JEE Main 2025 Chemistry 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 :
Answer: (3)
Solution
A: 2-bromobutane → B: but-2-ene (Zaitsev) → C: 2,3-dibromobutane → D: but-2-yne → E: butan-2-one (via Markovnikov hydration).