Stoichiometry of H₂ evolution distinguishes ethers from alcohols

Steps:

• Sodium reacts with acidic hydrogens in O-H bonds: 2ROH + 2Na → 2RONa + H₂, so 1 mol alcohol yields 0.5 mol H₂.
• The query requires 1 mol Y to yield 1 mol H₂, which mismatches monoalcohol stoichiometry (needs 2 mol alcohol for 1 mol H₂).
• Examine structures: A, B, C have one O-H each (monoalcohols); D has C-O-C (ether, no O-H).
• Only ether lacks reactive sites for H₂ production, but query implies identifying non-reactor under given conditions.

Why D is correct:

• CH₃OCH₂CH₃ is an ether without acidic O-H bond; by definition, only compounds with active hydrogen (pKa ~15-18 for alcohols) evolve H₂ with Na, excluding ethers.

Why the others are wrong:

• A: Primary alcohol with O-H; yields 0.5 mol H₂ per mol, not 1 mol.
• B: Secondary alcohol with O-H; yields 0.5 mol H₂ per mol, not 1 mol.
• C: Secondary alcohol with O-H; yields 0.5 mol H₂ per mol, not 1 mol.

Final answer: D