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Which of the following is more basic than aniline :-
A
Diphenylamine
B
Triphenylamine
C
p-Nitroaniline
D
Benzylamine
Detailed Solution
The basic strength of an amine depends on the availability of the lone pair on nitrogen for accepting a proton.
In aniline, $C_6H_5NH_2$, the lone pair on nitrogen is delocalised into the benzene ring by resonance ($+M$ effect), so it is less available.
In benzylamine, $C_6H_5CH_2NH_2$, the nitrogen is attached to a $CH_2$ group and not directly to the ring, so its lone pair is localised and cannot be delocalised into the ring.
Benzylamine therefore behaves like an aliphatic primary amine and is much more basic than aniline.
In diphenylamine and triphenylamine the lone pair is delocalised over two and three benzene rings respectively, so they are weaker bases than aniline.
In p-nitroaniline the electron withdrawing $-NO_2$ group pulls the lone pair further into the ring, so it is also weaker than aniline.
So the compound more basic than aniline is benzylamine.
In aniline, $C_6H_5NH_2$, the lone pair on nitrogen is delocalised into the benzene ring by resonance ($+M$ effect), so it is less available.
In benzylamine, $C_6H_5CH_2NH_2$, the nitrogen is attached to a $CH_2$ group and not directly to the ring, so its lone pair is localised and cannot be delocalised into the ring.
Benzylamine therefore behaves like an aliphatic primary amine and is much more basic than aniline.
In diphenylamine and triphenylamine the lone pair is delocalised over two and three benzene rings respectively, so they are weaker bases than aniline.
In p-nitroaniline the electron withdrawing $-NO_2$ group pulls the lone pair further into the ring, so it is also weaker than aniline.
So the compound more basic than aniline is benzylamine.