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Question
Arrange the following:
In increasing order of basic strength:
C6H5NH2, C6H5NHCH3, C6H5CH2NH2
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Solution
C6H5NHCH3 is more basic than C6H5NH2 due to the presence of an electron-donating −CH3 group in C6H5NHCH3.
Again, in C6H5NHCH3, −C6H5 group is directly attached to the N-atom. However, it is not so in C6H5CH2NH2. Thus, in C6H5NHCH3, the −R effect of −C6H5 group decreases the electron density over the N-atom. Therefore, C6H5CH2NH2 is more basic than C6H5NHCH3.
Hence, the increasing order of the basic strengths of the given compounds is as follows:
C6H5NH2 < C6H5NHCH3 < C6H5CH2NH2
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Write the structures of the main products of the following reactions:

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(I)

(II)

(III)

Account for the following:
Acylation of aniline is carried out in the presence of pyridine.
Which of the following statement is true about methyl amine?
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By the presence of a halogen atom in the ring, what is the effect of this on basic property of aniline?
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\[\ce{C6H5NH2, C6H5NHCH3, C6H5CH2NH2, CH3NH2, NH3}\]
The correct order of the increasing basic nature of Ammonia, Methylamine and Aniline is:
