In the electrophilic substitution reactions of aniline, the -NH2 grou…

The correct answer is Option (3) → As both Ortho and para directing group

The lone pair on the nitrogen of the –\( \ce{NH2} \) group is in resonance with the benzene ring. In the resonance structures of aniline a negative charge appears at the two ortho positions and at the para position, never at the meta positions. These positions become centres of high electron density, so an incoming electrophile attacks there. The –\( \ce{NH2} \) group is therefore ortho and para directing and a powerful activating group.

For example, bromine water at room temperature substitutes all three of these positions at once:

    \( \displaystyle \ce{C6H5NH2 + 3Br2} \xrightarrow{\ce{H2O}} \text{2,4,6-tribromoaniline} + \ce{3HBr} \)

So the –\( \ce{NH2} \) group directs to both the ortho and the para positions.

Why the other options are incorrect

• Option (1) As ortho directing group only — Incorrect. Resonance raises the electron density at the para position as much as at the ortho positions, so both are attacked.

• Option (2) As para directing group only — Incorrect. The para product is often the major one because the ortho positions are more crowded, but the ortho positions are activated too.

• Option (4) As meta directing group only — Incorrect. –\( \ce{NH2} \) itself never directs meta. In strongly acidic nitrating mixture aniline is protonated to the anilinium ion, \( \ce{C6H5NH3+} \), and it is this ion that is meta directing; that is why direct nitration gives a good deal of m-nitroaniline.

NCERT (2026-27 edition):
Electrophilic Substitution – “You have read earlier that aniline is a resonance hybrid of five structures. … Ortho- and para-positions to the –NH₂ group become centres of high electron density. Thus –NH₂ group is ortho and para directing and a powerful activating group.”