Match List-I (reagent or condition used with phenol) with List-II (ma…

The correct answer is Option (4) → A-III, B-IV, C-II, D-I

The –OH group strongly activates the ring of phenol and directs the incoming group to the ortho and para positions, so how far bromination goes depends on the conditions. Zinc dust removes the oxygen of phenol, while chromic acid oxidises phenol.

(A) Zinc dust, heat — This matches (III) Benzene. On heating with zinc dust, the –OH group of phenol is removed and benzene is left.

    \( \displaystyle \ce{C6H5OH + Zn} \xrightarrow{\Delta} \ce{C6H6 + ZnO} \)

(B) Bromine in CS₂ at low temperature — This matches (IV) o-Bromophenol and p-bromophenol. In a solvent of low polarity at low temperature only one bromine atom enters the ring, giving monobromophenols.

    \( \displaystyle \ce{C6H5OH + Br2} \xrightarrow[\text{low temperature}]{\ce{CS2}} \ce{BrC6H4OH + HBr} \)

(C) Chromic acid — This matches (II) Benzoquinone. Oxidation of phenol gives this conjugated diketone.

    \( \displaystyle \ce{C6H5OH + 2[O]} \xrightarrow{\text{chromic acid}} \ce{C6H4O2 + H2O} \)

(D) Bromine water — This matches (I) 2,4,6-Tribromophenol. In water all three ortho and para positions are brominated, and the product separates as a white precipitate.

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

NCERT (2026-27 edition):
7.4.4 Chemical Reactions – “(ii) 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.”

“(b) When phenol is treated with bromine water, 2,4,6-tribromophenol is formed as white precipitate.”

“4. Reaction of phenol with zinc dust: Phenol is converted to benzene on heating with zinc dust.”

“5. Oxidation: Oxidation of phenol with chromic acid produces a conjugated diketone known as benzoquinone.”