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By which of the following reaction is borazine prepared?

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$B_{2} H_{6} + {NH}_{3} (excess) \underset{High temperature}{\overset{(Ratio 2 {NH}_{3} : 1 B_{2} H_{6})}{⟶}}$

$B_{2} H_{6} + {NH}_{3} (excess) \overset{Low temperature}{⟶}$

None of the above

$B_{2} H_{6} + {NH}_{3} (excess) \overset{High temperature}{⟶}$

answer is C.

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Detailed Solution

Inorganic benzene(borazine ) is prepared by the reaction:

$3 B_{2} H_{6} + 6 {NH}_{3} \to 3 {[{BH}_{2} {({NH}_{3})}_{2}]}^{+} {[{BH}_{4}]}^{-} \overset{Heat}{⟶} 2 B_{3} N_{3} H_{6} + 12 H_{2}$

Hence, the correct option is C.

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