When phenol is treated with chloroform in the presence of Sodium hydr…

The correct answer is Option (3) → Salicylaldehyde

In the Reimer–Tiemann reaction, phenol is treated with chloroform in the presence of aqueous sodium hydroxide, and a –CHO group is introduced at the ortho position of the benzene ring. The phenoxide ion, whose ring is strongly activated, is attacked by dichlorocarbene, \( \ce{:CCl2} \), formed from chloroform and alkali. The substituted benzal chloride formed first is hydrolysed by the alkali, and acidification gives salicylaldehyde (2-hydroxybenzaldehyde):

    \( \displaystyle \ce{C6H5OH} \xrightarrow{\ce{CHCl3} + \text{aq. }\ce{NaOH}} o\text{-}\ce{NaOC6H4CHCl2} \xrightarrow{\text{aq. }\ce{NaOH}} o\text{-}\ce{NaOC6H4CHO} \xrightarrow{\ce{H+}} o\text{-}\ce{HOC6H4CHO} \)

So the product is salicylaldehyde.

Why the other options are incorrect

• Option (1) Salicylic acid — Incorrect. Salicylic acid (2-hydroxybenzoic acid) is made from sodium phenoxide and carbon dioxide by Kolbe’s reaction, not with chloroform.

      \( \displaystyle \ce{C6H5ONa} \xrightarrow[\text{(ii) }\ce{H+}]{\text{(i) }\ce{CO2},\ 400\,\text{K},\ 4\text{-}7\,\text{atm}} o\text{-}\ce{HOC6H4COOH} \)

• Option (2) Salicylate ion — Incorrect. The salicylate ion (2-hydroxybenzoate) is the anion of salicylic acid, formed in Kolbe’s reaction from sodium phenoxide and carbon dioxide. With chloroform and alkali the group put on the ring is –CHO, not –\( \ce{COO^-} \).

• Option (4) benzoquinone — Incorrect. Benzoquinone is the oxidation product of phenol with chromic acid. Chloroform and sodium hydroxide do not oxidise phenol; they add a –CHO group to the ring.

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

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