Question:

Which of the following compounds will form significant amount of meta product during mono-nitration reaction?

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In the strongly acidic nitrating mixture the \(\mathrm{-NH_2}\) group is protonated to \(\mathrm{-NH_3^+}\), which is a meta director.
Updated On: Nov 11, 2026
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The Correct Option is B

Solution and Explanation

Options: (1) phenyl acetate \(\mathrm{C_6H_5OCOCH_3}\), (2) aniline \(\mathrm{C_6H_5NH_2}\), (3) acetanilide \(\mathrm{C_6H_5NHCOCH_3}\), (4) phenol \(\mathrm{C_6H_5OH}\).
The substituents \(\mathrm{-OCOCH_3}\), \(\mathrm{-NHCOCH_3}\) and \(\mathrm{-OH}\) have lone pairs on the atom attached to the ring and are ortho/para directing.
Nitration is done with \(\mathrm{HNO_3 + H_2SO_4}\). In this strongly acidic medium aniline is protonated: \(\mathrm{C_6H_5NH_2 + H^+ \rightarrow C_6H_5NH_3^+}\) (anilinium ion).
The \(\mathrm{-NH_3^+}\) group is strongly electron-withdrawing (\(-I\)) and is a meta director, so a significant amount of meta-nitroaniline is obtained along with para and a little ortho product.
Answer: (2) aniline
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