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The Cannizaro Reaction

Compound X on...

Question

Compound X on reduction with $L i A l H_{4}$ gives a hydride Y containing 21.72% hydrogen and other products. The compound Y reacts explosively with air resulting in the formation of Boron Trioxide. What are X and Y ,respectively?

B C l_{3}, B_{2} H_{6}

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P C l_{3}, B_{2} H_{6}

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B_{2} H_{6}, B C l_{3}

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L i A l H_{4}, P C l_{3}

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Solution

The correct option is A $B C l_{3}, B_{2} H_{6}$
The reaction scheme is
From the reaction scheme and the properties, it is clear that the compound Y is boron hydride. The boron hydride that contains 21.72% hydrogen is $B_{2} H_{6} .$ So, Y is $B_{2} H_{6}$ (diborane). $B_{2} H_{6}$ is obtained by the reduction of $B C l_{3}$ by $L i A l H_{4} .$ So, X is $B C l_{3}$ .
Reactions : $4 B C l_{3} + 3 L I A l H_{4} \to 3 L i C l + 2 B_{2} H_{6} + 3 A l C l_{3}$
$B_{2} H_{6} + 3 O \to B_{2} O_{3} + 3 H_{2} O$

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