The reaction of chlorobezene with NaOH at. 626 K gives 'X' and in presence of `HNO_(3)` it gives 'Y. The 'X' and 'Y' are:

Options

A. X-Phenol, Y-p-nitrophenol
B. X-benzyl chloride, Y-benzyl alcohol
C. X-phenol, Y-o-nitrophenol
D. X-o-nitrophenol, Y-p-nitrophenol

Explanation

To solve the question regarding the reaction of chlorobenzene with NaOH at 626 K and the subsequent reaction with HNO3, we can break it down into steps:

### Step 1: Identify the reaction of chlorobenzene with NaOH at 626 K
Chlorobenzene (C6H5Cl) reacts with sodium hydroxide (NaOH) at high temperatures (626 K) through a process known as nucleophilic aromatic substitution. Under these conditions, chlorobenzene can undergo hydrolysis to form phenol (C6H5OH).

**Reaction:**
\[ \text{C}_6\text{H}_5\text{Cl} + \text{NaOH} \xrightarrow{626 K} \text{C}_6\text{H}_5\text{OH} + \text{NaCl} \]

Thus, the product 'X' is phenol (C6H5OH).

### Step 2: Identify the reaction of 'X' (phenol) with HNO3
When phenol is treated with nitric acid (HNO3), it undergoes nitration. The nitration of phenol can lead to the formation of ortho-nitrophenol and para-nitrophenol due to the activating effect of the hydroxyl group (-OH) on the aromatic ring.

**Reaction:**
\[ \text{C}_6\text{H}_5\text{OH} + \text{HNO}_3 \rightarrow \text{C}_6\text{H}_4(\text{NO}_2)\text{(ortho)} + \text{C}_6\text{H}_4(\text{NO}_2)\text{(para)} + \text{H}_2\text{O} \]

Since the question asks for 'Y', we can consider the major product formed, which is typically para-nitrophenol (C6H4(NO2) in para position).

### Final Answer:
- 'X' is phenol (C6H5OH).
- 'Y' is para-nitrophenol (C6H4(NO2)).