Acetoacetic Ester Synthesis

Concept: Concept: General Reactions

Video Transcript

In this video, I want to review the general reactions that acetoacetic esters can go through. This is like your cheat sheet for all the different reactions that you can do with acetoacetic ester synthesis. The first step for all of these is always going to be deprotonation and formation of an enolate. That enolate is going to selectively form right here because that the easiest proton to pull off. Remember, it’s the most acidic. We’re just going to go one by one down these reactions. Once you have your enolate, if you react that enolate with an alkyl halide, you’re going to get an alkylation. That just makes sense. We’ve done that plenty.
It turns out that usually you're going to have two hydrogens on this middle part, on this methylene carbon. Usually most of these acetoacetic esters start off with not just one hydrogen but two. If you wanted to, after you get the first alkylation on, you could deprotonate the other one with another equivalent of base and then react with another alkyl halide. That way, by reacting twice, you would remove both of these hydrogens one after the other. You would actually get now two R groups attached to the alpha-carbon. So this would be dialkylation. Just keep in mind, to do a dialkylation, you need to react with the base twice in order to get both protons off. It can't be two equivalents of base at the same time. It needs to be base, then alkyl halide, then base again, then alkyl halide again.
Then finally the last two steps, I’m skipping in all of these. Just so you guys know, I'm always assuming that at the end of this reaction, I’m going to do plus H3O+ and plus heat. Why am I assuming that? Why did I not draw it? I guess I could have drawn it but it happens every single time. It’s a waste of space. What I’m trying to show you is the stuff that changes. The stuff that stays the same that you always hydrolyze and you always decarboxylate. That's going to hold true for all of these.
Next. What if you use, instead of using a halide or an alkyl halide, what if you use a terminal alkyl dihalide? Then what you can do is you can actually cycloalkylate. What you can do is you can react with your base, react with one of your alkyl halide. Then react another cool enough base, then react with the other side of your alkyl halide. That’s going to form a ring around the alpha-carbon. You're going to get a ring with a size of n+1, where n is equal to the number of carbons in between in your chain. If you reacted with a three-carbon dihalide, you’re going to get a four-carbon chain. Why? Because these are the three carbons, and this is the alpha-carbon that you started with. It’s a four-member ring but only three of those new carbons came from the dihalide.
Finally, this one is easy – acylation. You know that you could attack here, kick out the Cl and you would get an acyl group after you do your H3O+ and your heat. I know this is a lot to go over. The only way you’re going to learn is just start applying it. Let's do this practice problem. Noticed it’s five steps. You’re going to have to figure out what's going on. Try to do this five-step product. Don’t worry about mechanisms. Just worry about the product, and then I’ll help out. By the way, I don’t want anyone to get stuck on molecule 2 so I’m going to give it to you. Notice that it says (BrCH2CH2)2O. What is that? That’s an ether. Let me draw it for you. That would look like this. O in the middle with these two groups coming off of both sides. It would be CH2CH2Br, CH2CH2Br. Now you have no excuse to get this one wrong because I'm actually giving hint to you. Go ahead and try to do it yourself and then I'll step in and show you how it's done.

Problem: Provide the major product for the following reaction


Concept: Practice 2: Acetoacetic Ester Retrosynthesis