0.40 g of an organic compound (A), (M.F.- C₅H₈O) reacts with x mole of CH₃MgBr to liberate 224 mL of a gas at STP. With excess of H₂, (A) gives pentan-1-ol. The correct structure of (A) is :

mol. mass of A = 12x5+ 8 + 16 = 84g
moles of $\mathrm{A}=\frac{0.40\mathrm{g}}{84\mathrm{g}}$ = 0.005moles
moles of gas released at STP = $\frac{224 \mathrm{mL}}{22.4 \mathrm{L}} = 0.01 \mathrm{moles}$
$\therefore 0.005\text{ moles of }\mathrm{A}\text{ release }0.01\text{ moles of gas. }$
i. e., ratio of 1 : 2.
Hence there must be two acidic H in the compound A. One acidic H is from alcohol the other must be terminal alkyne as only the H attached to terminal alkynes are acidic.

Hence option (d) is incorrect.