JEE Main 2021 · 24 February · morning · Single correct · +4 / −1
In the following reaction the reason why meta-nitro product also formed is :
- (a) –NH₂ group is highly meta-directive
- (b) –NO₂ substitution always takes place at meta-position
- (c) Formation of anilinium ion
- (d) low temperature
Answer and solution
Answer (c) Formation of anilinium ion
Solution
Chemistry ma’am · 4 steps
In strong acid, aniline grabs a proton first, and the ring it nitrates is no longer plain aniline.
- Free amine. In free aniline, the lone pair on N pushes electron density into the ring. That makes -NH₂ a strong activating group that directs to the ortho and para positions.
- Strong acid. The nitrating mixture of conc. HNO₃ and conc. H₂SO₄ is strongly acidic, and aniline is a base. A large part of it is protonated to the anilinium ion.
C₆H₅NH₂ + H⁺ → C₆H₅NH₃⁺
- Anilinium directs meta. In -NH₃⁺ the lone pair is used up in the bond to H⁺, and the positive charge pulls electrons out of the ring. This deactivates the ring and directs the incoming NO₂⁺ to the meta position.
- So. Free aniline gives the para product (51%) and a little ortho (2%), while the anilinium ion gives the meta product (47%). The meta product forms because of the formation of the anilinium ion, option (c).
The idea to keep. Always check what the group becomes under the reaction conditions. In strong acid, -NH₂ turns into -NH₃⁺, and an ortho/para director becomes a meta director.
Answer from the NTA final answer key. Solution by PrepWiser.