Consider the following reaction. Provide a stepwise mechanism ...

Question

Consider the following reaction. Provide a stepwise mechanism to explain the given transformation.

Reaction mechanism for intramolecular aldol condensation in organic chemistry.

Accepted Answer

Hey guys, let's take a look at the following question. So here I say, consider the following reaction. Provide a stepwise mechanism to explain the given transformation. What we have to really do here is first observe what is going on. Well, we have a dialdehyde, so we have a dicarbonyl. And what happens here is this carbonyl somehow reacts with itself to form a ring with a double bond in it. Here, this is our alpha carbon and this is our beta carbon. So we created a cyclic, alpha, beta, since they're double bonded, we use the word unsaturated aldehyde. So if we have a dicarbonyl and it somehow cyclizes into a ring, then we'd say that this is an example of an intramolecular aldol. Not just an aldol reaction because we made a double bond, it's an aldol condensation condensation reaction. The word condensation means that we form a double bond And unsaturated means double bond. Intramolecular means it's reacting with itself, closing in on itself to make a ring. Now, since we have to show the mechanism here, I'm going to take myself out of the image guys so we have more room to work with. What happens first is here is our alpha carbon and here is our alpha carbon. Since it's symmetrical, it doesn't matter which alpha carbon gets attacked. This is a base here, it's ionic, so it's not really together. It's really Na plus Oet negative. This base is going to shoot out, rip off this hydrogen, alpha carbon holds on to the electrons. So what we're going to get initially is we're going to have this carbon negatively charged because it has a lone pair. It's already connected to 1 carbonyl, but now it's attracted to this carbonyl carbon because it's partially positive. This is going to shoot out and attach to this carbonyl carbon kicking this bond up. Now, we count 1, 2, 3, 4, 5, 6. There'll be 6 elements that loop around each other to form a ring. That's why our ring is 6 carbons large. Here, we come over here. We have 6 carbons in there. We're going to say that this is 1, 2, 3, 4, 5, and 6. On carbon number 1, there's still this aldehyde. Carbon number 6 has an O negative. Remember, the Oh or the OET minus that basically ripped off an H could come back and do what? Protonate this oxygen here. So what we make initially is this and this is alpha, this is beta, so initially we make a beta hydroxyaldehide. Now, we need to make a double bond between the alpha and beta carbon. Now, what's going to happen next is here we could say, ETO negative comes back or maybe Oh- comes back because I put water here. Now, here's the thing. This comes back and what is it going to do? It's going to seek out a hydrogen to take off. It's going to take it from the alpha carbon. Now, here's the thing. It's going to shoot out and rip off this alpha hydrogen here which falls here to make a double bond which kicks out the Oh group. Now, there should be some bells going off in your heads because we've always heard that weak bases are good leaving groups. This is something we learned in Organic 1. Weak bases are good leaving groups. Now, Oh is a strong base, so it should be a bad leaving group. So, it should be able to leave. But for some reason, we're allowing it to leave in this case. The reason we're letting it leave is because we create in the process a conjugated aldehyde. Remember the word conjugated means you have So it's okay that we're letting not such a great, leaving So it's okay that we're letting not such a great, leaving group leave because it's kind of a strong base. It's okay because in the process of it leaving, we make a much more stable conjugated carbonyl group. So in this case, we will allow a strong base to leave. Here, this would be our mechanism to illustrate what's going on here. Remember, we're ripping off a hydrogen from the alpha carbon so that the Oh could get kicked out as we make the conjugated system. Now, if that alpha carbon didn't have a hydrogen to rip off, then we wouldn't be able to get rid of that Oh and we would have stopped at the beta hydroxy stage. Remember, this helps to make a conjugated carbonyl group. We call these conjugated carbonyl groups usually enones. Now, E means alkene but one means ketone. Now, why do we usually call them enones? That's because ketones really love to do this. Ketones hate being in the beta hydroxy form and will react further to help make a double bond between the alpha and beta carbon so that they become a conjugated ketone. Again, conjugation makes you more stable and that's what's going on here. This whole process, because it's making a cyclic ring, is intramolecular aldol condensation reaction. So remember that little things about this reaction help guide you to the final answer.

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Consider the following reaction. Provide a stepwise mechanism to explain the given transformation.

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