An optically active alkyl halide C 4 H 9 B r [A] reacts with hot KOH dissolved in ethanol and forms alkene [B] as major product which reacts with bromine to give dibromide

Step 1: Identify the optically active alkyl bromide (A)

Given molecular formula = C₄H₉Br and it is optically active.

Possible isomers:

  • CH₃CH₂CH₂CH₂Br (1-bromobutane) ❌
  • CH₃CHBrCH₂CH₃ (2-bromobutane) ✅ (chiral)
  • (CH₃)₂CHCH₂Br (isobutyl bromide) ❌
  • (CH₃)₃CBr (tert-butyl bromide) ❌

Hence,A=2-bromobutane​

Step 2: Alcoholic KOH (Hot)

Step 3: Addition of Br₂

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Step 4: Reaction with alcoholic NaNH₂

Vicinal dibromide + excess NaNH₂ undergoes double dehydrohalogenation.

Step 5: Hydration of alkyne

Conditions:

  • HgSO₄
  • Dil. H₂SO₄
  • 333 K

Hydration of an internal alkyne gives a ketone after keto-enol tautomerism.

Final Answer

(3) Butan-2-one​

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