Acetic acid on heating with NH_3 forms A. Whe | vedclass

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(B) $1$. Acetic acid $(CH_3COOH)$ reacts with $NH_3$ to form ammonium acetate,which upon heating undergoes dehydration to form acetamide $(A)$: $CH_3C

Vedclass · Accepted Answer

$CH_3CH_2NH_2, CH_3CH_2NC$

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(B) $1$. Acetic acid $(CH_3COOH)$ reacts with $NH_3$ to form ammonium acetate,which upon heating undergoes dehydration to form acetamide $(A)$: $CH_3COOH + NH_3$ $\rightarrow CH_3COONH_4$ $\xrightarrow{\Delta} CH_3CONH_2 (A) + H_2O$.$2$. Acetamide $(A)$ is reduced by $LiAlH_4$ to form ethylamine $(B)$: $CH_3CONH_2 \xrightarrow{LiAlH_4} CH_3CH_2NH_2 (B)$.$3$. Ethylamine $(B)$ reacts with chloroform $(CHCl_3)$ in the presence of $KOH$ (Carbylamine reaction) to form ethyl isocyanide $(C)$: $CH_3CH_2NH_2 + CHCl_3 + 3KOH \rightarrow CH_3CH_2NC (C) + 3KCl + 3H_2O$.$4$. Therefore,$B$ is $CH_3CH_2NH_2$ and $C$ is $CH_3CH_2NC$.

Acetic acid on heating with $NH_3$ forms $A$. When $A$ reacts with $LiAlH_4$ followed by hydrolysis,it gives $B$. When $B$ is heated with chloroform in $KOH$ medium,it gives $C$. What are $B$ and $C$ respectively?

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