The Hook: Why Fish Smell and Proteins Don’t
The Fishy Smell Mystery: Fresh fish is odorless, but decomposing fish releases trimethylamine [(CH₃)₃N] - a volatile, basic amine. That’s why squeezing lemon (citric acid) on fish neutralizes the smell!
But here’s the paradox: Proteins in our body contain thousands of amino groups (-NH₂), yet we don’t smell fishy. Why?
Answer: The amino groups in proteins are:
- Not free (part of peptide bonds)
- Protonated at physiological pH (-NH₃⁺)
- Not volatile (large molecules)
JEE Connection: Understanding amine properties explains:
- Why primary amines have higher boiling points than tertiary
- Why methylamine is water-soluble but aniline isn’t
- Why amines are bases but strength varies
Physical Properties
Boiling Points
General Trends
graph LR
A[Boiling Point Factors] --> B[Hydrogen Bonding]
A --> C[Molecular Mass]
A --> D[Van der Waals Forces]
B --> E[1° > 2° > 3°]
C --> F[Higher mass = Higher bp]For amines of similar mass:
$$\boxed{\text{Primary} > \text{Secondary} > \text{Tertiary}}$$Reason: Hydrogen bonding capability
| Amine Type | H-bonding ability | Boiling Point |
|---|---|---|
| Primary (RNH₂) | 2 N-H bonds → Strong H-bonding | Highest |
| Secondary (R₂NH) | 1 N-H bond → Moderate H-bonding | Medium |
| Tertiary (R₃N) | 0 N-H bonds → No H-bonding | Lowest |
Examples with Data
| Compound | Type | Molecular Mass | Boiling Point (°C) |
|---|---|---|---|
| CH₃NH₂ | 1° | 31 | -6 |
| (CH₃)₂NH | 2° | 45 | 7 |
| (CH₃)₃N | 3° | 59 | 3 |
| C₂H₅NH₂ | 1° | 45 | 17 |
| C₃H₇NH₂ | 1° | 59 | 48 |
Unexpected observation: (CH₃)₃N has higher mass but lower bp than (CH₃)₂NH!
Reason: Hydrogen bonding dominates over mass
JEE Formula:
$$\text{bp} \propto \text{H-bonding} > \text{Mass} > \text{Van der Waals}$$Memory trick: “Hydrogen bonding trumps mass!”
Comparison with Alcohols and Hydrocarbons
For similar mass:
$$\boxed{\text{Alcohols} > \text{Primary Amines} > \text{Hydrocarbons}}$$Example:
- CH₃CH₂OH (Ethanol): 78°C
- CH₃CH₂NH₂ (Ethylamine): 17°C
- CH₃CH₂CH₃ (Propane): -42°C
Why alcohols > amines?
- O is more electronegative than N
- O-H···O bonding stronger than N-H···N
- Oxygen pulls H more strongly
“Harry (H-bonding) Matters Most”
Boiling point order (similar mass): Alcohols > 1° Amines > 2° Amines > 3° Amines > Hydrocarbons
JEE Shortcut: Count N-H bonds:
- 2 N-H (primary) → Highest bp among amines
- 1 N-H (secondary) → Medium bp
- 0 N-H (tertiary) → Lowest bp
Related: Hydrogen bonding
Interactive Demo: Visualize Amine Reactions
See how amines react with various reagents and form different products.
Solubility in Water
General Trends
Lower aliphatic amines (C₁-C₄): Water-soluble
Higher aliphatic amines (>C₆): Water-insoluble
Aromatic amines: Generally insoluble (except in acidic solution)
Explanation
Why lower amines are soluble:
- Form H-bonds with water: R-NH₂···H-O-H
- Small hydrophobic (R) part
- Can accept and donate H-bonds
Why higher amines are insoluble:
- Large hydrophobic alkyl chain
- H-bonding not enough to overcome hydrophobic effect
Why aromatic amines (aniline) are insoluble:
- Large hydrophobic benzene ring
- NH₂ group can’t compensate for phenyl group
Solubility Order
In water (C₁-C₄):
$$\text{Primary} \approx \text{Secondary} > \text{Tertiary}$$Reason: Primary and secondary can both donate and accept H-bonds; tertiary can only accept
| Amine | Solubility in water | H-bonding |
|---|---|---|
| CH₃NH₂ | Highly soluble | Can donate (2 H) & accept |
| (CH₃)₂NH | Highly soluble | Can donate (1 H) & accept |
| (CH₃)₃N | Soluble | Can only accept |
| C₆H₅NH₂ | Insoluble | Hydrophobic phenyl ring |
Q: Aniline is insoluble in water but soluble in dilute HCl. Why?
