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 = 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).