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

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?


1 Answer 1


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.

  • $\begingroup$ So are you telling that a Br+ would attack in this case rather than a Br-. So what will the final product of this reaction be then? $\endgroup$ Commented Feb 16, 2017 at 14:34
  • $\begingroup$ No, $\ce{Br-}$ is present in solution and does attack the benzylic position. This is an $\mathrm{S_N1}$ reaction. $\endgroup$
    – ringo
    Commented Feb 16, 2017 at 14:40
  • $\begingroup$ So what is attacking the reactant then? $\endgroup$ Commented Feb 16, 2017 at 14:42
  • 1
    $\begingroup$ $\ce{H+}$ from $\ce{HBr}$ attacks first, $\ce{H2O}$ departs, and is replaced by $\ce{Br-}$. $\endgroup$
    – ringo
    Commented Feb 16, 2017 at 14:44
  • 1
    $\begingroup$ It does not and cannot replace it. It activates the alcohol oxygen to leave by forming water as a leaving group, which departs, then bromide attacks the same position to form 1-bromomethyl-4-methoxybenzene. $\endgroup$
    – ringo
    Commented Feb 16, 2017 at 15:01

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