I've been asked to draw the mechanism by which benzene with $\ce{Me}$ on position 1 and $\ce{OMe}$ on position 2 goes to benzene with $\ce{Me}$ on 1 and $\ce{OH}$ on 2 by reaction with concentrated $\ce{HBr}$.

I'm really unsure as to what is going on, I think its probably most likely that $\ce{Br}$ acts as a nucleophile and attacks the $\ce{C-O}$ $\sigma^*$ and then $\ce{OH}$ attacks the $\ce{C-Br}$ $\sigma^*$ but it doesn't seem like a likely pathway as $\ce{OMe}$ has a higher pKaH than $\ce{Br}$ and substitutions don't occur at $\mathrm{sp^2}$ centres.


1 Answer 1


Cleavage of ethers using strong acids (classically $\ce{HI}$ but $\ce{HBr}$ should work as well) can proceed by either an $S_N1$ or $S_N2$ mechanism. In this case the cation formed would be primary or phenyl, neither of which are stable, and so the reaction proceeds via an $S_N2$ mechanism.

enter image description here

The oxygen is protonated in an equilibrium reaction. The bromide ion then attacks at the least hindered carbon and the $\ce{C-O}$ bond breaks in a concerted $S_N2$ mechanism.



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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