`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 step by step, we will analyze the reactions involving C2H5Br (ethyl bromide) and the reagents AgCN and Zn-Hg/HCl.

### Step 1: Reaction of C2H5Br with AgCN
- **Reaction**: When ethyl bromide (C2H5Br) reacts with silver cyanide (AgCN), a nucleophilic substitution reaction occurs. The bromine atom is replaced by a cyanide group (CN).
- **Product (X)**: The product of this reaction is ethyl isocyanate (C2H5N=C=O), which can also be represented as C2H5NCO.

### Step 2: Reduction of X with Zn-Hg/HCl
- **Reaction**: The next step involves the reduction of ethyl isocyanate (X) using a reducing agent, Zn-Hg in the presence of hydrochloric acid (HCl). This is known as the Wolf-Kishner reduction.
- **Mechanism**: The isocyanate group (N=C=O) is reduced to an amine. In this case, the ethyl isocyanate will be converted to ethyl methyl amine.
- **Product (Y)**: The final product after reduction is ethyl methyl amine (C2H5NH2).

### Final Answer
Thus, the product Y is **ethyl methyl amine**.

### Summary of Steps:
1. **C2H5Br + AgCN → C2H5N=C=O (ethyl isocyanate)**
2. **C2H5N=C=O + Zn-Hg/HCl → C2H5NH2 (ethyl methyl amine)**