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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):

enter image description here

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.

enter image description here

What am I doing wrong?

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    $\begingroup$ Welcome to ChemSE. Compare the pKa of benzoic acid and phenol. What price does phenol have to pay vs. benzoic acid? $\endgroup$ – user55119 Oct 14 at 1:21
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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.

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  • $\begingroup$ ok, thanks! I'm not in a position to accept answers because I evaluate them. But if I understand you correctly, a charge shared between two oxygens' p orbitals is so favorable that an ion that can do that is more stable than a molecule that allow the electron density to occupy through a higher energy state in order to share it with the other oxygen. (I an a beginning chemistry student obviously so there might be something I'm saying incorrectly.) $\endgroup$ – Katie Oct 14 at 15:42
  • $\begingroup$ can't evaluate them $\endgroup$ – Katie Oct 14 at 17:16

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