectangle is more than a centimeter . it 's longer . you see , each rectangle is actually two centimeters . so , it 's longer than a centimeter . so it takes blank rectangles than centimeters to measure the line . well , since each rectangle is longer , it 's going to take fewer rectangles to measure the line . so we 'll select fewer right over there and you see that over here . it only took two rectangles to measure the line , but it took four centimeters . two is fewer than four . let 's see . let 's do one more of these . these are a lot of fun . alright , this is a rectangle . let 's measure the green line below . same idea . the line is four rectangles long . the line is seven centimeters long . the length of each rectangle is shorter , sorry , longer than a centimeter . you see that right here . each rectangle is more than one centimeter . so it takes fewer rectangles than centimeters to measure the line . so , we 've seen this , we 've seen this multiple times .
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now , in terms of centimeters , it is four , four centimeters long . four centimeters long . the length of each rectangle is blank than a centimeter .
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at 10 ; 17 , how is it 8 centimeters ?
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- this is a rectangle . and yes , this is a rectangle . let 's measure the green line below using both rectangles and a centimeter ruler . so , we have this green line and we 're going to measure it in terms of rectangles and we 're going to measure it on this ruler . and you immediately see the green line is equal to one rectangle long , but if we measure it in centimeters it 's equal to six centimeters long . so we can say the line is one rectangle long , but it 's six centimeters long and that makes sense because the rectangle is much more than a centimeter . actually , the length of a rectangle is six centimeters . so , the length of a rectangle is longer than a centimeter . you see that . this is the length of the rectangle , it 's the same as the line , while centimeter is this length . the rectangle is actually six centimeters long . so it takes blank rectangles , than centimeters , to measure the line . well , since the rectangle is longer , it takes fewer rectangles than centimeters to measure the line . so it 's fewer rectangles than centimeters and you see that . fewer rectangles , one is fewer than six . it took one , it only took one rectangle to measure the line because a rectangle 's so long and it took six centimeters to measure the line . so it took fewer rectangles than centimeters to measure the line . let 's keep going . alright , so we have . this is a rectangle . yes , that is a rectangle . let 's measure the blue line below using both a rectangle and a centimeter ruler . so it 's a similar idea . so , when we look at this blue line , the line is now two of these green rectangles long . so it is two rectangles long . now , in terms of centimeters , it is four , four centimeters long . four centimeters long . the length of each rectangle is blank than a centimeter . well , each rectangle is more than a centimeter . it 's longer . you see , each rectangle is actually two centimeters . so , it 's longer than a centimeter . so it takes blank rectangles than centimeters to measure the line . well , since each rectangle is longer , it 's going to take fewer rectangles to measure the line . so we 'll select fewer right over there and you see that over here . it only took two rectangles to measure the line , but it took four centimeters . two is fewer than four . let 's see . let 's do one more of these . these are a lot of fun . alright , this is a rectangle . let 's measure the green line below . same idea . the line is four rectangles long . the line is seven centimeters long . the length of each rectangle is shorter , sorry , longer than a centimeter . you see that right here . each rectangle is more than one centimeter . so it takes fewer rectangles than centimeters to measure the line . so , we 've seen this , we 've seen this multiple times .
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four centimeters long . the length of each rectangle is blank than a centimeter . well , each rectangle is more than a centimeter .
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why do people use inch , centimeter or length ?
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- this is a rectangle . and yes , this is a rectangle . let 's measure the green line below using both rectangles and a centimeter ruler . so , we have this green line and we 're going to measure it in terms of rectangles and we 're going to measure it on this ruler . and you immediately see the green line is equal to one rectangle long , but if we measure it in centimeters it 's equal to six centimeters long . so we can say the line is one rectangle long , but it 's six centimeters long and that makes sense because the rectangle is much more than a centimeter . actually , the length of a rectangle is six centimeters . so , the length of a rectangle is longer than a centimeter . you see that . this is the length of the rectangle , it 's the same as the line , while centimeter is this length . the rectangle is actually six centimeters long . so it takes blank rectangles , than centimeters , to measure the line . well , since the rectangle is longer , it takes fewer rectangles than centimeters to measure the line . so it 's fewer rectangles than centimeters and you see that . fewer rectangles , one is fewer than six . it took one , it only took one rectangle to measure the line because a rectangle 's so long and it took six centimeters to measure the line . so it took fewer rectangles than centimeters to measure the line . let 's keep going . alright , so we have . this is a rectangle . yes , that is a rectangle . let 's measure the blue line below using both a rectangle and a centimeter ruler . so it 's a similar idea . so , when we look at this blue line , the line is now two of these green rectangles long . so it is two rectangles long . now , in terms of centimeters , it is four , four centimeters long . four centimeters long . the length of each rectangle is blank than a centimeter . well , each rectangle is more than a centimeter . it 's longer . you see , each rectangle is actually two centimeters . so , it 's longer than a centimeter . so it takes blank rectangles than centimeters to measure the line . well , since each rectangle is longer , it 's going to take fewer rectangles to measure the line . so we 'll select fewer right over there and you see that over here . it only took two rectangles to measure the line , but it took four centimeters . two is fewer than four . let 's see . let 's do one more of these . these are a lot of fun . alright , this is a rectangle . let 's measure the green line below . same idea . the line is four rectangles long . the line is seven centimeters long . the length of each rectangle is shorter , sorry , longer than a centimeter . you see that right here . each rectangle is more than one centimeter . so it takes fewer rectangles than centimeters to measure the line . so , we 've seen this , we 've seen this multiple times .
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you see , each rectangle is actually two centimeters . so , it 's longer than a centimeter . so it takes blank rectangles than centimeters to measure the line .
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why do people use inch , centimeter , of lenght ?
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- this is a rectangle . and yes , this is a rectangle . let 's measure the green line below using both rectangles and a centimeter ruler . so , we have this green line and we 're going to measure it in terms of rectangles and we 're going to measure it on this ruler . and you immediately see the green line is equal to one rectangle long , but if we measure it in centimeters it 's equal to six centimeters long . so we can say the line is one rectangle long , but it 's six centimeters long and that makes sense because the rectangle is much more than a centimeter . actually , the length of a rectangle is six centimeters . so , the length of a rectangle is longer than a centimeter . you see that . this is the length of the rectangle , it 's the same as the line , while centimeter is this length . the rectangle is actually six centimeters long . so it takes blank rectangles , than centimeters , to measure the line . well , since the rectangle is longer , it takes fewer rectangles than centimeters to measure the line . so it 's fewer rectangles than centimeters and you see that . fewer rectangles , one is fewer than six . it took one , it only took one rectangle to measure the line because a rectangle 's so long and it took six centimeters to measure the line . so it took fewer rectangles than centimeters to measure the line . let 's keep going . alright , so we have . this is a rectangle . yes , that is a rectangle . let 's measure the blue line below using both a rectangle and a centimeter ruler . so it 's a similar idea . so , when we look at this blue line , the line is now two of these green rectangles long . so it is two rectangles long . now , in terms of centimeters , it is four , four centimeters long . four centimeters long . the length of each rectangle is blank than a centimeter . well , each rectangle is more than a centimeter . it 's longer . you see , each rectangle is actually two centimeters . so , it 's longer than a centimeter . so it takes blank rectangles than centimeters to measure the line . well , since each rectangle is longer , it 's going to take fewer rectangles to measure the line . so we 'll select fewer right over there and you see that over here . it only took two rectangles to measure the line , but it took four centimeters . two is fewer than four . let 's see . let 's do one more of these . these are a lot of fun . alright , this is a rectangle . let 's measure the green line below . same idea . the line is four rectangles long . the line is seven centimeters long . the length of each rectangle is shorter , sorry , longer than a centimeter . you see that right here . each rectangle is more than one centimeter . so it takes fewer rectangles than centimeters to measure the line . so , we 've seen this , we 've seen this multiple times .
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alright , this is a rectangle . let 's measure the green line below . same idea .
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what is the most lengths you can measure with different units at once ?
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- this is a rectangle . and yes , this is a rectangle . let 's measure the green line below using both rectangles and a centimeter ruler . so , we have this green line and we 're going to measure it in terms of rectangles and we 're going to measure it on this ruler . and you immediately see the green line is equal to one rectangle long , but if we measure it in centimeters it 's equal to six centimeters long . so we can say the line is one rectangle long , but it 's six centimeters long and that makes sense because the rectangle is much more than a centimeter . actually , the length of a rectangle is six centimeters . so , the length of a rectangle is longer than a centimeter . you see that . this is the length of the rectangle , it 's the same as the line , while centimeter is this length . the rectangle is actually six centimeters long . so it takes blank rectangles , than centimeters , to measure the line . well , since the rectangle is longer , it takes fewer rectangles than centimeters to measure the line . so it 's fewer rectangles than centimeters and you see that . fewer rectangles , one is fewer than six . it took one , it only took one rectangle to measure the line because a rectangle 's so long and it took six centimeters to measure the line . so it took fewer rectangles than centimeters to measure the line . let 's keep going . alright , so we have . this is a rectangle . yes , that is a rectangle . let 's measure the blue line below using both a rectangle and a centimeter ruler . so it 's a similar idea . so , when we look at this blue line , the line is now two of these green rectangles long . so it is two rectangles long . now , in terms of centimeters , it is four , four centimeters long . four centimeters long . the length of each rectangle is blank than a centimeter . well , each rectangle is more than a centimeter . it 's longer . you see , each rectangle is actually two centimeters . so , it 's longer than a centimeter . so it takes blank rectangles than centimeters to measure the line . well , since each rectangle is longer , it 's going to take fewer rectangles to measure the line . so we 'll select fewer right over there and you see that over here . it only took two rectangles to measure the line , but it took four centimeters . two is fewer than four . let 's see . let 's do one more of these . these are a lot of fun . alright , this is a rectangle . let 's measure the green line below . same idea . the line is four rectangles long . the line is seven centimeters long . the length of each rectangle is shorter , sorry , longer than a centimeter . you see that right here . each rectangle is more than one centimeter . so it takes fewer rectangles than centimeters to measure the line . so , we 've seen this , we 've seen this multiple times .
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now , in terms of centimeters , it is four , four centimeters long . four centimeters long . the length of each rectangle is blank than a centimeter .
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how many centimeters are in a inch ?
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- this is a rectangle . and yes , this is a rectangle . let 's measure the green line below using both rectangles and a centimeter ruler . so , we have this green line and we 're going to measure it in terms of rectangles and we 're going to measure it on this ruler . and you immediately see the green line is equal to one rectangle long , but if we measure it in centimeters it 's equal to six centimeters long . so we can say the line is one rectangle long , but it 's six centimeters long and that makes sense because the rectangle is much more than a centimeter . actually , the length of a rectangle is six centimeters . so , the length of a rectangle is longer than a centimeter . you see that . this is the length of the rectangle , it 's the same as the line , while centimeter is this length . the rectangle is actually six centimeters long . so it takes blank rectangles , than centimeters , to measure the line . well , since the rectangle is longer , it takes fewer rectangles than centimeters to measure the line . so it 's fewer rectangles than centimeters and you see that . fewer rectangles , one is fewer than six . it took one , it only took one rectangle to measure the line because a rectangle 's so long and it took six centimeters to measure the line . so it took fewer rectangles than centimeters to measure the line . let 's keep going . alright , so we have . this is a rectangle . yes , that is a rectangle . let 's measure the blue line below using both a rectangle and a centimeter ruler . so it 's a similar idea . so , when we look at this blue line , the line is now two of these green rectangles long . so it is two rectangles long . now , in terms of centimeters , it is four , four centimeters long . four centimeters long . the length of each rectangle is blank than a centimeter . well , each rectangle is more than a centimeter . it 's longer . you see , each rectangle is actually two centimeters . so , it 's longer than a centimeter . so it takes blank rectangles than centimeters to measure the line . well , since each rectangle is longer , it 's going to take fewer rectangles to measure the line . so we 'll select fewer right over there and you see that over here . it only took two rectangles to measure the line , but it took four centimeters . two is fewer than four . let 's see . let 's do one more of these . these are a lot of fun . alright , this is a rectangle . let 's measure the green line below . same idea . the line is four rectangles long . the line is seven centimeters long . the length of each rectangle is shorter , sorry , longer than a centimeter . you see that right here . each rectangle is more than one centimeter . so it takes fewer rectangles than centimeters to measure the line . so , we 've seen this , we 've seen this multiple times .
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- this is a rectangle . and yes , this is a rectangle .
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how many different measuring systems are there ?
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- this is a rectangle . and yes , this is a rectangle . let 's measure the green line below using both rectangles and a centimeter ruler . so , we have this green line and we 're going to measure it in terms of rectangles and we 're going to measure it on this ruler . and you immediately see the green line is equal to one rectangle long , but if we measure it in centimeters it 's equal to six centimeters long . so we can say the line is one rectangle long , but it 's six centimeters long and that makes sense because the rectangle is much more than a centimeter . actually , the length of a rectangle is six centimeters . so , the length of a rectangle is longer than a centimeter . you see that . this is the length of the rectangle , it 's the same as the line , while centimeter is this length . the rectangle is actually six centimeters long . so it takes blank rectangles , than centimeters , to measure the line . well , since the rectangle is longer , it takes fewer rectangles than centimeters to measure the line . so it 's fewer rectangles than centimeters and you see that . fewer rectangles , one is fewer than six . it took one , it only took one rectangle to measure the line because a rectangle 's so long and it took six centimeters to measure the line . so it took fewer rectangles than centimeters to measure the line . let 's keep going . alright , so we have . this is a rectangle . yes , that is a rectangle . let 's measure the blue line below using both a rectangle and a centimeter ruler . so it 's a similar idea . so , when we look at this blue line , the line is now two of these green rectangles long . so it is two rectangles long . now , in terms of centimeters , it is four , four centimeters long . four centimeters long . the length of each rectangle is blank than a centimeter . well , each rectangle is more than a centimeter . it 's longer . you see , each rectangle is actually two centimeters . so , it 's longer than a centimeter . so it takes blank rectangles than centimeters to measure the line . well , since each rectangle is longer , it 's going to take fewer rectangles to measure the line . so we 'll select fewer right over there and you see that over here . it only took two rectangles to measure the line , but it took four centimeters . two is fewer than four . let 's see . let 's do one more of these . these are a lot of fun . alright , this is a rectangle . let 's measure the green line below . same idea . the line is four rectangles long . the line is seven centimeters long . the length of each rectangle is shorter , sorry , longer than a centimeter . you see that right here . each rectangle is more than one centimeter . so it takes fewer rectangles than centimeters to measure the line . so , we 've seen this , we 've seen this multiple times .
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let 's see . let 's do one more of these . these are a lot of fun .
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how many sqare meter in one brase ?
