Acid-catalysed hydration of alkynes with $H_{2}S{O}_4$ and $HgS{O}_4$ adds water across the triple bond following Markovnikov's rule. The hydroxyl group attaches to the more substituted carbon, so with propyne, $C{H}_3-C\equiv CH$, water adds to give an enol in which OH sits on the central carbon, namely prop-1-en-2-ol, $C{H}_3-C(OH)=C{H}_2$. Enols are generally unstable and rapidly tautomerize to the more stable keto form by migration of a hydrogen and shift of the double bond, producing propan-2-one (acetone). Hence acetone is the isolable product. Propanal would arise only from anti-Markovnikov addition, which does not occur under these conditions, so it is wrong. Prop-1-en-2-ol is the genuine intermediate enol but it is not the final stable product because keto-enol tautomerism converts it, so it is incorrect as the answer. Propan-1-ol corresponds to simple alcohol reduction, not hydration of an alkyne, so it does not apply. The mercuric ion acts as a catalyst by coordinating to the pi system and making the triple bond susceptible to nucleophilic attack by water, after which loss of the metal regenerates the catalyst. The driving force for the final step is the much greater thermodynamic stability of the carbonyl group compared with the enol, which is why the equilibrium lies overwhelmingly toward acetone. This follows the NCERT description of alkyne hydration, where only ethyne gives an aldehyde. A consistency check: every terminal alkyne except ethyne yields a ketone, consistent with acetone here.