Question

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

Answer 1

We have to know that smaller amines on reaction with diborane gives diammonium salt but larger amines produce adduct.

Complete step by step answer:
We know that amines are basic chemical derivatives of ammonia in which the attached hydrogens are replaced by alkyl or aryl groups. We can classify amines into,
Primary amine: One organic group attached to nitrogen atoms.
Secondary amine: Two organic groups bonded to nitrogen atoms.
Tertiary amine: Three organic groups bonded to nitrogen atoms.
An example of primary amine is $C{H_3}N{H_2}$.
An example of secondary amine is ${\left( {C{H_3}} \right)_2}NH$.
An example of tertiary amine is ${\left( {C{H_3}} \right)_3}N$.
We can call $N{H_3}$, $C{H_3}N{H_2}$, ${\left( {C{H_3}} \right)_2}NH$ as smaller amines, whereas ${\left( {C{H_3}} \right)_3}N$ is a tertiary amine.
When smaller amines like $N{H_3}$, $C{H_3}N{H_2}$, ${\left( {C{H_3}} \right)_2}NH$ are reacted with diborane, an unsymmetrical cleavage of diborane occurs and the product formed will be an ionic compound. We can write the general reaction as,

.
Whereas, when larger amines like ${\left( {C{H_3}} \right)_3}N$ are reacted with diborane, asymmetric cleavage of diborane occurs and adduct is formed as product. We can write the general reaction as,

When $N{H_3}$ is reacted with diborane, the product formed would be ${\left[ {B{H_2}{{\left( {N{H_3}} \right)}_2}} \right]^ + }{\left[ {B{H_4}} \right]^ - }$. We can write the reaction as,
$2N{H_3} + {B_2}{H_6} \to {\left[ {B{H_2}{{\left( {N{H_3}} \right)}_2}} \right]^ + }B{H_4}^ -$
Therefore, the option (A) is correct.
When $C{H_3}N{H_2}$ is reacted with diborane, the product formed would be ${\left[ {B{H_2}{{\left( {C{H_3}N{H_2}} \right)}_2}} \right]^ + }{\left[ {B{H_4}} \right]^ - }$. We can write the reaction as,
$2C{H_3}N{H_2} + {B_2}{H_6} \to {\left[ {B{H_2}{{\left( {C{H_3}N{H_2}} \right)}_2}} \right]^ + }B{H_4}^ -$
Therefore, the option (B) is correct.
When ${\left( {C{H_3}} \right)_2}NH$ is reacted with diborane, the product formed would be ${\left[ {B{H_2}{{\left( {{{\left( {C{H_3}} \right)}_2}NH} \right)}_2}} \right]^ + }{\left[ {B{H_4}} \right]^ - }$. We can write the reaction as,
$2{\left( {C{H_3}} \right)_2}NH + {B_2}{H_6} \to {\left[ {B{H_2}{{\left( {{{\left( {C{H_3}} \right)}_2}NH} \right)}_2}} \right]^ + }{\left[ {B{H_4}} \right]^ - }$
Therefore, the option (C) is correct.
When ${\left( {C{H_3}} \right)_3}N$ is reacted with a diborane, the product formed would be adducted. We can write the reaction as,
$2{\left( {C{H_3}} \right)_3}N + {B_2}{H_6} \to 2\left[ {{{\left( {C{H_3}} \right)}_3}N \to B{H_3}} \right]$
Therefore, the option (D) is incorrect.

So, the correct answer is Option A,B,C.

Note:
As we know that primary amines have the highest melting points as they have polar nitrogen-hydrogen bonds present in them.
Secondary amines are said to have lower melting points than primary amines because the $N - H$ bond of a secondary amine is less polar than that of primary amine, the dipole-dipole attractions between the secondary amine molecules are lower. This is reflected in the lower melting points.
Tertiary amines have the least melting point because they do not have a $N - H$ bond, they cannot form any intermolecular hydrogen bonds with other tertiary amines, and this contributes to the least melting point of tertiary amine.
Diborane can be used as propellant in a rocket, used in production of borophosphosilicate, used as a reducing agent. It could act as catalyst and as rubber vulcanizer in polymerization reactions. It is used as a doping agent in manufacturing semiconductor devices.