Question

Why aniline on nitration forms a significant amount of m-nitroaniline?

Answer 1

In aniline, the NH₂ group acts as a bronsted base so it accepts H⁺ from HNO₃ to become Benzene-NH₃⁺ . Now in this NH₃⁺ does not have any +M effect. It has only -I effect, this means that addition of NO₂⁺ will be according to -I effect of NH₃⁺. At ortho position -I effect is maximum and at para position -I effect is minimum, so the cation NO₂⁺ is more likely to go to para and then meta rather than ortho. This is why m-Nitoaniline is very significant in amount.

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Answer 2

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Reason :-

Under strongly acidic conditions of nitration (in the presence of a mixture of conc. $HNO_3 + H_2SO_4$), aniline gets protonated and is converted into anilinium ion having —$NH_3^+$ group.

This group is deactivating group and m-directing. So, the nitration on aniline gives o, p-nitroaniline (mainly p-product) while the nitration of aniline ion gives m-nitroaniline.

Thus, nitration of aniline gives a substantial amount of m-nitroaniline due to protonation of aniline.