Identify the final product of the reaction $B C l_{3} + N H_{4} C l 140^{\circ} C - --- \to C_{6} H_{5} C l N a B H_{4} - --- \to$ ?

B_{3} N_{3} H_{6}

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B_{2} H_{6}

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N a B C l_{4}

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B N

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Solution

The correct option is A $B_{3} N_{3} H_{6}$
$3 B C l_{3} + 3 N H_{4} C l 140^{\circ} C - --- \to C_{6} H_{5} C l B_{3} N_{3} H_{3} C l_{3} N a B H_{4} - --- \to B_{3} N_{3} H_{6} borazine or inorganic benzene$

Option A is correct.

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