An aromatic compound 'A' of molecular formula \(C_7H_7ON\) undergoes a series of reaction as:
\(\underset{A}{C_7H_7ON} \overset{Br_2 / KOH}{\longrightarrow} B \overset{NaNO_2+ HCl}{\longrightarrow} C \overset{CH_3CH_2OH}{\longrightarrow} D\)
Identify the compound D:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(C_6H_6\)
An aromatic compound \( \ce{C7H7ON} \) that gives an amine with \( \ce{Br2} \)/KOH is an amide: benzamide, \( \ce{C6H5CONH2} \).
Hoffmann bromamide degradation (one carbon less): \( \ce{C6H5CONH2 + Br2 + 4KOH -> C6H5NH2 + K2CO3 + 2KBr + 2H2O} \) → B = aniline
Diazotisation (273-278 K): \( \ce{C6H5NH2 + NaNO2 + 2HCl -> C6H5N2^+Cl^- + NaCl + 2H2O} \) → C = benzenediazonium chloride
Reduction by ethanol (replacement by H): \( \ce{C6H5N2^+Cl^- + CH3CH2OH -> C6H6 + N2 + CH3CHO + HCl} \) → D = benzene
Why the other options are incorrect
Option (1) \(C_7H_5N\) — Incorrect. Benzonitrile; it would need the CN group (Sandmeyer with CuCN), and the carbon lost in the Hoffmann step cannot return.
Option (2) \(C_6H_6O\) — Incorrect. Phenol; formed when the diazonium salt is warmed with water, not with ethanol.
Option (3) \(C_6H_7N\) — Incorrect. Aniline; this is B, the intermediate, not the final product D.
NCERT (2026-27 edition):
Hoffmann Bromamide Degradation Reaction – “Hoffmann developed a method for preparation of primary amines by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide. In this degradation reaction, migration of an alkyl or aryl group takes place from carbonyl carbon of the amide to the nitrogen atom. The amine so formed contains one carbon less than that present in the amide.”
Reactions Involving Displacement of Nitrogen – “… 4. Replacement by H: Certain mild reducing agents like hypophosphorous acid (phosphinic acid) or ethanol reduce diazonium salts to arenes and themselves get oxidised to phosphorous acid and ethanal, respectively.”