`underset((A)) (PhCONH_2) underset(KOBr)rarr (B) underset(MeOH)rarr (C)`
Product is `( C)` is :

Options

A. `PhNH_2`
B. `PhNHCOOMe`
C. `PhNHCOOPh`
D. None

Explanation

To solve the given question step by step, we will analyze the reactions involved and identify the products formed at each stage.

### Step 1: Identify the Starting Material
The starting material (A) is **PhCONH2**, which is an amide (specifically, benzamide).

### Step 2: Reaction with KOBr
The first reaction involves treating benzamide (PhCONH2) with **KOBr** (potassium bromate). This reaction is known as the **Hofmann degradation** or **Hofmann rearrangement**.

In this reaction, the amide undergoes a rearrangement to form an isocyanate intermediate. The general reaction can be represented as:
\[ \text{PhCONH}_2 + \text{KOBr} \rightarrow \text{PhC(=O)N} \text{(isocyanate)} \]

### Step 3: Formation of Product B
After the rearrangement, the isocyanate can undergo hydrolysis or further reactions. In this case, the isocyanate can react with water or alcohols to form an **urea derivative**.

Thus, the product (B) after the reaction with KOBr will be:
\[ \text{PhC(=O)N} \text{(isocyanate)} \rightarrow \text{PhC(=O)NH} \text{(urea)} \]

### Step 4: Reaction with MeOH
Now, product (B) reacts with **MeOH** (methanol). This reaction involves the nucleophilic attack of methanol on the carbonyl carbon of the urea derivative.

The reaction can be represented as:
\[ \text{PhC(=O)NH} + \text{MeOH} \rightarrow \text{PhC(=O)OCH_3} + \text{NH}_2 \]

### Step 5: Formation of Product C
The final product (C) formed from the reaction of product (B) with MeOH will be:
\[ \text{PhC(=O)OCH_3} \]

This compound is a **methyl ester** of the corresponding carboxylic acid derived from the urea.

### Conclusion
Thus, the final product (C) is:
\[ \text{PhC(=O)OCH_3} \]

### Final Answer
The correct option for product (C) is:
**PhC(=O)OCH3** (which can also be written as Ph-CO-O-CH3).

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