What is the Hofmann Bromamide reaction?

Hofmann Bromamide reaction is the conversion of an amide to a primary amine with one carbon atom less. The reagent used is Bromine (Br2) and aqueous NaOH (or KOH). The reaction is: RCONH2 + Br2 + 4NaOH -> RNH2 + Na2CO3 + 2NaBr + 2H2O.

Amines JEE Notes 2026 | Structure, Basicity, Reactions, PYQs & Trends

1

Structure & Classification of Amines

Amines are derivatives of ammonia (NH₃) in which one or more hydrogen atoms are replaced by alkyl or aryl groups. They are the nitrogen analogs of alcohols and ethers, playing a vital role in biological systems and organic synthesis.

1.1 Classification of Amines

Primary (1°), Secondary (2°), and Tertiary (3°) Amines

Primary (1°): Nitrogen attached to one alkyl/aryl group (R-NH₂). Example: CH₃NH₂ (Methylamine).
Secondary (2°): Nitrogen attached to two alkyl/aryl groups (R₂NH). Example: (CH₃)₂NH (Dimethylamine).
Tertiary (3°): Nitrogen attached to three alkyl/aryl groups (R₃N). Example: (CH₃)₃N (Trimethylamine).

Note on Nomenclature: In secondary and tertiary amines, the alkyl/aryl groups may be same or different. The letter 'N' is used to indicate substitution on nitrogen (e.g., N-methylmethanamine for dimethylamine).

1.2 Structure of Amines

Important for JEE — Hybridization & Geometry

In amines, the nitrogen atom is sp³ hybridized. The four sp³ hybrid orbitals form three sigma bonds (with C or H) and hold one lone pair of electrons.

\text{Geometry: Pyramidal (like } NH_3 \text{)}\] \[\text{Bond Angle: } \approx 107^\circ \text{ (due to lone pair-bond pair repulsion)}

However, in quaternary ammonium salts (R₄N⁺), the nitrogen has no lone pair and is perfectly sp³ hybridized with tetrahedral geometry and a 109.5° bond angle.

1.3 Pyramidal Inversion (Umbrella Effect)

Amine molecules undergo rapid pyramidal inversion (like an umbrella turning inside out), where the nitrogen atom passes through a planar sp² transition state. This inversion interconverts the enantiomers of a chiral amine, making them optically inactive at room temperature.

PYQ — JEE Main 2024 (Jan 31 Shift 1)

Question: The hybridization and geometry of nitrogen in a tertiary amine (R₃N) and a quaternary ammonium salt (R₄N⁺Cl⁻) are respectively:

(a) sp³, Pyramidal and sp³, Tetrahedral

(b) sp², Planar and sp³, Tetrahedral

(c) sp³, Tetrahedral and sp³, Pyramidal

(d) sp³, Pyramidal and sp², Planar

Solution:

In R₃N (tertiary amine):

- Nitrogen has 3 bond pairs and 1 lone pair.

- Hybridization is sp³.

- Due to lone pair repulsion, geometry is Pyramidal.

In R₄N⁺Cl⁻ (quaternary ammonium salt):

- Nitrogen has 4 bond pairs and 0 lone pairs.

- Hybridization is sp³.

- With 4 identical regions of electron density, geometry is perfectly Tetrahedral.

\text{R}_3\text{N: sp}^3 \text{, Pyramidal} \quad \& \quad \text{R}_4\text{N}^+: \text{sp}^3 \text{, Tetrahedral}

Answer: (a) sp³, Pyramidal and sp³, Tetrahedral

2

Preparation of Amines

Amines can be prepared via several methods, including reduction of nitro compounds, nitriles, and amides, as well as specialized reactions like Gabriel synthesis and Hofmann bromamide degradation.

2.1 Reduction of Nitro Compounds

R-NO_2 \xrightarrow{Sn/HCl \text{ or } H_2/Pd \text{ or } LiAlH_4} R-NH_2 + H_2O

Key Points: Aromatic nitro compounds (like nitrobenzene) are easily reduced to aryl amines (like aniline) using Sn/HCl or catalytic hydrogenation.

