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?

Similar Questions

Which of the following is most reactive towards electrophile?

The sequence of reagents required to convert aniline to benzene nitrile is:

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Assertion : Acetamide reacts with $Br_2$ in presence of methanolic $CH_3ONa$ to form methyl $N$-methylcarbamate.
Reason : Methyl isocyanate is formed as an intermediate which reacts with methanol to form methyl $N$-methylcarbamate.

Which of the following products is a Schiff base?
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$C_2H_5NH_2 \xrightarrow{C_6H_5SO_2Cl} d$

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