Provide the mechanism for the following transformation.
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Question

Provide the mechanism for the following transformation.

Mechanism of aldol condensation showing reactants and products with hydroxide ion.

Accepted Answer

Provide the mechanism for the following transformation. Alright guys, so what we see is that we have a beta hydroxyketone that is reacting with base to form an alpha beta unsaturated ketone. So this is going to be a dehydration reaction, but remember that when you're dehydrating one of these aldol condensates, you're going to be using the e1cb mechanism. Let's write that down. E1cb, CB, which stands for e one conjugate base, which is a mechanism that you really shouldn't know anything about yet. We haven't focused on this mechanism a whole lot. It's a dehydration mechanism that actually uses base. Okay. So here we're going to show you I'm going to show you just like step by step how this works. Okay? But what we should know is 2 different facts about this. 1, since it's an e1, do you think that this should be a concerted mechanism or a 2 step mechanism? It has to be 2 step, right? Remember that your E2 reactions were concerted and your E1 reactions always had a carbocation, right? Well, the second thing is that this is in base, so should there be a carbocation? No, it should be a carbanion. So the intermediate should be a carbanion drawn like that, okay? Let's go ahead and draw the mechanism out. So the first step is that the base takes an alpha proton. So it's going to come over here. It's going to take away this proton and it's going to put a negative charge right here. So it's going to make an enolate. So this is really an enolate reaction to begin with. This is going to make an intermediate that looks like this. That methyl is still there, that methyl is still there, but now we have a negative charge here. Okay? So now we have a negative charge. This is our intermediate. This is our high energy intermediate. Now it wants to eliminate. How can we eliminate? By making a bond and breaking a bond. We're gonna go ahead and make a double bond and kick out the Oh, and what that's gonna give me is our dehydrated product. Cool? Now you guys, you might have a lot of questions going on here. So let me just go ahead and try to answer them 1 by 1. Your first question might be, okay, Johnny, we learned in orgo 1 that Oh is a bad leaving group. How are you just kicking out Oh? Don't you need to turn it into a better leaving group to make this happen? Well guys, great job if you were thinking that, but the reason that we can still kick out Oh as a bad leaving group is because we're reacting this in basic conditions. So there's already a lot of Oh present in the solution. So kicking out an Oh in basic conditions isn't that difficult to do if there's plenty of Oh around. Okay? The second reason is because the product is really really favored. It's not just a regular double bond like we have with a normal e one. It's actually a conjugated double bond that has resonance possible all throughout here. Okay? So Le Chatelier's principle is going to make this really favorite and once this forms, it's irreversible. So even if only a little bit of it gets dehydrated, it keeps forming this dehydration product shifting the equilibrium to the right. So basically even if you don't make a whole lot of this, the fact that it's irreversible means that it's eventually going to go completely to the right and make the dehydration product if left alone in base. Okay? Now the last question you might have is you might be thinking well Johnny, how did you know that it was going to take off that H? If this is an enolate question, question, why didn't it just take off this one instead and make a negative charge on the other side? Perfectly legitimate question. Great question. But that enolate is reversible. Notice that that enolate would just make a reversible enolate that could then be protonated again and there'd be no change in the product. Whereas the enolate in the middle is the one that has the ability to form the irreversible double bond and by having an irreversible product, it means that you can continue to shift the reaction to the right as long as you keep it reacting in base. Okay. Awesome guys. This is a reaction that's really unique to the Aldol Condensation. We don't use it a whole lot in organic chemistry, but it's a good mechanism for you to understand here when we're talking about condensation chemistry. Alright guys. Let's move on to the next video.

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Provide the mechanism for the following transformation.

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