Match List I with List II
| List I | List II |
| (A) Phenol + \(Br_2\) water | (I) Reimer-Tiemann Reaction |
| (B) Phenol + \(CO_2/NaOH\) | (II) Azo-dye test |
| (C) Phenol + \(CHCl_3 / KOH\) | (III) Kolbe's Reaction |
| (D) Phenol + Benzene diazonium chloride and \(NaOH\) | (IV) White precipitate |
Choose the correct answers from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-IV, B-III, C-I, D-II
These are the main reactions of phenol: the –OH group (or the phenoxide ion in alkali) makes the ring very reactive, so it is substituted at the ortho and para positions.
(A) Phenol + Br2 water — This matches (IV) White precipitate. Bromine water gives 2,4,6-tribromophenol as a white precipitate:
\( \ce{C6H5OH + 3Br2(aq) -> C6H2Br3OH v + 3HBr} \) (2,4,6-tribromophenol)
(B) Phenol + CO2/NaOH — This matches (III) Kolbe’s Reaction. Sodium phenoxide takes up \( \ce{CO2} \); acidification gives salicylic acid:
\( \ce{C6H5ONa + CO2} \xrightarrow[\text{(ii) }\ce{H+}]{\text{(i) 400 K, 4-7 atm}} \text{o-}\ce{HOC6H4COOH} \)
(C) Phenol + CHCl3/KOH — This matches (I) Reimer-Tiemann Reaction. A –CHO group is introduced at the ortho position:
\( \ce{C6H5OH + CHCl3 + 3NaOH} \xrightarrow[\text{(ii) }\ce{H+}]{\text{(i) 340 K}} \text{o-}\ce{HOC6H4CHO} \) (salicylaldehyde)
(D) Phenol + Benzene diazonium chloride and NaOH — This matches (II) Azo-dye test. Coupling at the para position gives the orange dye p-hydroxyazobenzene:
\( \ce{C6H5N2+Cl- + C6H5OH} \xrightarrow{\ce{OH-}} \text{p-}\ce{HOC6H4N=NC6H5 + Cl- + H2O} \)