Which of the following is the major product formed in Reimer-Tiemann reaction?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Salicylaldehyde
In the Reimer-Tiemann reaction, phenol is treated with chloroform (CHCl3) and aqueous sodium hydroxide. A –\( \ce{CHCl2} \) group is introduced at the ortho position of the phenoxide ion; the alkali hydrolyses it to –CHO, giving the sodium salt of salicylaldehyde. Acidification then gives salicylaldehyde (2-hydroxybenzaldehyde):
\( \displaystyle \ce{C6H5OH} \xrightarrow[\text{(ii) }\ce{H^+}]{\text{(i) }\ce{CHCl3}\text{, aq. NaOH}} \ce{C6H4(OH)CHO} \)
So the major product is salicylaldehyde.
Why the other options are incorrect
Option (2) Salicylic acid — Incorrect. It is the product of Kolbe’s reaction, in which sodium phenoxide is heated with \( \ce{CO2} \) under pressure and then acidified; an –COOH group, not –CHO, enters the ring:
\( \displaystyle \ce{C6H5OH} \xrightarrow[\text{(ii) }\ce{CO2}\text{, (iii) }\ce{H^+}]{\text{(i) NaOH}} \ce{C6H4(OH)COOH} \)
Option (3) Phenol — Incorrect. Phenol is the starting compound of the reaction, not its product.
Option (4) Benzoquinone — Incorrect. It is formed when phenol is oxidised with chromic acid (\( \ce{Na2Cr2O7} \)/\( \ce{H2SO4} \)):
\( \displaystyle \ce{C6H5OH} \xrightarrow{\ce{Na2Cr2O7}/\ce{H2SO4}} \underset{\text{benzoquinone}}{\ce{C6H4O2}} \)
NCERT (2026-27 edition):
Reimer-Tiemann Reaction
