Question

Arrange the following:

In decreasing order of the pK_b values:

C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂NH and C₆H₅NH₂

Answer 1

In C₂H₅NH₂, only one −C₂H₅ group is present while in (C₂H₅)₂NH, two −C₂H₅ groups are present. Thus, the +I effect is more in (C₂H₅)₂NH than in C₂H₅NH₂. Therefore, the electron density over the N-atom is more in (C₂H₅)₂NH than in C₂H₅NH₂. Hence, (C₂H₅)₂NH is more basic than C₂H₅NH₂.

Also, both C₆H₅NHCH₃ and C₆H₅NH₂ are less basic than (C₂H₅)₂NH and C₂H₅NH₂ due to the delocalization of the lone pair in the former two. Further, among C₆H₅NHCH₃ and C₆H₅NH₂, the former will be more basic due to the +I effect of −CH₃ group. Hence, the order of increasing basicity of the given compounds is as follows:

C₆H₅NH₂ < C₆H₅NHCH₃ < C₂H₅NH₂ < (C₂H₅)₂NH

We know that the higher the basic strength, the lower the pK_b values.

C₆H₅NH₂ > C₆H₅NHCH₃ > C₂H₅NH₂ > (C₂H₅)₂NH