Solution:
In water:
$$\text{C}_6\text{H}_5\text{NH}_2 \text{ (insoluble - hydrophobic)}$$In HCl:
$$\text{C}_6\text{H}_5\text{NH}_2 + \text{HCl} \rightarrow [\text{C}_6\text{H}_5\text{NH}_3^+]\text{Cl}^-$$Anilinium chloride is ionic → water-soluble!
Key concept: Amines become water-soluble as their salts
JEE Application: Separation of aniline from organic mixtures using acid extraction
Odor and Physical State
Smell characteristics:
| Amines | Odor | Example |
|---|---|---|
| Lower aliphatic (C₁-C₄) | Fishy, pungent | Methylamine, Trimethylamine |
| Higher aliphatic | Less pungent | Octylamine (weak odor) |
| Aromatic | Unpleasant | Aniline (oily, amine-like) |
Physical state at room temperature:
- C₁ (Methylamine): Gas (-6°C bp)
- C₂-C₃: Gases/volatile liquids
- C₄-C₁₈: Liquids
- >C₁₈: Solids
Chemical Properties
Basicity of Amines
Fundamental concept: Amines are Lewis bases (lone pair donors)
$$\text{R-NH}_2 + \text{H}^+ \rightleftarrows \text{R-NH}_3^+$$Base strength measured by:
- Kb (base dissociation constant)
- pKb (lower pKb = stronger base)
- pKa of conjugate acid (higher pKa = stronger base)
Detailed basicity comparison: See Basicity of Amines
Reactions of Amines
graph TD
A[Amine Reactions] --> B[With Acids]
A --> C[Acylation]
A --> D[Benzoylation]
A --> E[Alkylation]
A --> F[Carbylamine]
A --> G[Diazotization]
A --> H[Electrophilic Substitution]1. Reaction with Acids (Salt Formation)
General Reaction:
$$\boxed{\text{R-NH}_2 + \text{HCl} \rightarrow [\text{R-NH}_3^+]\text{Cl}^-}$$Examples:
Aliphatic amine:
$$\text{CH}_3\text{NH}_2 + \text{HCl} \rightarrow [\text{CH}_3\text{NH}_3^+]\text{Cl}^-$$(Methylammonium chloride)
Aromatic amine:
$$\text{C}_6\text{H}_5\text{NH}_2 + \text{HCl} \rightarrow [\text{C}_6\text{H}_5\text{NH}_3^+]\text{Cl}^-$$(Anilinium chloride)
Properties of amine salts:
- Ionic, crystalline solids
- Water-soluble (even if amine wasn’t)
- Non-basic (no lone pair)
- Can regenerate amine with base:
Problem: Separate aniline from an organic mixture
Solution:
Step 1: Add dilute HCl
- Aniline → Anilinium chloride (water-soluble)
- Other organics remain in organic layer
Step 2: Separate aqueous layer (contains anilinium chloride)
Step 3: Add NaOH
- Regenerates free aniline
- Extract with ether
Key concept: Acid-base extraction using salt formation
2. Acylation (Reaction with Acid Chlorides/Anhydrides)
Purpose: Replace H on nitrogen with acyl group (R-CO-)
Acylation with Acetyl Chloride
General Reaction:
$$\boxed{\text{R-NH}_2 + \text{CH}_3\text{COCl} \rightarrow \text{R-NH-CO-CH}_3 + \text{HCl}}$$Examples:
Primary amine:
$$\text{C}_6\text{H}_5\text{NH}_2 + \text{CH}_3\text{COCl} \rightarrow \text{C}_6\text{H}_5\text{-NH-CO-CH}_3 + \text{HCl}$$(Aniline → Acetanilide)
Secondary amine:
