Q30
1 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?

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 →(NaNO₂ + HCl, 273–278 K)→ C₆H₅N₂⁺Cl⁻ →(CuCN/KCN, warm)→ C₆H₅CN (benzonitrile)
Hinsberg reagentsecondary amineinsoluble in alkaliN-methylethanaminebenzenesulphonyl chloridediazotisationSandmeyer CuCNbenzonitrile

Marking Scheme

  • 11 mark: X = N-methylethanamine / ethylmethylamine (CH₃NHC₂H₅), a secondary amine (justification via alkali-insoluble sulphonamide accepted).
  • 2OR 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→benzonitrile, 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.

Analysis & 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 → sulphonamide with an acidic N–H → soluble in alkali.
  • 2° amine → N,N-disubstituted sulphonamide, no N–H → insoluble in alkali.
  • 3° amine → does not react.
  • 'Insoluble in alkali' therefore pins X as a secondary amine. The only secondary amine of formula C₃H₉N is ethylmethylamine, CH₃NHC₂H₅.

Aniline → 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 → benzenediazonium chloride.
  • Step 2 (Sandmeyer): treat with CuCN/KCN → 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.

Common Mistakes

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

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 → diazonium → nitrile route is a versatile branch point in aromatic synthesis.

Spotted a mistake or something unclear?

Tell us — we fix reported answers fast.

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.