Match List-I with List-II
| List-I Reactions | List-II Product |
(A)  | (I)  |
(B)  | (II)  |
(C)  | (III)  |
(D)  | (IV)  |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
The –OH group makes the benzene ring of phenol very electron rich. So phenol is easily oxidised, and it undergoes electrophilic substitution at the ortho and para positions much faster than benzene. The reagent and the conditions decide the product:
(A) Phenol with \( \ce{Na2Cr2O7} \)/\( \ce{H2SO4} \) — This matches structure III (benzoquinone). Chromic acid oxidises phenol to the conjugated diketone benzoquinone (p-benzoquinone).
\( \displaystyle \ce{C6H5OH} \xrightarrow{\ce{Na2Cr2O7}/\ce{H2SO4}} \ce{C6H4O2}\ \text{(benzoquinone)} \)
(B) Phenol with (i) NaOH (ii) \( \ce{CO2} \) (iii) \( \ce{H^{+}} \) — This matches structure I (salicylic acid). This is the Kolbe reaction: the phenoxide ion, even more reactive than phenol, is attacked by the weak electrophile \( \ce{CO2} \) at the ortho position; acidification gives 2-hydroxybenzoic acid.
\( \displaystyle \ce{C6H5OH} \xrightarrow{\ce{NaOH}} \ce{C6H5ONa} \xrightarrow[\text{(ii) }\ce{H^+}]{\text{(i) }\ce{CO2}} \text{salicylic acid} \)
(C) Phenol with \( \ce{Br2} \) in water — This matches structure IV (2,4,6-tribromophenol). In the polar solvent water, phenol partly ionises to the highly activated phenoxide ion, so all three ortho and para positions are brominated and a white precipitate forms.
\( \displaystyle \ce{C6H5OH + 3Br2} \xrightarrow{\ce{H2O}} \ce{C6H2Br3OH v + 3HBr} \)
(D) Phenol with conc. \( \ce{HNO3} \) — This matches structure II (2,4,6-trinitrophenol, picric acid). Concentrated nitric acid nitrates all three ortho and para positions.
\( \displaystyle \ce{C6H5OH + 3HNO3} \xrightarrow{\text{conc.}} \ce{C6H2(NO2)3OH + 3H2O} \)
Shortcut: The pair (B)–(I) appears only in Option (1).