Organic chemistry (Amines)
An organic compound (A) with the molecular formula C8H9NO when treated with bromine in aqueous solution of sodium hydroxide gives a new compound (B) having the molecular formula C7H9N. The compound (B) on heating with chloroform and ethanolic potassium hydroxide forms a compound (C) having foul-smelling. Further, the compound (B) gives an alkali soluble sulphonamide on reacting with benzenesulfonyl chloride.
Moreover, a thermally unstable compound (D) is obtained when the compound (B) reacts with sodium nitrite in concentrated hydrochloric acid at very low temperature. Further, the compound (D) on heating with water yields a compound which gives colour with neutral ferric chloride solution.
Answer the following questions on the basis of this passage.
Identify the compound (D):
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4-methylbenzenediazonium chloride
Work through the passage one clue at a time:
A (C8H9NO) → B (C7H9N) with Br2/NaOH: this is the Hofmann bromamide degradation. It turns an amide into a primary amine with one carbon atom fewer, so A is an amide and B is a primary amine.
\( \displaystyle \ce{CH3C6H4CONH2 + Br2 + 4NaOH -> CH3C6H4NH2 + Na2CO3 + 2NaBr + 2H2O} \)
B has 4 degrees of unsaturation: (2 × 7 + 2 + 1 − 9)/2 = 4, which fits a benzene ring. So B is a methyl-substituted aniline (toluidine).
Checks on B: it gives the carbylamine test and an alkali-soluble Hinsberg product, both of which confirm a primary amine.
B + NaNO2/HCl at low temperature → D: a primary aromatic amine gives an arenediazonium salt. This is stable only at 273–278 K and decomposes on warming, which is why the passage calls it “thermally unstable”.
D + warm water → a compound that gives a colour with neutral FeCl3: that compound is a phenol, which confirms D is an aromatic diazonium salt.
\( \displaystyle \ce{CH3C6H4NH2 + NaNO2 + 2HCl} \xrightarrow{\text{273-278 K}} \ce{CH3C6H4N2^+Cl^- + NaCl + 2H2O} \)
The passage does not say where the methyl group is, but every option shows the para isomer. So D is 4-methylbenzenediazonium chloride (p-toluenediazonium chloride).
Why the other options are incorrect
Option (1) 4-methylcyclohexanediazonium chloride — Incorrect. Aliphatic diazonium salts are too unstable to form even at low temperature; they lose \( \ce{N2} \) at once. Also, heating one with water would give an alcohol, which gives no colour with \( \ce{FeCl3} \).
Option (2) 4-methylphenol (p-cresol) — Incorrect. This is the product formed when D is warmed with water, not D itself.
Option (4) 4-methylaniline (p-toluidine) — Incorrect. This is compound B, the amine that is diazotised to give D.
NCERT (2026-27 edition):
Method of Preparation of Diazonium Salts
