What is Z in the following reaction sequence | vedclass

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(A) Step $1$: $C_3H_6$ (Propene) reacts with $Br_2$ in $CCl_4$ to form $X$,which is $1,2$-dibromopropane $(CH_3-CHBr-CH_2Br)$.Step $2$: $1,2$-dibromop

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Acetone

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(A) Step $1$: $C_3H_6$ (Propene) reacts with $Br_2$ in $CCl_4$ to form $X$,which is $1,2$-dibromopropane $(CH_3-CHBr-CH_2Br)$.Step $2$: $1,2$-dibromopropane reacts with alcoholic $KOH$ followed by $NaNH_2$ and heat (dehydrohalogenation) to form $Y$,which is $CH_3-C \equiv CH$ (Propyne).Step $3$: Propyne undergoes hydration in the presence of $Hg^{2+}$ and $H^+$ at $333 \ K$ (Kucherov reaction) to form an enol intermediate,which tautomerizes to form $Z$,which is $CH_3-CO-CH_3$ (Acetone).

List-$I$ (Reaction)	List-$II$ (Major product)
$A. CH_3-C \equiv CH \xrightarrow{H_2O, Hg^{2+}/H^+, 333K}$	$I. CH_3-CH(OH)-CH_3$
$B. CH_3COONa \xrightarrow{H_2O, Pt, \text{electrolysis}}$	$II. CH_3-CO-CH_3$
$C. CH_3-CH=CH_2 \xrightarrow{H_2O/H^+}$	$III. CH_3-CH(OH)-CH_2OH$
$D. CH_3-CH=CH_2 \xrightarrow{H_2O, \text{dil. } KMnO_4, 273K}$	$IV. CH_3-CH_3$
	$V. CH_4$

What is $Z$ in the following reaction sequence ?
$C_3H_6$ $\xrightarrow[CCl_4]{Br_2} \underset{\text{(colourless)}}{X}$ $\xrightarrow[\text{(ii) } NaNH_2, \Delta]{\text{(i) } KOH / \text{alcohol}} Y$ $\xrightarrow[\substack{Hg^{2+}, H^{+} \\ 333 \ K}]{H_2O} Z$

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What is $Z$ in the following reaction sequence ?$C_3H_6$ $\xrightarrow[CCl_4]{Br_2} \underset{\text{(colourless)}}{X}$ $\xrightarrow[\text{(ii) } NaNH_2, \Delta]{\text{(i) } KOH / \text{alcohol}} Y$ $\xrightarrow[\substack{Hg^{2+}, H^{+} \\ 333 \ K}]{H_2O} Z$