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so here i 'm gon na talk about the gradient . and in this video , i 'm only gon na describe how you compute the gradient , and in the next couple ones i 'm gon na give the geometric interpretation . and i hate doing this , i hate showing the computation before the geometric intuition since usually it should go the other way around , but the gradient is one of those weird things where the way that you compute it actually seems kind of unrelated to the intuition and you 'll see that . we 'll connect them in the next few videos . but to do that , we need to know what both of them actually are . so on the computation side of things , let 's say you have some sort of function . and i 'm just gon na make it a two-variable function . and let 's say it 's f of x , y , equals x-squared sine of y . the gradient is a way of packing together all the partial derivative information of a function . so let 's just start by computing the partial derivatives of this guy . so partial of f with respect to x is equal to , so we look at this and we consider x the variable and y the constant . well in that case sine of y is also a constant . as far as x is concerned , the derivative of x is 2x so we see that this will be 2x times that constant sine of y , sine of y . whereas the partial derivative with respect to y . now we look up here and we say x is considered a constant so x-squared is also considered a constant so this is just a constant times sine of y , so that 's gon na equal that same constant times the cosine of y , which is the derivative of sine . so now what the gradient does is it just puts both of these together in a vector . and specifically , maybe i 'll change colors here , you denote it with a little upside-down triangle . the name of that symbol is nabla , but you often just pronounce it del , you 'd say del f or gradient of f. and what this equals is a vector that has those two partial derivatives in it . so the first one is the partial derivative with respect to x , to x times sine of y . and the bottom one , partial derivative with respect to y x-squared cosine of y . and notice , maybe i should emphasize , this is actually a vector-valued function . so maybe i 'll give it a little bit more room here and emphasize that it 's got an x and a y . this is a function that takes in a point in two-dimensional space and outputs a two-dimensional vector . so you could also imagine doing this with three different variables . then you would have three partial derivatives , and a three-dimensional output . and the way you might write this more generally is we could go down here and say the gradient of any function is equal to a vector with its partial derivatives . partial of f with respect to x , and partial of f with respect to y . and in some sense , we call these partial derivatives . i like to think as the gradient as the full derivative cuz it kind of captures all of the information that you need . so a very helpful mnemonic device with the gradient is to think about this triangle , this nabla symbol as being a vector full of partial derivative operators . and by operator , i just mean like partial with respect to x , something where you could give it a function , and it gives you another function . so you give this guy the function f and it gives you this expression , this multi-variable function as a result . so the nabla symbol is this vector full of different partial derivative operators . and in this case it might just be two of them , and this is kind of a weird thing because it 's like what , this is a vector , it 's got like operators in it , that 's not what i thought vectors do . but you can kind of see where it 's going . it 's really just , you can think of it as a memory trick , but in some sense it 's a little bit deeper than that . and really when you take this triangle and you say ok let 's take this triangle and you can kind of imagine multiplying it by f , really it 's like an operator taking in this function and it 's gon na give you another function . it 's like you take this triangle and you put an f in front of it , and you can imagine , like this part gets multipled , quote unquote multiplied with f , this part gets quote unquote multiplied with f but really you 're just saying you take the partial derivative with respect to x and then with y , and on and on . and the reason for doing this , this symbol comes up a lot in other contexts . there are two other operators that you 're gon na learn about called the divergence and the curl . we 'll get to those later , all in due time . but it 's useful to think about this vector-ish thing of partial derivatives . and i mean one weird thing about it , you could say ok so this nabla symbol is a vector of partial derivative operators . what 's its dimension ? and it 's like how many dimensions do you got ? because if you had a three-dimensional function that would mean that you should treat this like it 's got three different operators as part of it . and you know i 'd kinda , finish this off down here , and if you had something that was 100-dimensional it would have 100 different operators in it and that 's fine . it 's really just again , kind of a memory trick . so with that , that 's how you compute the gradient . not too much too it , it 's pretty much just partial derivatives , but you smack em into a vector where it gets fun and where it gets interesting is with the geometric interpretation . i 'll get to that in the next couple videos . it 's also a super important tool for something called the directional derivative . so you 've got a lot of fun stuff ahead .
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as far as x is concerned , the derivative of x is 2x so we see that this will be 2x times that constant sine of y , sine of y . whereas the partial derivative with respect to y . now we look up here and we say x is considered a constant so x-squared is also considered a constant so this is just a constant times sine of y , so that 's gon na equal that same constant times the cosine of y , which is the derivative of sine .
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5 , when we take the derivative of f in respect to x , therefore take y = sin ( y ) as a constant , why does n't it disappear in the derivative ?
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so here i 'm gon na talk about the gradient . and in this video , i 'm only gon na describe how you compute the gradient , and in the next couple ones i 'm gon na give the geometric interpretation . and i hate doing this , i hate showing the computation before the geometric intuition since usually it should go the other way around , but the gradient is one of those weird things where the way that you compute it actually seems kind of unrelated to the intuition and you 'll see that . we 'll connect them in the next few videos . but to do that , we need to know what both of them actually are . so on the computation side of things , let 's say you have some sort of function . and i 'm just gon na make it a two-variable function . and let 's say it 's f of x , y , equals x-squared sine of y . the gradient is a way of packing together all the partial derivative information of a function . so let 's just start by computing the partial derivatives of this guy . so partial of f with respect to x is equal to , so we look at this and we consider x the variable and y the constant . well in that case sine of y is also a constant . as far as x is concerned , the derivative of x is 2x so we see that this will be 2x times that constant sine of y , sine of y . whereas the partial derivative with respect to y . now we look up here and we say x is considered a constant so x-squared is also considered a constant so this is just a constant times sine of y , so that 's gon na equal that same constant times the cosine of y , which is the derivative of sine . so now what the gradient does is it just puts both of these together in a vector . and specifically , maybe i 'll change colors here , you denote it with a little upside-down triangle . the name of that symbol is nabla , but you often just pronounce it del , you 'd say del f or gradient of f. and what this equals is a vector that has those two partial derivatives in it . so the first one is the partial derivative with respect to x , to x times sine of y . and the bottom one , partial derivative with respect to y x-squared cosine of y . and notice , maybe i should emphasize , this is actually a vector-valued function . so maybe i 'll give it a little bit more room here and emphasize that it 's got an x and a y . this is a function that takes in a point in two-dimensional space and outputs a two-dimensional vector . so you could also imagine doing this with three different variables . then you would have three partial derivatives , and a three-dimensional output . and the way you might write this more generally is we could go down here and say the gradient of any function is equal to a vector with its partial derivatives . partial of f with respect to x , and partial of f with respect to y . and in some sense , we call these partial derivatives . i like to think as the gradient as the full derivative cuz it kind of captures all of the information that you need . so a very helpful mnemonic device with the gradient is to think about this triangle , this nabla symbol as being a vector full of partial derivative operators . and by operator , i just mean like partial with respect to x , something where you could give it a function , and it gives you another function . so you give this guy the function f and it gives you this expression , this multi-variable function as a result . so the nabla symbol is this vector full of different partial derivative operators . and in this case it might just be two of them , and this is kind of a weird thing because it 's like what , this is a vector , it 's got like operators in it , that 's not what i thought vectors do . but you can kind of see where it 's going . it 's really just , you can think of it as a memory trick , but in some sense it 's a little bit deeper than that . and really when you take this triangle and you say ok let 's take this triangle and you can kind of imagine multiplying it by f , really it 's like an operator taking in this function and it 's gon na give you another function . it 's like you take this triangle and you put an f in front of it , and you can imagine , like this part gets multipled , quote unquote multiplied with f , this part gets quote unquote multiplied with f but really you 're just saying you take the partial derivative with respect to x and then with y , and on and on . and the reason for doing this , this symbol comes up a lot in other contexts . there are two other operators that you 're gon na learn about called the divergence and the curl . we 'll get to those later , all in due time . but it 's useful to think about this vector-ish thing of partial derivatives . and i mean one weird thing about it , you could say ok so this nabla symbol is a vector of partial derivative operators . what 's its dimension ? and it 's like how many dimensions do you got ? because if you had a three-dimensional function that would mean that you should treat this like it 's got three different operators as part of it . and you know i 'd kinda , finish this off down here , and if you had something that was 100-dimensional it would have 100 different operators in it and that 's fine . it 's really just again , kind of a memory trick . so with that , that 's how you compute the gradient . not too much too it , it 's pretty much just partial derivatives , but you smack em into a vector where it gets fun and where it gets interesting is with the geometric interpretation . i 'll get to that in the next couple videos . it 's also a super important tool for something called the directional derivative . so you 've got a lot of fun stuff ahead .
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so here i 'm gon na talk about the gradient . and in this video , i 'm only gon na describe how you compute the gradient , and in the next couple ones i 'm gon na give the geometric interpretation .
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so the output of the gradient is mutlidimensional ?
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so here i 'm gon na talk about the gradient . and in this video , i 'm only gon na describe how you compute the gradient , and in the next couple ones i 'm gon na give the geometric interpretation . and i hate doing this , i hate showing the computation before the geometric intuition since usually it should go the other way around , but the gradient is one of those weird things where the way that you compute it actually seems kind of unrelated to the intuition and you 'll see that . we 'll connect them in the next few videos . but to do that , we need to know what both of them actually are . so on the computation side of things , let 's say you have some sort of function . and i 'm just gon na make it a two-variable function . and let 's say it 's f of x , y , equals x-squared sine of y . the gradient is a way of packing together all the partial derivative information of a function . so let 's just start by computing the partial derivatives of this guy . so partial of f with respect to x is equal to , so we look at this and we consider x the variable and y the constant . well in that case sine of y is also a constant . as far as x is concerned , the derivative of x is 2x so we see that this will be 2x times that constant sine of y , sine of y . whereas the partial derivative with respect to y . now we look up here and we say x is considered a constant so x-squared is also considered a constant so this is just a constant times sine of y , so that 's gon na equal that same constant times the cosine of y , which is the derivative of sine . so now what the gradient does is it just puts both of these together in a vector . and specifically , maybe i 'll change colors here , you denote it with a little upside-down triangle . the name of that symbol is nabla , but you often just pronounce it del , you 'd say del f or gradient of f. and what this equals is a vector that has those two partial derivatives in it . so the first one is the partial derivative with respect to x , to x times sine of y . and the bottom one , partial derivative with respect to y x-squared cosine of y . and notice , maybe i should emphasize , this is actually a vector-valued function . so maybe i 'll give it a little bit more room here and emphasize that it 's got an x and a y . this is a function that takes in a point in two-dimensional space and outputs a two-dimensional vector . so you could also imagine doing this with three different variables . then you would have three partial derivatives , and a three-dimensional output . and the way you might write this more generally is we could go down here and say the gradient of any function is equal to a vector with its partial derivatives . partial of f with respect to x , and partial of f with respect to y . and in some sense , we call these partial derivatives . i like to think as the gradient as the full derivative cuz it kind of captures all of the information that you need . so a very helpful mnemonic device with the gradient is to think about this triangle , this nabla symbol as being a vector full of partial derivative operators . and by operator , i just mean like partial with respect to x , something where you could give it a function , and it gives you another function . so you give this guy the function f and it gives you this expression , this multi-variable function as a result . so the nabla symbol is this vector full of different partial derivative operators . and in this case it might just be two of them , and this is kind of a weird thing because it 's like what , this is a vector , it 's got like operators in it , that 's not what i thought vectors do . but you can kind of see where it 's going . it 's really just , you can think of it as a memory trick , but in some sense it 's a little bit deeper than that . and really when you take this triangle and you say ok let 's take this triangle and you can kind of imagine multiplying it by f , really it 's like an operator taking in this function and it 's gon na give you another function . it 's like you take this triangle and you put an f in front of it , and you can imagine , like this part gets multipled , quote unquote multiplied with f , this part gets quote unquote multiplied with f but really you 're just saying you take the partial derivative with respect to x and then with y , and on and on . and the reason for doing this , this symbol comes up a lot in other contexts . there are two other operators that you 're gon na learn about called the divergence and the curl . we 'll get to those later , all in due time . but it 's useful to think about this vector-ish thing of partial derivatives . and i mean one weird thing about it , you could say ok so this nabla symbol is a vector of partial derivative operators . what 's its dimension ? and it 's like how many dimensions do you got ? because if you had a three-dimensional function that would mean that you should treat this like it 's got three different operators as part of it . and you know i 'd kinda , finish this off down here , and if you had something that was 100-dimensional it would have 100 different operators in it and that 's fine . it 's really just again , kind of a memory trick . so with that , that 's how you compute the gradient . not too much too it , it 's pretty much just partial derivatives , but you smack em into a vector where it gets fun and where it gets interesting is with the geometric interpretation . i 'll get to that in the next couple videos . it 's also a super important tool for something called the directional derivative . so you 've got a lot of fun stuff ahead .
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so here i 'm gon na talk about the gradient . and in this video , i 'm only gon na describe how you compute the gradient , and in the next couple ones i 'm gon na give the geometric interpretation .
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what is the class code ?
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so here i 'm gon na talk about the gradient . and in this video , i 'm only gon na describe how you compute the gradient , and in the next couple ones i 'm gon na give the geometric interpretation . and i hate doing this , i hate showing the computation before the geometric intuition since usually it should go the other way around , but the gradient is one of those weird things where the way that you compute it actually seems kind of unrelated to the intuition and you 'll see that . we 'll connect them in the next few videos . but to do that , we need to know what both of them actually are . so on the computation side of things , let 's say you have some sort of function . and i 'm just gon na make it a two-variable function . and let 's say it 's f of x , y , equals x-squared sine of y . the gradient is a way of packing together all the partial derivative information of a function . so let 's just start by computing the partial derivatives of this guy . so partial of f with respect to x is equal to , so we look at this and we consider x the variable and y the constant . well in that case sine of y is also a constant . as far as x is concerned , the derivative of x is 2x so we see that this will be 2x times that constant sine of y , sine of y . whereas the partial derivative with respect to y . now we look up here and we say x is considered a constant so x-squared is also considered a constant so this is just a constant times sine of y , so that 's gon na equal that same constant times the cosine of y , which is the derivative of sine . so now what the gradient does is it just puts both of these together in a vector . and specifically , maybe i 'll change colors here , you denote it with a little upside-down triangle . the name of that symbol is nabla , but you often just pronounce it del , you 'd say del f or gradient of f. and what this equals is a vector that has those two partial derivatives in it . so the first one is the partial derivative with respect to x , to x times sine of y . and the bottom one , partial derivative with respect to y x-squared cosine of y . and notice , maybe i should emphasize , this is actually a vector-valued function . so maybe i 'll give it a little bit more room here and emphasize that it 's got an x and a y . this is a function that takes in a point in two-dimensional space and outputs a two-dimensional vector . so you could also imagine doing this with three different variables . then you would have three partial derivatives , and a three-dimensional output . and the way you might write this more generally is we could go down here and say the gradient of any function is equal to a vector with its partial derivatives . partial of f with respect to x , and partial of f with respect to y . and in some sense , we call these partial derivatives . i like to think as the gradient as the full derivative cuz it kind of captures all of the information that you need . so a very helpful mnemonic device with the gradient is to think about this triangle , this nabla symbol as being a vector full of partial derivative operators . and by operator , i just mean like partial with respect to x , something where you could give it a function , and it gives you another function . so you give this guy the function f and it gives you this expression , this multi-variable function as a result . so the nabla symbol is this vector full of different partial derivative operators . and in this case it might just be two of them , and this is kind of a weird thing because it 's like what , this is a vector , it 's got like operators in it , that 's not what i thought vectors do . but you can kind of see where it 's going . it 's really just , you can think of it as a memory trick , but in some sense it 's a little bit deeper than that . and really when you take this triangle and you say ok let 's take this triangle and you can kind of imagine multiplying it by f , really it 's like an operator taking in this function and it 's gon na give you another function . it 's like you take this triangle and you put an f in front of it , and you can imagine , like this part gets multipled , quote unquote multiplied with f , this part gets quote unquote multiplied with f but really you 're just saying you take the partial derivative with respect to x and then with y , and on and on . and the reason for doing this , this symbol comes up a lot in other contexts . there are two other operators that you 're gon na learn about called the divergence and the curl . we 'll get to those later , all in due time . but it 's useful to think about this vector-ish thing of partial derivatives . and i mean one weird thing about it , you could say ok so this nabla symbol is a vector of partial derivative operators . what 's its dimension ? and it 's like how many dimensions do you got ? because if you had a three-dimensional function that would mean that you should treat this like it 's got three different operators as part of it . and you know i 'd kinda , finish this off down here , and if you had something that was 100-dimensional it would have 100 different operators in it and that 's fine . it 's really just again , kind of a memory trick . so with that , that 's how you compute the gradient . not too much too it , it 's pretty much just partial derivatives , but you smack em into a vector where it gets fun and where it gets interesting is with the geometric interpretation . i 'll get to that in the next couple videos . it 's also a super important tool for something called the directional derivative . so you 've got a lot of fun stuff ahead .
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so you give this guy the function f and it gives you this expression , this multi-variable function as a result . so the nabla symbol is this vector full of different partial derivative operators . and in this case it might just be two of them , and this is kind of a weird thing because it 's like what , this is a vector , it 's got like operators in it , that 's not what i thought vectors do .
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how would you describe the way the total derivative works versus the full ( del operator ) derivative ?
