The reaction of H3N3B3Cl3 (A) with LiBH4 in tetrahydrofuran gives inorganic benzene (B). Furthur, the reaction of (A) with (C) leads to H3N3B3(Me)3. Compounds (B) and (C) respectively, are:
A
Boron nitride, MeBr
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B
Diborane, MeMgBr
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C
Borazine, MeBr
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D
Borazine, MeMgBr
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Solution
The correct option is D Borazine, MeMgBr B3N3H3Cl3+LiBH4→B3N3H6+LiCl+BCl3 B3N3H3Cl3+3CH3MgBr→B3N3H3(CH3)3+3MgBrCl