2.2 Gabriel Phthalimide Synthesis

Most Repeated PYQ Topic — Gabriel Synthesis

Gabriel synthesis is used to prepare primary aliphatic amines only. Phthalimide is treated with KOH, then alkyl halide (R-X), and finally hydrolyzed.

\text{Phthalimide} \xrightarrow{KOH} \text{Potassium Phthalimide} \xrightarrow{R-X} N\text{-Alkylphthalimide} \xrightarrow{NaOH/H^+} R-NH_2

Cannot prepare aromatic amines because aryl halides (Ar-X) do not undergo nucleophilic substitution (SN2) easily.
Gives pure 1° amines without 2° or 3° amine contamination.

2.3 Hofmann Bromamide Degradation

High PYQ Frequency — Hofmann Bromamide

Converts an amide to a primary amine with one carbon atom less (step-down reaction).

R-CONH_2 + Br_2 + 4NaOH \rightarrow R-NH_2 + Na_2CO_3 + 2NaBr + 2H_2O

Reagent: Bromine in aqueous NaOH (or KOH).
Product is always a 1° amine containing one carbon less than the amide.
Useful for descending the homologous series.

2.4 Reduction of Nitriles and Oximes

R-CN \xrightarrow{H_2/Ni \text{ or } LiAlH_4} R-CH_2-NH_2\] \[R-CH=NOH \xrightarrow{Na/C_2H_5OH \text{ or } LiAlH_4} R-CH_2-NH_2

PYQ — JEE Main 2023

Question: Which of the following statements is NOT correct regarding Gabriel phthalimide synthesis?

(a) It is used to prepare primary aliphatic amines.

(b) It involves the formation of an intermediate N-alkyl phthalimide.

(c) It can be used to prepare aniline.

(d) It gives a pure primary amine without secondary amine by-products.

Solution:

Gabriel phthalimide synthesis proceeds via an SN2 mechanism where an alkyl halide attacks the nitrogen of potassium phthalimide.

Aryl halides (like chlorobenzene) do not undergo SN2 reactions because the carbon-halogen bond has partial double bond character due to resonance, and the benzene ring is electron-rich, repelling the nucleophile.

Therefore, aniline cannot be prepared using Gabriel phthalimide synthesis.

\text{Aryl halides (Ar-X) do not undergo SN2} \rightarrow \text{No aniline}

Answer: (c) It can be used to prepare aniline.

3

Physical Properties & Basicity of Amines

3.1 Physical Properties

State & Odor: Lower aliphatic amines are gases/fishy odor; higher are solids. Aniline is a toxic oily liquid.
Boiling Point: 1° & 2° amines form intermolecular H-bonds → higher b.p. than 3° amines (which cannot form H-bonds, only accept). Order: 1° > 2° > 3° (for same alkyl group).
Solubility: Lower amines are soluble in water due to H-bonding. Solubility decreases with molecular mass.

3.2 Basicity of Amines

Most Repeated PYQ Topic — Basicity Order

Amines are basic due to the lone pair on nitrogen. In gas phase, basicity order is: 3° > 2° > 1° > NH₃ (pure +I effect).

In aqueous solution, basicity is determined by +I effect, steric hindrance, and solvation:

2^\circ > 1^\circ > 3^\circ > NH_3 \quad \text{(for methyl and ethyl amines)}

3.3 Basicity of Aromatic Amines

Important — Aniline vs Ammonia

Aniline is less basic than ammonia because the lone pair on nitrogen delocalizes into the benzene ring (resonance), making it less available for protonation.

\[NH_3 > C_6H_5NH_2 > (C_6H_5)_2NH > (C_6H_5)_3N\]

JEE Shortcut — Substituted Anilines

EDG (e.g., -OCH₃, -CH₃): Increase electron density on N → increases basicity (p-toluidine > aniline).
EWG (e.g., -NO₂, -Cl): Withdraws electron density → decreases basicity (p-nitroaniline < aniline).
Ortho effect: o-substituted anilines are generally less basic than para and meta due to steric inhibition of solvation.

