Concept: Concept: Claisen-Schmidt Reaction3m
Now we're going to talk about a very specific type of crossed aldol reaction called the 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 guys 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, let’s say this is molecule A and this is molecule B. 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. But what actually happens is that you only form the enolate on A. Why? Because B happens to be an aldehyde. It’s a group that is easier to do a 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 reagents and you’re still just going to get one product. Makes sense?
You guys got this. Go ahead and predict the major product of the following Claisen-Schmidt condensation. Notice I have a ketone and an aldehyde. Notice they’re both enolizable. Try to predict it and then I’ll show you guys what the answer is.
Concept: Example: Predict the Major Product3m
Alright, so which one did you choose as the enolate? Okay guys. So, 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, okay? So, 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 enolate going towards, towards the electrophile and 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, looks like this, enolate there.
Now, I have OH carbon, okay? Now, guys I am skipping a few steps like, for example, I skip the protonation step because this isn't a mechanism question, this is predict the product, okay? 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, okay? Now, we know that what these things like to do, your beta hydroxy carbonyl, what's wrong, guys it's going to be very difficult to keep this as a beta hydroxy carbonyl because we've got an irreversible dehydration hanging around, okay? Once you dehydrate it can't go back to the original ketone. So, what we're going to wind up getting is a dehydration product, okay? Awesome, okay. So, there you go, that's going to be pretty much my final product. Notice between the Alpha and the beta, this is my enone, right? And we're done, okay? So, that's it for claisen Schmidt. So, 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, okay? Awesome. So, let's move on to the next topic.
What two carbonyl compounds are required for the synthesis of morachalcone A (the aromatase inhibitor discussed in the box on page 874), via a Claisen–Schmidt condensation?
Predict the product for the following reaction.
What would be the major organic product of this base-catalyzed reaction?
Predict the product(s) in the following aldol condensation using aqueous sodium hydroxide.