$$(CH_3)_2\text{NH} + \text{CH}_3\text{COCl} \rightarrow (\text{CH}_3)_2\text{N-CO-CH}_3 + \text{HCl}$$(Dimethylamine → N,N-Dimethylacetamide)
Tertiary amine:
$$(\text{CH}_3)_3\text{N} + \text{CH}_3\text{COCl} \rightarrow \text{No reaction (no N-H to replace)}$$Acylation with Acetic Anhydride
$$\text{R-NH}_2 + (\text{CH}_3\text{CO})_2\text{O} \rightarrow \text{R-NH-CO-CH}_3 + \text{CH}_3\text{COOH}$$Example:
$$\text{C}_6\text{H}_5\text{NH}_2 + (\text{CH}_3\text{CO})_2\text{O} \rightarrow \text{C}_6\text{H}_5\text{-NH-CO-CH}_3 + \text{CH}_3\text{COOH}$$1. Protection of amino group:
- Prevents oxidation
- Protects during other reactions
- Can be removed later by hydrolysis
2. Reduces basicity:
- Amide is non-basic (lone pair delocalized)
- Useful for controlling reactivity
3. Controls electrophilic substitution:
Aniline (strongly activating):
- Polysubstitution, oxidation issues
Acetanilide (moderately activating):
- Controlled monosubstitution
- Ortho/para directing
- No oxidation
JEE Strategy:
- Direct halogenation of aniline → polysubstitution
- Acetylate first → controlled monosubstitution → deprotect
3. Benzoylation (Schotten-Baumann Reaction)
Reagent: Benzoyl chloride (C₆H₅COCl) + NaOH (aqueous)
General Reaction:
$$\boxed{\text{R-NH}_2 + \text{C}_6\text{H}_5\text{COCl} \xrightarrow{\text{NaOH}} \text{R-NH-CO-C}_6\text{H}_5 + \text{NaCl} + \text{H}_2\text{O}}$$Example:
$$\text{C}_6\text{H}_5\text{NH}_2 + \text{C}_6\text{H}_5\text{COCl} \xrightarrow{\text{NaOH}} \text{C}_6\text{H}_5\text{-NH-CO-C}_6\text{H}_5$$(Aniline → Benzanilide)
Why use NaOH?
- Neutralizes HCl formed
- Reaction proceeds in aqueous medium
Distinguishing amines:
| Amine Type | Product with C₆H₅COCl | Solubility in NaOH |
|---|---|---|
| 1° amine | R-NH-CO-C₆H₅ | Insoluble |
| 2° amine | R₂N-CO-C₆H₅ | Insoluble |
| 3° amine | No reaction | - |
Alternative to Hinsberg: Benzoylation can also distinguish amines
With C₆H₅COCl:
- 1° and 2° amines → Products (benzamides)
- 3° amine → No reaction
But Hinsberg (C₆H₅SO₂Cl) is better because:
- 1° product soluble in base (acidic H)
- 2° product insoluble in base
- 3° no reaction
Distinguishes all three!
Related: Classification of amines
4. Alkylation (Not Recommended)
Problem: Gives mixture of products
$$\text{R-NH}_2 + \text{R'-X} \rightarrow \text{R-NH-R'} + \text{R}_2\text{N-R'} + \text{R}_3\text{N} + \text{R}_4\text{N}^+\text{X}^-$$Why mixture?
- Primary amine more nucleophilic than NH₃
- Keeps reacting with R-X
- Over-alkylation occurs
Better method: Gabriel synthesis for pure primary amines
Related: Preparation of amines
5. Carbylamine Reaction (Isocyanide Test)
Test for: Primary amines ONLY
Reagent: CHCl₃ + alcoholic KOH (heat)
General Reaction:
$$\boxed{\text{R-NH}_2 + \text{CHCl}_3 + 3\text{KOH} \xrightarrow{\Delta} \text{R-N≡C} + 3\text{KCl} + 3\text{H}_2\text{O}}$$Product: Isocyanide (carbylamine) - offensive smell!
