Match List-I (Name reaction) with List-II corresponding substrates.LI…

The correct answer is Option (1) → A - III, B - I, C - IV, D - II

Each named reaction is recognised from its substrate and reagent. Two pairs look alike: the carbylamine and Reimer–Tiemann reactions both use chloroform and an alkali (the substrate decides: an amine or phenol), and the Reimer–Tiemann and Kolbe reactions both start from phenol and NaOH (the third reagent decides: \( \ce{CHCl3} \) or \( \ce{CO2} \)).

• A. Carbylamine reaction — This matches III. RNH₂ + CHCl₃ + 3 KOH. A primary amine heated with chloroform and alcoholic KOH gives a foul-smelling isocyanide; secondary and tertiary amines do not, so this is a test for primary amines.

      \( \displaystyle \ce{R-NH2 + CHCl3 + 3KOH} \xrightarrow{\Delta} \ce{R-NC + 3KCl + 3H2O} \)

• B. Hofmann bromamide degradation reaction — This matches I. RCONH₂ + Br₂ + 4 NaOH. A primary amide gives a primary amine with one carbon atom fewer; the carbonyl carbon leaves as carbonate.

      \( \displaystyle \ce{R-CONH2 + Br2 + 4NaOH -> R-NH2 + Na2CO3 + 2NaBr + 2H2O} \)

• C. Reimer- Tiemann reaction — This matches IV. Ph – OH + CHCl₃ + aq.NaOH. Phenol with chloroform and aqueous NaOH puts a –CHO group at the ortho position; acidification gives salicylaldehyde.

      \( \displaystyle \ce{C6H5OH} \xrightarrow[\text{(ii) }\ce{H+}]{\text{(i) }\ce{CHCl3}\text{ + aq. }\ce{NaOH}} \text{2-hydroxybenzaldehyde} \)

• D. Kolbe's reaction — This matches II. Ph-OH + NaOH + CO₂ + H⁺. Sodium phenoxide, made from phenol and NaOH, is attacked by the weak electrophile CO₂ at the ortho position; acidification gives salicylic acid.

      \( \displaystyle \ce{C6H5OH} \xrightarrow{\ce{NaOH}} \ce{C6H5ONa} \xrightarrow[\text{(ii) }\ce{H+}]{\text{(i) }\ce{CO2}} \text{2-hydroxybenzoic acid} \)

Shortcut: The pair A–III appears in Options (1) and (3), and B–I then leaves only Option (1).