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The P-Block Elements

Question
CBSEENCH12010267

Arrange the following in increasing order of their basic strength:

 

(i) C6H5 – NH2, C6H5 – CH2 – NH2, C6H5 – NH – CH3

 

Solution

(i) C6H5-NH2 < C6H5-NH-CH3 < C6H5-CH2-NH2
Reason:
C6H5-NH2 will be least basic because of the delocalization of the lone pair of electrons present on the N-atom over the benzene ring due to the ‒R effect of the C6H5 group. However, C6H5-CH2-NH2 will be more basic than C6H5-NH-CH3 because of the electron-releasing nature of the CH3- a group that increases the electron density on the N-atom, making the lone pair of electrons on the N-atom easily available for donation to a proton. The basicity of C6H5-NH-CH3 will be intermediate of C6H5-NH2 and C6H5-CH2-NH2 because the C6H5- the group will tend to pull the electron density from the N-atom. On the other hand, the CH3- group will tend to increase the electron density on the N-atom. Thus, the basic strength of the given amines will follow the above-mentioned order.


(ii) 

 

Reason:

The -CH3 group will increase the electron density on the benzene ring because of it is +I effect, while the NO2 group will decrease the electron density on the benzene ring because of its ‒I effect. Hence, the C6H5-NH2 molecule having - the CH3 group will be more basic than C6H5-NH2. Also, it will be more basic than the C6H5-NH2 molecule with the NO2 group.