Mechanism:
Step 1: CHCl₃ → Dichlorocarbene
$$\text{CHCl}_3 + \text{KOH} \rightarrow :\text{CCl}_2 + \text{KCl} + \text{H}_2\text{O}$$Step 2: Carbene insertion
$$\text{R-NH}_2 + :\text{CCl}_2 \rightarrow \text{R-NH-CCl}_2\text{H}$$Step 3: Elimination to isocyanide
$$\text{R-NH-CCl}_2\text{H} \xrightarrow{-2\text{HCl}} \text{R-N≡C}$$Examples:
Positive test (1° amines):
$$\text{CH}_3\text{NH}_2 \xrightarrow{\text{CHCl}_3/\text{KOH}} \text{CH}_3\text{-NC} + \text{foul smell}$$ $$\text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{CHCl}_3/\text{KOH}} \text{C}_6\text{H}_5\text{-NC} + \text{foul smell}$$Negative test (2° and 3° amines):
$$(CH_3)_2\text{NH} \xrightarrow{\text{CHCl}_3/\text{KOH}} \text{No reaction, no smell}$$Common mistake: “Carbylamine test distinguishes aliphatic from aromatic amines”
Wrong! It distinguishes 1° from 2° and 3° amines
Correct understanding:
- Both aliphatic and aromatic primary amines give positive test
- C₂H₅NH₂ (aliphatic 1°) → Positive ✓
- C₆H₅NH₂ (aromatic 1°) → Positive ✓
- (CH₃)₂NH (aliphatic 2°) → Negative
- C₆H₅NHCH₃ (aromatic 2°) → Negative
Test distinguishes: 1° vs (2° and 3°), NOT aliphatic vs aromatic
6. Reaction with Nitrous Acid (HNO₂)
Most important reaction for JEE!
Different products for 1°, 2°, and 3° amines
Primary Aliphatic Amines
Reaction: Forms alcohol + N₂ gas (via diazonium intermediate)
$$\text{R-NH}_2 + \text{HNO}_2 \xrightarrow{0-5°\text{C}} [\text{R-N}_2^+]\text{Cl}^- \xrightarrow{\text{unstable}} \text{R-OH} + \text{N}_2↑$$Example:
$$\text{CH}_3\text{CH}_2\text{NH}_2 \xrightarrow{\text{NaNO}_2/\text{HCl}} \text{CH}_3\text{CH}_2\text{OH} + \text{N}_2↑$$Observation: Evolution of nitrogen gas (brisk effervescence)
Mechanism:
- Nitrosation: R-NH₂ → R-NH-N=O
- Tautomerization: R-NH-N=O → R-N=N-OH
- Dehydration: R-N=N-OH → R-N₂⁺ (diazonium cation)
- Loss of N₂: R-N₂⁺ → R⁺ + N₂ (carbocation)
- Nucleophilic attack: R⁺ + H₂O → R-OH
Note: Aliphatic diazonium salts are unstable even at 0°C!
Primary Aromatic Amines
Reaction: Forms stable diazonium salt (0-5°C)
$$\text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{0-5°\text{C}} [\text{C}_6\text{H}_5\text{-N}_2^+]\text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O}$$Product: Benzenediazonium chloride (stable below 5°C)
Key difference: Aromatic diazonium salts are stable at low temperature!
Importance: Used for synthesis of many compounds!
Related: Diazonium salts
Secondary Amines (Aliphatic and Aromatic)
Reaction: Forms N-nitrosoamine (yellow oil)
$$\text{R}_2\text{NH} + \text{HNO}_2 \rightarrow \text{R}_2\text{N-N=O} + \text{H}_2\text{O}$$Example:
$$(\text{CH}_3)_2\text{NH} + \text{HNO}_2 \rightarrow (\text{CH}_3)_2\text{N-N=O}$$(N-Nitrosodimethylamine - yellow oil)
Observation: Formation of yellow/orange oily layer
Note: N-Nitrosamines are carcinogenic!
Tertiary Aliphatic Amines
Reaction: Forms nitrite salt (ionic)
$$\text{R}_3\text{N} + \text{HNO}_2 \rightarrow [\text{R}_3\text{NH}^+]\text{NO}_2^-$$Example:
$$(\text{CH}_3)_3\text{N} + \text{HNO}_2 \rightarrow [(\text{CH}_3)_3\text{NH}^+]\text{NO}_2^-$$(Trimethylammonium nitrite)
Observation: Water-soluble salt formed
Tertiary Aromatic Amines
Reaction: Electrophilic substitution at para position
$$\text{C}_6\text{H}_5\text{-N(CH}_3)_2 + \text{HNO}_2 \rightarrow \text{p-}\text{NO-C}_6\text{H}_4\text{-N(CH}_3)_2$$Product: p-Nitroso-N,N-dimethylaniline (green compound)
Observation: Green coloration
“One-Two-Three, Different as can be!”
