An organic halide is shaken with aqueous NaOH followed by the addition of dil `HNO_(3)` and silver nit rate solution gave white ppt The substane can be .

Options

A. `C_(6)H_(4)(CH_(3))`Br
B. `C_(6)H_(5)CH_(2)`Cl
C. `C_(6)H_(5)`Cl
D. None of these

Answer

To solve the question, we need to analyze the reactions that occur when an organic halide is treated with aqueous NaOH, followed by dilute HNO3 and silver nitrate solution, leading to the formation of a white precipitate.

### Step-by-Step Solution:

1. **Identify the Organic Halide**:
The organic halide can be represented as R-X, where R is an organic group and X is a halogen (Cl, Br, I).

2. **Reaction with Aqueous NaOH**:
When the organic halide (R-X) is shaken with aqueous NaOH, the hydroxide ion (OH-) acts as a nucleophile. The electronegative halogen (X) is displaced, leading to the formation of an alcohol (R-OH) and the corresponding sodium halide (NaX).
- Reaction:
\[ R-X + OH^- \rightarrow R-OH + Na^+ + X^- \]

3. **Formation of Sodium Halide**:
The sodium ion (Na+) combines with the halide ion (X-) to form sodium halide (NaX).

4. **Addition of Dilute Nitric Acid (HNO3)**:
The addition of dilute nitric acid does not directly affect the halide but helps in maintaining the acidic medium for the next step.

5. **Reaction with Silver Nitrate (AgNO3)**:
When sodium halide (NaX) is treated with silver nitrate (AgNO3), a precipitation reaction occurs, producing silver halide (AgX) and sodium nitrate (NaNO3).
- Reaction:
\[ NaX + AgNO3 \rightarrow AgX + NaNO3 \]
The silver halide (AgX) is the precipitate formed.

6. **Identify the Precipitate**:
The question specifies that a white precipitate is formed. The color of the precipitate depends on the halide:
- AgCl (silver chloride) forms a white precipitate.
- AgBr (silver bromide) forms a yellow precipitate.
- AgI (silver iodide) forms a slightly yellow precipitate.
Since we are looking for a white precipitate, it indicates that the halide must be chloride (Cl).

7. **Conclusion**:
The organic halide that would lead to the formation of a white precipitate (AgCl) upon the described reactions is likely benzyl chloride (C6H5CH2Cl). This is because it can easily form a carbocation that stabilizes and allows the release of the chloride ion.

### Final Answer:
The substance can be **C6H5CH2Cl (Benzyl Chloride)**.