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so here i 'm gon na talk about the gradient . and in this video , i 'm only gon na describe how you compute the gradient , and in the next couple ones i 'm gon na give the geometric interpretation . and i hate doing this , i hate showing the computation before the geometric intuition since usually it should go the other way around , but the gradient is one of those weird things where the way that you compute it actually seems kind of unrelated to the intuition and you 'll see that . we 'll connect them in the next few videos . but to do that , we need to know what both of them actually are . so on the computation side of things , let 's say you have some sort of function . and i 'm just gon na make it a two-variable function . and let 's say it 's f of x , y , equals x-squared sine of y . the gradient is a way of packing together all the partial derivative information of a function . so let 's just start by computing the partial derivatives of this guy . so partial of f with respect to x is equal to , so we look at this and we consider x the variable and y the constant . well in that case sine of y is also a constant . as far as x is concerned , the derivative of x is 2x so we see that this will be 2x times that constant sine of y , sine of y . whereas the partial derivative with respect to y . now we look up here and we say x is considered a constant so x-squared is also considered a constant so this is just a constant times sine of y , so that 's gon na equal that same constant times the cosine of y , which is the derivative of sine . so now what the gradient does is it just puts both of these together in a vector . and specifically , maybe i 'll change colors here , you denote it with a little upside-down triangle . the name of that symbol is nabla , but you often just pronounce it del , you 'd say del f or gradient of f. and what this equals is a vector that has those two partial derivatives in it . so the first one is the partial derivative with respect to x , to x times sine of y . and the bottom one , partial derivative with respect to y x-squared cosine of y . and notice , maybe i should emphasize , this is actually a vector-valued function . so maybe i 'll give it a little bit more room here and emphasize that it 's got an x and a y . this is a function that takes in a point in two-dimensional space and outputs a two-dimensional vector . so you could also imagine doing this with three different variables . then you would have three partial derivatives , and a three-dimensional output . and the way you might write this more generally is we could go down here and say the gradient of any function is equal to a vector with its partial derivatives . partial of f with respect to x , and partial of f with respect to y . and in some sense , we call these partial derivatives . i like to think as the gradient as the full derivative cuz it kind of captures all of the information that you need . so a very helpful mnemonic device with the gradient is to think about this triangle , this nabla symbol as being a vector full of partial derivative operators . and by operator , i just mean like partial with respect to x , something where you could give it a function , and it gives you another function . so you give this guy the function f and it gives you this expression , this multi-variable function as a result . so the nabla symbol is this vector full of different partial derivative operators . and in this case it might just be two of them , and this is kind of a weird thing because it 's like what , this is a vector , it 's got like operators in it , that 's not what i thought vectors do . but you can kind of see where it 's going . it 's really just , you can think of it as a memory trick , but in some sense it 's a little bit deeper than that . and really when you take this triangle and you say ok let 's take this triangle and you can kind of imagine multiplying it by f , really it 's like an operator taking in this function and it 's gon na give you another function . it 's like you take this triangle and you put an f in front of it , and you can imagine , like this part gets multipled , quote unquote multiplied with f , this part gets quote unquote multiplied with f but really you 're just saying you take the partial derivative with respect to x and then with y , and on and on . and the reason for doing this , this symbol comes up a lot in other contexts . there are two other operators that you 're gon na learn about called the divergence and the curl . we 'll get to those later , all in due time . but it 's useful to think about this vector-ish thing of partial derivatives . and i mean one weird thing about it , you could say ok so this nabla symbol is a vector of partial derivative operators . what 's its dimension ? and it 's like how many dimensions do you got ? because if you had a three-dimensional function that would mean that you should treat this like it 's got three different operators as part of it . and you know i 'd kinda , finish this off down here , and if you had something that was 100-dimensional it would have 100 different operators in it and that 's fine . it 's really just again , kind of a memory trick . so with that , that 's how you compute the gradient . not too much too it , it 's pretty much just partial derivatives , but you smack em into a vector where it gets fun and where it gets interesting is with the geometric interpretation . i 'll get to that in the next couple videos . it 's also a super important tool for something called the directional derivative . so you 've got a lot of fun stuff ahead .
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and i 'm just gon na make it a two-variable function . and let 's say it 's f of x , y , equals x-squared sine of y . the gradient is a way of packing together all the partial derivative information of a function .
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how the result was square of x multply cosy , it should be siny , shouldnt it ?
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so here i 'm gon na talk about the gradient . and in this video , i 'm only gon na describe how you compute the gradient , and in the next couple ones i 'm gon na give the geometric interpretation . and i hate doing this , i hate showing the computation before the geometric intuition since usually it should go the other way around , but the gradient is one of those weird things where the way that you compute it actually seems kind of unrelated to the intuition and you 'll see that . we 'll connect them in the next few videos . but to do that , we need to know what both of them actually are . so on the computation side of things , let 's say you have some sort of function . and i 'm just gon na make it a two-variable function . and let 's say it 's f of x , y , equals x-squared sine of y . the gradient is a way of packing together all the partial derivative information of a function . so let 's just start by computing the partial derivatives of this guy . so partial of f with respect to x is equal to , so we look at this and we consider x the variable and y the constant . well in that case sine of y is also a constant . as far as x is concerned , the derivative of x is 2x so we see that this will be 2x times that constant sine of y , sine of y . whereas the partial derivative with respect to y . now we look up here and we say x is considered a constant so x-squared is also considered a constant so this is just a constant times sine of y , so that 's gon na equal that same constant times the cosine of y , which is the derivative of sine . so now what the gradient does is it just puts both of these together in a vector . and specifically , maybe i 'll change colors here , you denote it with a little upside-down triangle . the name of that symbol is nabla , but you often just pronounce it del , you 'd say del f or gradient of f. and what this equals is a vector that has those two partial derivatives in it . so the first one is the partial derivative with respect to x , to x times sine of y . and the bottom one , partial derivative with respect to y x-squared cosine of y . and notice , maybe i should emphasize , this is actually a vector-valued function . so maybe i 'll give it a little bit more room here and emphasize that it 's got an x and a y . this is a function that takes in a point in two-dimensional space and outputs a two-dimensional vector . so you could also imagine doing this with three different variables . then you would have three partial derivatives , and a three-dimensional output . and the way you might write this more generally is we could go down here and say the gradient of any function is equal to a vector with its partial derivatives . partial of f with respect to x , and partial of f with respect to y . and in some sense , we call these partial derivatives . i like to think as the gradient as the full derivative cuz it kind of captures all of the information that you need . so a very helpful mnemonic device with the gradient is to think about this triangle , this nabla symbol as being a vector full of partial derivative operators . and by operator , i just mean like partial with respect to x , something where you could give it a function , and it gives you another function . so you give this guy the function f and it gives you this expression , this multi-variable function as a result . so the nabla symbol is this vector full of different partial derivative operators . and in this case it might just be two of them , and this is kind of a weird thing because it 's like what , this is a vector , it 's got like operators in it , that 's not what i thought vectors do . but you can kind of see where it 's going . it 's really just , you can think of it as a memory trick , but in some sense it 's a little bit deeper than that . and really when you take this triangle and you say ok let 's take this triangle and you can kind of imagine multiplying it by f , really it 's like an operator taking in this function and it 's gon na give you another function . it 's like you take this triangle and you put an f in front of it , and you can imagine , like this part gets multipled , quote unquote multiplied with f , this part gets quote unquote multiplied with f but really you 're just saying you take the partial derivative with respect to x and then with y , and on and on . and the reason for doing this , this symbol comes up a lot in other contexts . there are two other operators that you 're gon na learn about called the divergence and the curl . we 'll get to those later , all in due time . but it 's useful to think about this vector-ish thing of partial derivatives . and i mean one weird thing about it , you could say ok so this nabla symbol is a vector of partial derivative operators . what 's its dimension ? and it 's like how many dimensions do you got ? because if you had a three-dimensional function that would mean that you should treat this like it 's got three different operators as part of it . and you know i 'd kinda , finish this off down here , and if you had something that was 100-dimensional it would have 100 different operators in it and that 's fine . it 's really just again , kind of a memory trick . so with that , that 's how you compute the gradient . not too much too it , it 's pretty much just partial derivatives , but you smack em into a vector where it gets fun and where it gets interesting is with the geometric interpretation . i 'll get to that in the next couple videos . it 's also a super important tool for something called the directional derivative . so you 've got a lot of fun stuff ahead .
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so here i 'm gon na talk about the gradient . and in this video , i 'm only gon na describe how you compute the gradient , and in the next couple ones i 'm gon na give the geometric interpretation .
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please explain : why is area the opposite of gradient in calculus ?
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so here i 'm gon na talk about the gradient . and in this video , i 'm only gon na describe how you compute the gradient , and in the next couple ones i 'm gon na give the geometric interpretation . and i hate doing this , i hate showing the computation before the geometric intuition since usually it should go the other way around , but the gradient is one of those weird things where the way that you compute it actually seems kind of unrelated to the intuition and you 'll see that . we 'll connect them in the next few videos . but to do that , we need to know what both of them actually are . so on the computation side of things , let 's say you have some sort of function . and i 'm just gon na make it a two-variable function . and let 's say it 's f of x , y , equals x-squared sine of y . the gradient is a way of packing together all the partial derivative information of a function . so let 's just start by computing the partial derivatives of this guy . so partial of f with respect to x is equal to , so we look at this and we consider x the variable and y the constant . well in that case sine of y is also a constant . as far as x is concerned , the derivative of x is 2x so we see that this will be 2x times that constant sine of y , sine of y . whereas the partial derivative with respect to y . now we look up here and we say x is considered a constant so x-squared is also considered a constant so this is just a constant times sine of y , so that 's gon na equal that same constant times the cosine of y , which is the derivative of sine . so now what the gradient does is it just puts both of these together in a vector . and specifically , maybe i 'll change colors here , you denote it with a little upside-down triangle . the name of that symbol is nabla , but you often just pronounce it del , you 'd say del f or gradient of f. and what this equals is a vector that has those two partial derivatives in it . so the first one is the partial derivative with respect to x , to x times sine of y . and the bottom one , partial derivative with respect to y x-squared cosine of y . and notice , maybe i should emphasize , this is actually a vector-valued function . so maybe i 'll give it a little bit more room here and emphasize that it 's got an x and a y . this is a function that takes in a point in two-dimensional space and outputs a two-dimensional vector . so you could also imagine doing this with three different variables . then you would have three partial derivatives , and a three-dimensional output . and the way you might write this more generally is we could go down here and say the gradient of any function is equal to a vector with its partial derivatives . partial of f with respect to x , and partial of f with respect to y . and in some sense , we call these partial derivatives . i like to think as the gradient as the full derivative cuz it kind of captures all of the information that you need . so a very helpful mnemonic device with the gradient is to think about this triangle , this nabla symbol as being a vector full of partial derivative operators . and by operator , i just mean like partial with respect to x , something where you could give it a function , and it gives you another function . so you give this guy the function f and it gives you this expression , this multi-variable function as a result . so the nabla symbol is this vector full of different partial derivative operators . and in this case it might just be two of them , and this is kind of a weird thing because it 's like what , this is a vector , it 's got like operators in it , that 's not what i thought vectors do . but you can kind of see where it 's going . it 's really just , you can think of it as a memory trick , but in some sense it 's a little bit deeper than that . and really when you take this triangle and you say ok let 's take this triangle and you can kind of imagine multiplying it by f , really it 's like an operator taking in this function and it 's gon na give you another function . it 's like you take this triangle and you put an f in front of it , and you can imagine , like this part gets multipled , quote unquote multiplied with f , this part gets quote unquote multiplied with f but really you 're just saying you take the partial derivative with respect to x and then with y , and on and on . and the reason for doing this , this symbol comes up a lot in other contexts . there are two other operators that you 're gon na learn about called the divergence and the curl . we 'll get to those later , all in due time . but it 's useful to think about this vector-ish thing of partial derivatives . and i mean one weird thing about it , you could say ok so this nabla symbol is a vector of partial derivative operators . what 's its dimension ? and it 's like how many dimensions do you got ? because if you had a three-dimensional function that would mean that you should treat this like it 's got three different operators as part of it . and you know i 'd kinda , finish this off down here , and if you had something that was 100-dimensional it would have 100 different operators in it and that 's fine . it 's really just again , kind of a memory trick . so with that , that 's how you compute the gradient . not too much too it , it 's pretty much just partial derivatives , but you smack em into a vector where it gets fun and where it gets interesting is with the geometric interpretation . i 'll get to that in the next couple videos . it 's also a super important tool for something called the directional derivative . so you 've got a lot of fun stuff ahead .
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so the first one is the partial derivative with respect to x , to x times sine of y . and the bottom one , partial derivative with respect to y x-squared cosine of y . and notice , maybe i should emphasize , this is actually a vector-valued function .
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especially when diferentiating requires one value and and intergation requires 2values ?
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construct a regular hexagon inscribed inside the circle . so what i 'm going to do , first , i 'm going to draw a diameter of the circle . and actually , i 'm going to go beyond the diameter of the circle . i 'm just going to draw a line that goes through the center of the circle and just keeps on going . and i 'm going to make it flat , so it goes directly through , so this one right over here , goes directly through the center . and now , what i 'm going to do is i 'm going to construct a circle that 's the exact same dimensions of this circle that 's already been drawn . so let me put this one right over here . and let me make it the same dimensions . and now , what i 'm going to do is i 'm going to move it over , so that this new circle intersects the center of the old circle . and these circles are the same size . so notice , this center intersects the old circle . and the new circle itself intersects the center of the old circle . now , the reason why this is interesting , we already know that this distance , the distance between these two centers , this is equal to a radius . we also know -- we have a straight edge -- we also know that this distance right over here is equal to a radius . it 's equal to the radius of our new circle right over here . we also know that this distance right over here is equal to a radius of our old circle . and that they both have the same radius . so this is a radius between these two points . between these two points is a radius . and then , in between these two points is a radius . so now , i have constructed an equilateral triangle . and essentially , you have to just do this six times . and then i 'm going to have a hexagon inscribed inside the circle . let me do it again . so we 'll go from here to here . this is a radius of my new circle , which is the same as the radius of the old circle . and i could go from here to here . that 's the radius of my old circle . so i have another equilateral triangle . radius , radius , radius . another equilateral triangle . i 've just got to do this , i have to do this four more times . so let me go to my original . let 's see , let me make sure i can -- well , it 's actually going to be hard for me to -- let me just add another circle here , to do it on the other side . so if i put the center of it right -- i want to move it a little bit , right over -- i want to make it the same size . so oh , that 's close enough . and let 's see , that looks pretty close . that 's the same size . now , let me move it over here . now , i want this center to be on the circle , right like that . and now , i 'm ready to draw some more equilateral triangles . and really , i do n't have to even draw the inside of it . now , i see my six vertices for my hexagon , here , here , here , here , here , here . and i think you 're satisfied now that you could break this up into six equilateral triangles . so let 's do that . so this would be the base of one of those equilateral triangles . and actually , let me move these . let me move this one out of the way . i can move this one right over here . because i really just care about the hexagon itself . and i can move this right over here . but we know that these are all the lengths of the radius anyway . actually , i 'm not even having to change the length there . then , i have to just connect one more right down here . so let me add another straight edge , connect those two points . and i would have done it . i would have constructed my regular hexagon inscribed in the circle .
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radius , radius , radius . another equilateral triangle . i 've just got to do this , i have to do this four more times .
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how can a polygon be equilateral and not equiangular ?
