Question:

The correct order of decreasing basic strength of the given amines is:

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When evaluating the basicity of amines, remember that alkyl groups increase basicity by donating electron density, while aromatic rings decrease basicity by withdrawing electron density.
Updated On: May 4, 2025
  • N-ethylmethanamine $>$ ethanamine $>$ benzenamine $>$ N-methylaniline
  • N-ethylmethanamine $>$ ethanamine $>$ N-methylaniline $>$ benzenamine
  • N-methylaniline $>$ ethanamine $>$ benzenamine $>$ N-ethylmethanamine
  • N-methylaniline $>$ N-ethylmethanamine $>$ benzenamine $>$ ethanamine
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The Correct Option is B

Solution and Explanation

Basic strength of amines depends on the availability of the nitrogen lone pair for protonation. The basicity of amines can be influenced by electron-donating and electron-withdrawing groups attached to the nitrogen. The general trend is:
- Alkyl groups (like ethyl and methyl) donate electrons to the nitrogen, increasing its basicity.
- A benzene ring (as in benzenamine) is electron-withdrawing due to resonance, which decreases the basicity of the nitrogen.
Let’s analyze the amines in the options:
- N-ethylmethanamine (ethylamine): The ethyl group is electron-donating, making the nitrogen more basic than ethanamine.
- Ethanamine (ethylamine): The amine group is attached to an ethyl group, giving it good electron-donating effects, but the effect is weaker than in N-ethylmethanamine.
- N-methylaniline: The methyl group is electron-donating, but the aromatic ring in benzenamine withdraws electron density through resonance, making N-methylaniline less basic than ethanamine.
- Benzenamine (aniline): The aromatic ring has a strong electron-withdrawing effect on the nitrogen, making this the least basic of all the options.

Thus, the correct order of basicity is N-ethylmethanamine $>$ ethanamine $>$ N-methylaniline $>$ benzenamine.

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