Ethyne is treated with $\mathrm{NaNH}{}_2$ and then with $\mathrm{CH}{}_3\mathrm{I}$ to give A. A is treated with dilute $\mathrm{H}{}_2\mathrm{SO}{}_4$ containing $\mathrm{HgSO}{}_4$ to give B. B is

$\mathrm{HC}\equiv \mathrm{CNa}$ is methylated to propyne, $\mathrm{CH}{}_3\mathrm{C}\equiv \mathrm{CH}$ (A). Hydration of a terminal alkyne follows Markovnikov's rule to give an enol that tautomerises to the methyl ketone, $\mathrm{CH}{}_3\mathrm{COCH}{}_3$ (B). Only ethyne would give an aldehyde.