Question

In the reaction: $2X + {B_2}{H_6} \to {\left[ {B{H_2}{{(X)}_2}} \right]^ + }{\left[ {B{H_4}} \right]^ - }$, the amine(s) X is (are):
[This question has multiple correct options]
(A) $N{H_3}$
(B) $C{H_3}N{H_2}$
(C) ${(C{H_3})_2}NH$
(D) ${(C{H_3})_3}N$

Answer 1

Diborane is electron deficient and so is a lewis acid. So it can react with a lewis base easily. But if the lewis base does not have steric strain then the diborane molecule will undergo asymmetric cleavage and if there is large steric strain then the cleavage will be symmetric.

Complete answer:
-First of all let us see what ${B_2}{H_6}$ is.
${B_2}{H_6}$ is known as diborane and it consists of 4 hydrogen atoms and 2 borane atoms. This compound is unstable. The diborane is an electron deficient compound and hence acts as a lewis acid.
Being a lewis acid diborane has the ability to form additional products with lewis bases. Lewis bases are those compounds which have the ability to donate an unbounded pair of electrons. Hence from this discussion we can conclude that for any compound to react with diborane, the compound should have an unbounded pair of electrons to donate to diborane.
Also we should keep in mind that if the base is small in size then there will be no steric strain and unsymmetric cleavage will take place along with ions formation, but if the base is large in size then there will be large steric strain and symmetric cleavage takes place.
-So, now coming back to the question, we need to tell which all compounds given in the options can undergo the following reaction:
$2X + {B_2}{H_6} \to {\left[ {B{H_2}{{(X)}_2}} \right]^ + }{\left[ {B{H_4}} \right]^ - }$
For (A) $N{H_3}$: Ammonia has a lone pair of electrons in nitrogen and so can easily donate the electron pair and it is not bulky so there will be no steric strain. Hence, there will be unsymmetric cleavage as shown in the following reaction:
${B_2}{H_6} + N{H_3} \to {\left[ {B{H_2}{{(N{H_3})}_2}} \right]^ + }{\left[ {B{H_4}} \right]^ - }$
So, $N{H_3}$can be the amine asked in the question.
For (B) $C{H_3}N{H_2}$: Methylamine has a lone pair at nitrogen and since only one of its substituents is bulkyl ($- C{H_3}$) so there is no steric strain. Hence it can easily donate the electron pair and undergo unsymmetric cleavage according to the following reaction:
${B_2}{H_6} + C{H_3}N{H_2} \to {\left[ {B{H_2}{{(C{H_3}N{H_2})}_2}} \right]^ + }{\left[ {B{H_4}} \right]^ - }$
So, $C{H_3}N{H_2}$can be the amine asked in the question.
For (C) ${(C{H_3})_2}NH$: Dimethylamine also has a lone pair and is not completely bulky due to one hydrogen substituent, so there will be no steric strain. Hence it can also donate electron pair and undergo unsymmetric cleavage according to the following reaction:
${B_2}{H_6} + {(C{H_3})_2}NH \to {\left[ {B{H_2}{{({{(C{H_3})}_2}NH)}_2}} \right]^ + }{\left[ {B{H_4}} \right]^ - }$
So, ${(C{H_3})_2}NH$can be the amine asked in the question.
For (D) ${(C{H_3})_3}N$: Trimethylamine has a lone pair of electrons but is extremely bulkyl due to the 3 methyl groups ($- C{H_3}$). Hence there will be large steric strain and so the cleavage of diborane will be symmetric as shown in the reaction given below:
${B_2}{H_6} + 2N{(C{H_3})_3} \to \left[ {{{(C{H_3})}_3}{N^ + } - {B^ - }{H_3}} \right]$
Hence ${(C{H_3})_3}N$ cannot be the amine we are talking about in the question.
So, from the above discussion we can conclude that the correct options will be:
(A) $N{H_3}$
(B) $C{H_3}N{H_2}$
(C) ${(C{H_3})_2}NH$

Note:
Diborane is highly flammable and has the ability to release large amounts of energy when burnt in the presence of large amounts of energy. Because of this property of diborane it is used as a rocket propellant. It mostly acts as a reducing agent and can even be used as a doping agent while manufacturing semiconductor devices.