Question:
Why is the $pK_b$ of aniline greater than that of methylamine?
Why is the $pK_b$ of aniline greater than that of methylamine?
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In aromatic amines, the electron-withdrawing effect of the phenyl group makes the amine less basic than methylamine.
Updated On: Jun 18, 2025
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Solution and Explanation
- Methylamine has a $–CH_3$ group, which donates electrons and increases the electron density on nitrogen, making it a stronger base.
- Aniline has a phenyl group ($–C_6H_5$), which withdraws electrons through resonance, reducing the electron density on nitrogen and making it a weaker base. Hence, the $pK_b$ of aniline is greater than that of methylamine.
- Aniline has a phenyl group ($–C_6H_5$), which withdraws electrons through resonance, reducing the electron density on nitrogen and making it a weaker base. Hence, the $pK_b$ of aniline is greater than that of methylamine.
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