(C) The reaction proceeds as follows: $CH_3Br + KCN \to CH_3CN + KBr$ $X$ is $CH_3CN$ (acetonitrile or ethanenitrile). Reduction of $CH_3CN$ using $Na

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(C) The reaction proceeds as follows: $CH_3Br + KCN \to CH_3CN + KBr$ $X$ is $CH_3CN$ (acetonitrile or ethanenitrile). Reduction of $CH_3CN$ using $Na / C_2H_5OH$ (Mendius reduction) yields $CH_3CH_2NH_2$. $CH_3CN + 4[H] \xrightarrow{Na/C_2H_5OH} CH_3CH_2NH_2$ Thus,$Y$ is $CH_3CH_2NH_2$ (ethanamine).