When acetylene is passed through $\text{ dil. }{\mathrm{H}}_2{\mathrm{SO}}_4$ in the presence of the ${\mathrm{HgSO}}_4$, the compound formed is

An alkene is treated with mercury(II) acetate, $\mathrm{Hg}(\mathrm{OAc}{)}_2$, in an oxymercuration-demercuration reaction, and the product is treated with sodium borohydride. The end result is a Markovnikov addition product in which the OH group bonds to the alkene's more substituted carbon atom.

The general reaction here is "Oxymercuration-Demercuration." Acetaldehyde is the final product.

$\mathrm{HC}\equiv \mathrm{CH}\underset{{\mathrm{HgSO}}_4}{\stackrel{\mathrm{dil}.{\mathrm{H}}_2{\mathrm{SO}}_4}{\to }}\underset{\text{Vinyl alcohol}}{{\mathrm{H}}_2\mathrm{C}=\underset{\mathrm{OH}}{\underset{\mid }{\mathrm{CH}}}}\stackrel{\text{Rearrangement}}{\leftrightarrow }\underset{\text{Acetaldehyde}}{{\mathrm{CH}}_3\mathrm{CHO}}$

Therefore, the correct option is (D) Acetaldehyde.