Question

Hi, Could any of you help me solving this organic chemistry exercise. The first part is a Claisen condensation, I got that one right, I think :P, I can't follow what happens after that with the base, I know there's a decarboxylation after a saponification, what I don't know is what ester I get after treating B with LDA, and when the IMe its added.

Answer 1

here's the image

Answer 2

Here's my guess, I'll try to explain my thinking so that hopefully it seems at least somewhat plausible to you. :D



So we have two good choices of hydrogen for LDA to remove. But the one between the two carbonyls is the most acidic since it has the full resonance stabilization of both carbonyls while the other one drawn in red if you work it out you'll see it won't have any resonance forms involving the carbonyl from the ester.

Moving on, now we add the iodomethane:



Iodine is a good leaving group, so SN2 reaction occurs and we've given our molecule a methyl group. So now if this is right, somehow hydroxide will have to cut this molecule apart.

Here I tried to box all the steps to make it clearer what's going on. So first off, we add hyrdoxide, then it ends up causing this molecule to leave. I'll admit, I don't know if this is exactly what happens but I know that the thing that leaves looks like what you have as the reaction product so I can only assume this had to have happened. At the very least it's not entirely impossible because it does form an enolate anion which should stabilize it as a leaving group, but we could also have the ethoxide anion leave as well. I suppose that it's possible that if the ethoxide leaves, we could have another hydroxide attack and do the same thing again, causing the end product to have carbonate with ethanol as the extra stuff.

Like I said, this is just my guess at what happens, feel free to ask or suggest corrections if you think of anything I messed up haha.