Claisen-Schmidt Condensation - Video Tutorials & Practice Problems

Claisen-Schmidt Reaction

Hey everyone. Now we're going to talk about a very specific type of crossed aldol reaction called a Claisen Schmidt reaction. The Claisen Schmidt condensation is what happens when you have an enolizable aldehyde plus ketone. This means that we're kind of breaking the rules. Remember that I told you is if you have a crossed aldol, then you should have one non-enolizable carbonyl. But this is the reaction that happens when you have specifically an enolizable aldehyde and an enolizable ketone, both together. It turns out that you actually get kind of lucky if you mix an aldehyde and a ketone together that are enolizable because theoretically, they could both react and you would get a cross product. But what happens in real life is that one product does actually predominate. Why? Because it turns out that aldehydes are more susceptible to nucleophilic addition than ketones. It's going to favor one of the enolates over the other. Basically, what I'm trying to say here is that you've got your ketone, let's say, and you've got your aldehyde. Theoretically, both of these could form enolates. I could form an enolate on A. I could form an enolate on B and I would start getting a mess. I would get my Punnett square again. But what actually happens is that you only form the enolate on A. Why? Because B happens to be an aldehyde, so it's a group that is easier to do nucleophilic addition with. Then it would prefer to be attacked and be the electrophile instead of being the enolate. With Claisen-Schmidt, it's the one example where you actually can have enolizable cross products and or cross reagents and you're still just going to get one product. Make sense? You guys got this. Go ahead and predict the major product of the following Claisen Schmidt condensation. Notice I have a ketone and aldehyde. Notice they're both enolizable. Try to predict it and then I'll show you guys what the answer is.

Predict the Major Product

Which one did you choose as the enolate? By definition, if I have a mixed ketone and aldehyde, the ketone is going to want to be the enolate so that it can attack the aldehyde. I'm going to bring this ketone down to my left side and draw it as my enolate. Remember that you always want the anion to face towards the electrophile. I'm going to react that with my aldehyde which I'll keep drawn exactly the way it is because I want the H, the smallest group to face towards the negative charge. I'm going to do my reaction. I'm going to get a molecule that looks like this. My enolate is there. Now I have OH carbon. I am skipping a few steps like for example, I skipped the protonation step because this isn't a mechanism question. This is a predictive product. But you guys know that you could protonate that with the conjugate of your base because your base would have an extra proton. It pulled a proton off the ketone. Now we know that what these things like to do, your beta-hydroxycarbonyls, what's wrong? It's going to be very difficult to keep this as a beta-hydroxycarbonyl because we've got an irreversible dehydration hanging around. Once you dehydrate, it can't go back to the original ketone. What we're going to wind up getting is a dehydration product. Awesome. There you go. That's going to be pretty much my final product. Notice it's between the alpha and the beta. This is my enone and we're done. That's it for the Claisen-Schmidt. As you guys can see, how it fits into the idea of crossed aldol. It just happens to be lucky that the aldehyde is less willing to be an enolate than the ketone. Awesome. Let's move on to the next topic.