Oxidative cleavage of alkenes with hot KMnO4 yields carboxylic acids from =CH- carbons in symmetric alkenes.

Steps:

• Hot concentrated acidified KMnO4 cleaves the C=C bond, converting each =CH-R to R-COOH.
• 2-Methylpropanoic acid, (CH3)2CHCOOH, as the sole product indicates a symmetric alkene where both fragments are identical.
• The alkene must be (CH3)2CH-CH=CH-CH(CH3)2 to yield two molecules of (CH3)2CHCOOH.
• Name this structure: the longest chain is six carbons with methyl branches at positions 2 and 5, double bond at 3, so 2,5-dimethylhex-3-ene.

Why A is correct:

• 2,5-Dimethylhex-3-ene is (CH3)2CHCH=CHCH(CH3)2; cleavage gives two (CH3)2CHCOOH per the oxidative rule for =CH- groups.

Why the others are wrong:

• B: 2-Methylbut-2-ene yields acetone and acetic acid, not 2-methylpropanoic acid.
• C: 2-Methylpropene yields acetone and CO2, not 2-methylpropanoic acid.
• D: Oct-4-ene yields butanoic acid, not 2-methylpropanoic acid.

Final answer: A