Q32 / 180 · COMEDK 2025 (10 May Shift 2) · Chemistry
An Alkene " X " on reaction with hot acidified $\mathrm{KMnO}_4$ gave a mixture of Ethanoic acid and Propanone. Identify " X ".
- ABut-2-ene
- BPent-2-ene
- Correct answerC2- Methylbut-2-ene.
- D2,3 -Dimethylbut-2-ene
Solution
The reaction of an alkene with hot acidified $\mathrm{KMnO}_4$ 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 $\mathrm{CH}_3\mathrm{COOH}$, which corresponds to a terminal $\mathrm{CH}_3\mathrm{CH}=$ group after oxidation. Propanone is $\mathrm{CH}_3\mathrm{COCH}_3$, which corresponds to a $(\mathrm{CH}_3)_2\mathrm{C}=$ group after oxidation.
Combining these two fragments, the alkene must be $\mathrm{CH}_3\mathrm{CH}=\mathrm{C}(\mathrm{CH}_3)_2$.
The structure $\mathrm{CH}_3\mathrm{CH}=\mathrm{C}(\mathrm{CH}_3)_2$ is 2-methylbut-2-ene.
Answer: 2- Methylbut-2-ene.
Ethanoic acid is $\mathrm{CH}_3\mathrm{COOH}$, which corresponds to a terminal $\mathrm{CH}_3\mathrm{CH}=$ group after oxidation. Propanone is $\mathrm{CH}_3\mathrm{COCH}_3$, which corresponds to a $(\mathrm{CH}_3)_2\mathrm{C}=$ group after oxidation.
Combining these two fragments, the alkene must be $\mathrm{CH}_3\mathrm{CH}=\mathrm{C}(\mathrm{CH}_3)_2$.
The structure $\mathrm{CH}_3\mathrm{CH}=\mathrm{C}(\mathrm{CH}_3)_2$ is 2-methylbut-2-ene.
Answer: 2- Methylbut-2-ene.
Hence, the correct answer is option: 'C'.