Ether cleavage with conc HBr

Question

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.

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.

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.

(source)