Question 301 markSection D

$(b) A compound 'X' with molecular formula C₃H₉N reacts with Hinsberg reagent to give a product insoluble in alkali. Identify 'X'. OR (b) How can you convert aniline to benzonitrile?$

Q: (b) A compound 'X' with molecular formula C₃H₉N reacts with Hinsberg reagent to give a product insoluble in alkali. Identify 'X'. OR (b) How can you convert aniline to benzonitrile?

**Hinsberg version — identify X:** A product **insoluble in alkali** means X is a **secondary amine** (its N,N-disubstituted sulphonamide has no acidic N–H, so it does not dissolve in KOH). - **X = N-methylethanamine, CH₃–NH–C₂H₅** (ethylmethylamine), a secondary amine of formula C₃H₉N. **OR — Aniline to benzonitrile:** Diazotise aniline, then replace the diazonium group by –CN (Sandmeyer reaction). - C₆H₅NH₂ →(NaNO₂ + HCl, 273–278 K)→ C₆H₅N₂⁺Cl⁻ →(CuCN/KCN, warm)→ **C₆H₅CN (benzonitrile)**

Amines

Hinsberg Test / Aniline to Benzonitrile

Official answer

Hinsberg version — identify X: A product insoluble in alkali means X is a secondary amine (its N,N-disubstituted sulphonamide has no acidic N-H, so it does not dissolve in KOH). X = N-methylethanamine, CH₃-NH-C₂H₅ (ethylmethylamine), a secondary amine of formula C₃H₉N. OR — Aniline to benzonitrile: Diazotise aniline, then replace the diazonium group by -CN (Sandmeyer reaction). C₆H₅NH₂ \to(NaNO₂ + HCl, 273-278 K)\to C₆H₅N₂⁺Cl⁻ \to(CuCN/KCN, warm)\to C₆H₅CN (benzonitrile)

Hinsberg reagent

secondary amine

insoluble in alkali

N-methylethanamine

benzenesulphonyl chloride

diazotisation

Sandmeyer CuCN

benzonitrile

Marking scheme

1. $1 mark: X = N-methylethanamine / ethylmethylamine (CH₃NHC₂H₅), a secondary amine (justification via alkali-insoluble sulphonamide accepted).$
2. $OR 1 mark: correct two-step route — diazotisation (NaNO₂/HCl, 273-278 K) followed by CuCN/KCN (Sandmeyer) to give benzonitrile; ½ for partial route.$

Hint

$Hinsberg: alkali-insoluble product = secondary amine, so C₃H₉N must be CH₃NHC₂H₅; for aniline\tobenzonitrile, diazotise then use CuCN/KCN (Sandmeyer).$

Quick oral answer

$Since the Hinsberg product is insoluble in alkali, X is a secondary amine, namely N-methylethanamine CH₃NHC₂H₅; and aniline is converted to benzonitrile by diazotising it with NaNO₂/HCl at 0-5 °C and then treating the diazonium salt with CuCN/KCN in the Sandmeyer reaction.$

Common mistakes

1. $Identifying X as a primary amine (n-propylamine/isopropylamine) — those give alkali-soluble sulphonamides; the alkali-insoluble product indicates a secondary amine.$
2. $Choosing trimethylamine (a 3° amine) — it does not react with Hinsberg reagent at all, so it cannot give the described product.$
3. $For aniline\tobenzonitrile, jumping straight to CuCN without first diazotising, or running diazotisation above 5 °C where the diazonium salt decomposes.$

Analysis and explanation

A one-mark case sub-part, but it rewards the exact logic of the Hinsberg test described in the passage. Hinsberg reasoning: Hinsberg reagent is benzenesulphonyl chloride, C₆H₅SO₂Cl. 1° amine \to sulphonamide with an acidic N-H \to soluble in alkali . 2° amine \to N,N-disubstituted sulphonamide, no N-H \to insoluble in alkali . 3° amine \to does not react.'Insoluble in alkali' therefore pins X as a secondary amine . The only secondary amine of formula C₃H₉N is ethylmethylamine, CH₃NHC₂H₅ . Aniline \to benzonitrile (OR part): This uses the passage's hint that aryldiazonium salts let you swap -N₂⁺ for many nucleophiles. Step 1 (diazotisation): aniline + NaNO₂/HCl at 0-5 °C \to benzenediazonium chloride. Step 2 (Sandmeyer): treat with CuCN/KCN \to the -CN replaces -N₂⁺ giving benzonitrile. Exam trap: For the Hinsberg part, do not pick a primary amine (n-propylamine or isopropylamine) — those give alkali-soluble products; trimethylamine (3°) does not react at all.

Interesting facts

$Hinsberg reagent (benzenesulphonyl chloride) was introduced by Oscar Hinsberg in 1890 and is still the classic way to sort 1°, 2° and 3° amines in the lab.$

$The Sandmeyer reaction (1884), which uses copper(I) salts to replace a diazonium group, lets chemists install -Cl, -Br, -CN and other groups on an aromatic ring with position-perfect selectivity.$

$Benzonitrile made this way can be hydrolysed to benzoic acid or reduced to benzylamine, so the aniline \to diazonium \to nitrile route is a versatile branch point in aromatic synthesis.$

Question details

Q30 of 37 questions

Board: CBSE
Class: Class 12
Subject: Chemistry
Year: 2026
Section: Section D
Marks: 1
Chapter: Amines
Topic: $Hinsberg Test / Aniline to Benzonitrile$
Bloom's level: apply
Syllabus topic: $Unit: Amines — identification of amines (Hinsberg test) and diazonium salt reactions$
Author: CBSE
Reviewed by: Padhantu Academic Team

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Frequently asked questions

$Why does an alkali-insoluble Hinsberg product mean the amine is secondary?$

$With Hinsberg reagent (benzenesulphonyl chloride), a primary amine forms a sulphonamide that still has an acidic N-H and dissolves in alkali, whereas a secondary amine forms an N,N-disubstituted sulphonamide with no N-H, so it stays insoluble in alkali. A tertiary amine does not react at all. Hence the alkali-insoluble product identifies a secondary amine, here CH₃NHC₂H₅.$

$How is aniline converted to benzonitrile?$

$First diazotise aniline with NaNO₂ and HCl at 273-278 K (0-5 °C) to form benzenediazonium chloride. Then treat this salt with CuCN/KCN (the Sandmeyer reaction); the cyanide group replaces the diazonium group to give benzonitrile, C₆H₅CN.$