Past exam of the chemistry B course of the University of Cambridge / 2022 / Paper 1 / B1 / a / iii / Solution

The activated aromatic pi system polarizes bromine and attacks its electrophilic end while Br–Br cleaves. This makes a C–Br bond at C1 or C3 and gives a Wheland intermediate. Donation from phenolic oxygen stabilizes contributors with positive charge at its attached ring carbon. acetate or bromide then removes the proton from the attacked carbon; the C–H bond electrons restore the aromatic pi system. Repetition at the other activated ortho site gives the dibrominated product.

Each step replaces C–H by C–Br and restores aromaticity. No radical initiator or benzylic abstraction is needed, and no Lewis acid catalyst should be invented for this already activated phenolic substrate. The overall substitution balance is $\mathrm{Ar}{\mathrm{H}}_2+2B{\mathrm{r}}_2\to \mathrm{ArB}{\mathrm{r}}_2+2HBr$.