First slide

Diborane

Question

An alkali metal hydride (NaH) reacts with diborane in ‘A’ to give a tetrahedral compound ‘B’ which is extensively used as reducing agent in organic synthesis. The compounds ‘A’ and ‘B’ respectively are

Easy

A

CH₃COCH₃ and B₃N₃H₆
B

(C₂H₅)₂O and NaBH₄
C

C₂H₆ and C₂H₅Na
D

C₆H₆ and NaBH₄

Solution

$2NaH + \mathop {{B_2}{H_6}}\limits_{Diborane} \xrightarrow[A]{{Diethyl\,\,Ether}}\mathop {2NaB{H_4}}\limits_{Tetra\,\,hedral\,\,complex} \approx 2N{a^ + }{[B{H_4}]^ - }$
Diborane reacts with alkali metal hydride (NaH) in dry ether solvent to give sodium tetra hydrido borate.
→ sodium tetra hydrido borate is used as strong reducing agent in organic synthesis.

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