The product of the reaction is:

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(A) The reaction given is the Hofmann bromamide degradation reaction,where an amide is treated with $NaOBr$ (or $Br_2/NaOH$) to form a primary amine w

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$CH_3-C(Ph)(H)-NH_2$

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(A) The reaction given is the Hofmann bromamide degradation reaction,where an amide is treated with $NaOBr$ (or $Br_2/NaOH$) to form a primary amine with one carbon atom less than the original amide.In this reaction,the migration of the alkyl group from the carbonyl carbon to the nitrogen atom occurs with complete retention of configuration at the chiral center.Starting material: $2$-phenylpropanamide.Product: $1$-phenylethanamine.Since the configuration is retained,the product is $1$-phenylethanamine with the same stereochemistry as the reactant.

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