AAMC 7 orgo question

[rcmp1234]

Passage 1 question 4: The base in step 5 of the synthesis abstracts a proton from which of the labelled carbon atoms of compound 7?

I understand why c and d are acidic; but isnt the proton on a even more acidic since it is alpha to an ester which contains a carbonyl carbon that it can delocalize to and an electron withdrawing oxygen?

the answer is c by the way.

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[V5RED]

Are you wondering why the answer is C or whether someone can tell you whether or not A/B are more acidic than C?

 

[rcmp1234]

Both actually. I'm wondering why C is more acidic than A.

 

[sciencebooks]

Both actually. I'm wondering why C is more acidic than A.

The question doesn't ask which is more acidic; the question ask from where the proton is removed. If you look at the product of the reaction, no hydrogen is removed from A, B, or D. A hydrogen is only removed from position C. There were two hydrogens there before, now there is one.

 

[SomethingLatin]

Can you identify anything in the structure of reactant that would suggest (A) isn't as acidic as you think. And draw out the deprotonated structure before it undergoes an intermolecular cyclization try to figure out why its not as stable as you think. You mentioned inductance of the electron withdrawing group, now think about resonance, and the overlap of orbitals that could increase stability.

 

[Erythropoietin]

Passage 1 question 4: The base in step 5 of the synthesis abstracts a proton from which of the labelled carbon atoms of compound 7?

I understand why c and d are acidic; but isnt the proton on a even more acidic since it is alpha to an ester which contains a carbonyl carbon that it can delocalize to and an electron withdrawing oxygen?

the answer is c by the way.

Here's what I think... By looking at the reactant and the product, c is definitely the most acidic. When you remove a H from c, it will become a negatively charged nucleophile, which can then attack carbon 'b' to make the ring. Consider doing the same with 'd' ...if you remove a hydrogen from d and let it attack b, the whole molecule have to flip and probably break some bonds to make the d-b bond, which would cost a lot of energy and won't happen.
B is definitely the most acidic of all because it will form the easiest nucleophile to attack b and make the product, and it is also an alpha carbon.

 

[brownbaglunch]

I think that your logic is probably on the right track.

I don't think the acidity is much different on the two alpha carbons. Let's consider for a moment that either hydrogen can be and is removed so that both versions of the enolate form. When that is the case, the enolate that contributes to the less stable product would tend to just get reprotonated.

 

[SomethingLatin]

I wouldnt be so quick to dismiss the acidity A, you have increased s character, (sp2 vs sp3, stabilization of negative charge) conjugation(overlapping of p orbitals that delocalize charge), resonance with carbonyl and the inductive effect of the ester . You do form an allene in the resonance structure, which should ring some bells with acidity/stability of conjugate base. All this is beyond the scope of the question as a quick and dirty numbering of the carbons shows which proton was abstracted but the OP wanted some deeper understanding.