Question

Borazine is sometimes called inorganic benzene. Which of the following reactions is expected to give this compound?
A. ${B_2}{H_6} + N{H_3}\xrightarrow[{excessN{H_3}}]{{low\,temp}}$

B. ${B_2}{H_6} + N{H_3}\xrightarrow[{excessN{H_3}}]{{High\,temp}}$

C. ${B_2}{H_6} + N{H_3}\xrightarrow[{ratio\,2N{H_3}:1{B_2}{H_6}}]{{high\,temp}}$
D. All of these

Answer 1

Hint: The borazine is known as the inorganic benzene which is prepared by the Stock and Pohlands method by reacting three moles of diborane with six moles of ammonia to give two moles of borazine and twelve moles of hydrogen.

Complete step by step answer:
Borazine is called inorganic benzene due to its isoelectronic and isostructural similarities with benzene. Borazine is a colorless liquid like benzene and gives an aromatic smell.
Borazine is an aromatic compound as it follows the 4n+2 rule and the bond length between boron and nitrogen is the same.
The chemical formula of borazine is ${(BH)_3}{(NH)_3}$ where three B-H units are present and three N-H units are present in an alternate manner.
Borazine can be prepared by the Stock and Pohlands method.
Alfred Stock and Erich Pohlands gave a method for the preparation of borazine and reported in 1926.
The reaction for the preparation of borazine is shown below.
$3{B_2}{H_6} + 6N{H_3}\xrightarrow[{ratio\,2N{H_3}:1{B_2}{H_6}}]{{high\,temp}}2{B_3}{N_3}{H_6} + 12{H_2}$

In this reaction, three moles of diborane reacts with six moles of ammonia to give two moles of borazine and twelve moles of hydrogen.
For the preparation of borazine, ammonia and diborane are taken in 2:1 ration and heated at high temperatures.

Therefore, the correct option is C .

Note: In the Stock and Pohlands method, the yield of borazine is in very less quantity because of the formation of many by-products. Borazine can also be prepared by heating $BC{l_3}$ and $N{H_4}Cl$ at $140^\circ C$ and also by heating mixture of $LiB{H_4}$ and $N{H_4}Cl$ at $230^\circ C$.