Secondary alcohols oxidize to ketones retaining the carbon chain length Steps:

• Count carbons: A has 5 (2-methylbutane chain), B/C/D have 6 (pentane + methyl).
• Classify alcohols: Primary if OH on CH2OH, secondary if on CHOH, tertiary if on COH with three alkyl groups.
• Recall oxidation: Acidified K2Cr2O7 oxidizes secondary alcohols to ketones, primaries to carboxylic acids, tertiaries not at all.
• Identify the secondary alcohol with 6 carbons to yield a 6-carbon ketone.

Why C is correct:

• 3-Methylpentan-2-ol is secondary (OH on C2 attached to one H and two carbons); oxidation removes H from C2 and OH to form C=O, yielding 3-methylpentan-2-one (6 carbons), per standard alcohol oxidation rules.

Why the others are wrong:

• A: Primary (5 carbons); oxidizes to 2-methylbutanoic acid (no ketone).
• B: Tertiary (6 carbons); resists oxidation, no ketone formed.
• D: Primary (6 carbons); oxidizes to 3-methylpentanoic acid (no ketone).

Final answer: C