Aniline on direct nitration yields
Aniline on direct nitration yields
Options
Official answer
Correct option: (C) 51%-para, 47%-meta, 2%-ortho derivatives
In the strongly acidic nitrating mixture, aniline is largely protonated to the anilinium ion (C₆H₅NH₃⁺), which is meta-directing and deactivating.
- A large share of meta product (≈47%) results, alongside
- para (≈51%) from the small fraction of un-protonated aniline, and
- only ≈2% ortho.
Marking scheme
- 1.1 mark: correct option (C) 51% para, 47% meta, 2% ortho.
- 2.No partial credit for options with swapped percentages.
Hint
In acid, aniline becomes the meta-directing anilinium ion — expect a large meta fraction plus dominant para and very little ortho.
Quick oral answer
Because the acidic medium protonates aniline to the meta-directing anilinium ion, direct nitration gives about 51% para, 47% meta and only 2% ortho product.
Common mistakes
- 1.Assuming aniline behaves purely as an o/p director and giving no meta product.
- 2.Swapping the meta and para percentages.
- 3.Forgetting that acetylation is used to suppress meta and get mainly p-nitroaniline.
Analysis and explanation
Concept
Free –NH₂ is a powerful o/p-directing activator. But nitration needs a strongly acidic medium (HNO₃/H₂SO₄), which protonates most of the aniline to the anilinium ion.
Why so much meta
The positively charged –NH₃⁺ group is electron-withdrawing (−I), making the ring deactivated and meta-directing. Because a substantial amount of aniline exists as anilinium ion, an abnormally large meta fraction (~47%) forms — far more than in typical o/p directors.
Why para dominates overall
The small equilibrium amount of neutral aniline reacts fast (activated ring) mainly at the less-hindered para position, giving ~51% para; ortho is only ~2% due to steric hindrance near the bulky protonated nitrogen.
Exam solution / practice
To get clean p-nitroaniline, the –NH₂ is first acetylated (protected) so that the acetamido group stays a moderate o/p director; nitration then gives mainly para, and hydrolysis restores –NH₂.
Why the distractors are wrong
- (A), (B), (D) scramble the percentages — the correct experimental split is para 51% > meta 47% ≫ ortho 2%.
Interesting facts
The unusually high meta yield in aniline nitration is classic proof that the reacting species is the protonated anilinium ion, not free aniline.
p-Nitroaniline, made cleanly via the acetanilide route, is an important intermediate for azo dyes and pigments.
Question details
Q10 of 37 questions
- Board
- CBSE
- Class
- Class 12
- Subject
- Chemistry
- Year
- 2026
- Section
- Section A
- Marks
- 1
- Type
- MCQ
- Chapter
- Amines
- Topic
- Direct nitration of aniline
- Bloom's level
- understand
- Syllabus topic
- Unit: Amines — Electrophilic Substitution of Aromatic Amines: Nitration of Aniline
- Author
- CBSE
- Reviewed by
- Padhantu Academic Team
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Frequently asked questions
Why does aniline give so much meta product on nitration?
The strongly acidic nitrating mixture protonates aniline to the anilinium ion (–NH₃⁺), which is electron-withdrawing and meta-directing. Because a large fraction of aniline is protonated, roughly 47% meta product forms, which is unusual for an amine.
How can we obtain mainly p-nitroaniline from aniline?
First acetylate aniline to acetanilide, which stays a moderate ortho/para director. Nitration then gives predominantly the para product; subsequent acid or base hydrolysis removes the acetyl group to yield p-nitroaniline.