See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Step 1: Determine moles of compound (A). Molecular formula C5H8O gives MW = 5(12) + 8(1) + 16 = 84 g/mol. Moles of A = 0.40 / 84 = 1/210 mol. Step 2: Determine moles of gas liberated with CH3MgBr. At STP, 224 mL = 0.224 L. Moles of gas = 0.224 / 22.4 = 0.01 mol. Step 3: Find x (moles of CH3MgBr reacting per mole of A). x = moles of gas / moles of A = 0.01 / (1/210) = 0.01 × 210 = 2.1 ≈ wait, let me recalculate. Moles of A = 0.40/84 = 4.76 × 10^-3 mol. Moles of gas = 0.224/22.4 = 0.01 mol. x = 0.01 / (0.40/84) = 0.01 × 84/0.40 = 0.84/0.40 = 2.1. Hmm, this suggests x ≈ 2, meaning 2 equivalents of CH3MgBr react per molecule. This means (A) has 2 acidic hydrogens (active H atoms) that react with Grignard reagent to release CH4 gas. Step 4: Identify active hydrogens in the options. - Option (a): CH3-C≡C-CH2-CH2-OH — internal alkyne (no terminal ≡C-H) + one OH → only 1 active H. - Option (b): CH3-CH2-C≡C-CH2-OH — internal alkyne (no terminal ≡C-H) + one OH → only 1 active H. - Option (c): H-C≡C-CH2-CH2-CH2-OH — terminal alkyne (1 acidic ≡C-H) + one OH → 2 active H atoms. Option (c) has both a terminal alkyne C-H and a hydroxyl OH, giving 2 acidic protons, consistent with x = 2. Step 5: Verify hydrogenation gives pentan-1-ol. Option (c): H-C≡C-CH2-CH2-CH2-OH + 2H2 (excess) → CH3-CH2-CH2-CH2-CH2-OH = pentan-1-ol. ✓ Step 6: Why other options fail. - Option (a) and (b) each have only 1 active hydrogen (only the OH group), so they would give x = 1, not 2. - Option (a) hydrogenation: CH3-C≡C-CH2-CH2-OH + 2H2 → CH3-CH2-CH2-CH2-CH2-OH = pentan-1-ol (this does work for hydrogenation), but fails the x = 2 criterion. - Option (b) hydrogenation: CH3-CH2-C≡C-CH2-OH + 2H2 → CH3-CH2-CH2-CH2-CH2-OH = pentan-1-ol (also works), but also fails x = 2 criterion. - Only option (c) satisfies both conditions: x = 2 (terminal alkyne H + OH) and gives pentan-1-ol on hydrogenation. Therefore, the correct answer is C.