1. Identify compound (A)

The given compound is $H_3N_3B_3Cl_3$ This is trichloroborazine, a borazine ring in which the three boron atoms carry chlorine substituents.

Borazine itself is: $B_3N_3H_6$ called inorganic benzene.

2. Reaction of (A) with $LiBH_4$ in THF

$LiBH_4$ is a hydride donor and reduces the $B-Cl$ bonds to $B-H$ bonds.

So, $H_3N_3B_3Cl_3 \xrightarrow[THF]{LiBH_4} H_3N_3B_3H_3$ which is the same as $B_3N_3H_6$ This compound is borazine, known as inorganic benzene.

Hence, $B = \text{Borazine}$

3. Reaction of (A) to form $H_3N_3B_3(Me)_3$

We need to replace the three chlorine atoms attached to boron by methyl groups.

That means we need a reagent that can provide a nucleophilic methyl group, i.e. effectively $Me^-$. Such substitution on boron is done by Grignard reagent: $MeMgBr$

The reaction is: $H_3N_3B_3Cl_3 + 3MeMgBr \rightarrow H_3N_3B_3(Me)_3 + 3MgBrCl$

So, $C = MeMgBr$

4. Match with options

We found:

• $(B) =$ Borazine
• $(C) =$ MeMgBr

This corresponds to:

Option D: Borazine and MeMgBr

5. Comparison with stored correct answer

Stored correct answer = D

Our derived answer = D

So the answer agrees with the stored correct answer.