# 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:

1. p-nitrobenzyl alcohol
2. p-chlorobenzyl alcohol
3. benzyl alcohol
4. 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?

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.
• No, $\ce{Br-}$ is present in solution and does attack the benzylic position. This is an $\mathrm{S_N1}$ reaction. – ringo Feb 16 '17 at 14:40
• $\ce{H+}$ from $\ce{HBr}$ attacks first, $\ce{H2O}$ departs, and is replaced by $\ce{Br-}$. – ringo Feb 16 '17 at 14:44