Question

An alkali metal hydride (NaH) reacts with diborane in ‘A’ to give a tetrahedral compound ‘B’ which is extensively used as a reducing agent in organic synthesis. The compounds ‘A’ and ‘B’ respectively are:
A. $\text{C}{{\text{H}}_{\text{3}}}\text{COC}{{\text{H}}_{\text{3}}}$ and ${{\text{B}}_{\text{3}}}{{\text{N}}_{\text{3}}}{{\text{H}}_{\text{6}}}$
B. ${{\left( {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}} \right)}_{\text{2}}}\text{O}$ and $\text{NaB}{{\text{H}}_{\text{4}}}$
C. ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$ and ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{Na}$
D. ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$ and $\text{NaB}{{\text{H}}_{\text{4}}}$

Answer 1

The above reaction has metal hydride and diborane, they should be allowed to react in a nonpolar medium to reduce the possibility of side reactions as polar and protic solvents might give us unwanted products. So, a non-polar solvent must be chosen to carry out this reaction.

Complete Solution :
The chemical formula of diborane is ${{\text{B}}_{\text{2}}}{{\text{H}}_{\text{6}}}$ . On reacting with NaH, it forms $\text{NaB}{{\text{H}}_{\text{4}}}$.
The reaction should take place in a nonpolar medium. $\text{NaB}{{\text{H}}_{\text{4}}}$ is soluble in ethers. Hence, the solvent used is diethyl-ether.
- Thus, when an alkali metal hydride ($\text{NaH}$) react with di-borane (${{\text{B}}_{\text{2}}}{{\text{H}}_{\text{6}}}$​) in the presence of diethyl-ether (${{\left( {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}} \right)}_{\text{2}}}\text{O}$), a tetrahedral compound which is sodium tetrahydroborate or sodium borohydride ($\text{NaB}{{\text{H}}_{\text{4}}}$), is formed which acts as a good reducing agent in organic synthesis.
- $\text{NaB}{{\text{H}}_{\text{4}}}$ is comparatively less reactive than Lithium aluminium hydride,$\text{LiAl}{{\text{H}}_{\text{4}}}$,but is otherwise similar. It is only powerful enough to reduce aldehydes, ketones and acid chlorides to alcohols and esters, amides, acids and nitriles are largely unreactive. It can also behave as a nucleophile toward halides and epoxides in given situations.
$2NaH+{{B}_{2}}{{H}_{6}}\xrightarrow{{{({{C}_{2}}{{H}_{5}})}_{2}}O}\underset{sodium~borohydride}{\mathop{2NaB{{H}_{4}}}}\,$
Thus, A is ${{\left( {{C}_{2}}{{H}_{5}} \right)}_{2}}O$ and B is $NaB{{H}_{4}}$​.
So, the correct answer is “Option B”.

Note: Diborane is a chemical compound consisting of boron and hydrogen with the formula ${{B}_{2}}{{H}_{6}}$. It is a colourless and pyrophoric gas with a repulsively sweet odour. Diborane is a key boron compound with a variety of applications.
- Reducing agent is defined as an element or compound that loses an electron to an electron recipient participating in a redox chemical reaction. A reducing agent is hence oxidized when it loses electrons in the redox reaction. Reducing agents reduce oxidizing agents.