Electrophilic Substitution in Aniline
Nitrogen Compounds
8
JEE Qs
8%
Hard
75
min
Master the exceptions and specific conditions (protection/deprotection) required for electrophilic substitution in aniline due to its strong activating nature and basicity, as these are frequent testing points.
✅ Key Points for JEE
- 1The -NH2 group is strongly activating due to the lone pair of electrons on nitrogen participating in resonance, making the ortho and para positions highly electron-rich.
- 2Direct bromination of aniline with bromine water leads to 2,4,6-tribromoaniline as the major product due to the extremely high activation of the ring.
- 3For mono-substitution (e.g., para-bromoaniline), the amino group must be protected by acetylation (reaction with acetic anhydride) to form acetanilide, which is a weaker activating group.
- 4Direct nitration of aniline with nitrating mixture (conc. HNO3 + conc. H2SO4) yields significant amounts of *meta*-nitroaniline (in addition to ortho/para) because aniline forms anilinium ion (C6H5-NH3+) in acidic medium, which is *meta*-directing and deactivating.
- 5Friedel-Crafts alkylation and acylation reactions are not possible directly on aniline because the amino group, being a Lewis base, forms a salt with the Lewis acid catalyst (e.g., AlCl3), deactivating the ring and preventing the reaction.
⚠️ Common Mistakes
- ✕Attempting Friedel-Crafts reaction directly on aniline and expecting substitution.
- ✕Forgetting to protect the amino group for mono-substituted product formation, especially in bromination.
- ✕Not understanding the formation of *meta*-substituted product during direct nitration due to anilinium ion formation.
- ✕Ignoring the oxidative side reactions that can occur during direct nitration of aniline.
📝 Practice Questions
See allQ56.Given below are two statements : Consider the following reaction (1) Both Statement I and Statement II are false (2) Statement I is true but Statement II is false (3) Statement I is false but Statement II is true (4) Both Statement I and Statement II are true
Q56.Identify correct statements : (A) Primary amines do not give diazonium salts when treated with NaNO2 in acidic condition. (B) Aliphatic and aromatic primary amines on heating with CHCl3 and ethanolic KOH form 2025 (28 Jan Shift 2) JEE Main Previous Year Paper carbylamines. (C) Secondary and tertiary amines also give carbylamine test. (D) Benzenesulfonyl chloride is known as Hinsberg's reagent. (E) Tertiary amines reacts with benzenesulfonyl chloride very easily. Choose the correct answer from the options given below : (1) (A) and (B) only (2) (D) and (E) only (3) (B) and (D) only (4) (B) and (C) only
Q68.Match the Compounds (List - I) with the appropriate Catalyst/Reagents (List - II) for their reduction into corresponding amines. Choose the correct answer from the options given below : (1) (A)-(II), (B)-(I), (C)-(III), (D)-(IV) (2) (A)-(III), (B)-(II), (C)-(IV), (D)-(I) (3) (A)-(II), (B)-(IV), (C)-(III), (D)-(I) (4) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
Q58.The correct set of ions (aqueous solution) with same colour from the following is: 2025 (23 Jan Shift 1) JEE Main Previous Year Paper (1) Sc3+, Ti3+, Cr2+ (2) V2+, Cr3+, Mn3+ (3) Ti4+, V4+, Mn2+ (4) Zn2+, V3+, Fe3+
Q66. For reaction The correct order of set of reagents for the above conversion is : (1) Br2 ∣FeBr3, H2O(Δ), NaOH (2) H2SO4, Ac2O, Br2, H2O(Δ), NaOH (3) Ac2O, Br2, H2O(Δ), NaOH (4) Ac2O, H2SO4, Br2, NaOH
Q55.Total number of nucleophiles from the following is : ⊖ NH3, PhSH, (H3C)2 S, H2C = CH2, O H, H3O⊕, (CH3)2CO, ⇌NCH3 (1) 7 (2) 4 (3) 6 (4) 5
NCERT Chapters
- Class 12 Chemistry Part 2 Ch 13: Amines