A) Diphenylamine
B) Triphenylamine
C) p-nitroaniline
D) Benzyiamine
Correct Answer: D
Solution :
Benzyl amine \[{{C}_{6}}{{H}_{5}}C{{H}_{2}}\to \,-\overset{\centerdot \,\centerdot }{\mathop{N}}\,{{H}_{2}}\] is more basic than aniline because benzyl group \[{{C}_{6}}{{H}_{5}}C{{H}_{2}}-\] is electron donating group due to + I effect. So, it is able to increase the electron density of N of \[-N{{H}_{2}}\] group. Thus due to higher electron density rate of donation of free pair of electron is increased i.e., basic character is higher. Phenyl and nitro group are electron attractive groups, so they are able to decrease the electron density of N of\[-\overset{\centerdot \,\centerdot }{\mathop{N}}\,{{H}_{2}}\] group. Hence, they are less basic with aniline.
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