The acid-catalyzed hydration we learned here in Chapter 8 is r...

Question

The acid-catalyzed hydration we learned here in Chapter 8 is reversible.

Chemical reaction diagram showing the dehydration of an alcohol to form an alkene, with sulfuric acid as a catalyst.

(a) Propose a mechanism for the formation of an alkene from an alcohol.

Accepted Answer

All right. Hi, everyone. So for this question, let's propose a mechanism for the acid catalyzed dehydration of an alcohol shown below. Now, like the question implies, right, this is going to be an example of the acid catalyzed dehydration of an alcohol. Now, this reaction, if you recall follows an E one mechanism in which the hydroxy group of one carbon and the hydrogen of another are subsequently removed to create a pi bond between the two. So here, right, the first thing I'm going to do is I'm going to go ahead and redraw our starting material on the bottom so that we can illustrate the mechanism. So here's my alcohol and let's go ahead and begin with our first step, right. So the mechanism in question is of acid catalyzed dehydration. So what that means right is that our acid is going to react with our hydroxy group in the first step as a means of catalyzing it, right. So here I'm going to go ahead and draw the Lewis structure of sulfuric acid in blue because here's the thing, right, the fact that the hydroxy group has to be eliminated from the molecule in addition to the other hydrogen implies that the hydroxy group is the leaving group, right? It has to go ahead and leave the molecule. But the problem is that hydroxy groups do not make good leaving groups by themselves. Because when they detach themselves from carbon, they detach themselves and form hydroxide, which is a strong base. So what that means is that we have to somehow modify the hydroxy group to make it a better living group. And one thing we can do is proton it, which is exactly what we're going to do here. So the fact that we're reacting with an acid means that we're going to go ahead and proin or hydroxy hydrogen or oxygen. Excuse me. So because of that, we end up with an intermediate in which we have a pronated alcohol, use our original hydrogen. And now oxygen is going to have a positive charge because it has another bond through another hydrogen. So now this prop needed this pro needed alcohol here. This is an excellent leaving group, right? Because when it goes ahead and detaches itself from the carbon, it's attached to, right, it becomes water. And so when it attaches itself to become water, it's stable on its own. So that's why it's a good lead group. And as a consequence of that leaving right, we end up with this tertiary carbo. So now we can go ahead and talk about so called beta elimination because recall that an elimination reaction involves the removal of a beta proton. So here, right, the carbon that bears the positive charge is the alpha carbon and beta carbons are those that are adjacent to the alpha carbon. So here what you'll notice is that we actually have two non equivalent beta positions because yes, we have two beta carbons on the ring here. But because of the element of symmetry present in this molecule, they would result in the same product. So for ease of reference, I'm only going to go ahead and label one. But the methyl group on the top here it is another beta position. So because we have two non equivalent beta positions, right, we can have potentially two different products. But here's the thing, right, in the majority of cases, right, whenever the product isn't all keen like the one we have here, right. Reactions are going to follow what's called Zit's rule. And in Zit's rule, the more substituted double bond is what's going to be formed as the major product. So because of that, right, the beta protein that we're going to remove is from the beta carbon inside of the ring, right? Because that would form a tris substituted alkene, whereas our other beta position would form a di substituted al. So our tri substituted his favorite and to go ahead and remove that beta proton, right, we're going to go ahead and bring back the water molecule that we formed in the previous step. The one which has gone ahead and detached itself from carbon to form the Carbocaine in the first place. So now our beta proton from the ring is what's going to be removed. And the pair of electrons between the beta carbon and the beta proton is going to go ahead and move in between the alpha and beta carbon. And so that is what creates the pie bond between the alpha and beta carbons. So now our products are as follows, right? We now end up with a pie bond in between the alpha and beta carbons inside the ring. And we also end up with water as a byproduct and there you have it, this is going to be our complete mechanism for the acid catalyzed dehydration of an alcohol. And it's important to recognize here that because we're forming a carbo CAD ion during our mechanism, right. Rearrangement is going to be a concern. But the thing is that here, right, the Carbocaine that we formed was already tertiary and recall that tertiary Carbocaine are quite stable by themselves, right? So here rearrangement would not occur because the carbo CAD ion is already quite stable, but that's not always the case, right? So when you do have a carbo CAD ion, you should always check and make sure whether or not rearrangement is going to occur before you proceed with beta elimination. But with that being said, if you stuck around to the end, thank you so much for watching and I hope you found this helpful.

The acid-catalyzed hydration we learned here in Chapter 8 is reversible.

(a) Propose a mechanism for the formation of an alkene from an alcohol.

Key Concepts

Acid-Catalyzed Dehydration

Carbocation Stability

Elimination Reactions (E1 and E2)

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