`C_(2)H_(5)I overset(AgNO_(2))rarr underset(("major product"))("X. Here X")` is :

Options

A. `C_(2)H_(5) - overset(O)overset(||)(N) rarr O`
B. `C_(2)H_(5) - O - N = O`
C. `C_(2) H_(5) - N = O`
D. All of above

Explanation

To solve the question regarding the reaction of ethyl iodide (C2H5I) with silver nitrite (AgNO2), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants in this reaction are ethyl iodide (C2H5I) and silver nitrite (AgNO2).

2. **Understand the Reaction Type**:
The reaction involves a nucleophilic substitution mechanism (SN2). In this case, the iodide ion (I-) is a good leaving group, and the nitrite ion (NO2-) from silver nitrite acts as the nucleophile.

3. **Write the Reaction**:
The nucleophilic substitution can be represented as follows:
\[ C2H5I + AgNO2 \rightarrow C2H5NO2 + AgI \]
Here, the iodide ion (I-) is replaced by the nitrite ion (NO2-), resulting in the formation of ethyl nitrite (C2H5NO2) and silver iodide (AgI).

4. **Identify the Major Product**:
The major product of this reaction is ethyl nitrite (C2H5NO2). This is because the nitrite ion is a strong nucleophile and effectively replaces the iodide ion.

5. **Byproducts**:
Along with the major product, silver iodide (AgI) is also formed, which is typically a white precipitate.

6. **Final Answer**:
Therefore, the major product X formed in this reaction is:
\[ X = C2H5NO2 \]