A compound (A , `B_3N_3H_3Cl_3`) reacts with `LiBH_4` to form inorganic benzene (B). (A) reacts with (C) to form `B_3N_3H_3(CH_3)_3`. (B) and (C) are respectively

Options

A. Boron nitride, MeMgBR
B. Boron nitride, MeBr
C. Borazine, MeBr
D. Borazine, MeMgBr

Answer

To solve the problem, we need to identify the compounds A, B, and C based on the reactions described. Let's break it down step by step.

### Step 1: Identify Compound A
The compound A is given as `B3N3H3Cl3`. This compound is known as trichloroborazine.

### Step 2: Reaction of A with LiBH4
When compound A (trichloroborazine) reacts with lithium borohydride (`LiBH4`), it produces borazine, which is represented as `B3N3H6`. The reaction can be summarized as follows:
\[ \text{B}_3\text{N}_3\text{H}_3\text{Cl}_3 + \text{LiBH}_4 \rightarrow \text{B}_3\text{N}_3\text{H}_6 + \text{LiCl} + \text{BCl} \]

Thus, compound B is `B3N3H6`, which is borazine, also referred to as inorganic benzene.

### Step 3: Identify Compound C
Next, we need to determine what compound C is. The problem states that compound A reacts with compound C to form `B3N3H3(CH3)3`, which is trimethylborazine.

### Step 4: Determine the Nature of Compound C
To form trimethylborazine from trichloroborazine, we can consider the use of a Grignard reagent. The Grignard reagent that would be suitable for this reaction is methylmagnesium bromide (`MeMgBr`).

The reaction can be represented as:
\[ \text{B}_3\text{N}_3\text{H}_3\text{Cl}_3 + 3 \text{MeMgBr} \rightarrow \text{B}_3\text{N}_3\text{H}_3(\text{CH}_3)_3 + 3 \text{MgBrCl} \]

### Conclusion
Therefore, the final answer is:
- Compound B is borazine (`B3N3H6`).
- Compound C is the Grignard reagent (`MeMgBr`).

### Final Answer
B (borazine) and C (Grignard reagent).

---