8
$\begingroup$

I am studying acidity of carboxylic acids and I found that benzoic Acid is more acidic than p-methoxybenzoic acid. It is written in my book that this happens because methoxy group is an electron donating group. I can understand why electron donating groups can reduce acidity of an carboxylic acid but I can't understand why methoxy group acts as an electron donating group?

$\endgroup$
11
$\begingroup$

The lone pair on the oxygen delocalizes into the conjugated aromatic ring.

Edit (based on Jan's suggestion):

Notice that the answer is actually a lot more complicated. A methoxy substituent actually impacts the ring electronics via two competing effects. The oxygen's lone pair is well-placed to delocalize and increase electron density within the ring's conjugated system. This allows delocalization to better stabilize positive charges. So the methoxy is electron-donating from a resonance perspective.

On the other hand, because oxygen is quite electronegative, the methoxy group is electron-withdrawing in an inductive sense via the $\sigma$ bonds.

Take a look at a linear free energy relationship (LFER) plot. For acidity of benzoic acids, the methoxy group has a positive $\sigma_{\mathrm{meta}}$ value but a negative $\sigma_{\mathrm{para}}$ value.

The meta $\sigma$ value measures the effect when there is no resonance structure that delocalizes effectively to the carboxylic acid, and its positive value tells us that the methoxy is electron-withdrawing in that position. The negative para $\sigma$ value tells us that when there is effective conjugation into the carboxylic acid, the methoxy is electron-donating. Notice that it is still electron-withdrawing inductively, but the resonance donation is much stronger.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.