The product obtained by the reaction of ethyl | vedclass

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(A) The reaction proceeds in two steps: $1$. Ethyl chloride $(C_2H_5Cl)$ reacts with potassium cyanide $(KCN)$ to form ethyl cyanide (propanenitrile):

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Propylamine

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(A) The reaction proceeds in two steps: $1$. Ethyl chloride $(C_2H_5Cl)$ reacts with potassium cyanide $(KCN)$ to form ethyl cyanide (propanenitrile): $C_2H_5Cl + KCN \rightarrow C_2H_5CN + KCl$. $2$. The reduction of ethyl cyanide $(C_2H_5CN)$ with sodium $(Na)$ and alcohol $(C_2H_5OH)$ (Mendius reduction) yields propylamine ($C_2H_5CH_2NH_2$ or $CH_3CH_2CH_2NH_2$): $C_2H_5CN + 4[H] \xrightarrow{Na/C_2H_5OH} CH_3CH_2CH_2NH_2$.

The product obtained by the reaction of ethyl chloride with potassium cyanide is reduced with sodium and alcohol to give ...............

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