JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Hydrocarbons

Hydration of an alkyne with $\mathrm{HgS}{\mathrm{O}}_4$ and ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ adds water across the triple bond by Markovnikov's rule, so the hydroxyl group lands on the more substituted carbon and the enol first formed tautomerises to a carbonyl compound. Acetylene itself gives acetaldehyde, a terminal alkyne gives a methyl ketone, and an internal alkyne gives a dialkyl ketone, so butan-2-one, cyclohexyl methyl ketone and acetaldehyde are all available by this route. Propanal would need the hydroxyl group to arrive at the terminal carbon of propyne, which is anti-Markovnikov addition and does not occur under these conditions. The carbonyl compound that cannot be prepared this way is propanal.