The reaction of H3N3B3Cl3A with LiBH4 in tetrahydrofuran gives inorganic benzene (B). Further, the reaction of (A) with (C) leads to H3N3B3Me3.Compounds (B) and (C) respectively are:

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Borazine and MeMgBr

Boron nitride and MeBr

Borazine and MeBr

Diborane and MeMgBr

answer is A.

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Detailed Solution

2B3N3H3Cl3+6LiBH4→2B3N3H6+3B2H6+6LiCl B3N3H3Cl3+3CH3MgBr→B3N3H3CH33+3MgBrClBoth are nucleophilic displacement reactions

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