The product in the given reaction is:

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(B) $LiAlH_4$ (Lithium aluminium hydride) is a strong reducing agent that reduces aldehydes,ketones,carboxylic acids,and esters to their corresponding

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$C_6H_5-CH=CH-CH_2OH$

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(B) $LiAlH_4$ (Lithium aluminium hydride) is a strong reducing agent that reduces aldehydes,ketones,carboxylic acids,and esters to their corresponding alcohols. However,it does not reduce isolated carbon-carbon double bonds $(C=C)$. In the given reaction,the reactant is cinnamaldehyde $(C_6H_5-CH=CH-CHO)$. $LiAlH_4$ selectively reduces the aldehyde group $(-CHO)$ to a primary alcohol group $(-CH_2OH)$ while leaving the conjugated double bond $(C=C)$ intact. Therefore,the product formed is cinnamyl alcohol,which is $C_6H_5-CH=CH-CH_2OH$.

List-$I$	List-$II$
$A$. Grignard reagent	$1$. $H_2 / Pd-BaSO_4$
$B$. Clemmensen reduction	$2$. $N_2H_4 / KOH / (CH_2OH)_2$
$C$. Rosenmund reduction	$3$. $CH_3MgX$
$D$. Wolff-Kishner reduction	$4$. $Zn-Hg / \text{conc. } HCl$
	$5$. $H_2 / Ni$

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Match the following lists:
List-$I$ List-$II$
$A$. Grignard reagent $1$. $H_2 / Pd-BaSO_4$
$B$. Clemmensen reduction $2$. $N_2H_4 / KOH / (CH_2OH)_2$
$C$. Rosenmund reduction $3$. $CH_3MgX$
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$5$. $H_2 / Ni$

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