Why is the carboxyl proton in salicylic acid more acidic than the phenol proton?

Question

I know I'm just wrong about this but I'd appreciate some help seeing why. It seems to me like the phenol proton in salicylic acid should be more acidic than the carboxyl proton. When I try to draw resonance structures for the conjugate base, it looks to me like the charge gets distributed over two oxygen atoms and the whole ring (if anyone can tell me what software makes the nice pictures in some answers here I'd take it):

Then when I try to draw the conjugate base when the carboxyl group is deprotonated, as far as I can tell only the oxygens take the charge, and trying to distribute over the ring results in invalid resonance structures because you aren't supposed to add formal charges to already charged species.

What am I doing wrong?

Answer 1

More resonance structures does not necessarily mean more stabilization.

Phenolate has a negative charge at an oxygen atom which cannot be shared without "disturbing" the aromaticity of the ring. This is in part because in the resonance structures drawn, there are negative charged carbons within the aromatic ring, which either implies that:

The ring is still aromatic, so there must be a free electron pair in a p orbital (very high in energy).

Or:

The free electron pair is in an sp³ orbital, which means the ring is no longer aromatic.

In the case of the carboxylic acid, the charge is distributed between two atoms more electronegative than carbon and the ring retains its aromaticity.

Answer 2

In this case, when the carboxyl group is deprotoneted and we draw the resonating structures the resonating structures are equivalent. But in case of phenolic deprotonation the resonating structures are not equivalent. So the carboxyl anion will be more stable, although it have less number of resonating structures than phenolic anion. So the carboxyl H will be more acidic.