The product “C” for the following reaction will be

The correct answer is Option (1) → Salicylaldehyde

The sequence converts benzene into phenol and then introduces a –CHO group ortho to –OH. Each step is a standard reaction:

• Sulphonation (A): benzene heated with oleum (\( \ce{SO3} \) in conc. \( \ce{H2SO4} \)) gives benzenesulphonic acid.

      \( \displaystyle \ce{C6H6 + SO3} \xrightarrow{\text{conc. }\ce{H2SO4}\text{ (oleum)}} \ce{C6H5SO3H} \)

• Alkali fusion (B): benzenesulphonic acid fused with NaOH gives sodium phenoxide, which on acidification gives phenol. This was an early industrial method for phenol.

      \( \displaystyle \ce{C6H5SO3H + 3NaOH} \xrightarrow{\text{fuse}} \ce{C6H5ONa + Na2SO3 + 2H2O} \)

      \( \displaystyle \ce{C6H5ONa + HCl -> C6H5OH + NaCl} \)

• Reimer–Tiemann reaction (C): phenol treated with chloroform and aqueous NaOH gets a –\( \ce{CHCl2} \) group at the ortho position, which is hydrolysed by the alkali; acidification then gives salicylaldehyde (2-hydroxybenzaldehyde).

      \( \displaystyle \ce{C6H5OH + CHCl3 + 4NaOH -> o-C6H4(ONa)CHO + 3NaCl + 3H2O} \)

      \( \displaystyle \ce{o-C6H4(ONa)CHO + HCl -> o-C6H4(OH)CHO + NaCl} \)

So A is benzenesulphonic acid, B is phenol, and the product C is salicylaldehyde.

Why the other options are incorrect

• Option (2) Phenol — Incorrect. Phenol is the intermediate B, formed before the Reimer–Tiemann step.

• Option (3) o-Cresol — Incorrect. o-cresol has a –CH₃ group; chloroform and alkali introduce a –CHO group, not a methyl group.

• Option (4) Benzaldehyde — Incorrect. Benzaldehyde has no –OH group; it is made from benzene by the Gattermann–Koch reaction (CO, HCl, anhyd. \( \ce{AlCl3} \)) or from toluene by the Etard reaction.

NCERT (2026-27 edition):
Reimer-Tiemann Reaction