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(A) $NaBH_4$ (sodium borohydride) is a selective reducing agent that specifically reduces aldehydes and ketones to their corresponding alcohols. It do

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(A) $NaBH_4$ (sodium borohydride) is a selective reducing agent that specifically reduces aldehydes and ketones to their corresponding alcohols. It does not reduce amides,esters,or $C=C$ double bonds under standard conditions. In the given compound,there is a ketonic group and an amide group along with a $C=C$ double bond. Therefore,$NaBH_4$ will selectively reduce the ketone to a secondary alcohol while leaving the amide and the $C=C$ double bond unaffected. The product is the corresponding hydroxy-amide with the double bond intact.

List-$I$	List-$II$
$A. \ C_6H_5NHNH_2$	$I. \ \text{Schiff base}$
$B. \ NH_2OH$	$II. \ \text{Hydrazone}$
$C. \ C_6H_5NH_2$	$III. \ \text{Oxime}$
	$IV. \ \text{Phenyl hydrazone}$

The main reduction product of the following compound with $NaBH_4$ in methanol is

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Match the following reagents with the products obtained when they react with a ketone.
List-$I$ List-$II$
$A. \ C_6H_5NHNH_2$ $I. \ \text{Schiff base}$
$B. \ NH_2OH$ $II. \ \text{Hydrazone}$
$C. \ C_6H_5NH_2$ $III. \ \text{Oxime}$
$IV. \ \text{Phenyl hydrazone}$

Correct answer is:

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