`C_(2)H_(5)Br overset(AgCN)rarr X overset("Reduction")underset(Zn-Hg//HCI)rarr Y`, Here, `Y` is:-

Options

A. n-propylamine
B. Isopropylamine
C. Ethyl methyl amine
D. Ethylamine

Answer

To solve the question, we need to analyze the given reactions step by step.

### Step 1: Identify the starting compound
The starting compound is ethyl bromide, which is represented as \( C_2H_5Br \) or \( CH_3CH_2Br \).

### Step 2: Reaction with AgCN
When ethyl bromide reacts with silver cyanide (AgCN), the bromine atom acts as a leaving group. The nitrogen atom from the cyanide ion attacks the carbon atom, leading to the formation of isocyanide. The reaction can be represented as follows:

\[ C_2H_5Br + AgCN \rightarrow C_2H_5NC + AgBr \]

Here, \( C_2H_5NC \) is the product formed, which is isocyanide (or ethyl isocyanide). We will denote this product as \( X \).

### Step 3: Reduction of Isocyanide
The next step involves the reduction of isocyanide \( X \) using zinc amalgam (Zn-Hg) in the presence of hydrochloric acid (HCl). The reduction of isocyanides typically leads to the formation of primary amines. The reaction can be represented as:

\[ C_2H_5NC + Zn-Hg/HCl \rightarrow C_2H_5NH_2 \]

Here, \( C_2H_5NH_2 \) is ethylamine, which is the product \( Y \).

### Step 4: Naming the Product
The product \( Y \) is ethylamine, which is a primary amine.

### Final Answer
Thus, the final product \( Y \) is ethylamine.

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