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JEE Main20266 April 2026Evening ShiftChemistryHydrocarbonsActual

An optically active alkyl bromide C₄ H₉ Br, reacts with ethanolic KOH to form major compound [A] which reacts with bromine to give compound [B]. Compound [B] reacts with ethanolic KOH and sodamide to give compound [C]. One molecule of water adds to compound [C] on warming with mercuric sulphate and dilute sulphuric acid at 333 K to form compound [D]. The functional group in compound D will be confirmed by :

Options

  1. AHaloform test
  2. BLucas test
  3. CSilver mirror test
  4. DBenedict test

Correct answer

A. Haloform test

Step-by-step solution

The optically active alkyl bromide C ₄ H ₉ Br is 2 -bromobutane, CH ₃ CH ( Br ) CH ₂ CH ₃ . Reaction with ethanolic KOH undergoes dehydrohalogenation to form but- 2 -ene as the major product [A] (Zaitsev's rule). But- 2 -ene [A] reacts with Br ₂ to form 2,3 -dibromobutane [B], CH ₃ CH ( Br ) CH ( Br ) CH ₃ . Compound [B] undergoes double dehydrohalogenation with ethanolic KOH and NaNH ₂ to give but- 2 -yne [C], CH ₃ C CCH ₃ . Hydration of but- 2 -yne [C] with HgSO ₄ and dilute H ₂ SO ₄ yields butan- 2 -one [D], CH

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