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Diborane

Which of the ...

Question

Which of the following product(s) may be formed by the reaction of diborane with ammonia?

A

(B N)_{x}

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B

3 B_{2} H_{6} \cdot 2 N H_{3}

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C

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

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D

B_{2} H_{6} \cdot N H_{3}

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Solution

The correct option is C $B_{3} N_{3} H_{6}$
$B_{2} H_{6} + N H_{3} (excess) low temperature - -------- \to 3 B_{2} H_{6} .2 N H_{3} o r [H_{2} B (N H_{3})_{2}]^{+} [B H_{4}]^{-}$

$B_{2} H_{6} + N H_{3} (excess) high temperature - ----------------- \to Higher temperature (> 200^{o} C) (B N)_{X} boron nitride$

$B_{2} H_{6} + N H_{3} high temperature(200^oC) - --------------- \to r a t i o 2 N H_{3} : 1 B_{2} H_{6} B_{3} N_{3} H_{6} borazine$
Hence option option A, B and C are correct.
THEORY
Properties :
1. $B_{2} H_{6}$ is a colourless gas and highly toxic gas with boiling point 180 K.

2. Diborane catches fire spontaneously upon exposure to air.
$B_{2} H_{6} + 3 O_{2} ⟶ B_{2} O_{3} + 3 H_{2} O$
$Δ_{c} H^{o} = - 1976 k J m o l^{- 1}$
It burns in oxygen releasing an enormous amount of energy. Most of the higher boranes are also spontaneously flammable in air.

3. Reacts with water to give boric acid
$B_{2} H_{6} + 6 H_{2} O ⟶ 2 B (O H)_{3} + 6 H_{2}$
Diborane is also hydrolysed by weaker acids (e.g. alcohols) or aqueous alkali
$B_{2} H_{6} + 6 R O H ⟶ 2 B (O R)_{3} + 6 H_{2}$
$B_{2} H_{6} + 2 K O H + 2 H_{2} O ⟶ 2 K B O_{2} + 6 H_{2}$

4. Diborane undergoes cleavage reactions with Lewis bases(L) to give borane adducts, $B H_{3} \cdot L$
$B_{2} H_{6} + 2 N M e_{3} ⟶ 2 B H_{3} . N M e_{3}$
$B_{2} H_{6} + 2 C O ⟶ 2 B H_{3} . C O$
4. Reaction with ammonia
$3 B_{2} H_{6} + 6 N H_{3} ⟶ 6 B H_{3} . N H_{3}$
$3 [H_{2} B (N H_{3})_{2}]^{+} [B H_{4}]^{-} ↓ 2 B_{3} N_{3} H_{6} + 12 H_{2} (Inorganic benzene$

5. Reaction with ammonia
$B_{2} H_{6} + N H_{3} (excess) low temperature - -------- \to B_{2} H_{6} .2 N H_{3} o r [H_{2} B (N H_{3})_{2}]^{+} [B H_{4}]^{-}$

$B_{2} H_{6} + N H_{3} (excess) high temperature - ----------------- \to Higher temperature (> 200^{o} C) (B N)_{X} boron nitride$

$B_{2} H_{6} + N H_{3} high temperature(200^oC) - --------------- \to r a t i o 2 N H_{3} : 1 B_{2} H_{6} B_{3} N_{3} H_{6} borazine$

6. Reduction of diborane can be accomplished with Na or sodium borohydride
$2 B_{2} H_{6} + 2 N a ⟶ N a B H_{4} + N a B_{3} H_{8}$
$B_{2} H_{6} + N a B H_{4} ⟶ N a B_{3} H_{8} + H_{2}$

7. Li and Na react with $B_{2} H_{6}$ to give borohydrides.
$B_{2} H_{6} + 2 M H Dimethyl ether ⟶ 2 M^{+} [B H_{4}]^{-}$
M=Li, Na

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