CH_3-CH_2-I xrightarrow{NaCN} A xrightarrow[P | vedclass

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Question

(C) Step $1$: $CH_3-CH_2-I$ reacts with $NaCN$ to form $CH_3-CH_2-CN$ $(A)$ via nucleophilic substitution.Step $2$: Partial hydrolysis of $CH_3-CH_2-C

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$CH_3-CH_2-NH_2$

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(C) Step $1$: $CH_3-CH_2-I$ reacts with $NaCN$ to form $CH_3-CH_2-CN$ $(A)$ via nucleophilic substitution.Step $2$: Partial hydrolysis of $CH_3-CH_2-CN$ $(A)$ in the presence of $\Theta OH$ yields $CH_3-CH_2-CONH_2$ $(B)$ (propanamide).Step $3$: The reaction of $CH_3-CH_2-CONH_2$ $(B)$ with $Br_2$ and $NaOH$ is the Hofmann bromamide degradation reaction,which converts an amide into a primary amine with one carbon atom less.Step $4$: The product $C$ is $CH_3-CH_2-NH_2$ (ethanamine).

$CH_3-CH_2-I$ $\xrightarrow{NaCN} A$ $\xrightarrow[Partial \ hydrolysis]{\Theta OH} B$ $\xrightarrow[Br_2]{NaOH} C$. Product $C$ is

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