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COMEDK202510 May 2025Evening ShiftChemistryHydrocarbonsActual

An Alkene " X " on reaction with hot acidified KMnO ₄ gave a mixture of Ethanoic acid and Propanone. Identify " X ".

Options

  1. ABut-2-ene
  2. BPent-2-ene
  3. C2- Methylbut-2-ene.
  4. D2,3 -Dimethylbut-2-ene

Correct answer

C. 2- Methylbut-2-ene.

Step-by-step solution

The reaction of an alkene with hot acidified KMnO ₄ results in oxidative cleavage of the carbon-carbon double bond. The products formed depend on the substitution pattern at the double bond carbons. Ethanoic acid is CH ₃ COOH , which corresponds to a terminal CH ₃ CH = group after oxidation. Propanone is CH ₃ COCH ₃ , which corresponds to a ( CH ₃)₂ C = group after oxidation. Combining these two fragments, the alkene must be CH ₃ CH = C ( CH ₃)₂ . The structure CH ₃ CH = C ( CH ₃)₂ is 2-methylbut-2-ene. Answer: 2-

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