JEE Main 2022 · 28 June · evening · Single correct · +4 / −1
With respect to the following reaction, consider the given statements :
(A) o-Nitroaniline and p-nitroaniline are the predominant products.
(B) p-Nitroaniline and m-nitroaniline are the predominant products.
(C) HNO₃ acts as an acid.
(D) H₂SO₄ acts as an acid.
Choose the correct option.
- (a) (A) and (C) are correct statements.
- (b) (A) and (D) are correct statements.
- (c) (B) and (D) are correct statements.
- (d) (B) and (C) are correct statements.
Answer and solution
Answer (c) (B) and (D) are correct statements.
Solution
Chemistry ma’am · 4 steps
In strong acid, aniline becomes the anilinium ion, and that is why the meta product shows up in bulk.
- Anilinium ion forms. The -NH₂ group is basic, so the strongly acidic nitrating mixture protonates it to -NH₃⁺. This positive group pulls electrons out of the ring and directs the incoming group to the meta position.
- Product mix. Some aniline stays unprotonated and still directs ortho and para. The result is about 51% p-nitroaniline, 47% m-nitroaniline and only 2% o-nitroaniline. So statement (B) is correct and (A) is wrong.
HNO₃ + 2H₂SO₄ ⇌ NO₂⁺ + H₃O⁺ + 2HSO₄⁻
- Who is the acid. H₂SO₄ is the stronger acid and donates a proton to HNO₃. Here HNO₃ accepts that proton, so it acts as a base, then loses water to give the electrophile NO₂⁺. So (D) is correct and (C) is wrong.
- So. Statements (B) and (D) are correct, which is option (c).
The idea to keep. Nitrating aniline directly gives a large share of meta product because -NH₃⁺ is meta-directing. In the nitrating mixture, H₂SO₄ is the acid and HNO₃ is the base.
Answer from the NTA final answer key. Solution by PrepWiser.