1° Amine:
- Aliphatic → Alcohol + N₂ (gas evolution)
- Aromatic → Diazonium salt (stable at 0-5°C)
2° Amine:
- Both → Yellow oily N-nitrosamine
3° Amine:
- Aliphatic → Ionic salt (colorless)
- Aromatic → Green p-nitroso product
JEE Shortcut Table:
| Amine | Product | Observation |
|---|---|---|
| 1° aliphatic | R-OH + N₂ | Gas bubbles |
| 1° aromatic | Ar-N₂⁺Cl⁻ | Stable salt |
| 2° (both) | R₂N-N=O | Yellow oil |
| 3° aliphatic | R₃NH⁺NO₂⁻ | Salt |
| 3° aromatic | p-NO-Ar-NR₂ | Green color |
7. Electrophilic Aromatic Substitution in Aniline
Key concept: -NH₂ group is strongly activating and ortho-para directing
Why activating?
- Lone pair on nitrogen delocalizes into benzene ring
- Increases electron density (especially at ortho/para)
- Resonance stabilization of intermediate
Resonance structures:
NH₂ NH₂⁺ NH₂⁺ NH₂⁺
| || || ||
⟨⟩ ←→ ⟨⟩⁻ ←→ ⟨⟩⁻ ←→ ⟨⟩⁻
(ortho⁻) (meta) (para⁻)
Electron density: ortho ≈ para > meta
Halogenation of Aniline
Problem: Aniline is TOO reactive!
Without protection:
$$\text{C}_6\text{H}_5\text{NH}_2 + 3\text{Br}_2(\text{aq}) \rightarrow 2,4,6-\text{tribromoaniline} + 3\text{HBr}$$Product: 2,4,6-Tribromoaniline (white precipitate)
All three positions substituted! (Cannot control)
Solution: Acetylation → Bromination → Deacetylation
Step 1: Protect -NH₂ (acetylation)
$$\text{C}_6\text{H}_5\text{NH}_2 + (\text{CH}_3\text{CO})_2\text{O} \rightarrow \text{C}_6\text{H}_5\text{-NH-CO-CH}_3$$(Aniline → Acetanilide)
Step 2: Controlled bromination
$$\text{C}_6\text{H}_5\text{-NH-CO-CH}_3 + \text{Br}_2 \rightarrow \text{p-Br-C}_6\text{H}_4\text{-NH-CO-CH}_3$$(Mainly para product)
Step 3: Deprotection (hydrolysis)
$$\text{p-Br-C}_6\text{H}_4\text{-NH-CO-CH}_3 \xrightarrow{\text{H}_3\text{O}^+} \text{p-Br-C}_6\text{H}_4\text{-NH}_2 + \text{CH}_3\text{COOH}$$Result: p-Bromoaniline (controlled monosubstitution!)
Why protect amino group?
Free -NH₂:
- Too strongly activating
- Polysubstitution
- Oxidation occurs
- Loss of control
Acetylated -NH-CO-CH₃:
- Moderately activating (resonance with C=O)
- Monosubstitution (controlled)
- No oxidation
- Still ortho/para directing
General strategy:
Aniline → Acetanilide → Substituted acetanilide → Substituted aniline
(too reactive) (moderate) (controlled product) (desired product)
JEE loves this concept: Protection-deprotection is high-yield!
Nitration of Aniline
Problem: Direct nitration oxidizes aniline (HNO₃ is oxidizing)
Solution: Protect as acetanilide
Step 1: Acetylation
$$\text{C}_6\text{H}_5\text{NH}_2 + (\text{CH}_3\text{CO})_2\text{O} \rightarrow \text{C}_6\text{H}_5\text{-NH-CO-CH}_3$$Step 2: Nitration
$$\text{C}_6\text{H}_5\text{-NH-CO-CH}_3 \xrightarrow{\text{HNO}_3/\text{H}_2\text{SO}_4} \text{p-NO}_2\text{-C}_6\text{H}_4\text{-NH-CO-CH}_3$$Step 3: Hydrolysis
$$\text{p-NO}_2\text{-C}_6\text{H}_4\text{-NH-CO-CH}_3 \xrightarrow{\text{H}_2\text{O}/\text{H}^+} \text{p-NO}_2\text{-C}_6\text{H}_4\text{-NH}_2$$Product: p-Nitroaniline
Sulphonation of Aniline
Special case: React as anilinium salt (prevents oxidation)
Step 1: Form anilinium sulfate
$$\text{C}_6\text{H}_5\text{NH}_2 + \text{H}_2\text{SO}_4 \rightarrow [\text{C}_6\text{H}_5\text{NH}_3^+]\text{HSO}_4^-$$Step 2: Heat (sulphonation)
$$[\text{C}_6\text{H}_5\text{NH}_3^+]\text{HSO}_4^- \xrightarrow{180-200°\text{C}} \text{H}_2\text{N-C}_6\text{H}_4\text{-SO}_3\text{H}$$Product: Sulfanilic acid (p-aminobenzenesulfonic acid)
Why meta directing as salt?
