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🧪 Chemistry  ·  Amines  ·  NEET & JEE

Which of the following will form an insoluble product with Hinsberg's reagent that is also insoluble in NaOH?

Answer: Diethylamine (secondary amine).

  • A Diethylamine (secondary amine)
  • B Ethylamine (primary amine)
  • C Triethylamine (tertiary amine)
  • D Aniline

Correct answer: A. Diethylamine (secondary amine)

Explanation: Secondary amines give N,N-disubstituted sulfonamides with no N-H; this is insoluble in NaOH.

Why Aniline Is a Weaker Base Than MethylamineNCH₃lone pair fully availableMethylamineN donates electrons freely → STRONG baseNlone pair delocalised into ringAniline

In aniline, the nitrogen's lone pair is pulled into the benzene ring through resonance, leaving less electron density available to accept a proton - which is why aniline is a much weaker base than methylamine, where the lone pair is fully available.

Concept context

Nitrogen-containing organic compounds. Covers classification (primary, secondary, tertiary), basicity comparison, preparation methods, and reactions including diazotization and coupling, key for understanding dyes and pharmaceuticals.

Read the full Amines notes →