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construct a regular hexagon inscribed inside the circle . so what i 'm going to do , first , i 'm going to draw a diameter of the circle . and actually , i 'm going to go beyond the diameter of the circle . i 'm just going to draw a line that goes through the center of the circle and just keeps on going . and i 'm going to make it flat , so it goes directly through , so this one right over here , goes directly through the center . and now , what i 'm going to do is i 'm going to construct a circle that 's the exact same dimensions of this circle that 's already been drawn . so let me put this one right over here . and let me make it the same dimensions . and now , what i 'm going to do is i 'm going to move it over , so that this new circle intersects the center of the old circle . and these circles are the same size . so notice , this center intersects the old circle . and the new circle itself intersects the center of the old circle . now , the reason why this is interesting , we already know that this distance , the distance between these two centers , this is equal to a radius . we also know -- we have a straight edge -- we also know that this distance right over here is equal to a radius . it 's equal to the radius of our new circle right over here . we also know that this distance right over here is equal to a radius of our old circle . and that they both have the same radius . so this is a radius between these two points . between these two points is a radius . and then , in between these two points is a radius . so now , i have constructed an equilateral triangle . and essentially , you have to just do this six times . and then i 'm going to have a hexagon inscribed inside the circle . let me do it again . so we 'll go from here to here . this is a radius of my new circle , which is the same as the radius of the old circle . and i could go from here to here . that 's the radius of my old circle . so i have another equilateral triangle . radius , radius , radius . another equilateral triangle . i 've just got to do this , i have to do this four more times . so let me go to my original . let 's see , let me make sure i can -- well , it 's actually going to be hard for me to -- let me just add another circle here , to do it on the other side . so if i put the center of it right -- i want to move it a little bit , right over -- i want to make it the same size . so oh , that 's close enough . and let 's see , that looks pretty close . that 's the same size . now , let me move it over here . now , i want this center to be on the circle , right like that . and now , i 'm ready to draw some more equilateral triangles . and really , i do n't have to even draw the inside of it . now , i see my six vertices for my hexagon , here , here , here , here , here , here . and i think you 're satisfied now that you could break this up into six equilateral triangles . so let 's do that . so this would be the base of one of those equilateral triangles . and actually , let me move these . let me move this one out of the way . i can move this one right over here . because i really just care about the hexagon itself . and i can move this right over here . but we know that these are all the lengths of the radius anyway . actually , i 'm not even having to change the length there . then , i have to just connect one more right down here . so let me add another straight edge , connect those two points . and i would have done it . i would have constructed my regular hexagon inscribed in the circle .
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construct a regular hexagon inscribed inside the circle . so what i 'm going to do , first , i 'm going to draw a diameter of the circle .
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is there any way to construct a regular octagon inscribed in a circle ?
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construct a regular hexagon inscribed inside the circle . so what i 'm going to do , first , i 'm going to draw a diameter of the circle . and actually , i 'm going to go beyond the diameter of the circle . i 'm just going to draw a line that goes through the center of the circle and just keeps on going . and i 'm going to make it flat , so it goes directly through , so this one right over here , goes directly through the center . and now , what i 'm going to do is i 'm going to construct a circle that 's the exact same dimensions of this circle that 's already been drawn . so let me put this one right over here . and let me make it the same dimensions . and now , what i 'm going to do is i 'm going to move it over , so that this new circle intersects the center of the old circle . and these circles are the same size . so notice , this center intersects the old circle . and the new circle itself intersects the center of the old circle . now , the reason why this is interesting , we already know that this distance , the distance between these two centers , this is equal to a radius . we also know -- we have a straight edge -- we also know that this distance right over here is equal to a radius . it 's equal to the radius of our new circle right over here . we also know that this distance right over here is equal to a radius of our old circle . and that they both have the same radius . so this is a radius between these two points . between these two points is a radius . and then , in between these two points is a radius . so now , i have constructed an equilateral triangle . and essentially , you have to just do this six times . and then i 'm going to have a hexagon inscribed inside the circle . let me do it again . so we 'll go from here to here . this is a radius of my new circle , which is the same as the radius of the old circle . and i could go from here to here . that 's the radius of my old circle . so i have another equilateral triangle . radius , radius , radius . another equilateral triangle . i 've just got to do this , i have to do this four more times . so let me go to my original . let 's see , let me make sure i can -- well , it 's actually going to be hard for me to -- let me just add another circle here , to do it on the other side . so if i put the center of it right -- i want to move it a little bit , right over -- i want to make it the same size . so oh , that 's close enough . and let 's see , that looks pretty close . that 's the same size . now , let me move it over here . now , i want this center to be on the circle , right like that . and now , i 'm ready to draw some more equilateral triangles . and really , i do n't have to even draw the inside of it . now , i see my six vertices for my hexagon , here , here , here , here , here , here . and i think you 're satisfied now that you could break this up into six equilateral triangles . so let 's do that . so this would be the base of one of those equilateral triangles . and actually , let me move these . let me move this one out of the way . i can move this one right over here . because i really just care about the hexagon itself . and i can move this right over here . but we know that these are all the lengths of the radius anyway . actually , i 'm not even having to change the length there . then , i have to just connect one more right down here . so let me add another straight edge , connect those two points . and i would have done it . i would have constructed my regular hexagon inscribed in the circle .
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construct a regular hexagon inscribed inside the circle . so what i 'm going to do , first , i 'm going to draw a diameter of the circle .
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why would we even need to construct inscribed figures in real world problems ?
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construct a regular hexagon inscribed inside the circle . so what i 'm going to do , first , i 'm going to draw a diameter of the circle . and actually , i 'm going to go beyond the diameter of the circle . i 'm just going to draw a line that goes through the center of the circle and just keeps on going . and i 'm going to make it flat , so it goes directly through , so this one right over here , goes directly through the center . and now , what i 'm going to do is i 'm going to construct a circle that 's the exact same dimensions of this circle that 's already been drawn . so let me put this one right over here . and let me make it the same dimensions . and now , what i 'm going to do is i 'm going to move it over , so that this new circle intersects the center of the old circle . and these circles are the same size . so notice , this center intersects the old circle . and the new circle itself intersects the center of the old circle . now , the reason why this is interesting , we already know that this distance , the distance between these two centers , this is equal to a radius . we also know -- we have a straight edge -- we also know that this distance right over here is equal to a radius . it 's equal to the radius of our new circle right over here . we also know that this distance right over here is equal to a radius of our old circle . and that they both have the same radius . so this is a radius between these two points . between these two points is a radius . and then , in between these two points is a radius . so now , i have constructed an equilateral triangle . and essentially , you have to just do this six times . and then i 'm going to have a hexagon inscribed inside the circle . let me do it again . so we 'll go from here to here . this is a radius of my new circle , which is the same as the radius of the old circle . and i could go from here to here . that 's the radius of my old circle . so i have another equilateral triangle . radius , radius , radius . another equilateral triangle . i 've just got to do this , i have to do this four more times . so let me go to my original . let 's see , let me make sure i can -- well , it 's actually going to be hard for me to -- let me just add another circle here , to do it on the other side . so if i put the center of it right -- i want to move it a little bit , right over -- i want to make it the same size . so oh , that 's close enough . and let 's see , that looks pretty close . that 's the same size . now , let me move it over here . now , i want this center to be on the circle , right like that . and now , i 'm ready to draw some more equilateral triangles . and really , i do n't have to even draw the inside of it . now , i see my six vertices for my hexagon , here , here , here , here , here , here . and i think you 're satisfied now that you could break this up into six equilateral triangles . so let 's do that . so this would be the base of one of those equilateral triangles . and actually , let me move these . let me move this one out of the way . i can move this one right over here . because i really just care about the hexagon itself . and i can move this right over here . but we know that these are all the lengths of the radius anyway . actually , i 'm not even having to change the length there . then , i have to just connect one more right down here . so let me add another straight edge , connect those two points . and i would have done it . i would have constructed my regular hexagon inscribed in the circle .
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so i have another equilateral triangle . radius , radius , radius . another equilateral triangle .
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if you 're given the radius of the circle is there an easy formula to find the area of the hexagon without using trig ?
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construct a regular hexagon inscribed inside the circle . so what i 'm going to do , first , i 'm going to draw a diameter of the circle . and actually , i 'm going to go beyond the diameter of the circle . i 'm just going to draw a line that goes through the center of the circle and just keeps on going . and i 'm going to make it flat , so it goes directly through , so this one right over here , goes directly through the center . and now , what i 'm going to do is i 'm going to construct a circle that 's the exact same dimensions of this circle that 's already been drawn . so let me put this one right over here . and let me make it the same dimensions . and now , what i 'm going to do is i 'm going to move it over , so that this new circle intersects the center of the old circle . and these circles are the same size . so notice , this center intersects the old circle . and the new circle itself intersects the center of the old circle . now , the reason why this is interesting , we already know that this distance , the distance between these two centers , this is equal to a radius . we also know -- we have a straight edge -- we also know that this distance right over here is equal to a radius . it 's equal to the radius of our new circle right over here . we also know that this distance right over here is equal to a radius of our old circle . and that they both have the same radius . so this is a radius between these two points . between these two points is a radius . and then , in between these two points is a radius . so now , i have constructed an equilateral triangle . and essentially , you have to just do this six times . and then i 'm going to have a hexagon inscribed inside the circle . let me do it again . so we 'll go from here to here . this is a radius of my new circle , which is the same as the radius of the old circle . and i could go from here to here . that 's the radius of my old circle . so i have another equilateral triangle . radius , radius , radius . another equilateral triangle . i 've just got to do this , i have to do this four more times . so let me go to my original . let 's see , let me make sure i can -- well , it 's actually going to be hard for me to -- let me just add another circle here , to do it on the other side . so if i put the center of it right -- i want to move it a little bit , right over -- i want to make it the same size . so oh , that 's close enough . and let 's see , that looks pretty close . that 's the same size . now , let me move it over here . now , i want this center to be on the circle , right like that . and now , i 'm ready to draw some more equilateral triangles . and really , i do n't have to even draw the inside of it . now , i see my six vertices for my hexagon , here , here , here , here , here , here . and i think you 're satisfied now that you could break this up into six equilateral triangles . so let 's do that . so this would be the base of one of those equilateral triangles . and actually , let me move these . let me move this one out of the way . i can move this one right over here . because i really just care about the hexagon itself . and i can move this right over here . but we know that these are all the lengths of the radius anyway . actually , i 'm not even having to change the length there . then , i have to just connect one more right down here . so let me add another straight edge , connect those two points . and i would have done it . i would have constructed my regular hexagon inscribed in the circle .
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but we know that these are all the lengths of the radius anyway . actually , i 'm not even having to change the length there . then , i have to just connect one more right down here .
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if the length of ed is 12 , what is the length of fg ?
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construct a regular hexagon inscribed inside the circle . so what i 'm going to do , first , i 'm going to draw a diameter of the circle . and actually , i 'm going to go beyond the diameter of the circle . i 'm just going to draw a line that goes through the center of the circle and just keeps on going . and i 'm going to make it flat , so it goes directly through , so this one right over here , goes directly through the center . and now , what i 'm going to do is i 'm going to construct a circle that 's the exact same dimensions of this circle that 's already been drawn . so let me put this one right over here . and let me make it the same dimensions . and now , what i 'm going to do is i 'm going to move it over , so that this new circle intersects the center of the old circle . and these circles are the same size . so notice , this center intersects the old circle . and the new circle itself intersects the center of the old circle . now , the reason why this is interesting , we already know that this distance , the distance between these two centers , this is equal to a radius . we also know -- we have a straight edge -- we also know that this distance right over here is equal to a radius . it 's equal to the radius of our new circle right over here . we also know that this distance right over here is equal to a radius of our old circle . and that they both have the same radius . so this is a radius between these two points . between these two points is a radius . and then , in between these two points is a radius . so now , i have constructed an equilateral triangle . and essentially , you have to just do this six times . and then i 'm going to have a hexagon inscribed inside the circle . let me do it again . so we 'll go from here to here . this is a radius of my new circle , which is the same as the radius of the old circle . and i could go from here to here . that 's the radius of my old circle . so i have another equilateral triangle . radius , radius , radius . another equilateral triangle . i 've just got to do this , i have to do this four more times . so let me go to my original . let 's see , let me make sure i can -- well , it 's actually going to be hard for me to -- let me just add another circle here , to do it on the other side . so if i put the center of it right -- i want to move it a little bit , right over -- i want to make it the same size . so oh , that 's close enough . and let 's see , that looks pretty close . that 's the same size . now , let me move it over here . now , i want this center to be on the circle , right like that . and now , i 'm ready to draw some more equilateral triangles . and really , i do n't have to even draw the inside of it . now , i see my six vertices for my hexagon , here , here , here , here , here , here . and i think you 're satisfied now that you could break this up into six equilateral triangles . so let 's do that . so this would be the base of one of those equilateral triangles . and actually , let me move these . let me move this one out of the way . i can move this one right over here . because i really just care about the hexagon itself . and i can move this right over here . but we know that these are all the lengths of the radius anyway . actually , i 'm not even having to change the length there . then , i have to just connect one more right down here . so let me add another straight edge , connect those two points . and i would have done it . i would have constructed my regular hexagon inscribed in the circle .
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construct a regular hexagon inscribed inside the circle . so what i 'm going to do , first , i 'm going to draw a diameter of the circle .
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is there any other ways to draw a circle-inscribed regular hexagon ?
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construct a regular hexagon inscribed inside the circle . so what i 'm going to do , first , i 'm going to draw a diameter of the circle . and actually , i 'm going to go beyond the diameter of the circle . i 'm just going to draw a line that goes through the center of the circle and just keeps on going . and i 'm going to make it flat , so it goes directly through , so this one right over here , goes directly through the center . and now , what i 'm going to do is i 'm going to construct a circle that 's the exact same dimensions of this circle that 's already been drawn . so let me put this one right over here . and let me make it the same dimensions . and now , what i 'm going to do is i 'm going to move it over , so that this new circle intersects the center of the old circle . and these circles are the same size . so notice , this center intersects the old circle . and the new circle itself intersects the center of the old circle . now , the reason why this is interesting , we already know that this distance , the distance between these two centers , this is equal to a radius . we also know -- we have a straight edge -- we also know that this distance right over here is equal to a radius . it 's equal to the radius of our new circle right over here . we also know that this distance right over here is equal to a radius of our old circle . and that they both have the same radius . so this is a radius between these two points . between these two points is a radius . and then , in between these two points is a radius . so now , i have constructed an equilateral triangle . and essentially , you have to just do this six times . and then i 'm going to have a hexagon inscribed inside the circle . let me do it again . so we 'll go from here to here . this is a radius of my new circle , which is the same as the radius of the old circle . and i could go from here to here . that 's the radius of my old circle . so i have another equilateral triangle . radius , radius , radius . another equilateral triangle . i 've just got to do this , i have to do this four more times . so let me go to my original . let 's see , let me make sure i can -- well , it 's actually going to be hard for me to -- let me just add another circle here , to do it on the other side . so if i put the center of it right -- i want to move it a little bit , right over -- i want to make it the same size . so oh , that 's close enough . and let 's see , that looks pretty close . that 's the same size . now , let me move it over here . now , i want this center to be on the circle , right like that . and now , i 'm ready to draw some more equilateral triangles . and really , i do n't have to even draw the inside of it . now , i see my six vertices for my hexagon , here , here , here , here , here , here . and i think you 're satisfied now that you could break this up into six equilateral triangles . so let 's do that . so this would be the base of one of those equilateral triangles . and actually , let me move these . let me move this one out of the way . i can move this one right over here . because i really just care about the hexagon itself . and i can move this right over here . but we know that these are all the lengths of the radius anyway . actually , i 'm not even having to change the length there . then , i have to just connect one more right down here . so let me add another straight edge , connect those two points . and i would have done it . i would have constructed my regular hexagon inscribed in the circle .
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construct a regular hexagon inscribed inside the circle . so what i 'm going to do , first , i 'm going to draw a diameter of the circle .
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why is der no loci ( of angles ) on khan academy ?
