Question

Reaction of BCl$_3$ with NH$_4$Cl at 140 °C produces compound P. Further, P reacts with NaBH$_4$ to give a colorless liquid Q. The reaction of Q with H$_2$O at 100 °C produces compound R and a diatomic gas S. Among the following, the CORRECT statement is

• A. P is B$_3$N$_3$H$_6$
• B. R is [B(OH)NH]$_3$
• C. Q is [BCINH]$_3$
• D. S is Cl$_2$

Hint:

Borazine (B$_3$N$_3$H$_6$) is called inorganic benzene. It is formed by heating BCl$_3$ and NH$_4$Cl and reacts with NaBH$_4$ to give borane derivatives.

Answer 1

The Correct Option is B

Step 1: Formation of compound P.
Boron trichloride reacts with ammonium chloride at high temperature (140°C) to form **borazine (B$_3$N$_3$H$_6$)**, also called inorganic benzene. \[ 3 \text{BCl}_3 + 3 \text{NH}_4\text{Cl} \rightarrow \text{B}_3\text{N}_3\text{H}_6 + 9 \text{HCl} \] Step 2: Reaction with NaBH$_4$.
Borazine reacts with sodium borohydride to form a colorless liquid borane derivative (Q), typically **borazine hydride**.
Step 3: Hydrolysis of Q.
On hydrolysis, borazine hydride produces **boric acid** and hydrogen gas. \[ \text{Q} + \text{H}_2\text{O} \rightarrow \text{R (boric acid)} + \text{H}_2 \] Step 4: Conclusion.
Hence, compound P is B$_3$N$_3$H$_6$ (borazine).