Which product is formed when phenol is treated with bromine water?

The correct answer is Option (3) → 2,4,6-Tribromophenol

The –OH group strongly activates the benzene ring of phenol and directs incoming groups to the ortho and para positions. With bromine water, bromine enters at all three of these positions (2, 4 and 6), and the product separates as a white precipitate:

    \( \displaystyle \ce{C6H5OH + 3Br2} \xrightarrow{\ce{H2O}} \ce{C6H2Br3OH + 3HBr} \)

Why the other options are incorrect

Option (1) o-Bromophenol and p-bromophenol — Incorrect. The monobromophenols form only in a solvent of low polarity such as CS₂ or CHCl₃ at low temperature, not with bromine water.

Option (2) 2,4-Dibromophenol — Incorrect. In bromine water the substitution does not stop at two bromine atoms; the third free ortho/para position is brominated as well.

Option (4) 2,3,5-Tribromophenol — Incorrect. Positions 3 and 5 are meta to the –OH group; the –OH group sends bromine to the ortho and para positions, not the meta positions.

NCERT (2026-27 edition):
(ii) Halogenation – “(b) When phenol is treated with bromine water, 2,4,6-tribromophenol is formed as white precipitate.”