Oxidative cleavage of terminal alkene by hot acidified KMnO4 Steps:

• Identify functional groups: secondary alcohol at C2 and terminal alkene at C5-C6 in CH3-CH(OH)-CH2-CH2-CH=CH2.
• Oxidize secondary alcohol to ketone: CH3-C(O)-CH2-CH2-CH=CH2.
• Cleave terminal alkene: -CH=CH2 becomes -COOH (from C5) and CO2 (from C6), yielding CH3-C(O)-CH2-CH2-COOH.
• Excess oxidant degrades methyl ketone, shortening chain to CH3-CH2-CH2-COOH.

Why C is correct:

• Hot acidified KMnO4 fully oxidizes methyl ketones R-C(O)-CH3 to R-CH2-COOH by cleaving and reforming the chain (standard aliphatic ketone degradation rule).

Why the others are wrong:

• A: Ignores alcohol oxidation to ketone.
• B: Matches a structure with one fewer CH2 group.
• D: Assumes ketone cleavage to separate acids without chain adjustment.

Final answer: C