PYQ — JEE Main 2024 (Apr 4 Shift 2)

Question: The correct decreasing order of basicity in aqueous solution is:

(a) (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃

(b) CH₃NH₂ > (CH₃)₂NH > (CH₃)₃N > NH₃

(c) NH₃ > (CH₃)₃N > (CH₃)₂NH > CH₃NH₂

(d) (CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃

Solution:

In aqueous solution, three factors compete:

1. +I effect: 3° > 2° > 1° (favors 3°)

2. Solvation of conjugate acid: 1° > 2° > 3° (favors 1°, stability of ammonium ion)

3. Steric hindrance: 3° > 2° > 1° (disadvantages 3°)

For methyl amines, the combined effect makes 2° amine the most basic. The 3° amine suffers from high steric hindrance and poor solvation, pushing it below 1°.

\[(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3\]

Answer: (a) (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃

4

Chemical Reactions of Amines

4.1 Carbylamine Reaction

Most Repeated PYQ — Isocyanide Test

Primary amines (aliphatic or aromatic) when heated with chloroform (CHCl₃) and alcoholic KOH form carbylamines (isocyanides) with a foul smell.

R-NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} R-NC + 3KCl + 3H_2O

Key Points: Given only by 1° amines. 2° and 3° amines do not give this test. Used as a test for -NH₂ group.

4.2 Hinsberg's Test

Most Repeated PYQ — Hinsberg Reagent

Hinsberg's reagent is benzene sulfonyl chloride (C₆H₅SO₂Cl). It distinguishes 1°, 2°, and 3° amines.

1° Amine: Forms sulphonamide that is soluble in alkali (NaOH) because it has an acidic hydrogen attached to Nitrogen.
2° Amine: Forms sulphonamide that is insoluble in alkali (no acidic hydrogen on N).
3° Amine: Does not react with Hinsberg's reagent.

4.3 Reaction with Nitrous Acid (HNO₂)

Nitrous acid is prepared in situ from NaNO₂ + HCl.

1° Aliphatic: Forms alcohol and N₂ gas. \(R-NH_2 + HNO_2 \rightarrow R-OH + N_2 + H_2O\)
2° Aliphatic: Forms N-nitrosoamine (yellow oily liquid). \(R_2NH + HNO_2 \rightarrow R_2N-NO\)
3° Aliphatic: Forms soluble salts. \(R_3N + HNO_2 \rightarrow R_3NH^+ NO_2^-\)
1° Aromatic (Aniline): Forms diazonium salt at 0-5°C (Benzene diazonium chloride).

PYQ — JEE Main 2022

Question: An organic compound 'A' on treatment with CHCl₃/KOH gives a foul-smelling compound 'B'. The compound 'A' can be:

(a) CH₃NH₂

(b) (CH₃)₂NH

(c) (CH₃)₃N

(d) C₆H₅N(CH₃)₂

Solution:

The reaction with CHCl₃/KOH to give a foul-smelling compound is the Carbylamine reaction.

The product 'B' is an isocyanide (R-NC), known for its extremely foul odor.

This reaction is given only by primary amines (both aliphatic and aromatic).

Analysis of options:

(a) CH₃NH₂ is a 1° amine → gives carbylamine test ✓

(b) (CH₃)₂NH is a 2° amine → no carbylamine test

(c) (CH₃)₃N is a 3° amine → no carbylamine test

(d) C₆H₅N(CH₃)₂ is a 2° amine → no carbylamine test

CH_3NH_2 + CHCl_3 + 3KOH \rightarrow CH_3NC + 3KCl + 3H_2O

Answer: (a) CH₃NH₂

5

Electrophilic Substitution Reactions of Aniline

The -NH₂ group in aniline is a strongly activating group and is ortho/para directing. This makes aniline highly reactive towards electrophilic substitution.

5.1 Bromination

Due to high activation, aniline reacts with bromine water at room temperature to form a white precipitate of 2,4,6-tribromoaniline.

C_6H_5NH_2 + 3Br_2 \rightarrow C_6H_2Br_3NH_2 \downarrow + 3HBr

5.2 Friedel-Crafts Reaction (Exception)

Critical Exception — Aniline fails Friedel-Crafts

Aniline does not undergo Friedel-Crafts alkylation or acylation. Why?

