`underset((A))("PhCOOMe") overset("(i)" NH_2 OH) underset("(ii)" overset (Ө)O H "(iii)" EtOH)rarr (B)`
Product `(B)` is :

Options

A. `PhNH_2`
B. `MeNH_2`
C. `PhNHCOOEt`
D. `PhNHCOOMe`

Explanation

To solve the problem step by step, we will analyze the reaction of the compound PhCOOMe (product A) with the reagents provided in the question.

### Step 1: Identify Product A
Product A is PhCOOMe, which is methyl benzoate. The structure can be represented as:
- Ph: Phenyl group (C6H5)
- COOMe: Methyl ester group (COOCH3)

### Step 2: Reaction with NH2OH
The first reagent is hydroxylamine (NH2OH). When methyl benzoate reacts with hydroxylamine, it forms an oxime. The reaction can be represented as follows:
- Methyl benzoate + NH2OH → Benzamide oxime (PhC(=NOH)OCH3) + Methanol (MeOH)

### Step 3: Reaction with OH⁻
The next step involves the reaction of the oxime with hydroxide ion (OH⁻). This step leads to the elimination of a water molecule and forms a nitrile:
- Benzamide oxime + OH⁻ → PhC(=N)O + H2O

### Step 4: Rearrangement
The product from the previous step undergoes a rearrangement (likely a Beckmann rearrangement), which converts the oxime into an amide:
- PhC(=N)O → PhC(=O)NH2

### Step 5: Reaction with Ethanol (EtOH)
Finally, the amide reacts with ethanol. This reaction typically results in the formation of an ethyl carbamate:
- PhC(=O)NH2 + EtOH → PhC(=O)OEt + NH3

### Final Product B
The final product (B) is therefore:
- PhC(=O)OEt (Ethyl benzoate)

### Summary of Product B
Product B is PhNHCOOEt, which can be written as:
- Ethyl benzoate (PhC(=O)OEt)

### Final Answer
Thus, the final product (B) is PhNHCOOEt.

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