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🧪 Chemistry  ·  Amines  ·  NEET & JEE

Which amine is prepared by the Schmidt reaction from a carboxylic acid and hydrazoic acid?

Answer: Primary amine with one less carbon.

  • A Primary amine with one less carbon
  • B Secondary amine formed by double substitution at nitrogen
  • C Tertiary amine formed by exhaustive N-alkylation
  • D Aromatic amine formed by direct ring amination

Correct answer: A. Primary amine with one less carbon

Explanation: Schmidt reaction: RCOOH + HN<sub>3</sub> → RNH<sub>2</sub> (primary amine with one fewer carbon, via isocyanate intermediate).

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.

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