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(B) The reaction starts with $hex-5-yn-1-ol$. Hydrogenation with $H_2$ (typically over a catalyst like $Pd/BaSO_4$ or similar) reduces the alkyne to a

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Racemic

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(B) The reaction starts with $hex-5-yn-1-ol$. Hydrogenation with $H_2$ (typically over a catalyst like $Pd/BaSO_4$ or similar) reduces the alkyne to an alkene,forming $hex-5-en-1-ol$ (intermediate $A$).In the presence of an acid catalyst $(H^{\oplus})$,the alkene undergoes protonation to form a carbocation at the $C_5$ position. The oxygen atom of the hydroxyl group then attacks this $sp^2$-hybridized carbocation to form a cyclic ether (tetrahydropyran derivative).Since the carbon atom where the ring closure occurs becomes a chiral center and the attack can occur from either face of the planar carbocation,both $R$ and $S$ enantiomers are formed in equal amounts.Therefore,the end product is a racemic mixture.

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

The end product formed in the above reaction is:

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