Question

Consider the Hofmann ammonolysis reaction. If $R=C{{H}_{3}}$ the correct order of basic character of 1 amine, 2 amine, 3 amine and $N{{H}_{3}}$ in aqueous medium is:
$R-X+\ddot{N}{{H}_{3}}\to HX+RN{{H}_{2}}\xrightarrow[373K]{R-X}HX+{{R}_{2}}NH\xrightarrow[373K]{R-X}{{R}_{3}}N+HX\xrightarrow[373K]{R-X}{{R}_{4}}{{N}^{+}}{{X}^{-}}$
(a)- ${{1}^{\circ }}>{{2}^{\circ }}>{{3}^{\circ }}>N{{H}_{3}}$
(b)- ${{2}^{\circ }}>{{3}^{\circ }}>{{1}^{\circ }}>N{{H}_{3}}$
(c)- ${{2}^{\circ }}>{{1}^{\circ }}>{{3}^{\circ }}>N{{H}_{3}}$
(d)- ${{2}^{\circ }}>{{1}^{\circ }}>N{{H}_{3}}>{{3}^{\circ }}$

Answer 1

Hint: The basicity order of the amine is based on the positive inductive effect, extent of hydrogen bonding with the water molecule, and steric effect of the alkyl group.

Complete step by step answer:
The basicity of an amine in an aqueous solution primarily depends upon the stability of the ammonium cation or the conjugate acid formed by accepting a proton from water. The stability of the ammonium cation depends upon a combination of the following three factors:
(i)- + I-Effect of the alkyl group
(ii)- Extend of H-bonding with water molecules
(iii)- Steric effects of the alkyl groups.

The stability of the ammonium cation due to H-bonding depends upon the number of H-atoms present on the N-atom. So, the ${{1}^{\circ }}$ will form three hydrogen bonding because of three H atoms, ${{2}^{\circ }}$ will for two hydrogen bonding because of two hydrogen atom, and ${{3}^{\circ }}$ will form one hydrogen bonding because of one hydrogen atom.
However, in the case of ammonium cation derived from tertiary amines, there is some steric repulsion to H-bonding, and hence the stability further decreases.
From the above discussion, we can conclude that it is a combination of +I-Effect, H-bonding, and steric effects are factors for determining the stability of ammonium cation. All these factors are favorable for ${{2}^{\circ }}$ amine and hence the ${{2}^{\circ }}$amine is the most stable and the strongest base. If the alkyl group is small like $C{{H}_{3}}$ then there is no steric hindrance to H-bonding, hence the stability due to hydrogen bonding predominates over the stability due to +I-Effect of the $C{{H}_{3}}$ group hence ${{1}^{\circ }}$ is a stronger base than the ${{3}^{\circ }}$.

Hence the stability order or basicity order is:
${{2}^{\circ }}>{{1}^{\circ }}>{{3}^{\circ }}>N{{H}_{3}}$
So, the correct answer is “Option C”.

Note: When the alkyl group is other than $C{{H}_{3}}$ like ${{C}_{2}}{{H}_{5}}$ due to steric effect, +I-Effect dominates over the H-bonding and the ${{3}^{\circ }}$ is more stable than ${{1}^{\circ }}$. The order will be:
${{2}^{\circ }}>{{3}^{\circ }}>{{1}^{\circ }}>N{{H}_{3}}$