Question

\[{C_2}{H_5}Cl
+ KCN \to X \overset{hydrolysis}{\rightarrow} Y\]. X and Y are
A. \[{C_2}{H_6}\] and \[{C_2}{H_5}CN\]
B. \[{C_2}{H_5}CN\] and \[{C_2}{H_6}\]
C. . \[{C_2}{H_5}CN\] and \[{C_2}{H_5}C{H_2}N{H_2}\]
D. . \[{C_2}{H_5}CN\] and \[{C_2}{H_5}COOH\]

Answer 1

Hint: Haloalkanes are the compound where the halogen group is attached to an alkyl group. Haloalkanes undergo a nucleophilic substitution reaction where a nucleophile replaces the halogen group to form the product.

Complete Step by Step Solution:
It is given that chloro ethane reacts with potassium cyanide to form a compound which further hydrolysis gives the final product.
The reaction between chloroethane with potassium cyanide takes place in a solvent, mainly ethanol. The reaction involved is nucleophilic substitution reaction 2 also said as \[{S_N}2\] reaction. It is a one step reaction where the nucleophile replaces the leaving group, usually halide.

The reaction is shown below.
\[{C_2}{H_5}Cl + KCN \to {C_2}{H_5}CN + KCl\]
In the above reaction chloroethane reacts with potassium cyanide to form ethyl cyanide and potassium chloride.
Here, the cyanide ion replaces the chloride ion from chloroethane to form ethyl cyanide.
Further ethyl cyanide on hydrolysis forms a carboxylic acid.
Hydrolysis is a type of reaction where a water molecule is added to the reactant.
The reaction is shown below.
\[{C_2}{H_5}CN \overset{hydrlysis}{\rightarrow} {C_2}{H_5}COOH\]
In the above reaction, ethyl cyanide on hydrolysis gives propanoic acid.
The overall reaction is given as shown below.
\[{C_2}{H_5}Cl + KCN \to {C_2}{H_5}CN \overset{hydrolysis}{\rightarrow}
{C_2}{H_5}COOH\]
Therefore, X is ethyl cyanide and Y is propanoic acid.
Therefore, the correct option is D.

Note: It should be noted that cyanide (CN) is an ambident nucleophile which means that it can attack the carbocation from any site. It can attach in form of CN or NC. In the given reaction it is attached as CN by attacking the carbon side to form ethyl cyanide. When it attacks from another side, then isocyanide is formed.