Which one of the following is the product $(Z)$ formed at the end of the given reaction: $\begin{aligned} & \text { Propan-1-ol }+\mathrm{SOCl}_2 \rightarrow[\mathrm{X}] \\ & {[\mathrm{X}]+\mathrm{HC} \equiv \mathrm{C}-\mathrm{Na} \rightarrow[\mathrm{Y}]} \\ & {[\mathrm{Y}]+\mathrm{H}_2 \mathrm{SO}_4 / \mathrm{HgSO}_4, 330 \mathrm{~K} \rightarrow[\mathrm{Z}]} \end{aligned}$

The reaction of propan-1-ol with $\text{SOCl}_2$ follows the Darzens halogenation mechanism to produce 1-chloropropane as product [X].

$\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + \text{SOCl}_2 \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + \text{SO}_2 + \text{HCl}$

The reaction of 1-chloropropane [X] with sodium acetylide ($\text{HC} \equiv \text{C}-\text{Na}$) is a nucleophilic substitution ($\text{S}_N2$) reaction, yielding pent-1-yne as product [Y].

$\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + \text{HC} \equiv \text{C}-\text{Na} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{C} \equiv \text{CH} + \text{NaCl}$

The hydration of pent-1-yne [Y] in the presence of $\text{H}_2\text{SO}_4$ and $\text{HgSO}_4$ (Kucherov reaction) follows Markovnikov addition of water across the triple bond to form an enol intermediate, which tautomerizes to pentan-2-one as product [Z].

$\text{CH}_3\text{CH}_2\text{CH}_2\text{C} \equiv \text{CH} + \text{H}_2\text{O} \xrightarrow{\text{H}_2\text{SO}_4, \text{HgSO}_4} \text{CH}_3\text{CH}_2\text{CH}_2\text{C}(\text{OH})=\text{CH}_2 \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{COCH}_3$

Answer: Pentan-2-one.