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Diborane

In the reacti...

Question

In the reaction, $2 X + B_{2} H_{6} \to [B H_{2} (X)_{2}]^{+} [B H_{4}]^{-}$ the amine(s) X is/are

A

N H_{3}

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B

C H_{3} N H_{2}

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C

(C H_{3})_{2} N H

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D

(C H_{3})_{3} N

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Solution

The correct option is C $(C H_{3})_{2} N H$
Diborane $(B_{2} H_{6})$ undergoes unsymmetric cleavage with Small amines like $N H_{3}$ , primary and secondary amine while tertiary amine brings about symmetrical cleavage of $B_{2} H_{6}$ as:

Different cleavage as per given option
$B_{2} H_{6} + 2 N H_{3} \to [H_{2} B (N H_{3})_{2}]^{+} [B H_{4}]^{-}$
$B_{2} H_{6} + 2 C H_{3} N H_{2} \to [H_{2} B (C H_{3} N H_{2})_{2}]^{+} [B H_{4}]^{-}$
$B_{2} H_{6} + 2 (C H_{3})_{2} N H \to [H_{2} B ((C H_{3})_{2} N H)_{2}]^{+} [B H_{4}]^{-}$
Large amines such as $(C H_{3})_{3} N$ gives symmetrical cleavage of diborane
$B_{2} H_{6} + 2 N (C H_{3})_{3} \to 2 H_{3} B \leftarrow N (C H_{3})_{3}$
Hence the X is a, b and c.

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