I understand that we need to add $\ce{FeBr3}$ while brominating benzene in order to polarize the bromine molecule. However, we do not need to add such a Lewis Acid in the case of phenol, because it is activated by the $\ce{-OH}$ group, and so polarizes the bromine molecule on its own. Anisole is also highly activated, however we still add $\ce{FeBr3}$ in order to brominate it. Why? (An answer which compares bromination in phenol and anisole will be highly appreciated.)

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    $\begingroup$ Can you give a reference to the synthesis, please? $\endgroup$ Commented Jun 18, 2015 at 10:15

1 Answer 1


I don't think polarization is the issue. Adding $\ce{FeBr3}$ is just to reduce the $\ce{Br-Br}$ bond strength, speeding up the reaction (attack of the aromatic ring on the bromine is part of the rate determining step, which is slow in the bromination of benzene). It functions as a catalyst being recycled when the $\ce{Br-}$ grabs a $\ce{H+}$ to restore aromaticity in the ring. I'm pretty sure — at least I see no reason to doubt — that the catalyst requirements for the bromination of anisole and the bromination of phenol do not differ much, if at all. That is, no catalyst is required for the bromination of anisole.

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    $\begingroup$ Um, as far as I know, FeBr3 is actually used to generate an electrophile. It grabs one Br atom from the Br2 molecule and gives us Br+, which is an electrophile. This make it easier for it to under bromination. Refer to this : chemwiki.ucdavis.edu/Organic_Chemistry/Hydrocarbons/Aromatics/… Thanks for the answer though, it answered my question well. Cheers! $\endgroup$
    – agdhruv
    Commented Jun 18, 2015 at 17:12
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    $\begingroup$ @ag_dhruv I agree, but that is exactly what I said? The $\ce{Br+}$ is not present as a lose species, it would be too unstable. I'm also not that big of a fan of UC Davis website. It contains plenty of incomplete (and erroneous) material. Cheers :) $\endgroup$
    – Jori
    Commented Jun 18, 2015 at 18:20

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