Question

If methyl is alkyl group, then which order of basicity is correct
A. ${{R}_{2}}NH$> \[RN{{H}_{2}}\] > ${{R}_{3}}N$ > $N{{H}_{3}}$
B. ${{R}_{2}}NH$> ${{R}_{3}}N$ > \[RN{{H}_{2}}\] > $N{{H}_{3}}$
C. \[RN{{H}_{2}}\] > $N{{H}_{3}}$ > ${{R}_{2}}NH$ > ${{R}_{3}}N$
D. $N{{H}_{3}}$ > \[RN{{H}_{2}}\] > ${{R}_{2}}NH$ > ${{R}_{3}}N$

Answer 1

Hint: The basicity character is directly proportional to the electron donating groups present in the compound i.e. higher the number of electron donating groups higher is the basicity but for small substituents like $C{{H}_{3}}$ the order vary in case of amines.

Complete Step by Step Solution:
Stronger should be the equivalent amine as a base if the substituted ammonium cation is more stable. As a result, contrary to the order based on the inductive effect, the basicity of aliphatic amines should be primary > secondary > tertiary.

Secondly, there is no steric barrier to H-bonding when the alkyl group is tiny, like the $C{{H}_{3}}$ group. In the event that the alkyl group is larger than the methyl group, H-bonding will be hampered sterically. As a result, a change in the alkyl group's nature, such as from $C{{H}_{3}}$ to ${{C}_{2}}{{H}_{5}}$, causes a change in the basic strength's order. As a result, the basic strength of alkyl amines in aqueous solutions is determined by the intricate interaction of the inductive effect, solvation effect, and steric hindrance of the alkyl group.

Thus the correct option is a.${{R}_{2}}NH$> \[RN{{H}_{2}}\] > ${{R}_{3}}N$ > $N{{H}_{3}}$.

Note:The important thing to note here is the exception that arises in the case of the $C{{H}_{3}}$ group due to its less hindrance power as compared to other groups like ${{C}_{2}}{{H}_{5}}$, usually the number of alkyl groups increase, there is an increase in the basicity but the $C{{H}_{3}}$ case is an exception.