Directed Condensations - Video Tutorials & Practice Problems

Directed Condensations

Hopefully by now, the aldol condensation is starting to make a little bit more sense. But what happens when you have an asymmetrical ketone? That presents a problem. Whenever you're reacting an enolate mediated reaction on an asymmetrical ketone, 2 enolates may be possible. We're going to have to use a directed reaction. Directed reactions are what we use to pick the enolate that we want because if you only have one choice of enolate, then your enolate is going to hit your electrophile and you're done. But what if you have 2 possible enolates? Then which one is the one that attacks the electrophilic carbon? Who knows? That's why we have to use thermodynamic versus kinetic control. The thermodynamic enolate, we've learned this before, is the more substituted one. It's going to be favored by small bases, whereas the kinetic enolate is the less substituted one. It's the one that's easier to reach and it's favored by bulky bases. What that means is that if I want to run an aldol reaction, let's say on the right side of my ketone here, I only want to attack with the enolate on the right side, then I would use a small base. For example, NaOH or any other small base. However, if I wanted to react on the less substituted side of the ring, making my enolate on the left side and then having that attack an electrophile, then I definitely have to use a bulky base. For that, we've got 2 options. But the most popular for this chapter being LDA because of the fact that it's a non-nucleophilic base, we don't have to worry that it's going to actually add to anything. It's just going to remove a hydrogen. But also, tert-butoxide would be a possibility. Excellent, guys. Now go ahead and try to predict the product for the following reaction and then I'll jump in.

Predict the Products

Okay. What base was this? I hope you didn't get tripped up on that because I've drawn this for you already. This is just another way to represent LDA. LDA stands for Lithium Diisopropylamide. That's what we have up here. We've got the 2 isopropyl groups, your negatively charged N and your lithium associated with it. Where is my enolate going to form? By the way, another question. Let's back up. Hold up. We know this is in the aldol section, so you're obviously thinking along those lines. But how would you know that this is even an aldol reaction? We've got LDA and a ketone. How do I know what to do? Remember I told you guys that anytime that you have a base plus a ketone, let me rearrange that. Anytime that you have a carbonyl like an aldehyde or ketone, this is important, plus a base to make the enolate, plus no other electrophile. That's going to be an aldol because what that means is that you're going to form an enolate but you're not going to have an electrophile to react it with, so it's just going to have to react with itself. That's exactly what we have here. We have a ketone with LDA, strong base, going to make an enolate and no other electrophile. I would make my enolate on this side. That means that the less substituted side is a directed reaction.

I'm going to have to flip this thing in order to use my rules of how to set up an aldol. Let's go for it. Remember, I told you guys I always want the enolate facing the anion towards the right. I would twist that around so that my anion is facing what's going to be the electrophile. My electrophile, I want to draw with the smallest group facing towards the anions, so I would keep it as is. We're good to go.

We're good to do our first mechanism. My negative attacks the O, kicks up the electrons. I wind up getting something like this where I have a molecule that looks like this, carbon. Now that is attached to—I'm just trying to use green—that is attached to what? An O negative and this thing in a methyl group. What's going to happen here? Basically, the conjugate of LDA. Remember LDA deprotonated? The conjugate could protonate this. We're always going to get a protonation. You're never just going to be stuck at the tetrahedral intermediate. You can always use at least the hydrogen that you took to replace this. I'm just going to put H+ because I know that there's at least one hydrogen hanging around since I took it off. What this is going to give me—I ran out of room over here, by the way, these are all reversible arrows, guys. What this is going to give me is it's going to give me a molecule that looks like this. That is one of the final answers. We're done. Good. That's the beta-hydroxycarbonyl.

But I told you that I'm going to be spontaneously dehydrating these guys. Why? Because that just seems to be the thing to do. It's going to make a very stab