Bromination of anisole in the presence of ethanoic acid does not require the Lewis acid because
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → of activation of benzene ring by methoxy Group.
Bromination of benzene needs a Lewis acid (\( \ce{FeBr3} \)), which polarises \( \ce{Br2} \) into a strong electrophile, because the benzene ring on its own is not reactive enough.
In anisole (methoxybenzene), the oxygen of the –\( \ce{OCH3} \) group has lone pairs that are delocalised into the ring (+R effect). This increases the electron density of the ring, especially at the ortho and para positions. The ring is then reactive enough to attack \( \ce{Br2} \) directly, so no Lewis acid is needed. Ethanoic acid is used only as the solvent:
\( \displaystyle \ce{C6H5OCH3 + Br2} \xrightarrow{\text{ethanoic acid}} \ce{p-BrC6H4OCH3 + HBr} \)
The para isomer, 4-bromoanisole, is the main product, with a little 2-bromoanisole.
So the Lewis acid is not required because of the activation of the benzene ring by the methoxy group.
Why the other options are incorrect
Option (1) ethanoic acid is a non-polar solvent. — Incorrect. Ethanoic acid is a polar solvent, and the choice of solvent is not what removes the need for the catalyst.
Option (2) bromine dimerizes in the presence of ethanoic acid. — Incorrect. Bromine does not dimerise; it stays as \( \ce{Br2} \).
Option (4) the methoxy group acts as a deactivating group. — Incorrect. The –\( \ce{OCH3} \) group is activating; a deactivated ring would need a catalyst even more.
NCERT (2026-27 edition):
Halogenation
