$ ext{Arrange the following in the decreasing order of their base strength :}
(I) C_6H_5NH_2 (II) p-NO_2C_6H_4NH_2 (III) m-NO_2C_6H_4NH_2 (IV) p-CH_3OC_6H_4NH_2$

Q: $\text{Arrange the following in the decreasing order of their base strength :} (I) C_6H_5NH_2 (II) p-NO_2C_6H_4NH_2 (III) m-NO_2C_6H_4NH_2 (IV) p-CH_3OC_6H_4NH_2$

$\text{Arrange the following in the decreasing order of their base strength :} (I) C_6H_5NH_2 (II) p-NO_2C_6H_4NH_2 (III) m-NO_2C_6H_4NH_2 (IV) p-CH_3OC_6H_4NH_2$

Question

Accepted Answer

To determine the base strength of the given compounds, we need to consider the effect of substituents on the aniline (C₆H₅NH₂) structure. The base strength of an amine is influenced by the electron-donating or electron-withdrawing nature of the substituents attached to the aromatic ring.

Let's analyze each compound:

1. **C₆H₅NH₂ (Aniline)**: This is the parent compound with no substituents. It serves as a reference point for base strength.

2. **p-NO₂C₆H₄NH₂**: The nitro group (NO₂) is a strong electron-withdrawing group due to its -M (mesomeric) and -I (inductive) effects. When positioned para to the amino group, it significantly reduces the electron density on the nitrogen atom, thereby decreasing the base strength.

3. **m-NO₂C₆H₄NH₂**: The nitro group at the meta position primarily exerts an inductive effect, which is less effective in withdrawing electron density compared to the para position. Therefore, the base strength is slightly higher than the para-substituted nitroaniline but still lower than aniline.

4. **p-CH₃OC₆H₄NH₂**: The methoxy group (CH₃O) is an electron-donating group due to its +M effect, which increases the electron density on the nitrogen atom when positioned para to the amino group. This enhances the base strength compared to aniline.

Based on these analyses, the decreasing order of base strength is:

- **IV (p-CH₃OC₆H₄NH₂)**: The strongest base due to the electron-donating methoxy group.
- **I (C₆H₅NH₂)**: Aniline itself, with no substituents affecting its base strength.
- **III (m-NO₂C₆H₄NH₂)**: Weaker than aniline due to the electron-withdrawing nitro group at the meta position.
- **II (p-NO₂C₆H₄NH₂)**: The weakest base due to the strong electron-withdrawing effect of the nitro group at the para position.

Thus, the correct order of decreasing base strength is IV > I > III > II, which corresponds to Option 1.

**Explanation of Incorrect Options:**

- **Option 2 (IV > III > II > I)**: This option incorrectly places the base strength of m-nitroaniline (III) above aniline (I), which is not possible as the nitro group reduces base strength.

- **Option 3 (III > II > I > IV)**: This option suggests that m-nitroaniline (III) is the strongest base, which is incorrect. It also places p-methoxyaniline (IV) as the weakest, which contradicts the electron-donating effect of the methoxy group.

- **Option 4 (I > IV > II > III)**: This option incorrectly suggests that aniline (I) is stronger than p-methoxyaniline (IV), which is not true due to the electron-donating nature of the methoxy group enhancing the base strength of IV.

Therefore, Option 1 is the correct answer, as it accurately reflects the influence of substituents on the base strength of the compounds.