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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.

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

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