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The reaction of \mathrm{H_{3}N_{3}B_{3}Cl_{3}} (A) with \mathrm{LiBH_{4}} in tetrahydrofuran gives inorganic benzene (B). Further, the reaction of (A) with (C) leads to \mathrm{H_{3}N_{3}B_{3}(Me)_{3}}. Compounds (B) and (C) respectively, are :

Option: 1

Diborane and \mathrm{M e M g B r}


Option: 2

Borazine and \mathrm{MeBr}


Option: 3

Borazine and \mathrm{MeMgBr}


Option: 4

Boron nitride and \mathrm{MeBr}


Answers (1)

best_answer

The reactions are given as :

\mathrm{\underset{(A)}{B_{3}N_{3}H_{3}Cl_{3}}\: +\: LiBH_{4}\rightarrow \underset{(B)}{B_{3}N_{3}H_{6}}\: +LiCl+BCl_{3}}

\mathrm{\underset{(A)}{B_{3}N_{3}H_{3}Cl_{3}}\: +\: 3\underset{(C)}{MeMgBr}\rightarrow B_{3}N_{3}H_{3}(CH_{3})_{3}\: +3 MgBrCl}

Hence, the option number (3) is correct.

Posted by

jitender.kumar

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