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
- ABut-2-ene
- BPent-2-ene
- C2- Methylbut-2-ene.
- 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-