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

Chemistry ma’am · 4 steps

The three Cl in (A) sit on boron: swap them for H to get borazine, and for Me you need a carbanion.

1. Read compound A. H₃N₃B₃Cl₃ is B-trichloroborazine, a six-membered ring of alternating B and N. Each N carries an H and each B carries a Cl. The boron is electron-poor, so a nucleophile attacks it and pushes out Cl⁻.
2. Find B. LiBH₄ supplies hydride, H⁻, which replaces each Cl on boron. The product is B₃N₃H₆, borazine. It is called inorganic benzene because it is isoelectronic with benzene and has a similar ring structure.

   2B₃N₃H₃Cl₃ + 6LiBH₄ →(THF) 2B₃N₃H₆ + 6LiCl + 3B₂H₆

3. Find C. To put CH₃ on each boron, you need a reagent that delivers CH₃⁻. The Grignard reagent MeMgBr does this, giving H₃N₃B₃Me₃ along with MgBrCl.
4. So. B is borazine and C is MeMgBr, which is option (d).

The idea to keep. To replace a halogen on boron, pick a reagent that brings an anion: LiBH₄ or LiAlH₄ for H⁻, a Grignard or alkyllithium for R⁻. An alkyl halide brings no carbanion.

Answer from the NTA final answer key. Solution by PrepWiser.