The product (P) of the below reaction sequenc | vedclass

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(B) Step $1$: Aldol condensation of $CH_3CH_2CHO$ (propanal) in the presence of $NaOH$ gives $\beta$-hydroxy aldehyde,which upon heating $( \Delta, H^

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$CH_3CH_2CH_2CH(CH_3)CH_2OH$

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(B) Step $1$: Aldol condensation of $CH_3CH_2CHO$ (propanal) in the presence of $NaOH$ gives $\beta$-hydroxy aldehyde,which upon heating $( \Delta, H^{\oplus} )$ undergoes dehydration to form an $\alpha,\beta$-unsaturated aldehyde: $CH_3CH_2CH=C(CH_3)CHO$.Step $2$: The final step involves hydrogenation using $H_2/Ni$ at $573 \ K$. This reagent reduces both the carbon-carbon double bond and the aldehyde group to a primary alcohol.Thus,the final product $(P)$ is $CH_3CH_2CH_2CH(CH_3)CH_2OH$.

List-$I$	List-$II$
$1.$ Ammoniacal $AgNO_3$	$a.$ Primary amine
$2.$ $HIO_4$	$b.$ Aldehyde
$3.$ Alkaline $KMnO_4$	$c.$ Vicinal-$OH$
$4.$ Chloroform $NaOH$	$d.$ Double bond

The product $(P)$ of the below reaction sequence is:

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Match List-$I$ with List-$II$.
List-$I$ List-$II$
$1.$ Ammoniacal $AgNO_3$ $a.$ Primary amine
$2.$ $HIO_4$ $b.$ Aldehyde
$3.$ Alkaline $KMnO_4$ $c.$ Vicinal-$OH$
$4.$ Chloroform $NaOH$ $d.$ Double bond

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