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
- AHaloform test
- BLucas test
- CSilver mirror test
- 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