Sn1 carbocation rearrangement (advanced)

- [Instructor] Here's another SN1 carbocation rearrangement but this one's pretty challenging. If you look on the left this is our starting alkyl halide and we're heating this alkyl halide with water to produce this tertiary alcohol on the right. The first step of this mechanism should be loss of a leaving group. These electrons come off onto the bromine to form the bromide anion. When we do that, we're taking a bond away from this carbon in red so that gets a plus one formal charge. If I draw in my six-membered ring, the carbon in red is this one and it has a plus one formal charge, so we have a carbocation. And we put in these two methyl groups here. This is a secondary carbocation because the carbon in red is directly bonded to two other carbons so this is secondary. If we look at our product, we look at our product here we have a five-membered ring and not a six-membered ring, and we know our nucleophile would have to be water in this reaction. And so, the oxygen and water forms a bond in our product with this carbon. Let me mark that carbon in blue. This carbon in blue which means that must be the plus one formal charge. That must be the carbocation of how we form the product just based on what we know from our earlier examples. Let's sketch that in. We have our five-membered ring and the carbon in blue is this carbon right here. That one must have our plus one formal charge because our nucleophile would attack that carbon in blue. Let me go ahead and just draw in the water molecule here, our nucleophile attacking that. 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. That would form, let's draw that in here next. We have our five-membered ring. We have our carbon in blue which is right here and let's draw in a bond to our oxygen. Our oxygen is bonded to two other hydrogens. We still have a lone pair of electrons on this oxygen which gives this oxygen a plus one formal charge. The last step of this mechanism is just loss of a proton. So we have proton transfers. I'm just gonna write here minus H plus and a base like a water molecule. We come along and take one of those protons to give us our final product. We've seen that in lots of earlier videos. But now let's think to ourselves, how do we go from the carbocation on the left to the carbocation on the right? Notice the carbocation on the right is a tertiary carbocation. The carbon in blue is directly bonded to three other carbons so this one is tertiary. We must get some sort of rearrangement going from a secondary carbocation to a tertiary carbocation but this one's different from any rearrangement we've seen so far. Let's go to the video to make this a little bit more clear. Here's the model of our carbocation, the carbon, the plus one formal charge is this carbon. I left the hydrogen in on that carbon only just to make it easier to see. For our carbocation rearrangement, I'm gonna take these electrons and the rings, we're gonna break the ring, go from a six-membered ring to a five-membered ring with this carbocation rearrangement. And we change the hybridization states of two carbons. Let's get a different model so we can see better what the carbocation looks like. The carbon with the hydrogen is now SP3 hybridized and this carbon is SP2 hybridized and plainer. 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 with this carbon and 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. This carbon in green is this one over here. In the video, we took these electrons and we move them over to this carbon. When we're drawing our mechanism up here, we take these electrons and we move them over to carbon in red. And that forms a bond between the carbon in green and the carbon in red. So, let's show that up here on the drawing. That moves us to a five-membered ring and the carbon in red is this one, and the carbon in green is this one. And let's go ahead and make, let's make this carbon right here blue. This carbon in blue is still attached to the carbon in red so let me just sketch that in here and then we have two methyl groups coming off that carbon. The carbon in blue is this one now. When we move to, actually let's highlight the carbon in blue over here. This is the carbon in blue on this picture. Now, let's move to this central picture here. The carbon in green is this one, the carbon in red is this one and the carbon in blue is this one down here. This forms our tertiary carbocations. There's a plus one formal charge on the carbon in blue. And this is the same thing, right? This is just two different ways of drawing our carbocation, the one on the right is a little bit better in terms of how to draw it. But it's really the same picture and let's identify those carbons here. 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 picture all the way over here on the left. The carbon in red starting off with a plus one formal charge is SP2 hybridized. But notice when we move over here, we're moving to SP3 hybridization. Tetrahedral geometry around the carbon in red. The carbon in blue, right, is going from SP3 hybridization over here to SP2 hybridization and that's why I switched the model sets because this carbon in blue is now SP2 hybridized when it's a carbocation. Here's our carbon in blue. It has plainer geometry around it. Hopefully the models helped clear up this strange carbocation rearrangement or challenging I should say. Let's go back here and let's look at the entire mechanism. The first step is loss of our leaving group then we get our carbocation rearrangement going from a secondary carbocation to a tertiary carbocation. The next step is nucleophilic attack where the water molecule attacks our positive charge. And then finally, we have an acid base reaction. We remove one of the protons to form our product.