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construct a regular hexagon inscribed inside the circle . so what i 'm going to do , first , i 'm going to draw a diameter of the circle . and actually , i 'm going to go beyond the diameter of the circle . i 'm just going to draw a line that goes through the center of the circle and just keeps on going . and i 'm going to make it flat , so it goes directly through , so this one right over here , goes directly through the center . and now , what i 'm going to do is i 'm going to construct a circle that 's the exact same dimensions of this circle that 's already been drawn . so let me put this one right over here . and let me make it the same dimensions . and now , what i 'm going to do is i 'm going to move it over , so that this new circle intersects the center of the old circle . and these circles are the same size . so notice , this center intersects the old circle . and the new circle itself intersects the center of the old circle . now , the reason why this is interesting , we already know that this distance , the distance between these two centers , this is equal to a radius . we also know -- we have a straight edge -- we also know that this distance right over here is equal to a radius . it 's equal to the radius of our new circle right over here . we also know that this distance right over here is equal to a radius of our old circle . and that they both have the same radius . so this is a radius between these two points . between these two points is a radius . and then , in between these two points is a radius . so now , i have constructed an equilateral triangle . and essentially , you have to just do this six times . and then i 'm going to have a hexagon inscribed inside the circle . let me do it again . so we 'll go from here to here . this is a radius of my new circle , which is the same as the radius of the old circle . and i could go from here to here . that 's the radius of my old circle . so i have another equilateral triangle . radius , radius , radius . another equilateral triangle . i 've just got to do this , i have to do this four more times . so let me go to my original . let 's see , let me make sure i can -- well , it 's actually going to be hard for me to -- let me just add another circle here , to do it on the other side . so if i put the center of it right -- i want to move it a little bit , right over -- i want to make it the same size . so oh , that 's close enough . and let 's see , that looks pretty close . that 's the same size . now , let me move it over here . now , i want this center to be on the circle , right like that . and now , i 'm ready to draw some more equilateral triangles . and really , i do n't have to even draw the inside of it . now , i see my six vertices for my hexagon , here , here , here , here , here , here . and i think you 're satisfied now that you could break this up into six equilateral triangles . so let 's do that . so this would be the base of one of those equilateral triangles . and actually , let me move these . let me move this one out of the way . i can move this one right over here . because i really just care about the hexagon itself . and i can move this right over here . but we know that these are all the lengths of the radius anyway . actually , i 'm not even having to change the length there . then , i have to just connect one more right down here . so let me add another straight edge , connect those two points . and i would have done it . i would have constructed my regular hexagon inscribed in the circle .
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construct a regular hexagon inscribed inside the circle . so what i 'm going to do , first , i 'm going to draw a diameter of the circle .
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what steps are different from an inscribed square ?
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construct a regular hexagon inscribed inside the circle . so what i 'm going to do , first , i 'm going to draw a diameter of the circle . and actually , i 'm going to go beyond the diameter of the circle . i 'm just going to draw a line that goes through the center of the circle and just keeps on going . and i 'm going to make it flat , so it goes directly through , so this one right over here , goes directly through the center . and now , what i 'm going to do is i 'm going to construct a circle that 's the exact same dimensions of this circle that 's already been drawn . so let me put this one right over here . and let me make it the same dimensions . and now , what i 'm going to do is i 'm going to move it over , so that this new circle intersects the center of the old circle . and these circles are the same size . so notice , this center intersects the old circle . and the new circle itself intersects the center of the old circle . now , the reason why this is interesting , we already know that this distance , the distance between these two centers , this is equal to a radius . we also know -- we have a straight edge -- we also know that this distance right over here is equal to a radius . it 's equal to the radius of our new circle right over here . we also know that this distance right over here is equal to a radius of our old circle . and that they both have the same radius . so this is a radius between these two points . between these two points is a radius . and then , in between these two points is a radius . so now , i have constructed an equilateral triangle . and essentially , you have to just do this six times . and then i 'm going to have a hexagon inscribed inside the circle . let me do it again . so we 'll go from here to here . this is a radius of my new circle , which is the same as the radius of the old circle . and i could go from here to here . that 's the radius of my old circle . so i have another equilateral triangle . radius , radius , radius . another equilateral triangle . i 've just got to do this , i have to do this four more times . so let me go to my original . let 's see , let me make sure i can -- well , it 's actually going to be hard for me to -- let me just add another circle here , to do it on the other side . so if i put the center of it right -- i want to move it a little bit , right over -- i want to make it the same size . so oh , that 's close enough . and let 's see , that looks pretty close . that 's the same size . now , let me move it over here . now , i want this center to be on the circle , right like that . and now , i 'm ready to draw some more equilateral triangles . and really , i do n't have to even draw the inside of it . now , i see my six vertices for my hexagon , here , here , here , here , here , here . and i think you 're satisfied now that you could break this up into six equilateral triangles . so let 's do that . so this would be the base of one of those equilateral triangles . and actually , let me move these . let me move this one out of the way . i can move this one right over here . because i really just care about the hexagon itself . and i can move this right over here . but we know that these are all the lengths of the radius anyway . actually , i 'm not even having to change the length there . then , i have to just connect one more right down here . so let me add another straight edge , connect those two points . and i would have done it . i would have constructed my regular hexagon inscribed in the circle .
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construct a regular hexagon inscribed inside the circle . so what i 'm going to do , first , i 'm going to draw a diameter of the circle .
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if a circle is inscribed in an hexagon , which of the following must be true ?
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construct a regular hexagon inscribed inside the circle . so what i 'm going to do , first , i 'm going to draw a diameter of the circle . and actually , i 'm going to go beyond the diameter of the circle . i 'm just going to draw a line that goes through the center of the circle and just keeps on going . and i 'm going to make it flat , so it goes directly through , so this one right over here , goes directly through the center . and now , what i 'm going to do is i 'm going to construct a circle that 's the exact same dimensions of this circle that 's already been drawn . so let me put this one right over here . and let me make it the same dimensions . and now , what i 'm going to do is i 'm going to move it over , so that this new circle intersects the center of the old circle . and these circles are the same size . so notice , this center intersects the old circle . and the new circle itself intersects the center of the old circle . now , the reason why this is interesting , we already know that this distance , the distance between these two centers , this is equal to a radius . we also know -- we have a straight edge -- we also know that this distance right over here is equal to a radius . it 's equal to the radius of our new circle right over here . we also know that this distance right over here is equal to a radius of our old circle . and that they both have the same radius . so this is a radius between these two points . between these two points is a radius . and then , in between these two points is a radius . so now , i have constructed an equilateral triangle . and essentially , you have to just do this six times . and then i 'm going to have a hexagon inscribed inside the circle . let me do it again . so we 'll go from here to here . this is a radius of my new circle , which is the same as the radius of the old circle . and i could go from here to here . that 's the radius of my old circle . so i have another equilateral triangle . radius , radius , radius . another equilateral triangle . i 've just got to do this , i have to do this four more times . so let me go to my original . let 's see , let me make sure i can -- well , it 's actually going to be hard for me to -- let me just add another circle here , to do it on the other side . so if i put the center of it right -- i want to move it a little bit , right over -- i want to make it the same size . so oh , that 's close enough . and let 's see , that looks pretty close . that 's the same size . now , let me move it over here . now , i want this center to be on the circle , right like that . and now , i 'm ready to draw some more equilateral triangles . and really , i do n't have to even draw the inside of it . now , i see my six vertices for my hexagon , here , here , here , here , here , here . and i think you 're satisfied now that you could break this up into six equilateral triangles . so let 's do that . so this would be the base of one of those equilateral triangles . and actually , let me move these . let me move this one out of the way . i can move this one right over here . because i really just care about the hexagon itself . and i can move this right over here . but we know that these are all the lengths of the radius anyway . actually , i 'm not even having to change the length there . then , i have to just connect one more right down here . so let me add another straight edge , connect those two points . and i would have done it . i would have constructed my regular hexagon inscribed in the circle .
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and these circles are the same size . so notice , this center intersects the old circle . and the new circle itself intersects the center of the old circle . now , the reason why this is interesting , we already know that this distance , the distance between these two centers , this is equal to a radius .
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can you tell me the derivation of the chord of contact of a circle ?
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construct a regular hexagon inscribed inside the circle . so what i 'm going to do , first , i 'm going to draw a diameter of the circle . and actually , i 'm going to go beyond the diameter of the circle . i 'm just going to draw a line that goes through the center of the circle and just keeps on going . and i 'm going to make it flat , so it goes directly through , so this one right over here , goes directly through the center . and now , what i 'm going to do is i 'm going to construct a circle that 's the exact same dimensions of this circle that 's already been drawn . so let me put this one right over here . and let me make it the same dimensions . and now , what i 'm going to do is i 'm going to move it over , so that this new circle intersects the center of the old circle . and these circles are the same size . so notice , this center intersects the old circle . and the new circle itself intersects the center of the old circle . now , the reason why this is interesting , we already know that this distance , the distance between these two centers , this is equal to a radius . we also know -- we have a straight edge -- we also know that this distance right over here is equal to a radius . it 's equal to the radius of our new circle right over here . we also know that this distance right over here is equal to a radius of our old circle . and that they both have the same radius . so this is a radius between these two points . between these two points is a radius . and then , in between these two points is a radius . so now , i have constructed an equilateral triangle . and essentially , you have to just do this six times . and then i 'm going to have a hexagon inscribed inside the circle . let me do it again . so we 'll go from here to here . this is a radius of my new circle , which is the same as the radius of the old circle . and i could go from here to here . that 's the radius of my old circle . so i have another equilateral triangle . radius , radius , radius . another equilateral triangle . i 've just got to do this , i have to do this four more times . so let me go to my original . let 's see , let me make sure i can -- well , it 's actually going to be hard for me to -- let me just add another circle here , to do it on the other side . so if i put the center of it right -- i want to move it a little bit , right over -- i want to make it the same size . so oh , that 's close enough . and let 's see , that looks pretty close . that 's the same size . now , let me move it over here . now , i want this center to be on the circle , right like that . and now , i 'm ready to draw some more equilateral triangles . and really , i do n't have to even draw the inside of it . now , i see my six vertices for my hexagon , here , here , here , here , here , here . and i think you 're satisfied now that you could break this up into six equilateral triangles . so let 's do that . so this would be the base of one of those equilateral triangles . and actually , let me move these . let me move this one out of the way . i can move this one right over here . because i really just care about the hexagon itself . and i can move this right over here . but we know that these are all the lengths of the radius anyway . actually , i 'm not even having to change the length there . then , i have to just connect one more right down here . so let me add another straight edge , connect those two points . and i would have done it . i would have constructed my regular hexagon inscribed in the circle .
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construct a regular hexagon inscribed inside the circle . so what i 'm going to do , first , i 'm going to draw a diameter of the circle .
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how can i use a straightedge and compass to construct a regular pentagon ?
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construct a regular hexagon inscribed inside the circle . so what i 'm going to do , first , i 'm going to draw a diameter of the circle . and actually , i 'm going to go beyond the diameter of the circle . i 'm just going to draw a line that goes through the center of the circle and just keeps on going . and i 'm going to make it flat , so it goes directly through , so this one right over here , goes directly through the center . and now , what i 'm going to do is i 'm going to construct a circle that 's the exact same dimensions of this circle that 's already been drawn . so let me put this one right over here . and let me make it the same dimensions . and now , what i 'm going to do is i 'm going to move it over , so that this new circle intersects the center of the old circle . and these circles are the same size . so notice , this center intersects the old circle . and the new circle itself intersects the center of the old circle . now , the reason why this is interesting , we already know that this distance , the distance between these two centers , this is equal to a radius . we also know -- we have a straight edge -- we also know that this distance right over here is equal to a radius . it 's equal to the radius of our new circle right over here . we also know that this distance right over here is equal to a radius of our old circle . and that they both have the same radius . so this is a radius between these two points . between these two points is a radius . and then , in between these two points is a radius . so now , i have constructed an equilateral triangle . and essentially , you have to just do this six times . and then i 'm going to have a hexagon inscribed inside the circle . let me do it again . so we 'll go from here to here . this is a radius of my new circle , which is the same as the radius of the old circle . and i could go from here to here . that 's the radius of my old circle . so i have another equilateral triangle . radius , radius , radius . another equilateral triangle . i 've just got to do this , i have to do this four more times . so let me go to my original . let 's see , let me make sure i can -- well , it 's actually going to be hard for me to -- let me just add another circle here , to do it on the other side . so if i put the center of it right -- i want to move it a little bit , right over -- i want to make it the same size . so oh , that 's close enough . and let 's see , that looks pretty close . that 's the same size . now , let me move it over here . now , i want this center to be on the circle , right like that . and now , i 'm ready to draw some more equilateral triangles . and really , i do n't have to even draw the inside of it . now , i see my six vertices for my hexagon , here , here , here , here , here , here . and i think you 're satisfied now that you could break this up into six equilateral triangles . so let 's do that . so this would be the base of one of those equilateral triangles . and actually , let me move these . let me move this one out of the way . i can move this one right over here . because i really just care about the hexagon itself . and i can move this right over here . but we know that these are all the lengths of the radius anyway . actually , i 'm not even having to change the length there . then , i have to just connect one more right down here . so let me add another straight edge , connect those two points . and i would have done it . i would have constructed my regular hexagon inscribed in the circle .
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construct a regular hexagon inscribed inside the circle . so what i 'm going to do , first , i 'm going to draw a diameter of the circle .
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how to construct a regular pentagon inscribed in a circle ?
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so in this video we 're going to install the switch you see in the lower center of the screen . that 's the on off switch for bit-zee . and the first thing we need to do is just take the switch and push it through the hole that we drilled out and marked in a previous video . so we 're going to push the switch through . and the hole 's just about the same size as the thread , so it 's a pretty tight fit . takes a little wiggling to get the switch through there . so then what we 're going to do is we will put the nut on the switch that will hold it in place . so we 're just using our screwdriver to slide it over the end of the switch there . and we 'll hand tighten that . it takes a little bit of doing . we got ta push the switch through from the underside and then just turn it a little bit . once we 've got the nut lined up and on the shaft of the switch , we can tighten it down the rest of the way with our -- it 's taking it a little bit of time there . and you 'll probably have a similar experience . it 's tricky to get that on , especially with all the other parts that are around the bot . we 're really close to the edge . but there we go , we finally got it . ok , so when you take your needle nose pliers , you can tighten it down the rest of the way . all right , so now what we 're going to do is , we 're going to solder our wires to the switch . and we have the wire coming from the battery . this is the positive wire coming from the battery . and we 're going to cut it to the right length . and then we 're going to strip the end of it . and then we 're going to bend it in a u shape . and that u shape , we can wrap around the contact on our switch . so we want to try and get it in almost a perfect u . and once we have in that shape , we can put it around the contact . and once it 's around the contact , we can use our needle nose pliers to crimp it in place , and make sure that it 's nice and tight against that connection there . and then what we 're going to do is once the wire is firmly in place , we 'll get our soldering iron and we 'll solder the wire to the post in the switch . so what the switch does is basically just stops the flow of electricity by disconnecting the wires . ok , so we 've soldered that one on , and we 'll solder another wire on to the other side of the switch . the wire that 's on the other side will go directly to our motor controller . and that 's where the power is going to come from . the power from our batteries is going to go to the motor controller , and the switch will allow us to break that circuit and shut the bit-zee bot off . ok , so we 've routed our power wires from underneath , up through the holes , and over to the motor controller . so now we 're just stripping the wire clean there . this is our positive 12 volt wire . and it 's going to go to the pin in our terminal block on the far right side . the terminal block in the center there . and we 're going to just -- we 've already loosened the screws , and we 're going to feed the wire into the pin . and then we 'll tighten the screw down on it to make sure that it does n't come out . so that is the wire that came directly from our switch . and there 's our tightening down the terminal block there . the screw in the terminal block . so we have our negative wire , which has come directly from the battery . and we are going to run that to the screw that is on the far left hand side of the terminal block . that is our ground wire . and so we 've stripped the end off of that wire . and just push it in underneath the screw there , in the left side of the terminal block . and now we 're going to take our 22 gauge wire , and we 're going to run 22 gauge wire into the ground . this will allow us to ground other components on the board using the motor controller as our sort of junction box . so we 're going to push our 22 gauge wire into the same terminal block as our 18 gauge wire from our battery . and it 's not a lot of room in there , but we 're going to try and slide it in . sometimes you have to move one of the wires out of the way to get the other one in . and our needle nose pliers can help push the wires into place . and then we 'll take our screwdriver again and tighten down that terminal block to make sure the wires wo n't come out . and then we 're going to take a 22 gauge wire and run it to the center of that terminal block , which is going to provide us five volts out that we can use to power our digital camera and our arduino and other components on our board . so we 'll wire that to our bread board in the upcoming videos .
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so that is the wire that came directly from our switch . and there 's our tightening down the terminal block there . the screw in the terminal block . so we have our negative wire , which has come directly from the battery .
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is there an electrical reason for putting the wires in specific holes in the terminal block ?
