Acidity of substituted benzoic acids

[CrimpJiggler]

Heres meta methoxy benzoic acid:

and here's para methoxy benzoic acid:

m-methoxybenzoic acid is a stronger acid than its para substituted analogue. I'm a bit confused about why this is. From what I've read, its because at the meta position, only the methoxy groups inductive withdrawing effects apply whereas at the para position, it also acts as a mesomeric (resonance) donator. Why does the methoxy groups resonance effects only work when its at the para position? Why doesn't it act as a resonance donator when its at the meta position?

 

[Ygggdrasil]

Draw out the resonance structures for the deprotonated form of both acids. For the para isomer, you will see that you will have more resonance structures because the negative charge can move to the methoxy group. For the meta isomer, however, you cannot draw a resonance structure that moves the negative charge to the methoxy group.

 

[CrimpJiggler]

**** my organic chem is a bit rusty, I don't know how to draw this resonance structure. Heres my attempt for the meta isomer:
[PLAIN]http://img28.imageshack.us/img28/2921/resonance.png
what am I doing wrong there?

 

[Ygggdrasil]

First, I made a mistake in my post. None of the resonance structures should move the negative charge to the methoxy group. Rather, for the para-isomer, there are resonance structures that give a slight negative charge to certain positions of the ring, and this slight negative charge will destabilize the carboxylate.

As for your resonance structures, the first two are ok, but the thrid is wrong as it involves the movement of a proton. Below are some resonance structures for methoxybenzene.

When you draw the resonance structures for the para-isomer you should find a resonance structure that would disfavor dissociation of the proton from the carboxyl group. This resonance structure you would not be able to see with the meta-isomer.