Question

The reaction of H₃N₃B₃Cl₃(A) with LiBH₄ in tetrahydrofuran gives inorganic benzene (B). Further, the reaction of (A) with (C) leads to H₃N₃B₃(Me)₃. Compounds (B) and (C) respectively, are ______.

Options

• Borazine and MeBr

• Diborane and MeMgBr

• Boron nitride and MeBr

• Borazine and MeMgBr

Answer 1

The reaction of H₃N₃B₃Cl₃(A) with LiBH₄ in tetrahydrofuran gives inorganic benzene (B). Further, the reaction of (A) with (C) leads to H₃N₃B₃(Me)₃. Compounds (B) and (C) respectively, are Borazine and MeMgBr.

Explanation:

\[\ce{\underset{(A)}{H3N3B3Cl3} + 3LiBH4 ->[T.H.F.] \underset{(B)}{B3N3H6} + 3LiCl + 3BH3}\]

\[\ce{\underset{(A)}{H3N3B3Cl3} + 3MeMgBr -> \underset{(C)}{H3N3B3(Me)3} + 3MgBrCl}\]