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(A) $1$. The reaction of the acid chloride $(CH_3-C(H)(C_6H_4-D)-COCl)$ with $NH_3$ yields the corresponding amide,$CH_3-C(H)(C_6H_4-D)-CONH_2$. This

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(A) $1$. The reaction of the acid chloride $(CH_3-C(H)(C_6H_4-D)-COCl)$ with $NH_3$ yields the corresponding amide,$CH_3-C(H)(C_6H_4-D)-CONH_2$. This step does not involve the chiral center,so the configuration is retained.$2$. The subsequent reaction with $OH^-/Br_2$ is the Hoffmann Bromamide degradation. In this reaction,the carbonyl group $(CO)$ is removed,and the $NH_2$ group attaches directly to the chiral carbon atom.$3$. The Hoffmann Bromamide degradation proceeds with complete retention of configuration at the chiral center because the migrating group does not break its bond with the chiral carbon until the new bond with nitrogen is formed.$4$. Therefore,the final product is $CH_3-C(H)(C_6H_4-D)-NH_2$ with the same configuration as the starting material.

Predict the major product of the following reaction sequence:
$CH_3-C(H)(C_6H_4-D)-COCl$ $\xrightarrow{NH_3} \text{Intermediate}$ $\xrightarrow{OH^-/Br_2} \text{Product}$

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Predict the major product of the following reaction sequence:$CH_3-C(H)(C_6H_4-D)-COCl$ $\xrightarrow{NH_3} \text{Intermediate}$ $\xrightarrow{OH^-/Br_2} \text{Product}$