Acetoacetic Ester Synthesis - Online Tutor, Practice Problems & Exam Prep

Acetoacetic esters undergo various reactions starting with deprotonation to form an enolate. This enolate can react with alkyl halides for alkylation, allowing for dialkylation through sequential reactions. Cycloalkylation occurs with terminal alkyl dihalides, forming rings. Acylation can introduce acyl groups. Hydrolysis and decarboxylation are common final steps. Understanding these reactions is crucial for mastering acetoacetic ester synthesis and applying these concepts in organic chemistry.

General Reactions

In this video, I want to review the general reactions that acetylacetic 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 guys is always going to be deprotonation, formation of an enolate. That enolate is going to selectively form right here because that's the easiest proton to pull off. Remember, it's the most acidic. We're just going to go 1 by 1 down these reactions. Once you have your enolate, if you react the 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 2 hydrogens on this middle carbon, on this methylene carbon. Usually, most acetoacetic esters start off with not just 1 hydrogen but 2. If you wanted to, after you get the 1^st 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, and you would actually get now 2 R groups attached to the alpha carbons. 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 2 equivalents of base at the same time. It needs to be base, then alkyl halide, then base again, then alkyl halide again. Perfect. 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 what two reactions? I'm always assuming that 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, so 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, because what you can do is you can react with your base, react with one side of your alkyl halide, then react another equivalent of 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 3 carbon dihalide, you're going to get a 4 carbon ring. Why? Because these are the 3 carbons, 1, 2, 3. And this is the alpha carbon that you started with. It's a 4-membered ring but only 3 of those new carbons came from the dihalide.

Finally, this one's 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 to start applying it. Let's do this practice problem. I've noticed it's 5 steps, so you're going to have to figure out what's going on. Try to do this 5 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 BR CH₂ CH₂ O₂. What is that? That's an ether. Let me draw it for you. That would look like this: O in the middle with these 2 groups coming off of both sides. It would be CH₂ CH₂ BR, CH₂ CH₂ BR. Now you have no excuse to get this wrong because I'm actually giving it to you. Go ahead and try to do it yourself and then I'll step in and show you how it's done.

Predict the products

All right, guys. Let's start off by drawing this ester out because I hate working with condensed formulas. That's just not helpful. I'm going to draw this out like this. My ketone CH₂COOR. That's going to be O O E T. That is acetoacetic ester. The first reagent I'm reacting with is OC₂H₅. Is this a good base to use to form an enolate on this dicarbonyl? What do you think? Remember that we said that you want to make sure that your R group is the same between your ester and your oxide or you're going to get what? A transesterification. Is this a good choice? Yes, it is guys. Ethyl and C₂H₅ are the same thing. It's just different ways of writing it. Tricky. It's fine. This is a good base to use. This is exactly what we want to use. The first reaction is going to be that the O grabs one of the H's and forms an enolate here. Now that enolate, let me see if I can move this down a little bit. Bit. I can't. This enolate is now going to react with my alkyl halide. And my alkyl halide looks like this. Again, it's BrOB_r. This enolate is going to do a backside attack. That's not really a backside attack. Let's do a backside attack. On one of the bromines and it's going to give me this. Let me move it up a little bit. It's going to give me carbonyl, carbonyl O E T. Plus, I've got this huge chain attached. That's going to be C₁₂H₂₅Br. Now that whole thing is attached. Now look at my 3rd reagent. Usually, we would be thinking, let's hydrolyze and decarboxylate. But no, I'm reacting with a second equivalent of base. What's that going to do? I've got one more proton there that can react. I'm going to take my base and I'm going to make an enolate there. What's going to happen? That enolate can do another backside attack. What we're going to get is now a compound that looks like this. With now, what does it have? It's a ring. It's going to be a ring of what size? To count out the size of a ring, I always start from nucleophile to electrophile. I would say 1, 2, 3, 4, 5, 6. This is going to make a 6-membered ring coming off of the middle like this, where you guys just have to tell me if I'm missing anything. This is 1, 2, 3, 4, 5, 6. Are there any atoms I should change? Anything I should add? Should I add a BR? I heard loud and clear. I need to change the 4 to an oxygen. Let's do that. You'd be surprised how much I can hear over the interwebs. What else? How about the BR? The BR is just a leaving group, so I'm just going to put plus BR negative. I don't need that. Excellent. What's next? It says now we're going to use water and acid. What's that going to do? That's going to do a hydrolysis reaction on my ester. That's going to give me this. Carboxylic acid instead of ester. That's what my H₃O⁺ does. And then heat. What is heat going to do, guys? Heat is going to decarboxylate, meaning that I take that entire carboxylic acid off. If you're curious about the mechanism for that definitely search decarboxylation. And what you're going to get is that now that whole thing is gone. That means the 6th of our ring is just coming straight off this alpha carbon. That's it. It's going to be that plus CO₂. That's going to give us our final answer. Once again, this was actually what? This was a cycloalkylation reaction because I reacted with a dihalide and 2 equivalents of base, and I was able to get a ring. Let's move on to the next set of videos.

Provide the major product for the following reaction

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Here’s what students ask on this topic:

What is the first step in acetoacetic ester synthesis?

The first step in acetoacetic ester synthesis is the deprotonation of the ester to form an enolate. This occurs because the hydrogen on the methylene carbon (the carbon between the two carbonyl groups) is the most acidic and easiest to remove. The enolate formation is crucial as it sets the stage for subsequent reactions such as alkylation, cycloalkylation, and acylation.

How does dialkylation occur in acetoacetic ester synthesis?

Dialkylation in acetoacetic ester synthesis occurs through a two-step process. First, the enolate formed from deprotonation reacts with an alkyl halide to introduce one alkyl group. Then, the remaining hydrogen on the methylene carbon is deprotonated again with another equivalent of base, followed by reaction with a second alkyl halide. This sequential process results in the introduction of two alkyl groups on the alpha carbon.

What is cycloalkylation in the context of acetoacetic ester synthesis?

Cycloalkylation in acetoacetic ester synthesis involves the reaction of the enolate with a terminal alkyl dihalide. The enolate first reacts with one end of the dihalide, followed by deprotonation and reaction with the other end of the dihalide. This forms a ring structure around the alpha carbon. The size of the ring is determined by the number of carbons in the dihalide plus one.

What are the final steps in acetoacetic ester synthesis reactions?

The final steps in acetoacetic ester synthesis reactions are hydrolysis and decarboxylation. After the desired alkylation or acylation reactions, the ester is hydrolyzed to form a β-keto acid, which then undergoes decarboxylation upon heating to yield the final product. These steps are common to all acetoacetic ester synthesis reactions and are often assumed in reaction schemes.

How does acylation occur in acetoacetic ester synthesis?

Acylation in acetoacetic ester synthesis occurs when the enolate attacks an acyl chloride. The enolate nucleophilically attacks the carbonyl carbon of the acyl chloride, displacing the chloride ion and forming a β-keto ester. This reaction is followed by hydrolysis and decarboxylation to yield the final acylated product.