Match List I with List II List I List II (A) Kolbe's reaction (I) Alk…

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

These reactions belong to three families. Electrophilic substitution of phenol (through the very reactive phenoxide ion): with \( \ce{CO2} \) it gives o-hydroxybenzoic acid (Kolbe’s reaction), and with \( \ce{CHCl3} \)/NaOH it gives salicylaldehyde (Reimer-Tiemann reaction). Preparation of ethers: an alkyl halide with a sodium alkoxide (Williamson synthesis). Dehydrogenation of alcohols: vapours passed over heated copper at 573 K give an aldehyde (1°) or a ketone (2°).

• (A) Kolbe’s reaction — This matches (IV) Phenol to o-hydroxybenzoic acid. Sodium phenoxide is attacked by \( \ce{CO2} \), a weak electrophile, at the ortho position; acidification gives salicylic acid.

      \( \displaystyle \ce{C6H5ONa + CO2} \xrightarrow{\text{400 K, 4-7 atm}} \ce{HOC6H4COONa} \xrightarrow{\ce{H+}} \ce{HOC6H4COOH} \)

• (B) Williamson’s synthesis — This matches (I) Alkylhalide with sodium alkoxide. The alkoxide ion replaces the halogen (S_N2), giving symmetrical or unsymmetrical ethers.

      \( \displaystyle \ce{R-X + R'-ONa -> R-O-R' + NaX} \)

• (C) Conversion of secondary alcohol to ketone — This matches (III) copper at 573K. The vapour of a secondary alcohol passed over heated copper loses hydrogen and gives a ketone.

      \( \displaystyle \ce{CH3CH(OH)CH3} \xrightarrow{\ce{Cu},\ \text{573 K}} \ce{CH3COCH3 + H2} \)

• (D) Reimer Tiemann reaction — This matches (II) Phenol to salicylaldehyde. Phenol with chloroform and NaOH gets a –CHO group at the ortho position; the benzal chloride intermediate is hydrolysed by the alkali.

      \( \displaystyle \ce{C6H5OH + CHCl3 + 4NaOH -> NaOC6H4CHO + 3NaCl + 3H2O} \)

      \( \displaystyle \ce{NaOC6H4CHO + H+ -> HOC6H4CHO + Na+} \)

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