Question

From ${B_2}{H_6}$ all of the following can be prepared except:
A.${B_2}{O_3}$
B.${H_3}B{O_3}$
C.${B_2}{(C{H_3})_6}$
D.$NaB{H_4}$

Answer 1

Diborane or ${B_2}{H_6}$ is a colorless gas that gives off an offensive odor and is toxic upon inhalation. It is a very important organic compound and can be used in the synthesis of a wide variety of compounds. To answer this question we have to find out which one of these cannot be prepared using diborane by the process of elimination.

Complete answer:
Diborane can be treated with oxygen or burned in the [presence of air to obtain boric anhydride. This is represented by ${B_2}{O_3}$ and the reaction is as follows:
${B_2}{H_6} + 3{O_2} \to {B_2}{O_3} + 3{H_2}O + Heat$
Diborane can be reacted with water to form boric acid which is ${H_3}B{O_3}$ . The reaction is represented by:
${B_2}{H_6} + 6{H_2}O \to {H_3}B{O_3} + 6{H_2}$
Sodium borohydride or $NaB{H_4}$ can be prepared from diborane by treating it with sodium hydride. This will give us the $NaB{H_4}$ . This reaction proceeds as follows:
${B_2}{H_6} + 2NaH \to NaB{H_4}$
Since there are only 4 hydrogens in ${B_2}{H_6}$ that can be replaced or reacted with an alkyl group that is present on the given plane, therefore 4 methyl groups can only be substituted. But in the option given there are 6 methyl groups substituted. This is wrong.
So the correct answer is option (C).

Note:
In this question, we could easily find out the answer at the first glance by saying that option C is never possible because such a compound is unstable as only the terminal hydrogens can be replaced in diborane. So the maximum number of methyl groups that can be substituted is 4 and thus option C is impossible.