During bromination of phenol, Lewis acid is not necessarily required because of :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → +R effect of −OH group
In the bromination of benzene a Lewis acid (\( \ce{FeBr3} \)) is needed to polarise the Br–Br bond and produce the electrophile, because the benzene ring alone is not electron-rich enough to do it.
In phenol, a lone pair on the oxygen of −OH is delocalised into the ring (+R or +M effect). This raises the electron density of the ring, especially at the ortho and para positions, and the activated ring can polarise the bromine molecule by itself. So phenol is brominated even without a Lewis acid:
\( \displaystyle \ce{C6H5OH + 3Br2} \xrightarrow{\text{bromine water}} \text{2,4,6-}\ce{Br3C6H2OH v + 3HBr} \)
In a solvent of low polarity (\( \ce{CHCl3} \) or \( \ce{CS2} \)) at low temperature, only monobromophenols (o- and p-bromophenol) are formed.
Why the other options are incorrect
Option (1) Electromeric effect — Incorrect. This is a temporary shift of the π electrons of a multiple bond when a reagent attacks; it does not explain why the phenol ring is permanently electron-rich.
Option (2) Inductive effect of −OH group — Incorrect. Oxygen is more electronegative than carbon, so −OH has a −I (electron-withdrawing) effect, which would make the ring less reactive, not more.
Option (4) Deactivating effect of −OH group — Incorrect. −OH is a strongly activating, ortho- and para-directing group, not a deactivating one.
NCERT (2026-27 edition):
Halogenation – “On treating phenol with bromine, different reaction products are formed under different experimental conditions.
(a) When the reaction is carried out in solvents of low polarity such as CHCl3 or CS2 and at low temperature, monobromophenols are formed. The usual halogenation of benzene takes place in the presence of a Lewis acid, such as FeBr3 (Unit 6, Class XII), which polarises the halogen molecule. In case of phenol, the polarisation of bromine molecule takes place even in the absence of Lewis acid. It is due to the highly activating effect of –OH group attached to the benzene ring.”