Oxidation of hydrocarbons to carboxylic acids

Steps:

• Pentanoic acid is CH3(CH2)3COOH, a 5-carbon carboxylic acid formed by oxidation cleaving C-C bonds.
• Evaluate A: Propan-1-ol (3 carbons) oxidizes to propanoic acid (3 carbons).
• Evaluate B: Butan-2-one (4 carbons) with H2SO4 undergoes acid-catalyzed reactions but not full oxidation to pentanoic acid.
• Evaluate C: Pentane (5 carbons) with hot concentrated H2SO4/KMnO4 oxidizes to pentanoic acid via cleavage.

Why C is correct:

• Oxidative cleavage of pentane (C5H12) with H2SO4(aq) (implying KMnO4 conditions) breaks the chain to yield CH3(CH2)3COOH.

Why the others are wrong:

• A: Yields propanoic acid (CH3CH2COOH), a 3-carbon acid, due to primary alcohol oxidation stopping at carboxylic acid.
• B: Ketone hydration or no reaction; does not produce a 5-carbon carboxylic acid.
• D: No reaction or information provided.

Final answer: C