Question

Explain why m-nitroaniline is a stronger base than p-nitroaniline

Short answer

In m-nitro aniline, the nitro ($-N{O}_2$) group is electron-withdrawing.

The electron does not get attracted directly by $-N{O}_2$ in m-nitroaniline because no resonance structure is possible with the electron delocalized over $-N{O}_2$.

Hence, it is a stronger base than p-nitroaniline.

Step-by-step solution

Basicity of disubstituted aromatic amines 

In a disubstituted aromatic amine ring, two factors show a significant effect on basicity

• A strong electron-withdrawing group reduce the basicity of nearby amines through inductive effect

• When amine is adjacent to the pi acceptor group basicity is affected due to resonance.

Explanation

In m-nitro aniline, the nitro ($-N{O}_2$) group is electron-withdrawing, hence decreasing the basicity of amine.

Resonance structure of m-nitro aniline

The electron does not get attracted directly by $-N{O}_2$ in m-nitroaniline because no resonance structure is possible with electron delocalized over $-N{O}_2$.

Hence, it is more basic than p-nitroaniline.

In p-nitro aniline, the electron gets attracted directly by $-N{O}_2$ in p-nitroaniline because one additional resonance structure is possible with electron delocalization over $-N{O}_2$.

Resonance structure of p-nitro aniline

Hence, p-nitro aniline is less basic.