Question

Borazole is prepared by heating the product isolated by reacting

• A. boron with dinitrogen.
• B. diborane with ammonium nitrate.
• C. diborane with ammonia.
• D. boron with ammonia.

Hint:

Borazole ($\text{B}_3\text{N}_3\text{H}_6$) is structural and isoelectronic with benzene ($\text{C}_6\text{H}_6$). It is synthesized by the reaction of diborane ($\text{B}_2\text{H}_6$) with ammonia ($\text{NH}_3$). Another common laboratory synthesis involves reacting $\text{BCl}_3$ with $\text{NH}_4\text{Cl}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Borazole (also known as Borazine), with the formula \(B_3N_3H_6\), is often called "inorganic benzene" because it is isoelectronic and isostructural with benzene (\(C_6H_6\)).
Its preparation typically involves the reaction of boron hydrides with nitrogen-donating molecules like ammonia.
Step 2: Detailed Explanation:
Borazine is classically prepared using diborane (\(B_2H_6\)) and ammonia (\(NH_3\)):
1. Reaction: When diborane reacts with ammonia in a \(1:2\) molar ratio at low temperatures, it forms an addition product called diammoniate of diborane:
\[ B_2H_6 + 2NH_3 \rightarrow [BH_2(NH_3)_2]^+[BH_4]^- \]
2. Pyrolysis: When this "isolated product" is heated to higher temperatures (around \(200^\circ C\)), it decomposes and undergoes cyclization to yield borazine and hydrogen gas:
\[ 3[BH_2(NH_3)_2]^+[BH_4]^- \xrightarrow{\Delta} 2B_3N_3H_6 + 12H_2 \]
Alternatively, heating a mixture of diborane and ammonia directly in a specific molar ratio at high temperatures also yields borazine.
None of the other options (\(B+N_2\), \(B_2H_6+NH_4NO_3\), or \(B+NH_3\)) are standard or efficient laboratory routes to borazine.
Step 3: Final Answer:
Borazole is prepared from the reaction of diborane with ammonia. The correct option is (C).