Oxidative cleavage of alkenes by KMnO4 Steps:

• Hot concentrated acidified KMnO4 cleaves the C=C bond in alkenes, converting =CH-R to R-COOH, =CR1R2 to R1R2C=O, and =CH2 to CO2.
• Analyze each option's alkene structure to determine cleavage products.
• Symmetric alkenes yield identical carboxylic acids; unsymmetric internal alkenes yield different ones.
• Select the option producing two distinct R-COOH groups.

Why B is correct:

• B is an unsymmetric internal alkene (e.g., pent-2-ene), cleaving to CH3COOH and CH3CH2COOH per the rule that each =CH-R carbon forms a unique carboxylic acid.

Why the others are wrong:

• A: Symmetric alkene (e.g., but-2-ene) yields two identical CH3COOH molecules.
• C: Terminal alkene (e.g., propene) yields CO2 and one carboxylic acid, not two different ones.
• D: Disubstituted alkene (e.g., 2-methylpropene) yields a ketone and CO2, not two carboxylic acids.

Final answer: B