Consider the reaction, $\mathrm{X}\underset{{\mathrm{HgSO}}_4,{60}^{\mathrm{o}}\mathrm{C}}{\overset{\mathrm{dil}.{\mathrm{H}}_2{\mathrm{SO}}_4}{\to }}\mathrm{Ketone}.$ The compound ‘X’ in this reaction can never be

Acidic hydration of alkyne involves addition of water by Markonikoff's rule. The enol formed tautomerises to a stable ketone.

$\because$Ethyne contains only two carbon atoms, the -OH group will be attached to terminal carbon atom.

the enol formed tautomerises to an aldehyde.

$$\begin{gathered} \mathrm{CH}\equiv \mathrm{CH}\stackrel{\mathrm{dil}.{\mathrm{H}}_2{\mathrm{SO}}_4}{\to }\left[\underset{\stackrel{|}{\mathrm{H}}}{\mathrm{C}}\mathrm{H}=\underset{\stackrel{|}{\mathrm{OH}}}{\mathrm{C}}\mathrm{H}\right]\underset{\mathrm{tautomerism}}{\overset{\mathrm{Keto}-\mathrm{enol}}{\rightleftharpoons }}{\mathrm{CH}}_3-\mathrm{CH}=\mathrm{O} \\ \left(2\right) . . . {\mathrm{CH}}_3-\mathrm{CO}-{\mathrm{CH}}_2-{\mathrm{CH}}_3 \mathrm{is} \mathrm{formed} \\ \left(3\right) . . . {\mathrm{CH}}_3-\mathrm{CO}-{\mathrm{CH}}_3 \mathrm{is} \mathrm{formed} \\ \left(4\right) . . . {\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{CO}-{\mathrm{CH}}_3 \mathrm{is} \mathrm{formed} \end{gathered}$$