Question

Under the conditions mentioned for each reaction, the reaction(s) that would give borazine (B₃N₃H₆) as the major product is/are:

• A. LiBH₄ + NH₄Cl   → 230 °C
• B. B₂H₆ + 2 NH₃   → 180 °C
• C. NaBH₄ + (NH₄)₂SO₄   → THF, 40 °C
• D. BCl₃ + NH₄Cl   → chlorobenzene, 135 °C

Hint:

Borazine forms under high temperature from precursors that can yield B–N bonds and release small molecules like Hsubscript{2} or HCl. Check for conditions that promote cyclization and dehydrogenation.

Answer 1

The Correct Option is A, B

Borazine (B₃N₃H₆) is an inorganic compound analogous to benzene, and it can be synthesized by reactions that generate B–N bonds from boron and nitrogen precursors under suitable thermal conditions.

• (A) LiBH₄ + NH₄Cl at high temperature (230 °C) yields borazine as the major product due to thermal decomposition and cyclization involving boron and nitrogen atoms.
• (B) B₂H₆ + NH₃ at 180 °C leads to the formation of borazine via an intermediate aminoborane species which polymerizes and cyclizes to form B₃N₃H₆.
• (C) The reaction of NaBH₄ and (NH₄)₂SO₄ at low temperature in THF forms only simple boron-nitrogen compounds like borohydride–ammonia adducts and not borazine.
• (D) BCl₃ with NH₄Cl in chlorobenzene gives B–N adducts or oligomers, but not borazine under the given conditions.

\[ \boxed{\text{Correct options: (A), (B)}} \]