Primary alcohol oxidation to carboxylic acid with molar mass comparison

Steps:

• Carboxylic acid P forms via oxidation, which requires a primary alcohol starting material, as secondary alcohols yield ketones.
• X is labeled as an alcohol, so it must be primary to produce P.
• Y and Z are alternative precursors (e.g., aldehyde or alkene); compare molar masses (Mr) from diagram.
• Greatest ΔMr occurs for Z to P, as Z has the lowest Mr, leading to largest increase upon forming P.

Why B is correct:

• Primary alcohols oxidize to carboxylic acids (RCH₂OH → RCOOH, ΔMr = +14 via O gain, 2H loss), enabling X → P; Z → P shows max ΔMr per diagram.

Why the others are wrong:

• A: ΔMr greatest for Y to P, but diagram shows Z to P larger.
• C: X secondary cannot oxidize to carboxylic acid, violating formation rule.
• D: Combines secondary X error with wrong ΔMr path (Y instead of Z).

Final answer: B