Question

Arrange the following in the increasing order of their pKₓ value in aqueous phase:
C₆H₅NH₂, (CH₃)₂NH, NH₃, CH₃NH₂, (CH₃)₃N

Hint:

Basicity increases with electron-donating groups like methyl groups and decreases with electron-withdrawing groups like phenyl groups.

Answer 1

To compare pKₓ values, we consider the basicity of the compounds. A higher pKₓ indicates a weaker base.
- NH₃ has the lowest pKₓ because it is a primary amine and has a high basicity.
- CH₃NH₂ (methylamine) is more basic than NH₃.
- (CH₃)₂NH (dimethylamine) is more basic than CH₃NH₂.
- (CH₃)₃N (trimethylamine) is more basic than (CH₃)₂NH.
- C₆H₅NH₂ (aniline) has the highest pKₓ due to the electron-withdrawing effect of the phenyl group, making it the weakest base.
Thus, the increasing order of pKₓ is: \[ \text{C}_6\text{H}_5\text{NH}_2<\text{NH}_3<\text{CH}_3\text{NH}_2<(\text{CH}_3)_2\text{NH}<(\text{CH}_3)_3\text{N} \]