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Why p- methoxy aniline is more basic than p-chloro aniline but less basic than p-methyl aniline?
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Why p- methoxy aniline is more basic than p-chloro aniline but less ba...
  1. Anilines are 
    Bronsted bases
    Comparative strength of a series of Bronsted bases are always explained in terms of the
     c
    omparative stability of their corresponding conjugate acids
    . More stable the conjugate acid (lower 
    pka
    ), stronger the Bronsted base. Now looking at the following figure, which one do you think should be more basic?
  1. Anilines are also 
    Lewis bases
     
    by definition. Comparative strength of a series of Lewis bases are always explained in terms of the 
    comparative availability of non-bonding, lone pair of electrons over aniline Nitrogen centres
    . Greater the availability, stronger the Lewis base. Now looking at the following figure, which one do you think should be more basic?
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Why p- methoxy aniline is more basic than p-chloro aniline but less ba...
Introduction:

In organic chemistry, the basicity of a compound refers to its ability to accept a proton (H+) and form a conjugate acid. Basicity is influenced by various factors such as electron density, resonance, inductive effects, and steric hindrance. In the case of p-methoxyaniline, p-chloroaniline, and p-methylaniline, the presence of different substituents on the aniline ring affects their basicity.

Effect of Electron-Donating Substituents:

1. p-Methoxyaniline:
- In p-methoxyaniline, the methoxy (-OCH3) group acts as an electron-donating substituent.
- The presence of this group increases the electron density on the aniline nitrogen atom through resonance and inductive effects.
- The lone pair of electrons on the nitrogen atom becomes more available for protonation, making p-methoxyaniline more basic.

2. p-Methylaniline:
- In p-methylaniline, the methyl (-CH3) group also acts as an electron-donating substituent.
- However, the electron-donating effect of the methyl group is weaker compared to the methoxy group.
- The electron density on the aniline nitrogen atom is slightly increased, making p-methylaniline less basic than p-methoxyaniline.

Effect of Electron-Withdrawing Substituents:

3. p-Chloroaniline:
- In p-chloroaniline, the chloro (-Cl) group acts as an electron-withdrawing substituent.
- The presence of this group decreases the electron density on the aniline nitrogen atom through resonance and inductive effects.
- The lone pair of electrons on the nitrogen atom becomes less available for protonation, making p-chloroaniline less basic compared to p-methoxyaniline.

Summary:

- p-Methoxyaniline is more basic than p-chloroaniline due to the electron-donating effect of the methoxy group.
- p-Methoxyaniline is less basic than p-methylaniline due to the stronger electron-donating effect of the methoxy group compared to the methyl group.
- p-Chloroaniline is less basic than p-methoxyaniline due to the electron-withdrawing effect of the chloro group.

In summary, the basicity of p-methoxyaniline is influenced by the electron-donating substituent, which increases the electron density on the aniline nitrogen atom, making it more basic than p-chloroaniline. However, p-methoxyaniline is less basic than p-methylaniline due to the weaker electron-donating effect of the methoxy group compared to the methyl group.
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Why p- methoxy aniline is more basic than p-chloro aniline but less basic than p-methyl aniline?
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