CH_3CH_2Br xrightarrow{AgCN} A xrightarrow{H_ | vedclass

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Question

(C) The reaction proceeds as follows:$1$. $CH_3CH_2Br$ reacts with $AgCN$ to form an isocyanide (carbylamine) as the major product because $AgCN$ is a

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$CH_3CH_2NHCH_3$

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(C) The reaction proceeds as follows:$1$. $CH_3CH_2Br$ reacts with $AgCN$ to form an isocyanide (carbylamine) as the major product because $AgCN$ is a covalent compound. The reaction is: $CH_3CH_2Br + AgCN \rightarrow CH_3CH_2NC + AgBr$. Thus,$A$ is $CH_3CH_2NC$ (ethyl isocyanide).$2$. The reduction of isocyanide $(R-NC)$ with $H_2$ in the presence of $Ni$ catalyst yields a secondary amine. The reaction is: $CH_3CH_2NC + 2H_2 \xrightarrow{Ni} CH_3CH_2NHCH_3$. Thus,$B$ is $CH_3CH_2NHCH_3$ ($N$-methylethanamine).

$CH_3CH_2Br$ $\xrightarrow{AgCN} A$ $\xrightarrow{H_2, Ni} B$. The final product $(B)$ is

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