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 :
Solution
B<sub>3</sub>N<sub>3</sub>H<sub>3</sub>Cl<sub>3</sub>(A) + LiBH<sub>4</sub> $\to$ B<sub>3</sub>N<sub>3</sub>H<sub>6</sub> (B)
<br><br>B = inorganic
benzene or Borazine
<br><br>B<sub>3</sub>N<sub>3</sub>H<sub>3</sub>Cl +
MeMgBr(C) $\to$ B<sub>3</sub>N<sub>3</sub>H<sub>3</sub>(CH<sub>3</sub>)<sub>3</sub> + 3MgBrCl
About this question
Subject: Chemistry · Chapter: p-Block Elements · Topic: Group 13: Boron Family
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