- As salt (-NH₃⁺), the group is meta directing
- Positive charge deactivates ortho/para
- But high temperature overcomes this → para product
Oxidation of Amines
Amines are easily oxidized (lone pair is susceptible)
Oxidation of Aliphatic Amines
With K₂Cr₂O₇/H₂SO₄:
Primary:
$$\text{R-CH}_2\text{-NH}_2 \xrightarrow{[\text{O}]} \text{R-CHO or R-COOH}$$Secondary:
$$\text{R}_2\text{CH-NH}_2 \xrightarrow{[\text{O}]} \text{R}_2\text{C=O}$$Tertiary: Complex mixture (C-N bond cleavage)
Oxidation of Aromatic Amines
Aniline is easily oxidized (even by air!)
With K₂Cr₂O₇:
$$\text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{[\text{O}]} \text{Black tar (complex mixture)}$$With mild oxidizing agents (e.g., chromyl chloride):
$$\text{C}_6\text{H}_5\text{NH}_2 \rightarrow \text{Azobenzene, nitrosobenzene, etc.}$$Important: Aniline cannot be oxidized to nitrobenzene!
Common mistake: C₆H₅NH₂ → C₆H₅NO₂ (oxidation)
Reality: Oxidation gives complex mixtures, not nitrobenzene
To get nitrobenzene from aniline: Must use indirect methods (diazotization, etc.)
Fresh aniline: Colorless Old aniline (exposed to air): Brown/black (oxidation products)
Common Mistakes to Avoid
Wrong: “Tertiary amines have highest bp (largest mass)”
Correct: Primary > Secondary > Tertiary (H-bonding trumps mass)
Example:
- (CH₃)₂NH (45 amu, bp = 7°C) > (CH₃)₃N (59 amu, bp = 3°C)
Reason: Dimethylamine has one N-H for H-bonding; trimethylamine has none!
Wrong: C₆H₅NH₂ + Br₂ → p-Bromoaniline
Correct: C₆H₅NH₂ + 3Br₂ → 2,4,6-Tribromoaniline
For monosubstitution: Must protect as acetanilide first!
JEE pattern: Questions often ask for “monobromo product” - always acetylate first!
Wrong: “Carbylamine test distinguishes aliphatic from aromatic amines”
Correct: “Carbylamine test distinguishes 1° from 2° and 3° amines”
Both aliphatic AND aromatic primary amines give positive test!
- CH₃CH₂NH₂ → Positive (1° aliphatic)
- C₆H₅NH₂ → Positive (1° aromatic)
- (CH₃)₂NH → Negative (2°)
Practice Problems
Level 1: Foundation (NCERT)
Q: Arrange in increasing order of boiling points: C₂H₅OH, (CH₃)₂NH, C₂H₅NH₂
Solution:
Molecular masses:
- C₂H₅OH: 46
- (CH₃)₂NH: 45
- C₂H₅NH₂: 45
Similar masses → Compare H-bonding
H-bonding strength:
- C₂H₅OH: O-H···O (strongest, O most electronegative)
- C₂H₅NH₂: N-H···N (2 N-H bonds, moderate)
- (CH₃)₂NH: N-H···N (1 N-H bond, weaker)
Order: (CH₃)₂NH < C₂H₅NH₂ < C₂H₅OH
Answer: Dimethylamine (7°C) < Ethylamine (17°C) < Ethanol (78°C)
Q: Why is aniline insoluble in water but soluble in dilute HCl?
