Question

All the products formed in the oxidation of $NaB{H_4}$ by ${I_2}$ are:
A.${B_2}{H_6}{\text{ }} and {\text{ }}NaI$
B.${B_2}{H_6}{\text{,}}{{\text{H}}_2}{\text{ }} and {\text{ }}NaI$
C.$B{{\text{I}}_3}{\text{ }} and {\text{ }}NaH$
D.$NaB{I_4}{\text{ }} and {\text{ H}}I$

Answer 1

Hint:Iodine(${I_2}$) is a very good reducing agent and it is always ready to give one electron and it reacts vigorously with reducing materials. Whereas Sodium Borohydride(\[NaB{H_4}\]) is a selective reducing agent.

Complete step by step answer:
The reaction of Sodium Borohydride ($NaB{H_4}$) reacts with Iodine (${I_2}$) in presence of a solvent known as Diglyme first yields Borane, sodium iodide, hydrogen iodide. The hydrogen iodide again reacts with sodium borohydride to yield borane and Sodium iodide. So the fine products obtained in the process are Sodium iodide, Diborane, and Hydrogen. The chemical reaction is written as:
\[2NaB{H_4}(s) + {I_2}(s) \to {B_2}{H_6}(g) + 2NaI(s) + {H_2}(g)\]
The diborane generated in the process can be trapped to form borane Lewis base complexes and that can be used in organic transformations. In the above reaction, Borane is primarily responsible for reducing acid to alcohol.

Additional Information: Sodium borohydride is not very sensitive to moisture and is very stable if stored in ambient conditions and it reacts with steam or water to produce hydrogen. It is not toxic in nature. Iodine, though very stable in ambient conditions, is highly dangerous when heated as it emits toxic fumes of iodine.


Note:
The combination of sodium borohydride ($NaB{H_4}$) and Iodine (${I_2}$) is widely used in the reduction of amino acid to amino alcohols as it is an inexpensive and convenient method and provides a very good yield of amino alcohol. Sodium borohydride is a selective reducing agent that reduces aldehydes, ketones, and oximes, but the functional groups such as esters, amines, and halides are not reduced under ambient conditions.