Phenol reacts with bromine WATER (aqueous $\mathrm{Br_2}$) at room temperature to give

1. Your syllabus (Phenol reactions — electrophilic substitution): phenol is HIGHLY ACTIVATED toward electrophilic substitution because the $\mathrm{-OH}$ donates electrons strongly into the ring (resonance + inductive ortho/para director). 2. With $\mathrm{Br_2}$ in WATER, the polar medium ENHANCES the reactivity further (because the phenoxide form $\mathrm{C_6H_5O^-}$ is even more reactive than phenol itself). 3. Multiple bromine atoms add — all three ortho/para positions get brominated, giving 2,4,6-TRIBROMOPHENOL as a WHITE PRECIPITATE. 4. In contrast: with $\mathrm{Br_2}$ in $\mathrm{CS_2}$ (non-polar solvent), the reaction is milder and gives mainly monobrominated product (ortho or para). 5. Option A is the $\mathrm{CS_2}$-medium product (selectivity by reagent/solvent). Option C is what would form with $\mathrm{C_6H_6}$ and Fe catalyst, NOT phenol. Option D contradicts phenol's very high reactivity. _Source: syllabus Chemistry Part 2, Ch 7, §7.7 (Phenol — bromination), p. 18._