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Compounds A and B both react with LDA (lithium diisopropylamide, a strong base) to form the same anion C with the formula $\ce{C4H5O^-Li^+}$.

LDA reaction scheme

I am unfamiliar with LDA. My first thought was that the base would attack the H of the carbonyl C, I presumed that this H would have the lowest $\mathrm{p}K_\mathrm{a}$ because it is an $\mathrm{sp^2}$ C and the C is slightly positive from the carbonyl.

However, the structures this gave were confusing and didn't give the same product for both A and B. I tried to google LDA which showed that it was a big bulky base so it can't attack hindered H, but this shouldn't be a problem as the carbonyl C H isn't hindered. However, I found one mechanism for a different aldehyde that had the H taken from the $\mathrm{sp^3}$ C next to the carbonyl.

Where would LDA preferentially attack an aldehyde and why?

Additional note: I have now played around with the structures to find that the only way LDA could react with A and B to form the same anion C if it in both cases it removes and $\ce{H+}$ from the $\mathrm{sp^3}$ carbon in the structure, these seems counter-intuitive! But perhaps LDA only attacks $\mathrm{sp^3}$ C-H for some reason?

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    $\begingroup$ LDA is a very strong base and not fussy about whether it takes an alpha proton or a gamma proton because you will end up with the same extended enolate i.e. C, which is a stable species. $\endgroup$
    – Waylander
    Apr 17 '19 at 10:18
  • $\begingroup$ While this question has a good goal, I think it is a terrible question, if this is a homework or exam question, or just an incomplete question if you're asking it yourself. Which enolate should be produced under the circumstances? (E)- or (Z)- ? They are different, non-interconverting, and probably formed in a mixture whose composition is dependent on whether you start from A or B. $\endgroup$
    – Zhe
    Apr 17 '19 at 11:57
  • $\begingroup$ @Zhe The initial part of the question is an undergraduate past-paper question. However no one set me the past paper and I asked the question out of curiosity (and frustration at my own lack of understanding); so it's not a homework question in the strict sense. $\endgroup$
    – Mirte
    Apr 18 '19 at 17:38
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    $\begingroup$ Sorry, let me be clear: I am talking specifically about the quoted question, which, based on your comment, is an assigned question, and I find that specific question to be quite poor: C is a mixture of enolates! Your own questions are quite reasonable, and I have no problem with them. $\endgroup$
    – Zhe
    Apr 18 '19 at 18:28
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If electronegativity alone were the only factor in determining the acidity of a carbon-hydrogen bond, the base would in fact attack the aldehyde hydrogen. But that is not so, for delocalization of negative charge can also stabilize the anion. If the base attacks at the initially saturated carbon in either compound, you get a delocalized negative charge distributed throughout the carbon chain and onto the oxygen.

Such delocalization stabilizes the anion so strongly that it overrides the original electronegativity difference between $\mathrm{sp^2}$ and $\mathrm{sp^3}$ carbon atoms. So, the base will attack along the carbon chain to produce this delocalized charge anion, instead of at the carbonyl carbon which offers no such delocalization.

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  • $\begingroup$ Is there a way to predict this other than by trying all options? $\endgroup$
    – Mirte
    Apr 18 '19 at 17:34

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