The reaction of BCl₃, with NH₄Cl gives product A which upon reduction by NaBH₄ gives product B. Product B… · ABC26IN1286

Subject: Inorganic Chemistry · Chapter: Main Group Elements · Topic: Main Group Elements – General · Exam: CSIR NET DEC 2015 · Marks: 2 · Difficulty: Medium

The reaction of $\mathrm{BCl_3}$, with $\mathrm{NH_4Cl}$ gives product A which upon reduction by $\mathrm{NaBH_4}$ gives product B. Product B upon reacting with HCl affords compound C, which is
(a)$\mathrm{Cl}_{3} \mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{9}$
(b)$[\mathrm{ClBNH}]_{3}$
(c)$[\mathrm{HBNH}]_{3}$
(d)$(\mathrm{ClH})_{3} \mathrm{B}_{3} \mathrm{N}_{3}(\mathrm{ClH})_{3}$
Answer
Answer: A ✓ checked by 4AB · confidence high
Explanation
A = B-trichloroborazine [ClBNH]₃; B = borazine [HBNH]₃; borazine adds three HCl across its B–N bonds (Cl to B, H to N) to give the cyclic adduct B₃N₃H₉Cl₃ (Cl₃B₃N₃H₉).

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