If we look at the resonance structures of benzoate anion, we would see that some amount of positive charge is present on the ring and this charge can get delocalised on the ring, like this:

resonance structures of benzoate anion

Now, if we had an electron donating group (EDG) at the para position, then the EDG should stabilize the ring because it can donate pi electrons to the empty p orbital of carbocation (look at third resonance structure) and form a double bond, thus resulting in more resonance and more resonance implies more stability and this should imply that EDG increases acidity as conjugate base is getting more stabilized.

But, it is a well established fact that EDG decrease the acidity of the molecule (benzoic acid). For example: $\mathrm{p}K_\mathrm{a}$ (para methoxy benzoic acid) > $\mathrm{p}K_\mathrm{a}$ (benzoic acid)

Why am I facing such a contradiction?


You seem to pay too much importance to these individual structures. However, note that there are no individual resonance structure in existence, but only the resonance hybrid - a combination of all these resonance structures in proportion to their stability.

The final three structures that you've drawn have a really negligible contribution to the final resonance hybrid of the benzoate anion. Why, you ask? Because they exhibit charge separation: creation of an additional positive and a negative charge.

Hence, your conclusion is wrong, because it's based on the invalid premise that the third (and by extension, the fifth) resonance structures have a major role in the resonance hybrid. The resonance hybrid in fact has no appreciable positive charge on the ring carbons.

The correct way to explain this is that you should look at the resonance hybrid as a whole, which is negatively charged. Adding an electron donating group to the ring would increase the negative charge further, thereby destabilizing the anion.


If you look closely at your diagram, you will notice that once benzoic acid has released its proton, the conjugate base that is formed has 6 resonance structures out of which 4 have 2 more charges that offset each other. The overall charge of benzoate anion is -1 regardless.

The important part of this discussion revolves around the last 4 resonance structures. For the sake of the discussion, I will only refer to the 2 charges that offset each other. When a negative charge appears on the oxygen atom in the carboxyl group, a positive charge appears on the ring (the charge is spread across the nucleus).

At this point, we see that the conjugate base is quite stable because it has 6 resonance structures. The more resonance structures the conjugate base has, the stronger the acid is. This is how the acidity of benzoic acid is explained.

Now, if you placed a methoxy group in para position, it would doante electrons to the ring through a mesomeric effect. If that happens, the electron density around the nucleus increases, diminishing the positive charge. The methoxy group increases the electron density and so the positive charge on the nucleus is no longer “that positive” while the negative charge remains unaffected (“just as negative”). If the positive charge can’t properly balance the negative charge, the resonance structures involving charge separation are not advantageous to exist and so the strength of the acid decreases.

This happens only if the methoxy group is situated para relative to the carboxyl group. If it is situated meta relative to the carboxyl, something interesting happens.

We know that the methoxy (like the hydroxy group) exerts a -I inductive effect and a mesomeric (M) effect. That is why, if found para, it “donates” electron density through a +M effect (which manifests more than -I). When in meta, the +M effect is no longer a dominant effect and so, methoxy manifests -I effect mostly, withdrawing electron density from the nucleus. It is often said that hydroxy and methoxy groups are EDG, but only when it comes to electromeric effects. They also have -I inductive effects which makes them EWG.

Overall, the presence of a EDG in ortho or para destabilises the actual negative charge of the benzoate anion, thus making it less stable.

  • $\begingroup$ sorry, I can;t understand this part of your answer..."The methoxy group increases the electron density and so the positive charge on the nucleus is no ............ so the strength of the acid decreases".Before methoxy group was attached we had a ring with positive charge and once we attach methoxy group we have less positive charge on the ring.it should stabilize it right? the amount of negative charge,at the carboxyl(oxygen) remains same after attachment also .moreover what do you mean by positive charge not being able to balance the negative charge properly(line 14)? $\endgroup$ – Banchin Jul 13 '18 at 23:51
  • $\begingroup$ @Gaurang Tandon, how do you know the last 4 structure have a insignificant contribution ? Though, your answer is beautifully couched. $\endgroup$ – AndrewB Jul 14 '18 at 21:25
  • $\begingroup$ @AndrewB The last three structures have an insignificant contribution because of charge separation. Charge separated resonance structures have a lower contribution to the resonance hybrid. $\endgroup$ – Gaurang Tandon Jul 15 '18 at 3:00

Benzoic acid ($\mathrm{p}K_\mathrm{a}=4.2$, PubChem) is actually more acidic than p-methoxybenzoic acid ($\mathrm{p}K_\mathrm{a}=4.47$, PubChem).

An electron donating group essentially destabilizes the conjugate base of the acid, because of which this phenomenon occurs. Your explanation speaks of the carbocation formed in some resonance structures which has a double negative charge and a single positive charge.

The electron donating group would increase the stability of that cation but there are two negative charges that aren't stabilised but destabilised instead.

Because of this, the overall stability of the conjugate base is lower than in case of no electron donating group.

  • $\begingroup$ I do agree that paramethoxy benzoic acid is less acidic than benzoic acid intuitively,but from my explanation(as mentioned in the question) I find some contradiction and my question asks you to explain it... $\endgroup$ – Banchin Jul 14 '18 at 1:49

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