An optically active alkyl halide C4H9Br [A] reacts with hot KOH dissolved in ethanol and forms alkene [B] as major product which reacts with bromine to give dibromide [C]. The compound [C] is converte

Question

An optically active alkyl halide C4H9Br [A] reacts with hot KOH dissolved in ethanol and forms alkene [B] as major product which reacts with bromine to give dibromide [C]. The compound [C] is converted into a gas [D] upon reacting with alcoholic NaNH2. During hydration 18 gram of water is added to 1 mole of gas [D] on warming with mercuric sulphate and dilute acid at 333 K to form compound [E]. The IUPAC name of compound [E] is :

(A) Butan-2-one
(B) But-2-yne
(C) Butane-1-al
(D) Butan-2-ol

Prashant Kotian · Accepted Answer

Correct answer: A. O Active Br A alc. KOH B Br2 Br Br C NaNH2 CH3–CC–CH3 D HgSO4 H2SO4 CH3–C–CH2–CH3 O E | JEE(Main) 2025 | DATE : 02-04-2025 (SHIFT-1) | PAPER-1 | CHEMISTRY PAGE # 2

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