Which of the following para-substituted benzyl alcohols reacts fastest with HBr? I had a question in a recent test which asked me to pick the compound that would react fastest with $\ce{HBr}$ with the options being:


p-nitrobenzyl alcohol
p-chlorobenzyl alcohol
benzyl alcohol
p-methoxybenzyl alcohol


The answer was p-methoxybenzyl alcohol.

Now what I don't understand is that $\ce{HBr}$ would produce nucleophilic $\ce{Br-}$, which would be looking to attack a region of electron deficiency. In that case option (1) p-nitrobenzyl alcohol would the right answer because the nitro group is an electron withdrawing group, thus making the ring more electron deficient.

Could someone please explain as to why this logic fails?
Inductive effects will play much more of a role in electrophilic aromatic substitution. Here, positive charge on a benzylic position is being delocalized to the aromatic ring, and the question is asking which group will best stabilize this positive charge (via resonance):




Between $\ce{-H}$ (neutral), $\ce{-NO2}$ (which would destabilize the positive charge), $\ce{-Cl}$ (poor at stabilizing positive charge), and $\ce{-OCH3}$ (good at stabilizing positive charge), $\ce{-OCH3}$ is the clear winner.