Question

In the following sequence of reactions, $CH _{3}- Br \stackrel{ KCN }{\longrightarrow} A \stackrel{ H _{3} O ^{+}}{\longrightarrow} B \stackrel{ LiAlH _{4}}{\longrightarrow} C$, the end product C is

• A. Acetone
• B. Methane
• C. Acetaldehyde
• D. Ethyl alcohol

Answer 1

Answer: (D)

Ethyl alcohol

Answer 2

The correct answer is D:Ethyl alcohol
$CH _{3}- Br \stackrel{ KCN }{\longrightarrow} CH _{3}- CN \stackrel{ H _{3} O ^{+}}{\longrightarrow} CH _{3} COOH \stackrel{ LiAlH _{4}}{\longrightarrow} CH _{3} CH _{2} OH$
Methyl bromide undergoes a reaction with $KCN$ to produce acetonitrile, where the bromine atom is substituted with a cyanide group. The subsequent hydrolysis of the cyanide group results in the formation of a carboxylic group. Finally, the reduction of the carboxylic group leads to the formation of a hydroxyl group.