Secondary alcohol oxidation yields ketone with lower boiling point

Steps:

• Classify each alcohol: A and C are primary; B is tertiary; D is secondary.
• Primary alcohols oxidize to carboxylic acids under reflux with excess acidified K₂Cr₂O₇.
• Secondary alcohols oxidize to ketones; tertiary alcohols do not react.
• Ketones lack hydrogen bonding, so they have lower boiling points than the parent alcohols; carboxylic acids have higher boiling points due to dimerization.

Why D is correct:

• Pentan-2-ol (secondary) oxidizes to pentan-2-one, a ketone without O-H hydrogen bonding, resulting in lower boiling point (119°C to 102°C).

Why the others are wrong:

• A: 2-Methylbutan-1-ol (primary) oxidizes to 2-methylbutanoic acid, which has higher boiling point due to strong hydrogen bonding.
• B: 2-Methylbutan-2-ol (tertiary) does not oxidize, so boiling point remains the same.
• C: Pentan-1-ol (primary) oxidizes to pentanoic acid, which has higher boiling point (138°C to 186°C).

Final answer: D