The Lewis acid catalyst (AlCl₃) acts as an electrophile and attacks the lone pair on nitrogen.
This forms a complex salt: \(C_6H_5NH_2 + AlCl_3 \rightarrow C_6H_5NH_2-AlCl_3\).
The N now becomes positively charged, acting as a strong electron-withdrawing group (-M effect), making the ring highly deactivated.
Hence, electrophilic substitution does not occur.

PYQ — JEE Main 2021

Question: Aniline does not undergo Friedel-Crafts reaction because:

(a) Aniline is basic in nature

(b) Aniline forms a salt with AlCl₃ which deactivates the ring

(c) Aniline is less reactive than benzene

(d) AlCl₃ oxidizes aniline

Solution:

Aniline has a lone pair of electrons on the nitrogen atom.

AlCl₃ (the Lewis acid catalyst required for Friedel-Crafts) is electron-deficient.

AlCl₃ coordinates with the lone pair of aniline to form a complex:

C_6H_5-NH_2 + AlCl_3 \rightarrow C_6H_5-\overset{+}{N}H_2-AlCl_3^-

Due to this coordination, the -NH₂ group effectively becomes -NH₂⁺-AlCl₃, which is a powerful electron-withdrawing group.

The benzene ring becomes highly deactivated and rejects the incoming electrophile.

\text{Forms salt with AlCl}_3 \rightarrow \text{Deactivates the ring}

Answer: (b) Aniline forms a salt with AlCl₃ which deactivates the ring

6

Diazonium Salts & Their Reactions

Diazonium salts have the general formula \(ArN_2^+X^-\). They are highly versatile intermediates in organic synthesis, especially for introducing various functional groups into the benzene ring.

6.1 Preparation (Diazotization)

Important — Diazotization Conditions

Aniline reacts with nitrous acid (generated in situ from NaNO₂ + HCl) at low temperature (273-278 K or 0-5°C) to form benzene diazonium chloride.

C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{0-5^\circ C} C_6H_5N_2^+Cl^- + NaCl + 2H_2O

Why low temperature? At higher temperatures, diazonium salts decompose to form phenols and nitrogen gas.

6.2 Sandmeyer & Gattermann Reactions

High PYQ Frequency — Sandmeyer Reaction

Sandmeyer Reaction

ArN_2^+X^- \xrightarrow{CuCl/HCl} ArCl

ArN_2^+X^- \xrightarrow{CuBr/HBr} ArBr

ArN_2^+X^- \xrightarrow{CuCN/KCN} ArCN

Uses Cu(I) salts.

Gattermann Reaction

ArN_2^+X^- \xrightarrow{Cu/HCl} ArCl

ArN_2^+X^- \xrightarrow{Cu/HBr} ArBr

Uses Cu powder instead of Cu(I) salt.

6.3 Other Important Replacements

Reagent	Group Introduced	Reaction Name
HBF₄, then heat	-F	Balz-Schiemann Reaction
KI	-I	—
H₂O (Boiling)	-OH	Hydrolysis (forms Phenol)
H₃PO₂ (Hypophosphorous acid)	-H	Deamination

6.4 Coupling Reactions

Diazonium salts undergo electrophilic aromatic substitution with highly activated rings like phenol and aniline to form azo dyes.

With Phenol: Occurs in basic medium (pH ~ 9-10). Forms Orange Dye (p-hydroxyazobenzene).
With Aniline: Occurs in acidic medium (pH ~ 4-5). Forms Yellow Dye (p-aminoazobenzene).

PYQ — JEE Main 2023

Question: The reaction of benzenediazonium chloride with fluoroboric acid (HBF₄) followed by heating gives:

(a) Chlorobenzene

(b) Fluorobenzene

(c) Bromobenzene

(d) Iodobenzene

Solution:

When benzenediazonium chloride reacts with fluoroboric acid (HBF₄), it forms benzenediazonium fluoroborate, which is a stable salt.

C_6H_5N_2^+Cl^- + HBF_4 \rightarrow C_6H_5N_2^+BF_4^- \downarrow + HCl

Upon heating this dry salt, it decomposes to give fluorobenzene, nitrogen gas, and boron trifluoride.

C_6H_5N_2^+BF_4^- \xrightarrow{\Delta} C_6H_5F + N_2 \uparrow + BF_3

This specific reaction is known as the Balz-Schiemann reaction, and it is the primary industrial method to introduce a fluorine atom into a benzene ring.

\text{Product: Fluorobenzene (Balz-Schiemann Reaction)}

Answer: (b) Fluorobenzene

7

Name Reactions Summary & Formula Sheet

7.1 Complete Name Reactions Table

Name Reaction	Starting Material	Reagent	Product
Gabriel Synthesis	Phthalimide	KOH, R-X, NaOH	1° Aliphatic Amine
Hofmann Bromamide	RCONH₂	Br₂ / NaOH	1° Amine (one C less)
Carbylamine Test	R-NH₂	CHCl₃ / KOH	R-NC (isocyanide)
Hinsberg Test	1°, 2°, 3° Amines	C₆H₅SO₂Cl	Sulphonamides (solubility varies)
Sandmeyer	ArN₂⁺X⁻	CuCl/HCl, CuBr/HBr, CuCN	ArCl, ArBr, ArCN
Gattermann	ArN₂⁺X⁻	Cu / HCl or HBr	ArCl, ArBr
Balz-Schiemann	ArN₂⁺X⁻	HBF₄, then Δ	ArF
Diazotization	Aniline	NaNO₂ / HCl at 0-5°C	ArN₂⁺Cl⁻
Coupling (Phenol)	ArN₂⁺ + Phenol	Basic Medium (pH 9)	Orange Azo Dye
Coupling (Aniline)	ArN₂⁺ + Aniline	Acidic Medium (pH 4)	Yellow Azo Dye

Common Mistakes Summary

Mistake: Assuming aniline is more basic than ammonia. Fact: Aniline is LESS basic due to resonance delocalization of the lone pair.
Mistake: Using Gabriel synthesis to make aniline. Fact: ArX does not undergo SN2, so no aromatic amines.
Mistake: Heating diazonium salts above 5°C during preparation. Fact: They decompose into phenols at higher temperatures.
Mistake: Thinking aniline gives Friedel-Crafts products. Fact: Aniline forms a complex with AlCl₃ and deactivates the ring.
Mistake: Confusing Sandmeyer with Gattermann. Fact: Sandmeyer uses Cu(I) salts; Gattermann uses Cu powder.
Mistake: Basicity order in gas phase vs aqueous. Fact: Gas phase is 3° > 2° > 1°. Aqueous is 2° > 1° > 3° (for small alkyl groups).

Previous Year Questions Analysis

JEE Main (Last 10 Years)

✓ Basicity Order (Aliphatic & Aromatic): 25%
✓ Diazonium Salt Reactions: 20%
✓ Gabriel & Hofmann Preparation: 18%
✓ Chemical Tests (Carbylamine, Hinsberg): 15%
✓ Electrophilic Substitution of Aniline: 10%
✓ Structure & Hybridization: 12%

JEE Advanced (Last 10 Years)

✓ Multi-step Synthesis via Diazonium: 30%
✓ Identification of Unknown Amines: 25%
✓ Substituted Aniline Basicity: 20%
✓ Coupling Reactions (Azo dyes): 15%
✓ Hofmann Rearrangement Mechanism: 10%

Weightage Analysis

JEE Main: 4-8 marks (1-2 questions)

JEE Advanced: 8-12 marks (2-3 questions)

Difficulty Level: Moderate to Hard

Practice Problem Set

Level 1: Basic (JEE Main Standard)

Write the IUPAC name of (C₂H₅)₂NH.
What is the hybridization and geometry of nitrogen in N(CH₃)₃?
Arrange in increasing order of basic strength: C₆H₅NH₂, NH₃, (CH₃)₂NH.
What is the product of Hofmann bromamide reaction of propanamide (CH₃CH₂CONH₂)?
Which amine does not give Carbylamine test: CH₃NH₂, C₂H₅NH₂, (CH₃)₂NH?

Level 2: Intermediate (JEE Main/Advanced)

Predict the product 'A' in the sequence: Aniline \(\xrightarrow{NaNO_2/HCl, 0^\circ C}\) Salt \(\xrightarrow{CuCN/KCN}\) 'A'.
Why is aniline less basic than cyclohexylamine?
How will you distinguish between CH₃NH₂ and (CH₃)₂NH using a chemical test?
Arrange in decreasing order of basicity: p-toluidine, aniline, p-nitroaniline.
What happens when benzene diazonium chloride is treated with KI?

Level 3: Advanced (JEE Advanced/Olympiad)

A compound 'X' (C₃H₉N) reacts with Hinsberg's reagent to form a product soluble in NaOH. 'X' gives carbylamine test. Identify 'X' and write all reactions.
Convert aniline to fluorobenzene using diazonium chemistry. Mention the specific reaction name.
Predict the major product when aniline is reacted with acetic anhydride, then treated with Br₂/CS₂, and finally hydrolyzed. Explain the necessity of acetylation.
Compare the basicity of aniline, N-methylaniline, and N,N-dimethylaniline. Justify your order.

> FAQ

Frequently Asked Questions

What is the order of basicity of aliphatic amines in aqueous solution?

In aqueous solution, the order of basicity for methyl and ethyl amines is: 2° > 1° > 3° > NH3. This is due to the combined effect of inductive effect, steric hindrance, and solvation effect. The 3° amine is less basic than 1° and 2° because the steric hindrance and poor solvation of its conjugate acid outweigh the +I effect of three alkyl groups.

What is Gabriel Phthalimide Synthesis?

Gabriel Phthalimide Synthesis is a method to prepare primary aliphatic amines only. Phthalimide is treated with KOH to form potassium phthalimide, which then reacts with an alkyl halide (R-X). Subsequent alkaline hydrolysis yields the primary amine (R-NH2). Aromatic amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution easily.

What is the Carbylamine test?

The Carbylamine test (or isocyanide test) is a chemical test for primary amines (both aliphatic and aromatic). When a primary amine is heated with chloroform (CHCl3) and alcoholic potassium hydroxide (KOH), it forms an isocyanide (carbylamine) which has a very foul smell. Secondary and tertiary amines do not give this test.

Why is aniline less basic than ammonia?

Aniline is less basic than ammonia because the lone pair of electrons on the nitrogen atom in aniline is delocalized into the benzene ring via resonance. This makes the lone pair less available for protonation. In ammonia, there is no such resonance, so the lone pair is freely available.

What is Hinsberg's reagent and what is it used for?

Hinsberg's reagent is benzene sulfonyl chloride (C6H5SO2Cl). It is used to distinguish between primary (1°), secondary (2°), and tertiary (3°) amines. Primary amines form a sulphonamide that is soluble in alkali (due to acidic H). Secondary amines form a sulphonamide insoluble in alkali. Tertiary amines do not react.

What is the Sandmeyer reaction?

The Sandmeyer reaction is a method to replace the diazonium group (-N2+Cl-) in an aromatic diazonium salt with a nucleophile like Cl-, Br-, or CN- using copper(I) salts (CuCl, CuBr, CuCN). For example, Benzenediazonium chloride + CuCl/HCl gives Chlorobenzene.

Does aniline give Friedel-Crafts reaction?

No, aniline does not undergo Friedel-Crafts reaction (alkylation or acylation). The basic nitrogen of aniline reacts with the Lewis acid catalyst (AlCl3) to form a complex salt, which deactivates the benzene ring and prevents the electrophilic substitution.

What is the Balz-Schiemann reaction?

The Balz-Schiemann reaction is the conversion of an aryl diazonium salt to an aryl fluoride. When benzenediazonium chloride is treated with fluoroboric acid (HBF4), it forms diazonium fluoroborate, which upon heating decomposes to give fluorobenzene, nitrogen gas, and boron trifluoride.

How many questions come from amines in JEE Mains?

On average 1-2 questions (4-8 marks) appear from Amines in every JEE Main session. Key topics tested include basicity comparison, Gabriel/Hofmann synthesis, identification tests (Carbylamine, Hinsberg), and diazonium salt reactions.

