Why is the $\ce{-OAc}$ group o,p-directing in electrophilic aromatic substitution?

How can I compare the activating effect of $\ce{-OAc}$ with $\ce{-OH}$ and $\ce{O-}$? Is it to do with the number of lone pairs on oxygen?


Phenoxide is more reactive than phenol towards electrophilic substitution. However, the number of lone pairs isn't the factor that controls it, because it is not possible for both lone pairs to be simultaneously in conjugation with the benzene ring. If one lone pair is in conjugation with the benzene ring, the other lone pairs will not be oriented correctly to overlap with the pi system of the ring. So, you can have one, two, or three lone pairs - it does not matter because any lone pairs beyond the first are useless.

Only the availability of the lone pairs matters. That has to be discussed with inductive or resonance effects. With an $\ce{-OH}$ group you have a hydrogen atom fighting for some of the electrons (inductive effect), and with a $\ce{-O^-}$ group there is nothing else.

Moving on to the acetoxy group $\ce{-OC(O)CH3}$, the lone pair on oxygen is delocalised into the electron-withdrawing carbonyl group, via resonance. You can think of it as the $\ce{C=O}$ and the benzene ring both fighting over the oxygen lone pair, which makes it less available for donation to the pi system. Consequently, the acetoxy group is still o/p-directing (because of the presence of at least one lone pair), but is not as strongly activating as the hydroxy group (because the lone pair is less available).


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.