JEE MAIN - Chemistry (2020 - 9th January Evening Slot - No. 13)
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 MeBr
Diborane and MeMgBr
Boron nitride and MeBr
Borazine and MeMgBr
Explanation
B3N3H3Cl3(A) + LiBH4 $$ \to $$ B3N3H6 (B)
B = inorganic benzene or Borazine
B3N3H3Cl + MeMgBr(C) $$ \to $$ B3N3H3(CH3)3 + 3MgBrCl
B = inorganic benzene or Borazine
B3N3H3Cl + MeMgBr(C) $$ \to $$ B3N3H3(CH3)3 + 3MgBrCl
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