Amines notes, reactions & PYQs for JEE 2026

📑 Table of Contents

Structure & Classification of Amines

1.1 Classification of Amines

Primary (1°), Secondary (2°), and Tertiary (3°) Amines

1.2 Structure of Amines

⭐ Important for JEE — Hybridization & Geometry

1.3 Pyramidal Inversion (Umbrella Effect)

📝 PYQ — JEE Main 2024 (Jan 31 Shift 1)

Preparation of Amines

2.1 Reduction of Nitro Compounds

2.2 Gabriel Phthalimide Synthesis

🔥 Most Repeated PYQ Topic — Gabriel Synthesis

2.3 Hofmann Bromamide Degradation

🔥 High PYQ Frequency — Hofmann Bromamide

2.4 Reduction of Nitriles and Oximes

📝 PYQ — JEE Main 2023

Physical Properties & Basicity of Amines

3.1 Physical Properties

3.2 Basicity of Amines

🔥 Most Repeated PYQ Topic — Basicity Order

3.3 Basicity of Aromatic Amines

⭐ Important — Aniline vs Ammonia

💡 JEE Shortcut — Substituted Anilines

📝 PYQ — JEE Main 2024 (Apr 4 Shift 2)

Chemical Reactions of Amines

4.1 Carbylamine Reaction

🔥 Most Repeated PYQ — Isocyanide Test

4.2 Hinsberg's Test

🔥 Most Repeated PYQ — Hinsberg Reagent

4.3 Reaction with Nitrous Acid (HNO₂)

📝 PYQ — JEE Main 2022

Electrophilic Substitution Reactions of Aniline

5.1 Bromination

5.2 Friedel-Crafts Reaction (Exception)

⚠️ Critical Exception — Aniline fails Friedel-Crafts

📝 PYQ — JEE Main 2021

Diazonium Salts & Their Reactions

6.1 Preparation (Diazotization)

⭐ Important — Diazotization Conditions

6.2 Sandmeyer & Gattermann Reactions

🔥 High PYQ Frequency — Sandmeyer Reaction

6.3 Other Important Replacements

6.4 Coupling Reactions

📝 PYQ — JEE Main 2023

Name Reactions Summary & Formula Sheet

7.1 Complete Name Reactions Table

⚠️ Common Mistakes Summary

📝 Previous Year Questions Analysis

JEE Main (Last 10 Years)

JEE Advanced (Last 10 Years)

Top 8 Most Repeated Question Types

Weightage Analysis

🎯 Practice Problem Set

Level 1: Basic (JEE Main Standard)

Level 2: Intermediate (JEE Main/Advanced)

Level 3: Advanced (JEE Advanced/Olympiad)

Frequently Asked Questions

Related Chemistry Notes

Biomolecules

Aldehydes, Ketones & Carboxylic Acids

Haloalkanes & Haloarenes

Complete Amines Mastery Guide for JEE 2026

🎯 Why This Chapter is Important

Topic-wise Weightage Analysis

📐 Must-Remember Formulas & Values

Basicity Rules

Hofmann Bromamide

Carbylamine Test

Diazonium Reactions

⚡ 5-Step Exam Strategy

Amines — Complete JEE Study Guide

Table of Contents

Important for JEE — Hybridization & Geometry

PYQ — JEE Main 2024 (Jan 31 Shift 1)

Most Repeated PYQ Topic — Gabriel Synthesis

High PYQ Frequency — Hofmann Bromamide

PYQ — JEE Main 2023

Most Repeated PYQ Topic — Basicity Order

Important — Aniline vs Ammonia

JEE Shortcut — Substituted Anilines

PYQ — JEE Main 2024 (Apr 4 Shift 2)

Most Repeated PYQ — Isocyanide Test

Most Repeated PYQ — Hinsberg Reagent

PYQ — JEE Main 2022

Critical Exception — Aniline fails Friedel-Crafts

PYQ — JEE Main 2021

Important — Diazotization Conditions

High PYQ Frequency — Sandmeyer Reaction

PYQ — JEE Main 2023

Common Mistakes Summary

Previous Year Questions Analysis

Practice Problem Set

Why This Chapter is Important

Must-Remember Formulas & Values

5-Step Exam Strategy