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An optically active alkyl halide C4H9Br [A] reacts with hot KOH dissolved in ethanol and forms product which reacts…

Haloalkanes & Haloarenes · Class 12 · JEE Main Previous Year Question

Question

An optically active alkyl halide \ceC4H9Br\ce{C4H9Br} [A] reacts with hot KOH dissolved in ethanol and forms product which reacts with bromine to give dibromide [C]. The compound [C] is converted into a gas [D] upon treatment with alcoholic KOH. 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:

Options
  1. a

    But-2-yne

  2. b

    Butan-2-ol

  3. c

    Butan-2-one

    ✓
  4. d

    Butan-1-al

Correct Answerc

Butan-2-one

Detailed Solution

Step 1: Identify compound A

Optically active \ceC4H9Br\ce{C4H9Br} means it has a chiral center.

Possible structures:

  • 2-Bromobutane: \ceCH3−CHBr−CH2−CH3\ce{CH3-CHBr-CH2-CH3} ✓ (chiral at C-2)
  • 1-Bromobutane: \ceCH3−CH2−CH2−CH2Br\ce{CH3-CH2-CH2-CH2Br} (not chiral)

Compound A = 2-bromobutane

Step 2: Reaction with alcoholic KOH

2-Bromobutane + alcoholic KOH → E2 elimination → alkene

\ceCH3−CHBr−CH2−CH3+KOH(alc)−>CH3−CH=CH−CH3\ce{CH3-CHBr-CH2-CH3 + KOH(alc) -> CH3-CH=CH-CH3}

Product B = but-2-ene

Step 3: Bromination of but-2-ene

But-2-ene + \ceBr2\ce{Br2} → addition → dibromide

\ceCH3−CH=CH−CH3+Br2−>CH3−CHBr−CHBr−CH3\ce{CH3-CH=CH-CH3 + Br2 -> CH3-CHBr-CHBr-CH3}

Compound C = 2,3-dibromobutane

Step 4: Dehydrohalogenation with alcoholic KOH

2,3-Dibromobutane + alcoholic KOH (excess) → double elimination → alkyne

\ceCH3−CHBr−CHBr−CH3+2KOH(alc)−>CH3−C≡C−CH3\ce{CH3-CHBr-CHBr-CH3 + 2 KOH(alc) -> CH3-C≡C-CH3}

Gas D = but-2-yne (2-butyne)

Step 5: Hydration of but-2-yne

But-2-yne + \ceH2O\ce{H2O} with \ceHg2+/H2SO4\ce{Hg^{2+}/H2SO4} at 333 K → hydration → ketone

For internal alkynes: \ceR−C≡C−R+H2O−>[Hg2+/H2SO4]R−CO−CH2−R\ce{R-C≡C-R + H2O ->[Hg^{2+}/H2SO4] R-CO-CH2-R}

For but-2-yne: \ceCH3−C≡C−CH3+H2O−>[Hg2+/H2SO4]CH3−CO−CH2−CH3\ce{CH3-C≡C-CH3 + H2O ->[Hg^{2+}/H2SO4] CH3-CO-CH2-CH3}

Wait, this would give butan-2-one: \ceCH3−C≡C−CH3−>CH3−C(=O)−CH3\ce{CH3-C≡C-CH3 -> CH3-C(=O)-CH3}

Actually, for symmetric alkyne \ceCH3−C≡C−CH3\ce{CH3-C≡C-CH3}, hydration gives: \ceCH3−C≡C−CH3+H2O−>CH3−CO−CH3\ce{CH3-C≡C-CH3 + H2O -> CH3-CO-CH3}

But this is propanone (3 carbons), not butanone.

Let me reconsider. If the alkyne is but-2-yne (\ceCH3−C≡C−CH3\ce{CH3-C≡C-CH3}), hydration gives:

Actually, I made an error. But-2-yne is \ceCH3−C≡C−CH3\ce{CH3-C≡C-CH3} which has 4 carbons total.

Hydration: \ceCH3−C≡C−CH3+H2O−>CH3−CO−CH2−CH3\ce{CH3-C≡C-CH3 + H2O -> CH3-CO-CH2-CH3}

No wait, for a symmetric internal alkyne, both sides are equivalent, so we get: \ceCH3−C≡C−CH3+H2O−>CH3−C(OH)=CH−CH3−>CH3−CO−CH2−CH3\ce{CH3-C≡C-CH3 + H2O -> CH3-C(OH)=CH-CH3 -> CH3-CO-CH2-CH3}

This is butan-2-one (methyl ethyl ketone).

Compound E = Butan-2-one

Answer: (c)

Key Points:

  • Optically active alkyl halide must have chiral center
  • Alcoholic KOH → elimination → alkene
  • Double elimination of vicinal dibromide → alkyne
  • Alkyne hydration with Hg²⁺/H₂SO₄ → ketone (for internal alkynes)

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