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so in this video we 're going to install the switch you see in the lower center of the screen . that 's the on off switch for bit-zee . and the first thing we need to do is just take the switch and push it through the hole that we drilled out and marked in a previous video . so we 're going to push the switch through . and the hole 's just about the same size as the thread , so it 's a pretty tight fit . takes a little wiggling to get the switch through there . so then what we 're going to do is we will put the nut on the switch that will hold it in place . so we 're just using our screwdriver to slide it over the end of the switch there . and we 'll hand tighten that . it takes a little bit of doing . we got ta push the switch through from the underside and then just turn it a little bit . once we 've got the nut lined up and on the shaft of the switch , we can tighten it down the rest of the way with our -- it 's taking it a little bit of time there . and you 'll probably have a similar experience . it 's tricky to get that on , especially with all the other parts that are around the bot . we 're really close to the edge . but there we go , we finally got it . ok , so when you take your needle nose pliers , you can tighten it down the rest of the way . all right , so now what we 're going to do is , we 're going to solder our wires to the switch . and we have the wire coming from the battery . this is the positive wire coming from the battery . and we 're going to cut it to the right length . and then we 're going to strip the end of it . and then we 're going to bend it in a u shape . and that u shape , we can wrap around the contact on our switch . so we want to try and get it in almost a perfect u . and once we have in that shape , we can put it around the contact . and once it 's around the contact , we can use our needle nose pliers to crimp it in place , and make sure that it 's nice and tight against that connection there . and then what we 're going to do is once the wire is firmly in place , we 'll get our soldering iron and we 'll solder the wire to the post in the switch . so what the switch does is basically just stops the flow of electricity by disconnecting the wires . ok , so we 've soldered that one on , and we 'll solder another wire on to the other side of the switch . the wire that 's on the other side will go directly to our motor controller . and that 's where the power is going to come from . the power from our batteries is going to go to the motor controller , and the switch will allow us to break that circuit and shut the bit-zee bot off . ok , so we 've routed our power wires from underneath , up through the holes , and over to the motor controller . so now we 're just stripping the wire clean there . this is our positive 12 volt wire . and it 's going to go to the pin in our terminal block on the far right side . the terminal block in the center there . and we 're going to just -- we 've already loosened the screws , and we 're going to feed the wire into the pin . and then we 'll tighten the screw down on it to make sure that it does n't come out . so that is the wire that came directly from our switch . and there 's our tightening down the terminal block there . the screw in the terminal block . so we have our negative wire , which has come directly from the battery . and we are going to run that to the screw that is on the far left hand side of the terminal block . that is our ground wire . and so we 've stripped the end off of that wire . and just push it in underneath the screw there , in the left side of the terminal block . and now we 're going to take our 22 gauge wire , and we 're going to run 22 gauge wire into the ground . this will allow us to ground other components on the board using the motor controller as our sort of junction box . so we 're going to push our 22 gauge wire into the same terminal block as our 18 gauge wire from our battery . and it 's not a lot of room in there , but we 're going to try and slide it in . sometimes you have to move one of the wires out of the way to get the other one in . and our needle nose pliers can help push the wires into place . and then we 'll take our screwdriver again and tighten down that terminal block to make sure the wires wo n't come out . and then we 're going to take a 22 gauge wire and run it to the center of that terminal block , which is going to provide us five volts out that we can use to power our digital camera and our arduino and other components on our board . so we 'll wire that to our bread board in the upcoming videos .
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so in this video we 're going to install the switch you see in the lower center of the screen . that 's the on off switch for bit-zee .
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what was the tool and ?
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so how do we take our products here that we 've taken apart -- except for the bottle , which is just a recycled good -- and we want to see if we can create value by building something new out of the parts that we 've taken out of all these different components here . so what we 've done is we 've designed a new product that 's kind of designed to get kids excited about math and science , and in order for all these products to work for us , we need to incorporate some electronics . so we have an arduino . it 's a basically programmable chip that you can easily wire to other components . it 's great for prototyping . and we have a l298 chip , which is this guy , on a motor control . it 's set up as a motor controller here , and the board that you can see here comes as a kit . so we put that together , and that 'll allow us to control the speed and direction of motors so that we can move our little bot around . and then we have a little sound chip here , and so that allows us to record and playback sound . and so we 'll control everything with the arduino . the motor controller will let us drive motors around , and then this guy will help us to record and play back sound . and we 're going to use some of the parts from these products to make something cool . so after three weeks of development , we 've come up with bit-zee bot . bit-zee is a little programmable robot that drives around and has a little personality . you can control bit-zee with a universal remote control -- the one that we took apart , actually . and that sends a signal right here to the ir sensor . and then that communicates with the arduino board and tells the arduino board that it just got a signal . and the arduino can turn on and off different functions , so we can make bit-zee do a little dance , make its lights come on . and bit-zee has a couple of switches here . there 's a switch here and a switch here , and those came out of the radio . and then you can that bit-zee 's nose came from a bottle cap , and its eyes came from a bottle cap . and the power for bit-zee is just some batteries that you can get from radio shack in a battery holder , and the motors came from the hair dryer . so here 's one of the hair dryer motors . and basically , we use that as a wheel to push bit-zee around . bit-zee 's also got the ability to record and play back sound , so we use that , and you can use the remote control to trigger that function . we 've also got a motor controller here , and that , of course , allows us to control the motors , the speed , and direction . the camera that we took apart , we mounted to the board here , and that came out of this camera . and you can use that to take photos of things . and again , the arduino controls that , and we control the arduino with the remote control . and you can see there 's what 's called a bread board here , and that 's where all the wires go to basically to allow us to connect more things to the arduino and we have a number of resistors and transistors here . the resistors limit current flow , and the transistors are functioning as switches , so they allow us to control different devices , again , with our arduino here . so this is bit-zee , and bit-zee has got a personality . you can see bit-zee lights up and does different things when it gets its bumper pressed or when it bumps into something . so we 're going to show you more about bit-zee in the next few sessions .
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and that sends a signal right here to the ir sensor . and then that communicates with the arduino board and tells the arduino board that it just got a signal . and the arduino can turn on and off different functions , so we can make bit-zee do a little dance , make its lights come on .
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is the arduino board mentioned at :and the l-298 chip described reusable ?
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so how do we take our products here that we 've taken apart -- except for the bottle , which is just a recycled good -- and we want to see if we can create value by building something new out of the parts that we 've taken out of all these different components here . so what we 've done is we 've designed a new product that 's kind of designed to get kids excited about math and science , and in order for all these products to work for us , we need to incorporate some electronics . so we have an arduino . it 's a basically programmable chip that you can easily wire to other components . it 's great for prototyping . and we have a l298 chip , which is this guy , on a motor control . it 's set up as a motor controller here , and the board that you can see here comes as a kit . so we put that together , and that 'll allow us to control the speed and direction of motors so that we can move our little bot around . and then we have a little sound chip here , and so that allows us to record and playback sound . and so we 'll control everything with the arduino . the motor controller will let us drive motors around , and then this guy will help us to record and play back sound . and we 're going to use some of the parts from these products to make something cool . so after three weeks of development , we 've come up with bit-zee bot . bit-zee is a little programmable robot that drives around and has a little personality . you can control bit-zee with a universal remote control -- the one that we took apart , actually . and that sends a signal right here to the ir sensor . and then that communicates with the arduino board and tells the arduino board that it just got a signal . and the arduino can turn on and off different functions , so we can make bit-zee do a little dance , make its lights come on . and bit-zee has a couple of switches here . there 's a switch here and a switch here , and those came out of the radio . and then you can that bit-zee 's nose came from a bottle cap , and its eyes came from a bottle cap . and the power for bit-zee is just some batteries that you can get from radio shack in a battery holder , and the motors came from the hair dryer . so here 's one of the hair dryer motors . and basically , we use that as a wheel to push bit-zee around . bit-zee 's also got the ability to record and play back sound , so we use that , and you can use the remote control to trigger that function . we 've also got a motor controller here , and that , of course , allows us to control the motors , the speed , and direction . the camera that we took apart , we mounted to the board here , and that came out of this camera . and you can use that to take photos of things . and again , the arduino controls that , and we control the arduino with the remote control . and you can see there 's what 's called a bread board here , and that 's where all the wires go to basically to allow us to connect more things to the arduino and we have a number of resistors and transistors here . the resistors limit current flow , and the transistors are functioning as switches , so they allow us to control different devices , again , with our arduino here . so this is bit-zee , and bit-zee has got a personality . you can see bit-zee lights up and does different things when it gets its bumper pressed or when it bumps into something . so we 're going to show you more about bit-zee in the next few sessions .
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the resistors limit current flow , and the transistors are functioning as switches , so they allow us to control different devices , again , with our arduino here . so this is bit-zee , and bit-zee has got a personality . you can see bit-zee lights up and does different things when it gets its bumper pressed or when it bumps into something .
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ie once you make the bit-zee bot can you take it apart and use these components to create something else ?
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so how do we take our products here that we 've taken apart -- except for the bottle , which is just a recycled good -- and we want to see if we can create value by building something new out of the parts that we 've taken out of all these different components here . so what we 've done is we 've designed a new product that 's kind of designed to get kids excited about math and science , and in order for all these products to work for us , we need to incorporate some electronics . so we have an arduino . it 's a basically programmable chip that you can easily wire to other components . it 's great for prototyping . and we have a l298 chip , which is this guy , on a motor control . it 's set up as a motor controller here , and the board that you can see here comes as a kit . so we put that together , and that 'll allow us to control the speed and direction of motors so that we can move our little bot around . and then we have a little sound chip here , and so that allows us to record and playback sound . and so we 'll control everything with the arduino . the motor controller will let us drive motors around , and then this guy will help us to record and play back sound . and we 're going to use some of the parts from these products to make something cool . so after three weeks of development , we 've come up with bit-zee bot . bit-zee is a little programmable robot that drives around and has a little personality . you can control bit-zee with a universal remote control -- the one that we took apart , actually . and that sends a signal right here to the ir sensor . and then that communicates with the arduino board and tells the arduino board that it just got a signal . and the arduino can turn on and off different functions , so we can make bit-zee do a little dance , make its lights come on . and bit-zee has a couple of switches here . there 's a switch here and a switch here , and those came out of the radio . and then you can that bit-zee 's nose came from a bottle cap , and its eyes came from a bottle cap . and the power for bit-zee is just some batteries that you can get from radio shack in a battery holder , and the motors came from the hair dryer . so here 's one of the hair dryer motors . and basically , we use that as a wheel to push bit-zee around . bit-zee 's also got the ability to record and play back sound , so we use that , and you can use the remote control to trigger that function . we 've also got a motor controller here , and that , of course , allows us to control the motors , the speed , and direction . the camera that we took apart , we mounted to the board here , and that came out of this camera . and you can use that to take photos of things . and again , the arduino controls that , and we control the arduino with the remote control . and you can see there 's what 's called a bread board here , and that 's where all the wires go to basically to allow us to connect more things to the arduino and we have a number of resistors and transistors here . the resistors limit current flow , and the transistors are functioning as switches , so they allow us to control different devices , again , with our arduino here . so this is bit-zee , and bit-zee has got a personality . you can see bit-zee lights up and does different things when it gets its bumper pressed or when it bumps into something . so we 're going to show you more about bit-zee in the next few sessions .
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and bit-zee has a couple of switches here . there 's a switch here and a switch here , and those came out of the radio . and then you can that bit-zee 's nose came from a bottle cap , and its eyes came from a bottle cap .
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what was the radio used for ?
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so how do we take our products here that we 've taken apart -- except for the bottle , which is just a recycled good -- and we want to see if we can create value by building something new out of the parts that we 've taken out of all these different components here . so what we 've done is we 've designed a new product that 's kind of designed to get kids excited about math and science , and in order for all these products to work for us , we need to incorporate some electronics . so we have an arduino . it 's a basically programmable chip that you can easily wire to other components . it 's great for prototyping . and we have a l298 chip , which is this guy , on a motor control . it 's set up as a motor controller here , and the board that you can see here comes as a kit . so we put that together , and that 'll allow us to control the speed and direction of motors so that we can move our little bot around . and then we have a little sound chip here , and so that allows us to record and playback sound . and so we 'll control everything with the arduino . the motor controller will let us drive motors around , and then this guy will help us to record and play back sound . and we 're going to use some of the parts from these products to make something cool . so after three weeks of development , we 've come up with bit-zee bot . bit-zee is a little programmable robot that drives around and has a little personality . you can control bit-zee with a universal remote control -- the one that we took apart , actually . and that sends a signal right here to the ir sensor . and then that communicates with the arduino board and tells the arduino board that it just got a signal . and the arduino can turn on and off different functions , so we can make bit-zee do a little dance , make its lights come on . and bit-zee has a couple of switches here . there 's a switch here and a switch here , and those came out of the radio . and then you can that bit-zee 's nose came from a bottle cap , and its eyes came from a bottle cap . and the power for bit-zee is just some batteries that you can get from radio shack in a battery holder , and the motors came from the hair dryer . so here 's one of the hair dryer motors . and basically , we use that as a wheel to push bit-zee around . bit-zee 's also got the ability to record and play back sound , so we use that , and you can use the remote control to trigger that function . we 've also got a motor controller here , and that , of course , allows us to control the motors , the speed , and direction . the camera that we took apart , we mounted to the board here , and that came out of this camera . and you can use that to take photos of things . and again , the arduino controls that , and we control the arduino with the remote control . and you can see there 's what 's called a bread board here , and that 's where all the wires go to basically to allow us to connect more things to the arduino and we have a number of resistors and transistors here . the resistors limit current flow , and the transistors are functioning as switches , so they allow us to control different devices , again , with our arduino here . so this is bit-zee , and bit-zee has got a personality . you can see bit-zee lights up and does different things when it gets its bumper pressed or when it bumps into something . so we 're going to show you more about bit-zee in the next few sessions .
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you can control bit-zee with a universal remote control -- the one that we took apart , actually . and that sends a signal right here to the ir sensor . and then that communicates with the arduino board and tells the arduino board that it just got a signal .
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what is an ir sensor , and what does it do ?
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so how do we take our products here that we 've taken apart -- except for the bottle , which is just a recycled good -- and we want to see if we can create value by building something new out of the parts that we 've taken out of all these different components here . so what we 've done is we 've designed a new product that 's kind of designed to get kids excited about math and science , and in order for all these products to work for us , we need to incorporate some electronics . so we have an arduino . it 's a basically programmable chip that you can easily wire to other components . it 's great for prototyping . and we have a l298 chip , which is this guy , on a motor control . it 's set up as a motor controller here , and the board that you can see here comes as a kit . so we put that together , and that 'll allow us to control the speed and direction of motors so that we can move our little bot around . and then we have a little sound chip here , and so that allows us to record and playback sound . and so we 'll control everything with the arduino . the motor controller will let us drive motors around , and then this guy will help us to record and play back sound . and we 're going to use some of the parts from these products to make something cool . so after three weeks of development , we 've come up with bit-zee bot . bit-zee is a little programmable robot that drives around and has a little personality . you can control bit-zee with a universal remote control -- the one that we took apart , actually . and that sends a signal right here to the ir sensor . and then that communicates with the arduino board and tells the arduino board that it just got a signal . and the arduino can turn on and off different functions , so we can make bit-zee do a little dance , make its lights come on . and bit-zee has a couple of switches here . there 's a switch here and a switch here , and those came out of the radio . and then you can that bit-zee 's nose came from a bottle cap , and its eyes came from a bottle cap . and the power for bit-zee is just some batteries that you can get from radio shack in a battery holder , and the motors came from the hair dryer . so here 's one of the hair dryer motors . and basically , we use that as a wheel to push bit-zee around . bit-zee 's also got the ability to record and play back sound , so we use that , and you can use the remote control to trigger that function . we 've also got a motor controller here , and that , of course , allows us to control the motors , the speed , and direction . the camera that we took apart , we mounted to the board here , and that came out of this camera . and you can use that to take photos of things . and again , the arduino controls that , and we control the arduino with the remote control . and you can see there 's what 's called a bread board here , and that 's where all the wires go to basically to allow us to connect more things to the arduino and we have a number of resistors and transistors here . the resistors limit current flow , and the transistors are functioning as switches , so they allow us to control different devices , again , with our arduino here . so this is bit-zee , and bit-zee has got a personality . you can see bit-zee lights up and does different things when it gets its bumper pressed or when it bumps into something . so we 're going to show you more about bit-zee in the next few sessions .
