\[C{H_3} - C{H_2} - Br \overset{alcKOH}{\rightarrow} C{H_3}C{H_2}CN \overset{HOH}{\rightarrow} X\]. In this reaction, product X is
A. Acetic acid
B. Propionic acid
C. Butyric acid
D. Formic acid

Hint: The chemical name of \[C{H_3}C{H_2}Br\] is Bromo ethane. It is an example of haloalkanes. Haloalkanes generally undergo nucleophilic substitution reactions where a nucleophile replaces the halogen group to form the main product.

Complete Step by Step Solution:
It is given that Bromo ethane reacts with alcoholic potassium cyanide to form ethyl cyanide which further hydrolysis and gives the final product X.

The reaction between bromoethane with potassium cyanide takes place in solvent 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 ions.

The reaction is shown below: \[{C_2}{H_5}Br + KCN \to {C_2}{H_5}CN + KBr\]

In the above reaction, bromoethane reacts with potassium cyanide to form ethyl cyanide and potassium bromide. Here, the cyanide ion replaces the bromide ion from bromoethane 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:
\[{C_2}{H_5}CN \overset{HOH}{\rightarrow} {C_2}{H_5}COOH\]

In the above reaction, ethyl cyanide on hydrolysis gives propanoic acid.

The overall reaction is given as shown below.
\[{C_2}{H_5}Br + KCN \to
{C_2}{H_5}CN \overset{HOH}{\rightarrow} {C_2}{H_5}COOH\]
Therefore, X is propanoic acid.
Therefore, the correct option is B.

Note: Ambident nucleophiles are the type of nucleophile which can attack the carbocation from either side. Cyanide ion is an example of an ambident nucleophile. It can attack both NC and CN. When it attacks from another side then isocyanide is formed.