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Arrange the following nitrogen compounds in decreasing order of basicity: (i) Aniline (Ph–NH2), (ii) Benzylamine (Ph–CH2–NH2), (iii) p-Nitroaniline (p-NO2–C6H4–NH2), (iv) Benzamide (Ph–C(=O)–NH2). Which compound has the highest basicity?

1. Aniline (benzene ring–NH2)
2. Benzylamine (benzene ring–CH2–NH2)
3. Nitroaniline (benzene ring with NO2 and NH2 substituents)
4. Benzamide (benzene ring–C(=O)–NH2)

Correct answer: Benzylamine (benzene ring–CH2–NH2)

Solution

In benzylamine the lone pair on N is not conjugated with the aromatic ring (separated by sp3 CH2), giving the highest basicity. Aniline has its lone pair partially delocalized into the ring. p-Nitroaniline is further deactivated by the NO2 group. Benzamide has the lone pair conjugated with the C=O, making it almost non-basic. Decreasing order: Benzylamine > Aniline > p-Nitroaniline > Benzamide.

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