The hydration of propyne in the presence of ${\mathrm{HgSO}}_4/{\mathrm{H}}_2{\mathrm{SO}}_4$ produces

The reaction is,

${\mathrm{CH}}_3\mathrm{C}=\mathrm{CH}\frac{{\mathrm{HgSO}}_4/{\mathrm{H}}_2{\mathrm{SO}}_4}{+{\mathrm{H}}_2\mathrm{O}}$$$\begin{gathered} {\mathrm{CH}}_3\mathrm{C}=\mathrm{CHc} \\ \overline{) } \\ \mathrm{OH} \end{gathered}$$$\frac{{\mathrm{HgSO}}_4/{\mathrm{H}}_2{\mathrm{SO}}_4}{{\mathrm{H}}_2\mathrm{O}}{\mathrm{CH}}_3{\mathrm{COCH}}_3$$$\begin{gathered} OH \\ \stackrel{-{\mathrm{H}}_2\mathrm{O}}{\leftarrow }{\mathrm{CH}}_3\underset{\overline{) }}{\overset{\overline{) }}{\mathrm{C}}}-{\mathrm{CH}}_3 \\ \mathrm{OH} \end{gathered}$$

Thus, The hydration of propyne in the presence of ${\mathrm{HgSO}}_4/{\mathrm{H}}_2{\mathrm{SO}}_4$ produces ${\mathrm{CH}}_3{\mathrm{COCH}}_3$.