When phenol is treated with chloroform in the presence of NaOH , the…

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) 2-Hydroxybenzoic acid — Incorrect. 2-Hydroxybenzoic acid (salicylic 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) Anisole — Incorrect. Anisole (methoxybenzene) is made by the Williamson synthesis, from sodium phenoxide and a methyl halide. Chloroform does not methylate the oxygen; it supplies the carbon of a –CHO group on the ring.

      \( \displaystyle \ce{C6H5ONa + CH3I -> C6H5OCH3 + NaI} \)

• Option (4) Methoxy toluene — Incorrect. Methoxytoluene carries a –\( \ce{OCH3} \) group and a ring –\( \ce{CH3} \) group. In the Reimer–Tiemann reaction the –OH group stays free and the new group on the ring is –CHO, not –\( \ce{CH3} \).

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: