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(C) $1$. The starting material is $Ph-CH(OH)-C \equiv C-CH=CH-OMe$. $2$. Treatment with $H_2 / Pd-BaSO_4$ (Lindlar's catalyst) selectively reduces the

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$Ph-(CH=CH)_2-CHO$

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(C) $1$. The starting material is $Ph-CH(OH)-C \equiv C-CH=CH-OMe$. $2$. Treatment with $H_2 / Pd-BaSO_4$ (Lindlar's catalyst) selectively reduces the alkyne to a cis-alkene,forming intermediate $(A)$: $Ph-CH(OH)-CH=CH-CH=CH-OMe$. $3$. Subsequent treatment with $H_3O^+$ leads to the hydrolysis of the enol ether. The protonation of the double bond followed by resonance and water attack leads to the formation of an aldehyde. $4$. The final product $(B)$ is $Ph-CH=CH-CH=CH-CHO$,which can be written as $Ph-(CH=CH)_2-CHO$.

List-$I$	List-$II$
$A$. The reaction of $1,6$-dibromohexane with $Zn$.	$i$. $H_3C-C \equiv CH$
$B$. Reaction of ethanol with concentrated $H_2SO_4$ at $443 \ K$.	$ii$. $H_2C=CH_2$
$C$. Major product in the reaction of propene with $HBr$ in the presence of benzoyl peroxide.	$iii$. $CH_3-CH_2-CH_2-Br$
$D$. The reaction of $1,1$-dibromopropane with $NaNH_2$ at $433 \ K$.	$iv$. Cyclohexane

Product $(B)$ is

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