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Consider the following four amine compounds: I = aniline (C6H5-NH2), II = N-methylaniline (C6H5-NH-CH3), III = diphenylamine ((C6H5)2NH), IV = triphenylamine ((C6H5)3N). Choose the correct statement(s).

1. The order of basicity is II > I > III > IV.
2. The magnitude of pKb difference between I and II is more than that between III and IV.
3. Resonance effect is more in III than in IV.
4. Steric effect makes compound IV more basic than III.

Correct answer: The order of basicity is II > I > III > IV.

Solution

N-methylaniline (II) is most basic because the methyl group donates electrons to nitrogen. Adding more phenyl rings (III, IV) progressively delocalises the lone pair via resonance, reducing basicity. Hence basicity order: II > I > III > IV.

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