Compound 'A' yield benzylamine on reaction with `LiAIH_(4)` following by hydrolysis. The reaction of 'A' with NaOH and `Br_(2)` will yield

Options

A. benzamine
B. aniline
C. nitrobenzene
D. N-bromoanile

Explanation

To solve the problem, we need to analyze the reactions involving compound 'A' and understand the transformations it undergoes. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify Compound 'A'
Given that compound 'A' yields benzylamine when reacted with LiAlH4 followed by hydrolysis, we can infer that compound 'A' is likely an amide. This is because LiAlH4 is a strong reducing agent that can reduce amides to amines.

**Hint:** Consider the structure of benzylamine and the type of compound that can be reduced to yield it.

### Step 2: Structure of Benzylamine
Benzylamine has the structure C6H5CH2NH2. When an amide is reduced by LiAlH4, the carbonyl group (C=O) is converted to a methylene group (CH2), resulting in the formation of benzylamine.

**Hint:** Visualize the reduction process of an amide to an amine.

### Step 3: Reaction of Compound 'A' with NaOH and Br2
The next part of the question involves the reaction of compound 'A' with NaOH and Br2. This reaction is known as the Hofmann bromamide reaction, where an amide reacts with bromine in the presence of a base to yield an amine with one less carbon atom.

**Hint:** Recall the characteristics of the Hofmann bromamide reaction and what it produces.

### Step 4: Determine the Product of the Hofmann Reaction
Since we established that compound 'A' is an amide (specifically, benzamide, C6H5CONH2), we can apply the Hofmann bromamide reaction. The reaction will remove one carbon atom from the amide, resulting in the formation of an amine.

1. The structure of benzamide is C6H5CONH2.
2. Upon reaction with NaOH and Br2, the carbonyl carbon is lost, resulting in the formation of aniline (C6H5NH2).

**Hint:** Focus on how the Hofmann reaction modifies the carbon chain of the amide.

### Step 5: Write the Final Reaction Equation
The final products of the reaction of compound 'A' (benzamide) with NaOH and Br2 will be:
- Aniline (C6H5NH2)
- By-products such as sodium carbonate (Na2CO3) and sodium bromide (NaBr).

**Hint:** Ensure to balance the reaction and identify all products formed.

### Conclusion
The reaction of compound 'A' with NaOH and Br2 yields aniline as the main product.

**Final Answer:** Aniline (C6H5NH2) is the product formed when compound 'A' reacts with NaOH and Br2.