Question

Identify the product '$B$' in the following sequence of reactions. $CH_{3}Br \xrightarrow{KCN} A \xrightarrow{Na/C_{2}H_{5}OH} B$

• A. Methyl cyanide
• B. Ethylamine
• C. Methylamine
• D. Ethyl cyanide

Hint:

Chemistry Tip: The reaction with $KCN$ is a classic "step-up" reaction, uniquely useful for extending a carbon chain by exactly one carbon atom.
• $Na/C_2H_5OH$ or $LiAlH_4$ are standard reagents for reducing nitriles completely to primary amines.

Answer 1

The Correct Option is B

Concept: Chemistry (Organic Chemistry) - Nucleophilic Substitution and Mendius Reduction.

Step 1: Analyze the first reaction step to find product A. The first step is the reaction of methyl bromide ($CH_3Br$) with potassium cyanide ($KCN$). This is a classic nucleophilic substitution ($S_N2$) reaction.

Step 2: Determine the structure of product A. The cyanide nucleophile ($CN^-$) displaces the bromide leaving group ($Br^-$). This forms methyl cyanide, also known as acetonitrile. Reaction: $CH_3Br + KCN \rightarrow CH_3CN + KBr$. Thus, intermediate $A$ is $CH_3CN$.

Step 3: Analyze the second reaction step to find product B. The second step involves treating product A ($CH_3CN$) with sodium metal in ethanol ($Na/C_2H_5OH$). This specific reagent combination is a strong reducing agent.

Step 4: Identify the specific named reaction. The reduction of an alkyl cyanide (nitrile) using sodium and ethanol is known as the Mendius reduction. This process systematically reduces the carbon-nitrogen triple bond ($-C\equiv N$) to a primary amine group ($-CH_2NH_2$).

Step 5: Determine the final structure and name of product B. During the reduction, four hydrogen atoms are added across the nitrile group. Reaction: $CH_3-C\equiv N + 4[H] \xrightarrow{Na/C_2H_5OH} CH_3-CH_2-NH_2$.
The resulting molecule contains a two-carbon chain attached to an amino group. The IUPAC name for this primary amine is ethanamine, and its common name is ethylamine. $$ \therefore \text{The product B is Ethylamine.} $$