What are X and Y respectively in the followin | vedclass

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(B) The conversion of benzamide $(C_6H_5CONH_2)$ to aniline $(C_6H_5NH_2)$ involves the removal of a carbonyl group,which is achieved by the Hoffmann

Vedclass · Accepted Answer

$NaOH + Br_2 ; (i) \ LiAlH_4 \ (ii) \ H_2O$

Vedclass · Answer

(B) The conversion of benzamide $(C_6H_5CONH_2)$ to aniline $(C_6H_5NH_2)$ involves the removal of a carbonyl group,which is achieved by the Hoffmann bromamide degradation reaction using $NaOH + Br_2$. Thus,$X = NaOH + Br_2$.The conversion of benzamide $(C_6H_5CONH_2)$ to benzylamine $(C_6H_5CH_2NH_2)$ involves the reduction of the amide group to an amine group,which is achieved using a strong reducing agent like $LiAlH_4$ followed by hydrolysis. Thus,$Y = (i) \ LiAlH_4 \ (ii) \ H_2O$.Therefore,$X = NaOH + Br_2$ and $Y = (i) \ LiAlH_4 \ (ii) \ H_2O$.

What are $X$ and $Y$ respectively in the following reactions?

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