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the resistors limit current flow , and the transistors are functioning as switches , so they allow us to control different devices , again , with our arduino here . so this is bit-zee , and bit-zee has got a personality . you can see bit-zee lights up and does different things when it gets its bumper pressed or when it bumps into something .
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hi , how would you view the pictures you take with the camera mounted to bit-zee ?
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so how do we take our products here that we 've taken apart -- except for the bottle , which is just a recycled good -- and we want to see if we can create value by building something new out of the parts that we 've taken out of all these different components here . so what we 've done is we 've designed a new product that 's kind of designed to get kids excited about math and science , and in order for all these products to work for us , we need to incorporate some electronics . so we have an arduino . it 's a basically programmable chip that you can easily wire to other components . it 's great for prototyping . and we have a l298 chip , which is this guy , on a motor control . it 's set up as a motor controller here , and the board that you can see here comes as a kit . so we put that together , and that 'll allow us to control the speed and direction of motors so that we can move our little bot around . and then we have a little sound chip here , and so that allows us to record and playback sound . and so we 'll control everything with the arduino . the motor controller will let us drive motors around , and then this guy will help us to record and play back sound . and we 're going to use some of the parts from these products to make something cool . so after three weeks of development , we 've come up with bit-zee bot . bit-zee is a little programmable robot that drives around and has a little personality . you can control bit-zee with a universal remote control -- the one that we took apart , actually . and that sends a signal right here to the ir sensor . and then that communicates with the arduino board and tells the arduino board that it just got a signal . and the arduino can turn on and off different functions , so we can make bit-zee do a little dance , make its lights come on . and bit-zee has a couple of switches here . there 's a switch here and a switch here , and those came out of the radio . and then you can that bit-zee 's nose came from a bottle cap , and its eyes came from a bottle cap . and the power for bit-zee is just some batteries that you can get from radio shack in a battery holder , and the motors came from the hair dryer . so here 's one of the hair dryer motors . and basically , we use that as a wheel to push bit-zee around . bit-zee 's also got the ability to record and play back sound , so we use that , and you can use the remote control to trigger that function . we 've also got a motor controller here , and that , of course , allows us to control the motors , the speed , and direction . the camera that we took apart , we mounted to the board here , and that came out of this camera . and you can use that to take photos of things . and again , the arduino controls that , and we control the arduino with the remote control . and you can see there 's what 's called a bread board here , and that 's where all the wires go to basically to allow us to connect more things to the arduino and we have a number of resistors and transistors here . the resistors limit current flow , and the transistors are functioning as switches , so they allow us to control different devices , again , with our arduino here . so this is bit-zee , and bit-zee has got a personality . you can see bit-zee lights up and does different things when it gets its bumper pressed or when it bumps into something . so we 're going to show you more about bit-zee in the next few sessions .
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the resistors limit current flow , and the transistors are functioning as switches , so they allow us to control different devices , again , with our arduino here . so this is bit-zee , and bit-zee has got a personality . you can see bit-zee lights up and does different things when it gets its bumper pressed or when it bumps into something .
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will they ever make another bit-zee bot like a friend for bit-zee bot ?
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so how do we take our products here that we 've taken apart -- except for the bottle , which is just a recycled good -- and we want to see if we can create value by building something new out of the parts that we 've taken out of all these different components here . so what we 've done is we 've designed a new product that 's kind of designed to get kids excited about math and science , and in order for all these products to work for us , we need to incorporate some electronics . so we have an arduino . it 's a basically programmable chip that you can easily wire to other components . it 's great for prototyping . and we have a l298 chip , which is this guy , on a motor control . it 's set up as a motor controller here , and the board that you can see here comes as a kit . so we put that together , and that 'll allow us to control the speed and direction of motors so that we can move our little bot around . and then we have a little sound chip here , and so that allows us to record and playback sound . and so we 'll control everything with the arduino . the motor controller will let us drive motors around , and then this guy will help us to record and play back sound . and we 're going to use some of the parts from these products to make something cool . so after three weeks of development , we 've come up with bit-zee bot . bit-zee is a little programmable robot that drives around and has a little personality . you can control bit-zee with a universal remote control -- the one that we took apart , actually . and that sends a signal right here to the ir sensor . and then that communicates with the arduino board and tells the arduino board that it just got a signal . and the arduino can turn on and off different functions , so we can make bit-zee do a little dance , make its lights come on . and bit-zee has a couple of switches here . there 's a switch here and a switch here , and those came out of the radio . and then you can that bit-zee 's nose came from a bottle cap , and its eyes came from a bottle cap . and the power for bit-zee is just some batteries that you can get from radio shack in a battery holder , and the motors came from the hair dryer . so here 's one of the hair dryer motors . and basically , we use that as a wheel to push bit-zee around . bit-zee 's also got the ability to record and play back sound , so we use that , and you can use the remote control to trigger that function . we 've also got a motor controller here , and that , of course , allows us to control the motors , the speed , and direction . the camera that we took apart , we mounted to the board here , and that came out of this camera . and you can use that to take photos of things . and again , the arduino controls that , and we control the arduino with the remote control . and you can see there 's what 's called a bread board here , and that 's where all the wires go to basically to allow us to connect more things to the arduino and we have a number of resistors and transistors here . the resistors limit current flow , and the transistors are functioning as switches , so they allow us to control different devices , again , with our arduino here . so this is bit-zee , and bit-zee has got a personality . you can see bit-zee lights up and does different things when it gets its bumper pressed or when it bumps into something . so we 're going to show you more about bit-zee in the next few sessions .
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the resistors limit current flow , and the transistors are functioning as switches , so they allow us to control different devices , again , with our arduino here . so this is bit-zee , and bit-zee has got a personality . you can see bit-zee lights up and does different things when it gets its bumper pressed or when it bumps into something .
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i 'm just wondering why did n't the people that made bit-zee but a cover over all of the wires ?
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so how do we take our products here that we 've taken apart -- except for the bottle , which is just a recycled good -- and we want to see if we can create value by building something new out of the parts that we 've taken out of all these different components here . so what we 've done is we 've designed a new product that 's kind of designed to get kids excited about math and science , and in order for all these products to work for us , we need to incorporate some electronics . so we have an arduino . it 's a basically programmable chip that you can easily wire to other components . it 's great for prototyping . and we have a l298 chip , which is this guy , on a motor control . it 's set up as a motor controller here , and the board that you can see here comes as a kit . so we put that together , and that 'll allow us to control the speed and direction of motors so that we can move our little bot around . and then we have a little sound chip here , and so that allows us to record and playback sound . and so we 'll control everything with the arduino . the motor controller will let us drive motors around , and then this guy will help us to record and play back sound . and we 're going to use some of the parts from these products to make something cool . so after three weeks of development , we 've come up with bit-zee bot . bit-zee is a little programmable robot that drives around and has a little personality . you can control bit-zee with a universal remote control -- the one that we took apart , actually . and that sends a signal right here to the ir sensor . and then that communicates with the arduino board and tells the arduino board that it just got a signal . and the arduino can turn on and off different functions , so we can make bit-zee do a little dance , make its lights come on . and bit-zee has a couple of switches here . there 's a switch here and a switch here , and those came out of the radio . and then you can that bit-zee 's nose came from a bottle cap , and its eyes came from a bottle cap . and the power for bit-zee is just some batteries that you can get from radio shack in a battery holder , and the motors came from the hair dryer . so here 's one of the hair dryer motors . and basically , we use that as a wheel to push bit-zee around . bit-zee 's also got the ability to record and play back sound , so we use that , and you can use the remote control to trigger that function . we 've also got a motor controller here , and that , of course , allows us to control the motors , the speed , and direction . the camera that we took apart , we mounted to the board here , and that came out of this camera . and you can use that to take photos of things . and again , the arduino controls that , and we control the arduino with the remote control . and you can see there 's what 's called a bread board here , and that 's where all the wires go to basically to allow us to connect more things to the arduino and we have a number of resistors and transistors here . the resistors limit current flow , and the transistors are functioning as switches , so they allow us to control different devices , again , with our arduino here . so this is bit-zee , and bit-zee has got a personality . you can see bit-zee lights up and does different things when it gets its bumper pressed or when it bumps into something . so we 're going to show you more about bit-zee in the next few sessions .
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the camera that we took apart , we mounted to the board here , and that came out of this camera . and you can use that to take photos of things . and again , the arduino controls that , and we control the arduino with the remote control .
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is there a simpler robot that can do all the same things ?
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so how do we take our products here that we 've taken apart -- except for the bottle , which is just a recycled good -- and we want to see if we can create value by building something new out of the parts that we 've taken out of all these different components here . so what we 've done is we 've designed a new product that 's kind of designed to get kids excited about math and science , and in order for all these products to work for us , we need to incorporate some electronics . so we have an arduino . it 's a basically programmable chip that you can easily wire to other components . it 's great for prototyping . and we have a l298 chip , which is this guy , on a motor control . it 's set up as a motor controller here , and the board that you can see here comes as a kit . so we put that together , and that 'll allow us to control the speed and direction of motors so that we can move our little bot around . and then we have a little sound chip here , and so that allows us to record and playback sound . and so we 'll control everything with the arduino . the motor controller will let us drive motors around , and then this guy will help us to record and play back sound . and we 're going to use some of the parts from these products to make something cool . so after three weeks of development , we 've come up with bit-zee bot . bit-zee is a little programmable robot that drives around and has a little personality . you can control bit-zee with a universal remote control -- the one that we took apart , actually . and that sends a signal right here to the ir sensor . and then that communicates with the arduino board and tells the arduino board that it just got a signal . and the arduino can turn on and off different functions , so we can make bit-zee do a little dance , make its lights come on . and bit-zee has a couple of switches here . there 's a switch here and a switch here , and those came out of the radio . and then you can that bit-zee 's nose came from a bottle cap , and its eyes came from a bottle cap . and the power for bit-zee is just some batteries that you can get from radio shack in a battery holder , and the motors came from the hair dryer . so here 's one of the hair dryer motors . and basically , we use that as a wheel to push bit-zee around . bit-zee 's also got the ability to record and play back sound , so we use that , and you can use the remote control to trigger that function . we 've also got a motor controller here , and that , of course , allows us to control the motors , the speed , and direction . the camera that we took apart , we mounted to the board here , and that came out of this camera . and you can use that to take photos of things . and again , the arduino controls that , and we control the arduino with the remote control . and you can see there 's what 's called a bread board here , and that 's where all the wires go to basically to allow us to connect more things to the arduino and we have a number of resistors and transistors here . the resistors limit current flow , and the transistors are functioning as switches , so they allow us to control different devices , again , with our arduino here . so this is bit-zee , and bit-zee has got a personality . you can see bit-zee lights up and does different things when it gets its bumper pressed or when it bumps into something . so we 're going to show you more about bit-zee in the next few sessions .
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the motor controller will let us drive motors around , and then this guy will help us to record and play back sound . and we 're going to use some of the parts from these products to make something cool . so after three weeks of development , we 've come up with bit-zee bot .
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what are key parts to the robot ?
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so how do we take our products here that we 've taken apart -- except for the bottle , which is just a recycled good -- and we want to see if we can create value by building something new out of the parts that we 've taken out of all these different components here . so what we 've done is we 've designed a new product that 's kind of designed to get kids excited about math and science , and in order for all these products to work for us , we need to incorporate some electronics . so we have an arduino . it 's a basically programmable chip that you can easily wire to other components . it 's great for prototyping . and we have a l298 chip , which is this guy , on a motor control . it 's set up as a motor controller here , and the board that you can see here comes as a kit . so we put that together , and that 'll allow us to control the speed and direction of motors so that we can move our little bot around . and then we have a little sound chip here , and so that allows us to record and playback sound . and so we 'll control everything with the arduino . the motor controller will let us drive motors around , and then this guy will help us to record and play back sound . and we 're going to use some of the parts from these products to make something cool . so after three weeks of development , we 've come up with bit-zee bot . bit-zee is a little programmable robot that drives around and has a little personality . you can control bit-zee with a universal remote control -- the one that we took apart , actually . and that sends a signal right here to the ir sensor . and then that communicates with the arduino board and tells the arduino board that it just got a signal . and the arduino can turn on and off different functions , so we can make bit-zee do a little dance , make its lights come on . and bit-zee has a couple of switches here . there 's a switch here and a switch here , and those came out of the radio . and then you can that bit-zee 's nose came from a bottle cap , and its eyes came from a bottle cap . and the power for bit-zee is just some batteries that you can get from radio shack in a battery holder , and the motors came from the hair dryer . so here 's one of the hair dryer motors . and basically , we use that as a wheel to push bit-zee around . bit-zee 's also got the ability to record and play back sound , so we use that , and you can use the remote control to trigger that function . we 've also got a motor controller here , and that , of course , allows us to control the motors , the speed , and direction . the camera that we took apart , we mounted to the board here , and that came out of this camera . and you can use that to take photos of things . and again , the arduino controls that , and we control the arduino with the remote control . and you can see there 's what 's called a bread board here , and that 's where all the wires go to basically to allow us to connect more things to the arduino and we have a number of resistors and transistors here . the resistors limit current flow , and the transistors are functioning as switches , so they allow us to control different devices , again , with our arduino here . so this is bit-zee , and bit-zee has got a personality . you can see bit-zee lights up and does different things when it gets its bumper pressed or when it bumps into something . so we 're going to show you more about bit-zee in the next few sessions .
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so after three weeks of development , we 've come up with bit-zee bot . bit-zee is a little programmable robot that drives around and has a little personality . you can control bit-zee with a universal remote control -- the one that we took apart , actually .
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what are the principles of spider robot science kit ?
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so how do we take our products here that we 've taken apart -- except for the bottle , which is just a recycled good -- and we want to see if we can create value by building something new out of the parts that we 've taken out of all these different components here . so what we 've done is we 've designed a new product that 's kind of designed to get kids excited about math and science , and in order for all these products to work for us , we need to incorporate some electronics . so we have an arduino . it 's a basically programmable chip that you can easily wire to other components . it 's great for prototyping . and we have a l298 chip , which is this guy , on a motor control . it 's set up as a motor controller here , and the board that you can see here comes as a kit . so we put that together , and that 'll allow us to control the speed and direction of motors so that we can move our little bot around . and then we have a little sound chip here , and so that allows us to record and playback sound . and so we 'll control everything with the arduino . the motor controller will let us drive motors around , and then this guy will help us to record and play back sound . and we 're going to use some of the parts from these products to make something cool . so after three weeks of development , we 've come up with bit-zee bot . bit-zee is a little programmable robot that drives around and has a little personality . you can control bit-zee with a universal remote control -- the one that we took apart , actually . and that sends a signal right here to the ir sensor . and then that communicates with the arduino board and tells the arduino board that it just got a signal . and the arduino can turn on and off different functions , so we can make bit-zee do a little dance , make its lights come on . and bit-zee has a couple of switches here . there 's a switch here and a switch here , and those came out of the radio . and then you can that bit-zee 's nose came from a bottle cap , and its eyes came from a bottle cap . and the power for bit-zee is just some batteries that you can get from radio shack in a battery holder , and the motors came from the hair dryer . so here 's one of the hair dryer motors . and basically , we use that as a wheel to push bit-zee around . bit-zee 's also got the ability to record and play back sound , so we use that , and you can use the remote control to trigger that function . we 've also got a motor controller here , and that , of course , allows us to control the motors , the speed , and direction . the camera that we took apart , we mounted to the board here , and that came out of this camera . and you can use that to take photos of things . and again , the arduino controls that , and we control the arduino with the remote control . and you can see there 's what 's called a bread board here , and that 's where all the wires go to basically to allow us to connect more things to the arduino and we have a number of resistors and transistors here . the resistors limit current flow , and the transistors are functioning as switches , so they allow us to control different devices , again , with our arduino here . so this is bit-zee , and bit-zee has got a personality . you can see bit-zee lights up and does different things when it gets its bumper pressed or when it bumps into something . so we 're going to show you more about bit-zee in the next few sessions .