Solution:
In water:
- Aniline (C₆H₅NH₂) has large hydrophobic phenyl group
- Small -NH₂ cannot overcome hydrophobic effect
- Insoluble
In dilute HCl:
$$\text{C}_6\text{H}_5\text{NH}_2 + \text{HCl} \rightarrow [\text{C}_6\text{H}_5\text{NH}_3^+]\text{Cl}^-$$- Forms anilinium chloride (ionic compound)
- Ionic compounds are water-soluble
- Soluble
Key concept: Amines form water-soluble salts with acids
Level 2: JEE Main
Q: How will you distinguish between the following pairs using a chemical test? (a) Aniline and N-methylaniline (b) Aniline and benzylamine
Solution:
(a) Aniline vs N-methylaniline:
Test: Carbylamine reaction (CHCl₃ + KOH, heat)
Aniline (1° amine):
$$\text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{CHCl}_3/\text{KOH}} \text{Foul smell} \text{ (positive)}$$N-methylaniline (2° amine):
$$\text{C}_6\text{H}_5\text{-NH-CH}_3 \xrightarrow{\text{CHCl}_3/\text{KOH}} \text{No smell} \text{ (negative)}$$(b) Aniline vs Benzylamine:
Test: Azo dye test (diazotization + coupling)
Aniline (aromatic 1°):
$$\text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{NaNO}_2/\text{HCl}, 0-5°\text{C}} \text{Diazonium salt (stable)}$$Then couple with phenol → Orange-red dye
Benzylamine (aliphatic 1°):
$$\text{C}_6\text{H}_5\text{CH}_2\text{NH}_2 \xrightarrow{\text{NaNO}_2/\text{HCl}} \text{Alcohol + N}_2 \text{ (no stable diazonium)}$$No dye formation
Related: Diazonium salts
Q: Predict the products:
(a) (CH₃)₂NH + HNO₂ →
(b) C₆H₅N(CH₃)₂ + HNO₂ →
Solution:
(a) Secondary aliphatic amine + HNO₂:
$$(\text{CH}_3)_2\text{NH} + \text{HNO}_2 \rightarrow (\text{CH}_3)_2\text{N-N=O} + \text{H}_2\text{O}$$Product: N-Nitrosodimethylamine (yellow oily liquid)
(b) Tertiary aromatic amine + HNO₂:
$$\text{C}_6\text{H}_5\text{N(CH}_3)_2 + \text{HNO}_2 \rightarrow \text{p-NO-C}_6\text{H}_4\text{-N(CH}_3)_2$$Product: p-Nitroso-N,N-dimethylaniline (green compound)
Observation: Green coloration appears
Level 3: JEE Advanced
Q: How will you prepare p-bromoaniline from aniline?
Solution:
Cannot do: Direct bromination (gives 2,4,6-tribromoaniline)
Strategy: Acetylation → Bromination → Deacetylation
Step 1: Protect amino group
$$\text{C}_6\text{H}_5\text{NH}_2 + (\text{CH}_3\text{CO})_2\text{O} \rightarrow \text{C}_6\text{H}_5\text{-NH-CO-CH}_3 + \text{CH}_3\text{COOH}$$(Acetanilide - moderately activating)
Step 2: Controlled bromination
$$\text{C}_6\text{H}_5\text{-NH-CO-CH}_3 + \text{Br}_2 \rightarrow \text{p-Br-C}_6\text{H}_4\text{-NH-CO-CH}_3 + \text{HBr}$$(p-Bromoacetanilide - major product)
Step 3: Deprotect (hydrolyze)
$$\text{p-Br-C}_6\text{H}_4\text{-NH-CO-CH}_3 \xrightarrow{\text{H}_3\text{O}^+ \text{ or NaOH}} \text{p-Br-C}_6\text{H}_4\text{-NH}_2 + \text{CH}_3\text{COOH}$$Final product: p-Bromoaniline
Why this works:
- Acetyl group reduces reactivity (prevents polysubstitution)
- Still ortho/para directing (mainly para due to steric factors)
- Acetyl group easily removed
Q: An organic compound (A) with molecular formula C₇H₉N gives positive carbylamine test. When A reacts with HNO₂, stable diazonium salt (B) is formed. Identify A and B.
