Empirical formula and oxidation product identification Steps:

• From percentages in 100 g Y: C = 54.54 g (4.545 mol), H = 9.10 g (9.10 mol), O = 36.36 g (2.273 mol).
• Ratio by dividing by smallest (2.273): C:H:O = 2:4:1, so empirical formula C₂H₄O (Mᵣ 44).
• Molecular formula is 2 × empirical (88/44 = 2), giving C₄H₈O₂.
• C₄H₈O₂ matches butanoic acid (CH₃CH₂CH₂COOH), oxidation product of primary alcohol butan-1-ol.

Why D is correct:

• Butan-1-ol (CH₃CH₂CH₂CH₂OH) oxidizes to butanoic acid (C₄H₈O₂, Mᵣ 88) via loss of 2 H and gain of O, matching exact composition.

Why the others are wrong:

• A: Propan-2-ol oxidizes to acetone (C₃H₆O, Mᵣ 58, 62.1% C), wrong formula and percentages.
• B: Propan-1-ol oxidizes to propanoic acid (C₃H₆O₂, Mᵣ 74, 48.6% C), wrong carbon count and percentages.
• C: Butan-2-ol oxidizes to butan-2-one (C₄H₈O, Mᵣ 72, 66.7% C), lacks second oxygen atom.

Final answer: D