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so what we 've done is we 've designed a new product that 's kind of designed to get kids excited about math and science , and in order for all these products to work for us , we need to incorporate some electronics . so we have an arduino . it 's a basically programmable chip that you can easily wire to other components .
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where is the code for the arduino ?
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so how do we take our products here that we 've taken apart -- except for the bottle , which is just a recycled good -- and we want to see if we can create value by building something new out of the parts that we 've taken out of all these different components here . so what we 've done is we 've designed a new product that 's kind of designed to get kids excited about math and science , and in order for all these products to work for us , we need to incorporate some electronics . so we have an arduino . it 's a basically programmable chip that you can easily wire to other components . it 's great for prototyping . and we have a l298 chip , which is this guy , on a motor control . it 's set up as a motor controller here , and the board that you can see here comes as a kit . so we put that together , and that 'll allow us to control the speed and direction of motors so that we can move our little bot around . and then we have a little sound chip here , and so that allows us to record and playback sound . and so we 'll control everything with the arduino . the motor controller will let us drive motors around , and then this guy will help us to record and play back sound . and we 're going to use some of the parts from these products to make something cool . so after three weeks of development , we 've come up with bit-zee bot . bit-zee is a little programmable robot that drives around and has a little personality . you can control bit-zee with a universal remote control -- the one that we took apart , actually . and that sends a signal right here to the ir sensor . and then that communicates with the arduino board and tells the arduino board that it just got a signal . and the arduino can turn on and off different functions , so we can make bit-zee do a little dance , make its lights come on . and bit-zee has a couple of switches here . there 's a switch here and a switch here , and those came out of the radio . and then you can that bit-zee 's nose came from a bottle cap , and its eyes came from a bottle cap . and the power for bit-zee is just some batteries that you can get from radio shack in a battery holder , and the motors came from the hair dryer . so here 's one of the hair dryer motors . and basically , we use that as a wheel to push bit-zee around . bit-zee 's also got the ability to record and play back sound , so we use that , and you can use the remote control to trigger that function . we 've also got a motor controller here , and that , of course , allows us to control the motors , the speed , and direction . the camera that we took apart , we mounted to the board here , and that came out of this camera . and you can use that to take photos of things . and again , the arduino controls that , and we control the arduino with the remote control . and you can see there 's what 's called a bread board here , and that 's where all the wires go to basically to allow us to connect more things to the arduino and we have a number of resistors and transistors here . the resistors limit current flow , and the transistors are functioning as switches , so they allow us to control different devices , again , with our arduino here . so this is bit-zee , and bit-zee has got a personality . you can see bit-zee lights up and does different things when it gets its bumper pressed or when it bumps into something . so we 're going to show you more about bit-zee in the next few sessions .
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so how do we take our products here that we 've taken apart -- except for the bottle , which is just a recycled good -- and we want to see if we can create value by building something new out of the parts that we 've taken out of all these different components here . so what we 've done is we 've designed a new product that 's kind of designed to get kids excited about math and science , and in order for all these products to work for us , we need to incorporate some electronics .
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9 , what is the spelling of that item ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ?
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why attack 2-methyl pent-2-ene with hydronium instead of water ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond .
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just to clarify , the first molecule could 've been calleded z-2-methylpent-2-ene as well right ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon .
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so what makes the pi bond between the two carbons less stable/prone to elicit these reactions ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium .
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is a pi bond easier to break than a sigma bond ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ?
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why does n't attack oh- the carbocation at 05 ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch .
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i mean in the end of the theory sal tells us there is a free proton left and this will bind to the oh- anyway , right ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon .
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isnt one of the carbons a tertiary and the other one a secondary carbon ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton .
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as in why is the hydrogen moved to another water molecule to form hydronium ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton .
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why does the name of the 1st molecule have 2 before the ene ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons .
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why does oxygen have a positive charge , with 1 lone pair left should n't it have a negative charge ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons .
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why does sal not put the positive charge on the oxygen atom ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ?
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if two oxygen electrons are shared , then why is n't a double bond formed ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water .
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what 's the name of the final product of the addition ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule .
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one of the methyl groups was an ether ) , would we expect that carbon to be r or s once the hydrogen leaves the water group forming the final product ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons .
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a carboncation is formed with a +1 charge , but should n't it have a +2 charge ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ?
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in this video , sal draws the molecule of water with two pairs of free electrons and two more electrons bonded to two hydrogen atoms , but should n't there be another pair of electrons because oxygen has 6 electrons in it 's last shell and needs two more to complete an octet ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond .
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so that means the new h should bind to the right c. but here sal says that the c-c bonding is what 's important ... can someone explain this to me please ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons .
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so the carbocation will always be the most `` substituted '' carbon ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive .
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would the rate determining step of this reaction be the first step of pi bond attacking hydronium ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton .
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why do the h and oh attach onto the same side of the molecule ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon .
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how do we apply markavnikov 's rule if both carbons are bonded to same number of carbons or functional groups ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene .
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is there a difference between having the h2o on top of the arrow or bottom of the arrow ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene .
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what is the name of the resulting compound ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon .
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if you have two molecules that are both 2-butene , one is z-isomer and the other is e-isomer , which is is more reactive ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond .
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also , if a molecule has more hydrogens , it is more stable than those with less right ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium .
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will the catalizer always be the first part of a mechanism ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water .
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would the final product be called 1-butanol ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again .
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what were we trying to obtain as the final result ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water .
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is this mechanism syn addition or trans addition ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons .
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why the oh^ ( -1 ) is not just bonds to the positively charged carbon ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged .
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at 2.52 sal states that in the h3o+ , the o wants the electron back from hydrogen , but why ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ?
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why does n't it take the electrons from the other hydrogens as well then ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule .
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what is the significance of the h-oh above the reaction arrow in the original equation , since the h3o is what is reacting ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here .
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i understand that there is a newly formed carbocation that is reactive because it is positively charged and thus is in search of an electron , but the water molecule does n't make a good electron donor , right ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons .
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sal , we only talk about situations in which the carbon becomes a carboncation , are there any moments in which it becomes a carbon anion ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around .
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since when h2o attacks the carbocation attack from both sides is equally hindered..so why is the cis product shown in particular ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge .
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if oxygen is electronegative in nature then why would oxygen will give its electron to hydrogen in the formation of h3o+ ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons .
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shouldn '' t sal have put a positive charge on the oxygen atom ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule .
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how would you usse this in reverse to determine the starting product ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ?
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why is 2-methyl pent-2-ene and not 2 bimethyl hex 2 ene ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water .
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does markovnikov 's rule work for all addition reactions apart for free radical addition ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium .
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i did n't get why that oxygen has positive charge ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond .
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so is water the solvent or the reagent ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond .
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would this type of similar mechanism be used to do for example : reaction of ethylene with water to produce ethanol ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond .
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why isnt there hydrogen bonding between oxygen of oh and the hydrogen attached to the other carbon ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond .
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why isnt there hydrogen bonding between oxygen of oh and the hydrogen attached to the other carbon ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ?
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can this reaction also be named hydration ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene .
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what is the name of the newly formed compound ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule .
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would this be considered a hydration reaction ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon .
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the carbon is acting as a base and taking the hydrogen away from hydronium , right ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond .
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how can we know that the following catalyst is better for a specific reaction ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons .
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do you have to draw the arrow coming from the lone electron of the carbon atom to the hydrogen ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond .
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how do we know that a particular equation is in equilibrium ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add .
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how can we generalize the idea of equilibrium while solving such problems ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge .
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for the last part of the mechanism , is n't it supposed to be that the h-o-h bonded to another h should be a positive charge ( making it a hydronium ion ) since the o lost one of its electrons which it shared with the h coming from the h-o-h bond from the entire pentene molecule ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond .
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it has a complete octate ) , so how can the carbocation ( formed above ) attack and extract an electon from the water molecule ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again .
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and is n't oxygen the second most electonegative element in the periodic table , so wo n't it attract the bond electrons towards itself ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons .
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why does the oxygen in the substituent group have a positive charge ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton .
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oxygen already has 8 valence electrons , 1 lone pair and 3 pairs sharing , so should n't the charge be neutral ?
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anytime you 're trying to come up with a mechanism for a reaction , it 's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different . so what we 're starting with , we could call this one , two , three , four , five , so this is , let 's see , we have methyl group on the number two carbon , it is a pentene , and that is double-bond between the number two and number three carbons , so this is two-methyl-pent-2-ene . so that 's what we start with , we 're in the presence , we 're in an acidic environment , we 've got what 's gon na be catalyzed by our hydronium here , and we end up with this , and how is our product different from what we started with ? well the double bond is now gone , the number three carbon gains this hydrogen , and now the number two carbon gains a hydroxyl group . so one way to think about this is , in the presence of an acid , it 's acid-catalyzed , we have gained two hydrogens and an oxygen , which is what we 've gained , what could be used to make a water . and this is actually called an acid-catalyzed addition of water . the water is n't sitting on one part of the molecule , but if you take the hydrogen we added , and the hydroxyl we added , if you combine them , that 's what you need to make a water . so let 's think about how we can , how this actually happens in the presence of our hydronium . so let me redraw this molecule right over here . so we copy and paste it , so that 's not exactly it yet , that is just with the single bond . so let me draw , woops , wrong tool . let me draw the double bond there . and now let me put it in the presence of some hydronium . alright , so we have an oxygen bonded , two . so this would just be water , as oxygen has two lone pairs , but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton , thus making the entire molecule positive , because the hydrogen proton is positive . so there you go , this now has a positive charge . and this can be pretty reactive , 'cause we know that oxygen is quite electronegative , it lives to keep its electrons . so what is there was a way , what if there 's a way for the oxygen to take back the electrons in this bond right over here , the two electrons in this bond . well what if one of these carbons , especially the ones that have the double bonds , what if some of the electrons from this double bond could be used to snab , to take that hydrogen proton , and then oxygen can hog its electrons again . and you might say , `` oh that 's reasonable , but which of these carbons would actually do it ? '' and to think about which of those carbons would do it , we have to turn to markovnikov 's rule . markovnikov 's rule tells us that , look , if you have a reaction like this , and alkene reaction , the carbon that already has , that 's already attached to more hydrogens is more likely to gain more hydrogens , the carbon that 's attached to more functional groups is more likely to gain more functional groups . another way to think about it is think about , well that is the order of the carbons ? because the higher order of carbon , the more stable it will be if it forms some type of cation . so if you look at this carbon right over here , our number two carbon , let me circle it , our number two carbon is bonded to one , two , three carbons . so this is a tertiary carbon . this one right over here , this carbon on the other side of the double bond is only bonded to one , two carbons . so this is a secondary carbon . so the tertiary carbon is going to be more stable as a carbocation , you can think of it as it can spread the charge a little bit . so it would be more likely to lose the electrons in one of these , in one of these bonds , and so the way that we can think about this mechanism , and it might be a little bit clearer when we form the carbocation , is let 's have , going to do this in blue . let 's have these two electrons that form this bond , well now they form a bond with that hydrogen , and now the oxygen can take back these two electrons , and what are we going , what is going to result ? and i 'm drawing it in equilibrium , remember all of these things are going back and forth depending on how things bump into each other , but what are we left with ? so let me copy and paste this again and i 'm copying and pasting in a way that , just so i , this is the backbone , and i 'll add what i need to add . so once this happens , we have this carbon , the number three carbon , now , woops i keep using the wrong tool . the number three carbon now forms a bond with this hydrogen just like that . this carbon , our number two carbon , has lost an electron , it 's no longer sharing this bond . and so now it is going to have a positive charge , it is a carbocation and once again is a tertiary carbocation , it is bonded to one , two , three carbons . that is stable , more stable than if we did it the other way around , if this one grabbed the hydrogen somehow , then this would be a secondary carbocation , it 'd be harder for it to spread that positive charge around . and what about our , what about our molecule up here ? let 's see what it looks like now . we have our oxygen bonded to the two hydrogens , it had one of those lone pairs , and now the electrons in this bond are now going to form another lone pair . so it took back an electron , or you can think of it , it gave away a hydrogen proton . and so this is now just neutral water , and we see that we have a conservation of charge here , this was positive in charge , now our original molecule is positively charged . and what feels good about this is we 're getting , we 're getting close to our end product , at least on our number three carbon , we now have , we now have this hydrogen . now we need to think about , well how do we get a hydroxyl group added right over here ? well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here . and you can imagine , if they just pump into each other in just the right way , this is , water is a polar molecule , it has a partially negative end near the oxygen because the oxygen likes to hog the electrons , and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged , so you can imagine the oxygen end might be attracted to this tertiary carbocation , and so just bumping it in just the right way , it might form a bond . so let me say these two electrons right over here , let 's say they form a bond with this , with that number two carbon , and then what is going to result ? so let me draw , so what is , what is going to result , let me scroll down a little bit , and let me paste , whoops , let me copy and paste our original molecule again . so , here we go . so what could happen ? this is the one we constructed actually , so we have the hydrogen there . we have the hydrogen , now this character , so we have the water molecule , so oxygen bonded to two hydrogens , you have this one lone pair that is n't reacting , but then you have the lone pair that does do the reacting . and so it now forms a bond . woops , let me do it in that orange color . it now , it now forms an actual bond . and we 're really close to our final product , we have our hydrogen on the number three carbon , we have more than we want on our number two carbon , we just want a hydroxyl group , now we have a whole water bonded to the carbon . so somehow we have to get one of these other hydrogens swiped off of it , well that could happen with just another water molecule . so let 's draw that . so another water molecule someplace , i 'll do the different color just to differentiate , although as we know water , well it 's hard to see what color is water if you 're looking at the molecular scale . so here we go , and we 're really in the home stretch at this point . you have another water molecule , let 's say , let me pick a color . so let 's say these electrons right over here , maybe they form a bond with that hydrogen proton , and then these , the electrons in that bond can go back to form a lone pair on that oxygen , and then what are we left with ? and this really is the home stretch . so we are in equilibrium with , so let me draw my five carbons , so let 's see , i have ... h3c , carbon , carbon , ch2 , ch3 , i have a ch3 , i say h3c instead of ch3 , i wrote it that way just so it 's clear that the carbons are bonded to the carbons , you have the original hydrogen right over there , you have the one that we just added as part of this mechanism , you have this orange bond to now , this hydroxyl group , the hydroxyl group , and it had one lone pair before , it had one lone pair before , but bound to both of the electrons from this bond to form another lone pair . to form another lone pair , which i am depicting in pink , and then this water is now , this water molecule is now a hydronium molecule . so let me draw that , so this is now , oxygen , hydrogen , had one lone pair that did n't react , and it had one lone pair that i put in blue that is reacting with this hydrogen proton . with that hydrogen , just like this , and so since it got the hydrogen proton it 's giving its , sharing its electrons now , now this has a positive , this has a positive charge , just like that . i have to be very careful , in the last step i forgot to draw the positive charge ! we always wan na make sure that your charge is being conserved , we started off with the positive charge on the hydronium , then we have the positive charge on the tertiary carbocation right over here on our number two carbon , and now we have the positive charge , would be right over here , because that oxygen , what we saw before that oxygen , which this water molecule was neutral , but you could say , you could view it is `` well now it 's going to be , it 's now sharing these two electrons instead of keeping them , '' so you can view it is maybe it 's giving away an electron , and so now it becomes positive . and then , and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium . but just like that we are done . we have added a hydroxyl group and a hydrogen combined , that 's a water , so that 's why we call it addition of water , and it was catalyzed by acid , so it 's an acid-catalyzed addition of water .
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well we have all this water , we have all this water floating around , let me , i could use this water molecule but the odds of it being the exact same water molecule , we do n't know . but there 's all sorts of water molecules , we 're in an aqueous solution , so let me draw another water molecule here . so the water molecules are all equivalent , but let me draw another water molecule here .
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is the solution always aqueous ?
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