`overset(NaBH_(4))(leftarrow)CH_(3)N_(3)overset(LAH)(rarr)(A)`
the products (A) and (B) are:

Options

A. a.`CH_(3)NH_(2), CH_(3)NH_(2)`
B. b.`CH_(3)NH_(2)`, No reaction
C. c.No reaction, `CH_(3)NH_(2)`
D. d.No reaction, No reaction

Explanation

To solve the problem, we need to analyze the reactions of the azide compound (CH₃N₃) with two different reducing agents: sodium borohydride (NaBH₄) and lithium aluminum hydride (LiAlH₄).

### Step-by-Step Solution:

1. **Identify the Reactants:**
- The azide compound is CH₃N₃ (methyl azide).
- The two reducing agents are NaBH₄ and LiAlH₄.

2. **Understanding the Reducing Agents:**
- **Sodium Borohydride (NaBH₄):** A mild reducing agent that can reduce azides to primary amines.
- **Lithium Aluminum Hydride (LiAlH₄):** A stronger reducing agent that can also reduce azides to primary amines.

3. **Reaction with Sodium Borohydride (NaBH₄):**
- When CH₃N₃ reacts with NaBH₄, the azide group (N₃) is reduced to an amine group (NH₂).
- The reaction can be summarized as:
\[ \text{CH}_3\text{N}_3 + \text{NaBH}_4 \rightarrow \text{CH}_3\text{NH}_2 + \text{by-products} \]
- The product formed (A) is methylamine (CH₃NH₂).

4. **Reaction with Lithium Aluminum Hydride (LiAlH₄):**
- Similarly, when CH₃N₃ reacts with LiAlH₄, the azide group is also reduced to an amine group (NH₂).
- The reaction can be summarized as:
\[ \text{CH}_3\text{N}_3 + \text{LiAlH}_4 \rightarrow \text{CH}_3\text{NH}_2 + \text{by-products} \]
- The product formed (B) is also methylamine (CH₃NH₂).

5. **Final Products:**
- Both reactions yield the same product:
- Product (A) from NaBH₄: CH₃NH₂ (methylamine)
- Product (B) from LiAlH₄: CH₃NH₂ (methylamine)

### Conclusion:
The products (A) and (B) from the reactions are both methylamine (CH₃NH₂).