Match List I with List II List I List II A. Reimer-Tiemann reaction I…

The correct answer is Option (2) → A-IV, B-III, C-II, D-I

Phenol has an –OH group that activates the ring towards electrophilic substitution at the ortho and para positions; in alkali the phenoxide ion is even more reactive. Reimer-Tiemann and Kolbe’s reactions put a –CHO or –COOH group at the ortho position, nitration puts in –\( \ce{NO2} \) groups, and zinc dust removes the –OH group altogether.

• A. Reimer-Tiemann reaction — This matches IV. Salicylaldehyde. Phenol with chloroform in aqueous NaOH introduces a –CHO group at the ortho position; the substituted benzal chloride formed is hydrolysed by the alkali and acidification gives salicylaldehyde.

      \( \displaystyle \ce{C6H5OH} \xrightarrow[\text{(ii) }\ce{H+}]{\text{(i) }\ce{CHCl3}\text{, aq. }\ce{NaOH}} \ce{\mathit{o}-HOC6H4CHO} \)

• B. Kolbe’s reaction — This matches III. Salicylic acid. Sodium phenoxide reacts with carbon dioxide, a weak electrophile, and acidification gives ortho-hydroxybenzoic acid.

      \( \displaystyle \ce{C6H5ONa} \xrightarrow[\text{(ii) }\ce{H+}]{\text{(i) }\ce{CO2}} \ce{\mathit{o}-HOC6H4COOH} \)

• C. Nitration of phenol with conc. HNO₃ — This matches II. Picric acid. Concentrated nitric acid nitrates all three activated positions, giving 2,4,6-trinitrophenol.

      \( \displaystyle \ce{C6H5OH + 3HNO3} \xrightarrow{\text{conc.}} \ce{C6H2(NO2)3OH + 3H2O} \)

• D. Reaction of phenol with Zn dust — This matches I. Benzene. On heating, zinc dust takes the oxygen and phenol is reduced to benzene.

      \( \ce{C6H5OH + Zn -> C6H6 + ZnO} \)

Shortcut: A-IV (Reimer-Tiemann gives the aldehyde) appears only in Option (2).

Note: The same/similar question has been asked in different years by NTA as per the details given below: