Chapter Overview & Weightage
Amines is one of those chapters that rewards understanding over memorisation. The reactions follow a clear logic — if you understand why amines are basic and nucleophilic, most reactions become predictable rather than a list to memorise.
Amines typically contributes 1 question in JEE Main (4 marks) and occasionally appears in JEE Advanced as part of organic synthesis problems. Over 2019–2024, it has appeared in roughly 60% of JEE Main sessions — treat it as a near-certain question.
| Year | JEE Main Questions | Topic Tested |
|---|---|---|
| 2024 | 1 | Basicity comparison, diazonium coupling |
| 2023 | 1 | Gabriel synthesis, Hofmann bromamide |
| 2022 | 1 | Carbylamine test, classification |
| 2021 | 2 | Basicity order, Hinsberg test |
| 2020 | 1 | Diazonium salt reactions |
| 2019 | 1 | Basicity, amide vs amine |
Weightage verdict: 3–4% of chemistry marks. One focused week is enough to lock this chapter.
Key Concepts You Must Know
Prioritised by how often each concept appears in PYQs:
Tier 1 — Appears almost every session:
- Basicity order of amines (this is the single most-tested concept)
- Classification: primary, secondary, tertiary amines vs quaternary ammonium salts
- Carbylamine reaction (Hofmann isocyanide test) — identifying 1° amines
- Diazonium salt formation and its reactions (coupling, Sandmeyer, Balz–Schiemann)
Tier 2 — Appears frequently:
- Gabriel phthalimide synthesis (prepares pure 1° amines)
- Hofmann bromamide degradation (1° amine with one fewer carbon)
- Hinsberg test for distinguishing 1°, 2°, 3° amines
- Reaction with nitrous acid (NaNO₂ + HCl) — different products for 1°, 2°, 3°
Tier 3 — Appears in Advanced or tricky mains:
- Coupling reactions with phenols and aromatic amines
- Acylation, benzoylation (Schotten–Baumann conditions)
- Reduction of nitro compounds, amides, nitriles to amines
Important Formulas
In aqueous solution: inductive effect (electron donation) increases basicity, but steric hindrance and solvation both matter. The 2° amine wins because it has two alkyl groups pushing electrons yet isn’t sterically overwhelmed. The 3° amine has three alkyl groups but poor solvation — so it drops behind even 1° amine.
The lone pair on nitrogen in aniline delocalises into the benzene ring — this reduces availability for protonation. Aniline’s pKb ≈ 9.4 vs methylamine’s pKb ≈ 3.4.
Produces isocyanide (foul smell). Only primary amines give this — 2° and 3° amines do not. This is a confirmatory test for 1° amines.
Temperature is critical: above 5°C, the diazonium salt decomposes. This is a favourite trap in MCQs.
Use Sandmeyer when you need Cl, Br, or CN on a benzene ring at a position where direct substitution is difficult (meta directing would otherwise block it).
Coupling occurs at para position preferentially. With aniline: weakly acidic medium. With phenol: alkaline medium.
The product amine has one carbon fewer than the amide. This is the standard method to “go down the homologous series.”
Solved Previous Year Questions
PYQ 1 — Basicity Order (JEE Main 2021, February Session)
Question: Arrange the following in increasing order of basicity: Aniline (I), p-Nitroaniline (II), p-Methylaniline (III), Diphenylamine (IV)
Solution:
The key is understanding what increases or decreases availability of the lone pair on nitrogen.
- Aniline — lone pair delocalised into ring, moderate basicity
- p-Nitroaniline (II) — nitro group is electron-withdrawing, further pulls the lone pair — least basic
- p-Methylaniline (III) — methyl is electron-donating via hyperconjugation — more basic than aniline
- Diphenylamine (IV) — lone pair delocalised into two rings — even less available than aniline
Increasing order: II < IV < I < III
Students often rank diphenylamine as more basic because it has two phenyl groups “pushing” electrons. Wrong — both rings pull the lone pair through resonance, making diphenylamine less basic than aniline. Only alkyl groups push electrons; aryl groups pull via resonance.
PYQ 2 — Reaction Identification (JEE Main 2023, April Shift 2)
Question: Which of the following reagents will convert benzamide to aniline?
(A) Br₂/KOH (B) LiAlH₄ (C) H₂/Ni (D) Zn/HCl
Solution:
Benzamide is C₆H₅–CO–NH₂ (an amide). We need to get to aniline, C₆H₅–NH₂.
- Option A (Br₂/KOH): This is Hofmann bromamide degradation — gives C₆H₅–NH₂ (aniline) with loss of one carbon (CO₂ leaves). ✓
- Option B (LiAlH₄): Reduces amide to amine, but retains all carbons — gives C₆H₅–CH₂–NH₂ (benzylamine), not aniline. ✗
- Options C, D: Won’t cleave the C–N amide bond selectively. ✗
Answer: A
The mechanism — Br₂ replaces one NH hydrogen, KOH causes rearrangement (nitrene intermediate), CO₂ is lost.
Whenever the question asks to go from an amide (R–CO–NH₂) to an amine with one fewer carbon (R–NH₂), think Hofmann immediately. LiAlH₄ gives you one more carbon in the chain, not one fewer.
PYQ 3 — Diazonium Salt Reactions (JEE Main 2024, January Shift 1)
Question: The product formed when benzenediazonium chloride reacts with HBF₄ followed by heating is:
(A) Fluorobenzene (B) Chlorobenzene (C) Benzene (D) Phenol
Solution:
This is the Balz–Schiemann reaction — specifically designed for introducing F onto a benzene ring.
Step 1: C₆H₅–N₂⁺Cl⁻ + HBF₄ → C₆H₅–N₂⁺BF₄⁻ (diazonium tetrafluoroborate)
Step 2: On dry heating → C₆H₅–F + N₂ + BF₃
Answer: A — Fluorobenzene
Why Balz–Schiemann and not Sandmeyer for fluorine? Because CuF is unstable. The tetrafluoroborate salt is thermally stable enough to isolate, then decomposes cleanly to give ArF.
Sandmeyer: Cl, Br, CN. Balz–Schiemann: F. Gattermann: Cl and Br (uses Cu powder instead of CuCl/CuBr). This distinction appears repeatedly — a single question can test all three by listing different reagents.
Difficulty Distribution
For JEE Main Amines questions (based on 2019–2024 analysis):
| Level | % of Questions | What It Tests |
|---|---|---|
| Easy | 40% | Classification, carbylamine test, direct basicity comparison |
| Medium | 45% | Basicity with substituents, identifying synthesis routes, reaction conditions |
| Hard | 15% | Multi-step synthesis using diazonium/Gabriel, Advanced-level mechanism |
JEE Advanced context: In Advanced, Amines rarely appears as a standalone question — it shows up as a step in organic synthesis problems. The key skill there is recognising when to use Gabriel/Hofmann to insert a specific amine into a synthesis.
Expert Strategy
Week 1 — Build the logic tree (3 hours): Start with basicity. Understand the three competing factors — inductive effect, steric hindrance, resonance. Work through the standard comparison: aliphatic 2° > 1° > 3° > NH₃ > ArNH₂. Don’t memorise — derive it once fully, then it sticks.
Week 2 — Master the named reactions (2 hours): There are exactly five named reactions in this chapter that JEE tests: Gabriel, Hofmann, Carbylamine, Sandmeyer/Balz–Schiemann, and Coupling. For each, write down: starting material → reagent → product → what it’s used for. That last column is what saves you in synthesis problems.
A clean way to remember diazonium reactions: group them by what replaces N₂. Sandmeyer (Cl, Br, CN), Balz–Schiemann (F), direct decomposition (OH from warm water, H from H₃PO₂), coupling (no replacement — N₂ unit stays as –N=N– in azo dye).
Revision approach: Practice basicity comparison questions from PYQs — at least 15 questions. These are pattern-recognisable once you’ve seen enough variations. The substituent effect questions follow strict rules; they don’t surprise you if your fundamentals are right.
Do not spend more than 20% of your time on mechanisms. JEE Main tests products and conditions, not arrow-pushing.
Common Traps
Trap 1 — Aliphatic 3° amine basicity in non-aqueous vs aqueous: In gas phase (or non-aqueous solvents), basicity order is strictly 3° > 2° > 1° > NH₃ (only inductive effect matters). In aqueous solution, solvation matters and 3° drops below 1°. JEE questions are almost always in aqueous context — but read carefully.
Trap 2 — Gabriel synthesis cannot give secondary or tertiary amines: Gabriel phthalimide only gives primary amines. If an MCQ option says Gabriel synthesis gives 2° amine, it’s wrong. The nitrogen in phthalimide has both bonds tied up; after hydrolysis, you only get R–NH₂.
Trap 3 — Carbylamine test is for primary amines only, but aliphatic AND aromatic: Students sometimes believe carbylamine test is only for aliphatic amines. Wrong — aniline (aromatic 1° amine) also gives positive carbylamine test. The test is for primary, not aliphatic.
Trap 4 — Diazotisation temperature: The reaction requires 0–5°C. If the temperature is raised, the diazonium salt decomposes to phenol. Questions will sometimes ask what happens on warming the diazonium salt solution — the answer is phenol (+ N₂), not a coupling product.
Trap 5 — Hofmann product vs LiAlH₄ product from amide: Hofmann (Br₂/KOH) gives R–NH₂ from R–CO–NH₂, with R losing one carbon. LiAlH₄ gives R–CH₂–NH₂ from R–CO–NH₂, with the chain extended by keeping all carbons. This is a classic “one carbon up or down” trap in synthesis questions.
The shortcut for substituent effect on aniline basicity: Electron-donating groups (–CH₃, –OCH₃, –OH at para/ortho) increase basicity of aniline. Electron-withdrawing groups (–NO₂, –CN, –COOH, –CHO) decrease basicity. For ortho substituents, also consider steric hindrance reducing solvation — even electron-donating ortho groups can sometimes reduce basicity compared to the para isomer.