Arrange the following:
(i) In decreasing order of the pKb values:C2H5NH2,C6H5NHCH3,(C2H5)2NH,C6H5NH2
(ii) In increasing order of basic strength: C6H5NH2,C6H5N(CH3)2,(C2H5)2NH,CH3NH2
Arrange the following:
(i) In decreasing order of the pKb values:C2H5NH2,C6H5NHCH3,(C2H5)2NH,C6H5NH2
(ii) In increasing order of basic strength: C6H5NH2,C6H5N(CH3)2,(C2H5)2NH,CH3NH2

(i) C2 HNH consists of one –CH5 group.

(C2 H5 )2NH consists of two –C2 H5 groups.

As a result, the +I effect in ( C2H5)2NH is greater than that in C2H5NH2 and ( C2H5)2NH has a higher electron density over the N-atom than C2H5NH2. This implies that (C2H5)2NH has higher basicity than C2H5NH2.

Due to the delocalization of the lone pair in the former two, both C6H5NHCH3 and C6H5NH2 are less basic than ( C2H5)2NH and C2H5NH2. Furthermore, due to the +I effect of the –CH3 group, the former will be more basic than the latter. Thus, the following is the sequence in which the basicity of the supplied compounds increases:

We know that the higher the basic strength, the lower is the p Kb values.

(ii) The presence of the + I impact of two –CH3 groups in C6H5N(CH3)2 makes it more basic than C6H5NH2. In addition, CH3NH2 has one CH3 group, whereas (C2H5 )2NH has two C2H5 groups. As a result, (C2H5 )2NH has higher basicity than C2H5NH2.

Because of the – R impact of the – C6H5 group, C6H5N(CH3)2 is now less basic than CH3NH2.

As a result, the following is the ascending order of the basic strengths of the provided compounds: