Aldol Condensation: Study with Video Lessons, Practice Problems & Examples

Aldol condensation is a key reaction where an enolate ion, formed from a ketone or aldehyde, reacts with another carbonyl compound to produce a beta hydroxy carbonyl, known as an aldol. The mechanism involves nucleophilic addition, where the enolate attacks the electrophile, leading to a tetrahedral intermediate. This intermediate can spontaneously dehydrate to form a stable alpha-beta unsaturated carbonyl compound, or enone. Understanding the proper alignment of carbonyls and the stability of the products is crucial for accurate predictions in organic synthesis.

General Features

In this video, I want to focus on a specific type of condensation reaction called an aldol condensation. From prior videos, we know that an enolate is a negatively charged species and it can attack electrophiles. But what we didn't fully realize up until this point is that these enolates can actually react with themselves to condense. Specifically, if it's a ketone or aldehyde, it's going to condense into this category of molecule called a beta-hydroxy carbonyl. The final products are called aldols. That's the reason that we have that name because they're part aldehyde and they're part alcohol. Hence the name aldol. What I'm going to do here is I'm going to show you the full mechanism for an aldol condensation. One of the most important things to realize about any condensation reaction is that a big part of getting these questions right isn't just knowing the mechanism. It's knowing how to line up your carbonyls because if you line them up incorrectly, you're going to be doing a lot more work and you're probably going to get it wrong. I'm going to teach you not only just the mechanism but I'm going to use my years of teaching experience to try to engrain this method into you of how is the best way to line up your carbonyls so that you can always visualize the product the easiest. The first step guys is super easy. All it is is that you're using some kind of base, usually OH negative, to pull off a proton and make your enolate. Remember that enolates have two different resonance structures. There's also a resonance structure with the negative on the O but we're not going to use that one. For this mechanism, you always want to be using the enolate on the carbon because that's going to help predict what our product looks like. The reason that we would get an aldol condensation is when you have a base, but you have no other electrophile. I'd be wondering how would I know that something is an aldol and how would I know that something is let's say a reaction from another chapter because there's a lot of reagents that you know now. But guys, if you are forming an enolate without an electrophile to react with, then you know this is going to be a self-condensation, which is exactly what we're going to draw. We've got our enolate. Notice that my enolate, I'm drawing it on the left-hand side. I'm always going to draw the enolate on the left. I don't have an electrophile on the right. I don't have a halide or an acyl chloride, whatever. That means I'm forced to react with just another carbonyl. How does that work? I like to call them; there's the enolate and there's the electrophile. Why? The enolate is the one with the negative charge. The electrophile is the one that you're going to attack the partial positive of. This one should not be in an enolate form. You should just draw this the normal way that it was before the base reacted. A few other pointers here. You want to make sure that your enolate is drawn on the left, but that your anion is toward the electrophile because you want to have the electrophile and the anion as close to each other as possible. Here, I'm drawing the enolate on the left but I want to make sure my negative charge is as close to the electrophile as possible. Also, any R groups on it face downwards. This is going to come up more later. But if I had bulky R groups on that CH2, I should actually face them facing down to clear the way for the electrophile that I'm about to attack. My electrophile, you've actually got some special rules about that one. The electrophile you draw on the right. Always draw on the right and you should draw your smallest group toward anion. The reason is because we're going to have to attack with the anion and it's going to be easiest to visualize if I have my small group close to the anion and my large group away from it. I think that's enough to get going. This mechanism is going to look like follows. This is a nucleophilic addition mechanism. My negative charge is going to attack my partial positive. I'm going to break a bond and I'm going to make a tetrahedral intermediate. It's a big tetrahedral intermediate. I understand. But you still have an O negative with now what groups? Now you've got the extra CH2 that's attached here. We still have that H. The H that was here is still here. It's just I'm not drawing it. Then there's the CH3. This is called a nucleophilic addition mechanism because we're going to protonate the O at the end. We're not going to kick out a leaving group. We use water or acid to protonate and to give us our beta-hydroxycarbonyl because I have a beta-hydroxy on my carbonyl. The reason I call it carbonyl is because it could be either a ketone or an aldehyde, just depending on what you started with. That's really it. It's a really easy mechanism. It's nuc...

What product can be isolated from the following aldol condensation reaction? 

• A.
• B.
• C.
• D.

Provide the mechanism for the following transformation.

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 are 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 one by one. 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 an 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. The second reason is that the product is really really favored. It's not just a regular double bond like we have with a normal E1; 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 favored, 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. 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 reaction, why didn't it just take off this one instead and make a negative charge on the other side? A 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. 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.

Here’s what students ask on this topic:

What is the mechanism of aldol condensation?

The mechanism of aldol condensation involves several steps. First, a base deprotonates the α-carbon of a ketone or aldehyde to form an enolate ion. This enolate ion then attacks the carbonyl carbon of another molecule of the ketone or aldehyde, forming a tetrahedral intermediate. This intermediate is then protonated to form a β-hydroxy carbonyl compound, known as an aldol. The aldol can further undergo dehydration to form an α,β-unsaturated carbonyl compound, or enone, which is highly stable due to conjugation.

What are the products of aldol condensation?

The primary product of aldol condensation is a β-hydroxy carbonyl compound, commonly referred to as an aldol. This compound can further undergo dehydration to form an α,β-unsaturated carbonyl compound, also known as an enone. The enone is particularly stable due to conjugation, making the dehydration process highly favored.

Why is the dehydration step in aldol condensation often spontaneous?

The dehydration step in aldol condensation is often spontaneous because the resulting α,β-unsaturated carbonyl compound (enone) is highly stable due to conjugation. This stability lowers the energy barrier for the dehydration process, allowing it to occur without the need for strong acids or heat, unlike typical alcohol dehydration reactions.

How do you identify an aldol condensation reaction?

You can identify an aldol condensation reaction by looking for the formation of an enolate ion from a ketone or aldehyde, which then reacts with another carbonyl compound to form a β-hydroxy carbonyl compound (aldol). If the reaction conditions favor dehydration, the final product will be an α,β-unsaturated carbonyl compound (enone). Key indicators include the presence of a base and the absence of other electrophiles.

What is the role of the base in aldol condensation?

The base in aldol condensation plays a crucial role by deprotonating the α-carbon of a ketone or aldehyde to form an enolate ion. This enolate ion is a nucleophile that can attack the carbonyl carbon of another molecule, leading to the formation of the β-hydroxy carbonyl compound (aldol). Common bases used include hydroxide ions (OH⁻).