JEE Main 2021 · 18 March · evening · Single correct · +4 / −1
Consider the given reaction, percentage yield of :
- (a) B > C > A
- (b) A > C > B
- (c) C > A > B
- (d) C > B > A
Answer and solution
Answer (d) C > B > A
Solution
Chemistry ma’am · 4 steps
In nitrating acid, much of the aniline is protonated first, so meta product forms in large amounts.
- Two species present. In the strongly acidic HNO₃/H₂SO₄ mixture, much of the aniline is protonated to the anilinium ion, C₆H₅NH₃⁺. A small amount of free aniline, C₆H₅NH₂, remains in equilibrium.
- Opposite directing effects. Free -NH₂ is a strong activator and an ortho/para director. The -NH₃⁺ group has no lone pair and a positive charge, so it deactivates the ring and directs to meta.
Product Formed from Approximate yield C (para-nitroaniline) free C₆H₅NH₂, the para position is unhindered about 51% B (meta-nitroaniline) anilinium ion C₆H₅NH₃⁺ about 47% A (ortho-nitroaniline) free C₆H₅NH₂, but the ortho position is crowded about 2% - Why ortho is least. Free aniline is present only in small amounts, and the position next to -NH₂ is sterically hindered. Most of its nitration therefore goes to para.
- So. The yields fall in the order para > meta > ortho, that is C > B > A, which is option (d).
The idea to keep. Direct nitration of aniline gives about 47% meta product because the acid turns -NH₂ into -NH₃⁺. To get mainly para, protect the amine as acetanilide first.
Answer from the NTA final answer key. Solution by PrepWiser.