What would be the product X in the following chemical reaction?
\( \displaystyle \text{Aniline} \xrightarrow[\text{ii) }\ce{H2O}\text{, Warm}]{\text{i) }\ce{NaNO2 + HCl}} \text{A} \xrightarrow[\ce{H3O+}]{\ce{CHCl3} + \text{aq. }\ce{NaOH}} \text{X} \)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Salicylaldehyde
This is a two-stage conversion: aniline is first changed to phenol through its diazonium salt, and phenol then undergoes the Reimer–Tiemann reaction, which puts a –CHO group at the ortho position.
First, aniline is diazotised with nitrous acid (from \( \ce{NaNO2 + HCl} \)) in the cold, giving benzenediazonium chloride:
\( \displaystyle \ce{C6H5NH2 + NaNO2 + 2HCl} \xrightarrow{273-278\ \mathrm{K}} \ce{C6H5N2^+Cl^- + NaCl + 2H2O} \)
On warming with water, the diazonium salt is hydrolysed: N2 escapes and –OH takes its place. So A is phenol:
\( \displaystyle \ce{C6H5N2^+Cl^- + H2O} \xrightarrow{\text{warm}} \ce{C6H5OH + N2 ^ + HCl} \)
Next, phenol is treated with chloroform and aqueous NaOH. The phenoxide ion is attacked at the ortho position, and the substituted benzal chloride (–CHCl2) formed as the intermediate is hydrolysed by the alkali to –CHO. The product is present as its sodium phenoxide salt:
\( \displaystyle \ce{C6H5OH + CHCl3 + 4NaOH -> 2-NaOC6H4CHO + 3NaCl + 3H2O} \)
Acidification (\( \ce{H3O+} \)) converts the phenoxide back to the free –OH group:
\( \displaystyle \ce{2-NaOC6H4CHO + H3O^+ -> 2-HOC6H4CHO + Na^+ + H2O} \)
So X is 2-hydroxybenzaldehyde, that is, salicylaldehyde.
Why the other options are incorrect
Option (1) Salicylic acid — Incorrect. Salicylic acid (2-hydroxybenzoic acid) has –COOH, not –CHO, at the ortho position. It is the product of Kolbe’s reaction, in which sodium phenoxide reacts with CO2 and the product is acidified:
\( \displaystyle \ce{C6H5ONa} \xrightarrow[\text{(ii) }\ce{H3O+}]{\text{(i) }\ce{CO2}} \ce{2-HOC6H4COOH} \)
Option (2) Benzoquinone — Incorrect. Benzoquinone is a conjugated diketone formed when phenol is oxidised with chromic acid (\( \ce{Na2Cr2O7/H2SO4} \)). Chloroform and NaOH do not oxidise phenol.
Option (4) 4-Hydroxybenzoic acid — Incorrect. This has the wrong group in the wrong place: a –COOH group at the para position. The Reimer–Tiemann reaction with chloroform introduces –CHO, and mainly at the ortho position.
NCERT (2026-27 edition):
From diazonium salts – “A diazonium salt is formed by treating an aromatic primary amine with nitrous acid (NaNO2 + HCl) at 273-278 K. Diazonium salts are hydrolysed to phenols by warming with water or by treating with dilute acids (Unit 9, Class XII).”
Reimer-Tiemann reaction – “On treating phenol with chloroform in the presence of sodium hydroxide, a –CHO group is introduced at ortho position of benzene ring. This reaction is known as Reimer - Tiemann reaction.
The intermediate substituted benzal chloride is hydrolysed in the presence of alkali to produce salicylaldehyde.”