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JEE Main 2025 Apr 2 Shift 1, Chemistry Q2: Preparation of Alkynes

JEE Main2025Apr 2, Shift 1Chemistry
Q.

An optically active alkyl halide [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 . 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 :

  1. A

    But-2-yne

  2. B

    Butan-2-ol

  3. C

    Butan-2-one

  4. D

    Butan-1-al

Solution

The optically active is sec-butyl bromide (2-bromobutane), the only chiral isomer.

Step 1 (A B): Alcoholic KOH induces E2 elimination. By Saytzeff's rule the major alkene is but-2-ene.

Step 2 (B C): Addition of across the double bond gives 2,3-dibromobutane.

Step 3 (C D): Alcoholic carries out a double dehydrohalogenation, producing the alkyne. Since only 1 mole of water (18 g) adds to 1 mole of D, the alkyne is terminal: but-1-yne, .

Step 4 (D E): Hydration of a terminal alkyne with at 333 K follows Markovnikov addition and gives a methyl ketone via the enol intermediate.

Compound (E) is butan-2-one. Option (3) is correct.

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