Determine molecular formula of Y and match to oxidation product of primary alcohol X

Steps:

• From percentages in 100 g Y: C = 54.54/12 = 4.545 mol, H = 9/1 = 9 mol, O = 36/16 = 2.25 mol.
• Divide by smallest (2.25): C ≈ 2, H = 4, O = 1, so empirical formula C₂H₄O (mass 44).
• Molecular mass 88 ÷ 44 = 2, so molecular formula C₄H₈O₂ (butanoic acid).
• Primary alcohols oxidize to carboxylic acids (CₙH₂ₙ₊₂O → CₙH₂ₙO₂); C₄H₈O₂ from C₄H₉OH (butan-1-ol, option D).

Why D is correct:

• Butan-1-ol (primary) oxidizes to butanoic acid (C₄H₈O₂, M=88, matching composition via formula C₄H₈O₂ %C=54.55).

Why the others are wrong:

• A: Propan-1-ol oxidizes to propanoic acid (C₃H₆O₂, M=74, %C=40%).
• B: Propan-2-ol (secondary) oxidizes to propanone (C₃H₆O, M=58, %C=61.9%).
• C: Butan-2-ol (secondary) oxidizes to butan-2-one (C₄H₈O, M=72, %C=66.7%).

Final answer: D