The mechanism of the acid-catalyzed hydration of an alkene to ...

Question

The mechanism of the acid-catalyzed hydration of an alkene to make an alcohol is shown. If this reaction is reversible, draw a mechanism that justifies formation of the alkene from the alcohol under similar conditions. How do you know this mechanism is correct?

Accepted Answer

Welcome back, everyone. Let's take a look at our next problem. The mechanism of the acid catalyzed hydration of an alkene to make an alcohol is shown below sketch, the mechanism of the reverse reaction. So we have this mechanism and the alkene we're starting with is cyclops. It's reacting with a hydro num ion. So H 30 plus the electrons from one of those cc bonds are detonating or hydro ion with the electrons from the oh bond going on to the oxygen to generate water. Then we see the next step we've generated a Carbocaine. The double bond has been broken and one of the carbons in that cyclops has a positive charge. This reacts to the molecule of water. The water is the Nile, it's one of its lone pairs bonds to that positively charged carbon. So we now have a cyclops with an 02 plus substituent. Another molecule of water detonates that 02 plus the electrons from the ho bond go on to that positively charged oxygen. And we've generated our water and regenerated a hydro ion. So now we need the reverse reaction. So we need to go from and alcohol two and keen and we need to be a similar reaction with similar reagent. So we'll be doing acid catalyzed just as we had an acid catalyst for our other first reaction to an acid catalyzed dehydration. So it's kind of interesting that you can use the same reagent to drive two opposite reactions. So we need to think about the mechanism of this. Well, it's an elimination reaction. So, is it E one or E two? Well, we have, what, what is our nuclear file? In this case, we're going to have water as a nucleo removing the hydrogen to create the double bond. So our Nile is water which is weak. So we have a weak nuclei. And then what kind of alcohol do we have? Well, our alcohol has bonds to two carbons in the ring. So we have a secondary alcohol and when we have a secondary or tertiary alcohol and a weak nuclei, our secondary alcohol can generate a stable carbo Kamm. So that will be an E one reaction since it generates that stable Carbocaine intermediate. So just as the initial reaction did, this will go through Carbocaine intermediate, but we can't just draw our reactions in reverse. So let's think about how this will work. We'll draw our alcohol here. So we've got our cyclops ring and our oh group with the two lone pairs on the oxygen, we need the alcohol group to leave. But of course, alcohol is a very poor leaving group. So we need to have protonation to convert the alcohol to a good leaving group. And we are in an acid acidic solution. So that's very easy to proton, eat something. And again, we don't have a specific acid here because any acidic solution will generate these hydro ions. So we've got plenty of them floating around. So again, we have 03 plus, it's gonna write it like that and it has a lone pair actually. Excuse me, it's tricky. We've got this reverse reaction. I don't want to write it that way because I need a proton to be going to my water. So I'm going to draw it as H 02 plus, get a lone pair in the water. But what we're interested in is the lone pair on our alcohol. So that lone pair on that oxygen comes on over and snags that proton from our hydro hydro ion and then the electrons from the ho bond go to the oxygen generating water. And now we have in a reversible reaction, we now have converted our alcohol to a very good leaving group of 02 plus. So we've got water as a leaving group, very excellent since that will generate that neutral water molecule. So as an excellent leaving group, the electrons from that co bond will go up to the oxygen, it will leave all by itself and generate that carbo Cathy. So we'll draw our ring again. And now we just have a positive charge on that top carbon note that we go through the exact same carbine intermediate that we did in the original reaction, I'm going to draw the hydrogen on our beta carbon since I'm doing this elimination reaction. So I'll need to have that indicated for my mechanism and my molecule is symmetrical. So it doesn't matter which of those two beta carbons, doesn't matter which I use, they'll generate the same result. So now as I said, I will have water as my nucleo file. So I'll write that as 02, it's got its two lone pairs. So the alone pair on the water molecule will grab a proton from that beta carbon and then the electrons from ach bond go down to form the new double bond to eliminate that positive charge of my carbon C I am. So my final step, my elimination reaction has taken place. So I now have gener generated my cyclo Pantene since I know how that cc double bond, it's now a neutral molecule. And as the problem says, it's a acid or as I've said, it's an acid catalyzed reaction. And so we've also regenerated our hydro ion in that last step. So we have 03 plus there. So here we go, we have our reverse reaction. It's acid catalyzed just as the original reaction was going through that Carbocaine intermediate, going through that two step elimination process. We have the protonation of our alcohol to create a good leaving group that water molecule leaves to generate the carbon cion and then water acts as a nucleo to grab the proton which then generates our alkene. See you in the next video.

The mechanism of the acid-catalyzed hydration of an alkene to make an alcohol is shown. If this reaction is reversible, draw a mechanism that justifies formation of the alkene from the alcohol under similar conditions. How do you know this mechanism is correct?

Key Concepts

Acid-Catalyzed Hydration

Reversibility of Reactions

Mechanistic Justification

Watch next