# Ether cleavage with conc HBr

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.

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.