Question

Given, ${C_2}{H_5}Br\xrightarrow{{KCN}}\left( A \right)\xrightarrow[{Hydrolysis}]{{HCl}}\left( B \right)$ the compound (B) in the reaction is:

Answer 1

Alkyl bromides on treatment with potassium cyanide, the bromide ion is replaced by cyanide, as cyanide is more electron withdrawing group. Further treatment with hydrogen chloride and hydrolysis, the alkyl cyanides undergo hydrolysis to form carboxylic acids.

Complete answer:
Given a molecule is ethyl bromide, the alkyl group is attached to the electronegative atom i.e.., bromine atom and the alkyl group is ethyl group.
When this ethyl bromide is treated with potassium cyanide, the bromide ion in the alkyl bromide is replaced by a cyanide ion, leading to the formation of ethyl cyanide.
Further, ethyl cyanide on treatment with hydrogen chloride and hydrolysis, the cyanide group undergoes hydrolysis to form carboxylic acids.
Thus, ethyl cyanide forms propanoic acid. Generally, alkyl cyanides or nitriles upon hydrolysis form carboxylic acids.
Thus, the compound A is ethyl cyanide and the compound B is propanoic acid.
The carboxylic acid with three carbons can be named as propanoic acid. Thus, the compound B in the reaction is propionic acid.
${C_2}{H_5}Br\xrightarrow{{KCN}}{C_2}{H_5}CN\xrightarrow[{Hydrolysis}]{{HCl}}{C_2}{H_5}COOH$
Propionic acid has the molecular formula of ${C_2}{H_5}COOH$, as it is an acid the root word for the three carbon atoms is prop and the suffix of the acid functional group must be -oic acid. Thus, the product can be named as propionic acid.

Note:
The alkyl halides are the compounds consisting of alkyl part and halogen atom, the halogen atom may be substituted by other groups like bromine can be replaced by cyanide group. Cyanide are also known as nitriles, upon hydrolysis forms carboxylic acids as the products.