Consider the following reaction sequence:B xl | vedclass

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(B) $NaBH_4$ is a selective reducing agent that reduces aldehydes and ketones to alcohols but does not reduce carbon-carbon double bonds $(C=C)$.There

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$CH_3-CH_2-CH_2OH, CH_2=CH-CH_2OH$

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(B) $NaBH_4$ is a selective reducing agent that reduces aldehydes and ketones to alcohols but does not reduce carbon-carbon double bonds $(C=C)$.Therefore,$CH_2=CH-CHO \xrightarrow{NaBH_4} CH_2=CH-CH_2OH$ ($B$ is allyl alcohol).$H_2/Pt$ is a strong reducing agent that reduces both aldehydes/ketones and carbon-carbon double bonds $(C=C)$.Therefore,$CH_2=CH-CHO \xrightarrow{H_2/Pt} CH_3-CH_2-CH_2OH$ ($A$ is propan$-1-$ol).Thus,$A$ is $CH_3-CH_2-CH_2OH$ and $B$ is $CH_2=CH-CH_2OH$.

List-$I$ (Reagents)	List-$II$ (Reaction Name)
$A$. $NH_2-NH_2, KOH$	$I$. Tollen's Test
$B$. $[Ag(NH_3)_2]OH$	$II$. Clemmensen Reduction
$C$. Aq. $CuSO_4$,Sodium Potassium tartarate,$KOH$	$III$. Wolff-Kishner Reduction
$D$. $Zn-Hg, HCl$	$IV$. Fehling's Test

Consider the following reaction sequence:
$B \xleftarrow{NaBH_4} CH_2=CH-CHO \xrightarrow{H_2/Pt} A$
Identify $A$ and $B$.

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Consider the following reaction sequence:$B \xleftarrow{NaBH_4} CH_2=CH-CHO \xrightarrow{H_2/Pt} A$Identify $A$ and $B$.