Secondary alcohol oxidizes to ketone under given conditions

Steps:

• X (C4H10O) is oxidized to Y (C4H8O) with excess acidified K2Cr2O7 under reflux, indicating loss of 2H without forming carboxylic acid (which would be C4H8O2).
• Y forms orange crystals with 2,4-DNPH, confirming Y is an aldehyde or ketone (carbonyl group).
• Y produces H2 with Na metal, but given formula and conditions, this likely distinguishes from acid (standard test: ketones do not, acids do); thus Y is ketone.
• Only secondary alcohols yield ketones (C4H8O) under these conditions; primary yield acids, tertiary do not react.

Why A is correct:

• A is butan-2-ol (secondary alcohol), oxidizes to butan-2-one (C4H8O ketone), matching Y's formula and tests.

Why the others are wrong:

• B (butan-1-ol, primary) oxidizes to butanoic acid (C4H8O2), not C4H8O.
• C (2-methylpropan-2-ol, tertiary) does not oxidize, remains C4H10O.
• D (2-methylpropan-1-ol, primary) oxidizes to 2-methylpropanoic acid (C4H8O2), not C4H8O.

Final answer: A