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(A) $1$. The first reaction is the Hoffmann bromamide degradation reaction. In this reaction,an amide $(R-CONH_2)$ reacts with $Br_2$ in the presence

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$3-$bromobenzenamine and $3-$bromobenzylamine

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(A) $1$. The first reaction is the Hoffmann bromamide degradation reaction. In this reaction,an amide $(R-CONH_2)$ reacts with $Br_2$ in the presence of a strong base (like $KOH$) to form a primary amine $(R-NH_2)$ with one carbon atom less than the original amide. Thus,$3$-bromobenzamide reacts to form $3$-bromobenzenamine $(A)$: $m-Br-C_6H_4-CONH_2 \xrightarrow{KOBr} m-Br-C_6H_4-NH_2$$2$. The second reaction is the reduction of an amide using $LiAlH_4$ followed by $H_3O^+$. $LiAlH_4$ is a strong reducing agent that reduces an amide $(R-CONH_2)$ to a primary amine $(R-CH_2NH_2)$ with the same number of carbon atoms. Thus,$3$-bromobenzamide reacts to form $3$-bromobenzylamine $(B)$:$m-Br-C_6H_4-CONH_2 \xrightarrow{LiAlH_4, H_3O^+} m-Br-C_6H_4-CH_2NH_2$Therefore,the products $A$ and $B$ are $3$-bromobenzenamine and $3$-bromobenzylamine respectively.

In the above reactions,product $A$ and product $B$ respectively are:

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