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Arrange the following aromatic amines in decreasing order of basicity: (I) Aniline (C6H5-NH2) (II) p-Nitroaniline (O2N-C6H4-NH2) (III) p-Methoxyaniline (CH3O-C6H4-NH2) (IV) p-Methylaniline (CH3-C6H4-NH2)

1. III > IV > I > II
2. III > IV > II > I
3. I > II > III > IV
4. IV > III > II > I

Correct answer: III > IV > I > II

Solution

The -OCH3 group donates electrons to the ring and increases electron density at N via resonance and induction, making p-methoxyaniline the most basic. The -NO2 group strongly withdraws electrons via resonance, delocalizing the N lone pair and making p-nitroaniline the least basic.

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