`C_(6) H_(5) - CH_(2) - I underset(Delta) overset(NaN_3) (to)` Products
Reaction is assumed to involve nitrene as intermediate, then various possible products are :

Options

A. `C_6 H_5 CH_2 NH_2`
B. `C_6 H_5 N = CH_2`
C. `C_6 H_5 CH = NH`
D.

Explanation

To solve the question regarding the reaction of C₆H₅-CH₂-I with NaN₃, we will follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- C₆H₅-CH₂-I (Benzyl iodide)
- NaN₃ (Sodium azide)

### Step 2: Understand the Role of Nitrene
The problem states that the reaction involves nitrene as an intermediate. Nitrenes are reactive species containing a nitrogen atom with a valence of three, typically represented as R-N: (where R is a carbon chain).

### Step 3: Reaction Mechanism
1. **Nucleophilic Substitution**: The iodide (I) in benzyl iodide is a good leaving group. When NaN₃ is added, the azide ion (N₃⁻) will attack the carbon atom bonded to the iodine, resulting in the formation of an intermediate.

\[ C₆H₅-CH₂-I + NaN₃ \rightarrow C₆H₅-CH₂-N₃ + NaI \]

2. **Formation of Nitrene**: The azide group (N₃) can undergo a rearrangement to form a nitrene intermediate. This nitrene can then lose nitrogen gas (N₂) to yield a more stable product.

### Step 4: Elimination of Nitrogen Gas
The nitrene intermediate can eliminate nitrogen gas (N₂), leading to the formation of a double bond between the carbon and nitrogen.

### Step 5: Final Product Formation
The final product after the elimination of nitrogen gas will be:
\[ C₆H₅-CH=NH \]
This product is known as a benzyl imine.

### Conclusion
The major product formed from the reaction of C₆H₅-CH₂-I with NaN₃, assuming nitrene as an intermediate, is:
\[ C₆H₅-CH=NH \]