Identify X and Y in the following reaction.
$B C l_{3} + N H_{4} C l 140^{O} C - --- \to C_{6} H_{5} C l X N a B H_{4} - --- \to Y$

X = N a B O_{2}, Y = B_{2} O_{3}

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X = N a_{2} B_{4} O_{7}, Y = H_{3} B O_{3}

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X = B N, Y = [N H_{4}]^{+} [B C l_{4}]^{-}

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X = B_{3} N_{3} H_{3} C l_{3}, Y = B_{3} N_{3} H_{6}

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Solution

The correct option is D $X = B_{3} N_{3} H_{3} C l_{3}, Y = B_{3} N_{3} H_{6}$
Answer:- (D) $X = B_{3} N_{3} H_{3} {C l}_{3}, Y = B_{3} N_{3} H_{6}$
Ammonium chloride on reaction with boron chloride at a temperature of $140 - 150 ° C, (c a t a l y s t - C_{6} H_{6}, C_{6} H_{5} C l)$ produces trichloridoborazine.
$3 {B C l}_{3} + 3 N H_{4} C l 140 ° C - --- \to C_{6} H_{5} C l B_{3} N_{3} H_{3} {C l}_{3} + 9 H C l ↑$
$∴ X = B_{3} N_{3} H_{3} {C l}_{3}$
Trichloridoborazine on further reaction with sodium borohydride produces borazine.
$2 B_{3} N_{3} H_{3} {C l}_{3} + 6 N a [B H_{4}] ⟶ 2 B_{3} N_{3} H_{6} + 6 N a C l + 3 B_{2} H_{6}$
$∴ Y = B_{3} N_{3} H_{6}$

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