The product obtained by the reaction of ethyl | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

Propylamine

Vedclass · Answer

(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 ...............

Similar Questions

Hofmann's hypobromite reaction affords a method of

$CH_3-COOH$ $\xrightarrow{NH_3} X$ $\xrightarrow{\Delta} Y$ $\xrightarrow{LiAlH_4} Z$; $Z$ is:

What is the number of moles of hydrogen atoms needed to obtain one mole of ethanamine from acetamide using $LiAlH_4 / \text{ether}$ as a reducing agent?

The sequence from the following that would result in giving predominantly $3,4,5-$Tribromoaniline is $:$

Which of the following reagents is used to convert a $-CN$ group to a $-CH_2NH_2$ group?

Vedclass Test Series

Exam Paper Generator

Online Exam Module