# LDA and an aldehyde

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

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?

• 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. – Waylander Apr 17 at 10:18
• 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. – Zhe Apr 17 at 11:57
• @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. – Mirte Apr 18 at 17:38
• 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. – Zhe Apr 18 at 18:28

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.