I was given just this statement at face value, but I think it refers to nucleophilic reactions where the carbonyl is the electrophile.

From what I understand, the $\ce{-OR}$ in ester is more electron donating inductively than the $\ce{-OH}$ in $\ce{-COOH}$ (due to the alkyl groups in $\ce{-OR}$). So this should make the carbonyl carbon less positive, making it a weaker electrophile, making it less reactive no?

The resources I read say that it's because the $\ce{-OR}$ is a better leaving group than $\ce{-OH}$. But alkoxide ($\ce{RO-}$) is a stronger base than hydroxide ($\ce{OH-}$), so why would it leave first?


1 Answer 1


Remember that a carboxylic acid is an ACID. It donates a proton, forming an anion and will protonate anionic and basic nucleophiles rendering them non-nucleophilic. If there is an excess of the nucleophile, this means that it is attempting to attack a species that is negatively charged; this is not a favourable interaction. Should it succeed the leaving group is not -OH it is O-


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