Which one of the following carbonyl compounds cannot be prepared by addition of water on an alkyne in the presence of $$HgSO_4$$ and $$H_2SO_4$$?

When an alkyne undergoes hydration in the presence of $$HgSO_4$$ and $$H_2SO_4$$, the addition of water follows Markovnikov's rule. The $$-OH$$ group preferentially adds to the more substituted (internal) carbon of the triple bond, and the resulting enol tautomerises to a carbonyl compound.

Consider acetylene ($$HC \equiv CH$$): since it is a symmetric alkyne, the hydration gives the enol $$CH_2 = CHOH$$, which tautomerises to acetaldehyde ($$CH_3CHO$$). So option (3), acetaldehyde, can be prepared by this method.

Consider 1-butyne ($$CH_3CH_2C \equiv CH$$): by Markovnikov addition, the $$-OH$$ adds to the internal carbon ($$C_2$$ of the triple bond), giving the enol $$CH_3CH_2C(OH) = CH_2$$. This tautomerises to methyl ethyl ketone ($$CH_3COCH_2CH_3$$). So option (2) can be prepared.

Cyclohexanone can be obtained by the acid-catalysed hydration of cyclohex-1-yne. The triple bond in this cyclic alkyne undergoes Markovnikov hydration to give the corresponding enol, which tautomerises to cyclohexanone. So option (4) can be prepared.

Now consider propanal ($$CH_3CH_2CHO$$): this is an aldehyde with the carbonyl at the terminal position ($$C_1$$). The only three-carbon alkyne is propyne ($$CH_3C \equiv CH$$). By Markovnikov hydration, the $$-OH$$ adds to the internal carbon ($$C_2$$), forming the enol $$CH_3C(OH) = CH_2$$, which tautomerises to acetone ($$CH_3COCH_3$$), a ketone, not propanal. In general, Markovnikov hydration of any terminal alkyne gives a methyl ketone, and only acetylene (being symmetric) can give an aldehyde. Since there is no alkyne whose Markovnikov hydration yields propanal, it cannot be prepared by this method.

The correct answer is Option (1): $$CH_3CH_2CHO$$ (propanal).