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Current time:0:00Total duration:7:24

Video transcript

here's another sn1 carbocation rearrangement but this one's pretty challenging so they look on the left this is our starting alkyl halides and we're heating this alkyl halide with water to produce this tertiary alcohol on the right so the first step of this mechanism should be loss of the leading group so these electrons come off onto the bromine to form the bromide anion so when we do that we're taking a bond away from this carbon in red so that gets a +1 formal charge so if I draw in my six membered ring the carbon in red is this one and it has a plus 1 formal charge so we have a carbo cation and we put in these two methyl groups here this is a secondary carbo cation because the carbon in red is directly bonded to two other carbons so this is secondary if we look at our product look at our product here we have a five membered ring not a six membered ring and we know our nucleophile would have to be water in this reaction and so the oxygen and water forms a bond in our products with this carbon so let me mark that carbon in blue so this carbon in blue which means that must be the plus one formal charge that must be the carbo cation of how we form the product just based on what we know from earlier examples so let's sketch that in so we have our five membered ring and the carbon in blue is this carbon right here and that one must have our plus one formal charge because our nucleophile would attack that carbon and blues let me do I just draw in the water molecule here our nucleophile attacking that so here's our water molecule two lone pairs of electrons on the oxygen and our nucleophile attacks our electrophile to form a bond between the oxygen and that carbon so that would form let's draw that in here next we have our five membered ring we have our carbon in blue which is right here and let's draw in a bond to our oxygen oxygen is bonded to two other hydrogen's we still have a lone pair of electrons on this oxygen which gives us oxygen a plus 1 formal charge so the last step of this mechanism is just loss of a proton to leave a proton transfer so I was going to write here minus h plus and a base like a water molecule would come along and take one of those protons to give us our final product so we've seen that in lots of earlier videos but now let's think to ourselves how do we go from the carbo cation on the left to the carbo cation on the right notice the carbo cation on the right is a tertiary carbo cation the carbon in blue is directly bonded to three other carbons so this one is tertiary we must get some sort of rearrangement going from a secondary carbo cation to a tertiary carbyl going but this one is different from any rearrangement we've seen it so far so let's go to the video to make this a little bit more clear here's the model of our carbo cation the carbon to +1 formal charge is this carbon I left the hydrogen in on that carbon only just to make it easier to see so for our carbo cation rearrangement I'm going to take these electrons in the ring so we're going to break the ring go from a six membered ring to a five membered ring with this carbo cation rearrangement and we change the hybridization states of two carbons so let's get a different model so we can see better what the carbo cation looks like so the carbon with the hydrogen is now sp3 hybridized and this carbon is sp2 hybridized and planar took some images from the video to help us understand this tricky rearrangement let's start with this picture the carbon in red that I marked above what this carbon we Circle it one more time here that is this carbon let me highlight another carbon on here and I'll do this one in green so this carbon in green is this one over here so in the video we took these electrons and we move them over to this carbon so when we're drawing our mechanism up here we take these electrons we move them over to the carbon in red and that forms the bond between the carbon in green and the carbon in red so let's show that up here on the drawing so that moves us to a five membered ring and the carbon in red is this one and the carbon in green is this one and let's let's go ahead and make let's make this carbon right here blue so this carbon in blue it's still attached to the carbon in red so let me just sketch that in here and then we have two methyl groups coming off that carbon so the carbon in blue is this one now when we move to actually let's highlight the carbon in blue over here so this is the carbon in blue on this picture so now let's move to this central picture here so the carbon in green is this one the carbon in red is this one and the carbon in blue is this one down here so this forms our tertiary carbo cations there's a plus one formal charge on the carbon in blue and this is the same thing right this is just two different ways of drawing our carbo cation the one on the right is a little bit better in terms of how to draw it but it's really the same picture and let's identify those carbons here so the carbon in red is this one and on our picture it is this one here the carbon in green is this one which is this carbon and finally the carbon in blue right here is this carbon the carbon in red goes let's go back to the to the picture all the way over here on the left the carbon in red starting off with a plus one formal charge is sp2 hybridized but notice we move over here moving to sp3 hybridization so tetrahedral geometry around the carbon in red the carbon in blue right is going from sp3 hybridization over here to SP at to hybridize a and that's why I switch the model sets because this carbon in blue is now sp2 hybridized 1 it's a carbo cations and here's our carbon in blue so it has planar geometry around it so hopefully the the models helps clear up this strange carbo cation rearrangements I should say so let's go back here and let's look at the entire mechanism so the first step is loss of our leaving group then we get a carbo cation rearrangements secondary carbo cation to a tertiary carbo cation the next step is nucleophilic attack where the water molecule attacks our positive charge and then finally we have an acid-base reaction we remove one of the protons to form our products