If aniline is treated with a 1:1 mixture of concentrated $HNO_3$ and concentrated $H_2SO_4$, p-nitroaniline and m-nitroaniline are formed nearly in equal amounts. This is due to

Question

Protonation of the - $NH_2$ group, which causes deactivation of the benzene ring

m-directing property of the - $NH_2$ group

Isomerization of some p-nitroaniline into m-nitroaniline

m- and p-directing properties of the $NH_2$ group

Solution

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