Arrange the following in the increasing order of their basicities. I. p - Toluidine II. N, N - Dimethyl - p - toluidine III. p - Nitroaniline IV. Aniline
A
III < IV < I < II
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B
II < IV < I < III
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C
III < II < I < IV
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D
I < IV < III < II
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Solution
The correct option is A III < IV < I < II N,N-Dimethyl p-toluidine is most basic whereas p-nitroaniline is least basic. II +I effect of two Me groups of N atom and Me group at p position (+I and hyperconjugative. effects) are responsible for highest basicity. I +I and H.C. effects of (Me) group at p position makes p-toluidine more basic than aniline. IV (Standard) (Aniline) III -I and -R effect of (-NO2) group at p position are responsible for least basicity.