Solution:
Analysis:
Clue 1: Positive carbylamine test → Primary amine
Clue 2: Stable diazonium salt → Aromatic primary amine (Aliphatic diazonium salts decompose immediately)
Molecular formula: C₇H₉N
Possible aromatic 1° amines with C₇H₉N:
- C₆H₅-NH-CH₃ (N-methylaniline) - But this is 2° amine ✗
- CH₃-C₆H₄-NH₂ (Toluidine) - 1° amine ✓
Compound A: CH₃-C₆H₄-NH₂ (Toluidine - ortho, meta, or para)
Most common: p-Toluidine (4-methylaniline)
Reaction with HNO₂:
$$\text{CH}_3\text{-C}_6\text{H}_4\text{-NH}_2 \xrightarrow{\text{NaNO}_2/\text{HCl}, 0-5°\text{C}} [\text{CH}_3\text{-C}_6\text{H}_4\text{-N}_2^+]\text{Cl}^-$$Compound B: 4-Methylbenzenediazonium chloride
Verification:
- C₇H₉N ✓
- Carbylamine positive (1° amine) ✓
- Stable diazonium (aromatic) ✓
Quick Revision Box
| Property/Reaction | Key Points | JEE Formula/Observation |
|---|---|---|
| Boiling Point | H-bonding dominant | 1° > 2° > 3° (similar mass) |
| Solubility | Lower amines soluble | C₁-C₄ water-soluble |
| Salt Formation | With acids | R-NH₂ + HCl → R-NH₃⁺Cl⁻ |
| Acylation | Protection/deactivation | 1°, 2° react; 3° no reaction |
| Benzoylation | Schotten-Baumann | With C₆H₅COCl/NaOH |
| Carbylamine | 1° amines only | Foul smell (isocyanide) |
| With HNO₂ | Different for each type | 1° aromatic → stable diazonium |
| Halogenation | Aniline too reactive | Acetylate first for mono-substitution |
| Oxidation | Aniline → black tar | Easily oxidized (store carefully) |
Decision Tree: Identifying Amines
Given amine, determine type?
│
├─ Add HNO₂ (NaNO₂/HCl)
│ ├─ Stable diazonium salt → 1° aromatic
│ ├─ N₂ gas evolution → 1° aliphatic
│ ├─ Yellow oily product → 2° amine
│ ├─ Ionic salt (colorless) → 3° aliphatic
│ └─ Green color → 3° aromatic
│
├─ Carbylamine test (CHCl₃/KOH)
│ ├─ Foul smell → 1° amine
│ └─ No smell → 2° or 3° amine
│
└─ Hinsberg test (C₆H₅SO₂Cl)
├─ Product soluble in KOH → 1° amine
├─ Product insoluble in KOH → 2° amine
└─ No reaction → 3° amine
Connection to Other Topics
Prerequisites:
- Classification of Amines - Structure, nomenclature
- Preparation of Amines - Synthesis methods
- Chemical Bonding - H-bonding, hybridization
Related Topics:
- Basicity of Amines - Detailed basicity comparison
- Diazonium Salts - From aromatic amines
- Coupling Reactions - Azo dye formation
Applications:
- Haloalkanes - Substitution reactions
- Aromatic Compounds - Electrophilic substitution
Summary
1. Physical Properties:
Boiling Points:
- 1° > 2° > 3° (H-bonding effect)
- Alcohols > Amines > Hydrocarbons
- H-bonding dominates over molecular mass
Solubility:
- Lower amines (C₁-C₄) water-soluble
- All amines soluble as salts
2. Chemical Properties:
Salt Formation:
- R-NH₂ + HCl → R-NH₃⁺Cl⁻ (water-soluble)
Acylation/Benzoylation:
- Protects amino group
- Reduces reactivity
- 1° and 2° react; 3° doesn’t
3. Identification Tests:
Carbylamine (CHCl₃/KOH):
- Only 1° amines → Foul smell
Nitrous Acid (HNO₂):
- 1° aliphatic → Alcohol + N₂
- 1° aromatic → Stable diazonium salt
- 2° → Yellow N-nitrosamine
- 3° aliphatic → Salt
- 3° aromatic → Green p-nitroso
Hinsberg (C₆H₅SO₂Cl):
- 1° → Soluble in KOH
- 2° → Insoluble
- 3° → No reaction
4. Aromatic Substitution:
Direct halogenation: Polysubstitution (2,4,6-tribromo)
For monosubstitution:
- Acetylate (protect)
- Halogenate (controlled)
- Deacetylate (regenerate)
5. JEE Strategy:
| To distinguish | Use test |
|---|---|
| 1° from 2°, 3° | Carbylamine (foul smell) |
| 1° aliphatic from 1° aromatic | HNO₂ (N₂ vs stable salt) |
| All three types | Hinsberg or HNO₂ |
“Amines show unique behavior based on degree - from boiling points to chemical tests, the number of hydrogens on nitrogen matters!”
Next, dive deep into basicity of amines to understand why different amines have different base strengths!