Main content
Current time:0:00Total duration:6:50

Video transcript

- [instructor] On the left, we have a tertiary alcohol. The carbon that's bonded to the OH would be the alpha carbon, and that carbon is directly bonded to three other carbons, which is why this is tertiary. This carbon, let's call beta one. This carbon, let's say this is the beta two carbon. And finally, this would be the beta three carbon. If you react a tertiary alcohol with sulfuric acid, and you heat up your reaction mixture, this is gonna be an E1 mechanism, and we'll talk about the regiochemistry for this reaction, and why this is a regioselective reaction in a few minutes. First, let's go through the mechanism. We know that with an alcohol, the alcohol we've protonated by the sulfuric acid. So instead of drawing out the dot structure for sulfuric acid, I'll just write H+ here. So sulfuric acid is a source of protons. And one of the lone pairs of electrons and oxygen picks up that proton. So our first step is a proton transfer, and let me draw that in here. So now this oxygen would be bonded to two hydrogens with one lone pair of electrons and a plus one formal charge on the oxygen. So the loan pair, let's say this lone pair here in magenta, picks up a proton from sulfuric acid to form this bond, and that gives us water as a leaving group. And we know water is a good leaving group. The electrons in this bond can come off onto the oxygen to form H2O. And when that happens, we take a bond away from this carbon in red. So we're gonna form a carbocation. So we take away a bond from the carbon in red. Let me go ahead and draw in our carbocation here. So the carbon in red would be this carbon. That carbon would have a plus one formal charge. This is a tertiary carbocation, because the carbon in red is directly bonded to three other carbons. So this one, this one, and this one. So this is a stable carbocation. Next, let's think about the next step of an E1 mechanism. A base is gonna come along and take a proton from one of the beta carbons. And let's start with beta two. So let's thinking about a proton on the carbon, so let me draw on in here. And our weak base comes along and takes this proton, which would leave this electrons to move into here to form a double bond. So let's draw that product. So we would have our double bond forming right here. Let me draw in the rest of the molecule. So our electrons in, let me make these blue here. So the electrons in light blue are going to move in to form our double bond, and so we would get this out alkene. So that's beta two. Let's think about what would happen if we took a proton away from beta one. So let's draw that one in next. So I'm gonna draw the carbocation again. Let me draw that in here. And beta one would be up here, so let me put in a proton on beta one. You think about a weak base coming along, and taking this proton, so our base is probably water. And these electrons would move into here this time. So if that happens, let's draw that product. Our double bond would form up here, and let me draw in the rest of this molecule. So let me use red for those electrons. So electrons in red are going to move into here to form this alkene. Notice that the two alkenes that we just drew is really the same molecule. This is the same compound. So we haven't formed two different products. If you take a proton away from the beta one or the beta two carbon, you're gonna make the same alkene. But what about the beta three carbon? So that's our last example. And let me go ahead. I forgot to in a plus one formal charge on our carbocation. Let me draw one more carbocation, the same one at tertiary carbocation. The difference is this time we're gonna take a proton away from our beta three carbon. And so, let me draw in a proton there. And we think about a weak base coming along and taking that proton, so I'll draw in my weak base here. So it takes this proton and these electrons move into here, so let's draw this product. So we would have our alkene that looks like this. So let's follow those electrons. I'll make them dark blue. Electrons in this bond move into here to form our double bond. And so, now we've gone through the complete mechanism, and we have two products. So let me circle our two products, so this is really just one product, and then this would be our second product. For this reaction, we actually get 90% of the alkene on the right and 10% of the alkene on the left. And so, let's look at the degree of substitution of our two products, and let's start with the one on the right. So, let me use red for this. If we think about the degree of substitution for the alkene on the right, by drawing my hydrogen right here, it makes it a little bit easier to see we have three alkyl groups, so this one, this one, and this one. So this would be a trisubstituted alkene. So the one on the right is a trisubstituted alkene, and the one on the left, so this one right here, would be a disubstituted alkene. These are the two carbons across our double bond. We have two hydrogens on this carbon, and the carbon on the right has two alkyl groups bonded to it. So this one is a disubstituted alkene. Now we've gone through the whole E1 mechanism, and we've seen that we get a disubstituted product, and a trisubstituted. Now let's think about regiochemistry. For this reaction, it's the region of the molecule where the double bond forms. For the disubstituted product, the double bond formed in this region of the molecule, and for the trisubstituted product, the double bond formed in this region. The trisubstituted product is the major product, and it's also the more stable alkene. So remember, from the video in alkene stability, the more substituted your alkene is, the more stable it is, so this product is more stable, and that's why we form more of it. And the more stable products or the more substituted product is called the Zaitsev product. So we say that this E1 reaction is regioselective because it has a preference to form the more stable product, the more substituted product, which we call the Zaitsev product.