Basicity of Amines: Order and Factors

Complete guide to amine basicity - comparison of aliphatic, aromatic amines, effect of substituents, and factors affecting base strength for JEE

15 min read

The Hook: Why Soap Feels Slippery

Connect: Real Life → Chemistry

Ever wondered why soap feels slippery? Soap molecules contain carboxylate anions (weak bases) that react with water to make it slightly basic. This alkalinity dissolves grease and oils.

Similarly, ammonia (NH₃) in cleaning products makes them basic - it’s why window cleaners often have a pungent smell!

But here’s the JEE puzzle:

  • NH₃ (ammonia) is basic
  • CH₃NH₂ (methylamine) is more basic than NH₃
  • C₆H₅NH₂ (aniline) is less basic than NH₃

Why does basicity change when we simply replace H with alkyl or aryl groups?

Understanding amine basicity helps explain:

  • Why dimethylamine is used in pharmaceuticals (stronger base)
  • Why aniline doesn’t dissolve in water but dissolves in acid
  • How to predict relative base strengths in complex molecules

The Core Concept

What Makes Amines Basic?

Amines are Lewis bases because they have a lone pair of electrons on nitrogen that can accept a proton (H⁺).

General reaction:

$$\boxed{\text{R-NH}_2 + \text{H}_2\text{O} \rightleftarrows \text{R-NH}_3^+ + \text{OH}^-}$$

Base strength depends on:

  1. Availability of lone pair on nitrogen
  2. Stability of conjugate acid (R-NH₃⁺)
JEE Weightage
Basicity of Amines: 3-4 questions in JEE Main, 2-3 in JEE Advanced High-yield areas: Basicity order (aliphatic vs aromatic), effect of substituents, solvent effect

Measuring Basicity

Base Dissociation Constant (Kb)

$$\text{R-NH}_2 + \text{H}_2\text{O} \rightleftarrows \text{R-NH}_3^+ + \text{OH}^-$$ $$K_b = \frac{[\text{R-NH}_3^+][\text{OH}^-]}{[\text{R-NH}_2]}$$

Larger Kb → Stronger base

pKb Scale

$$\text{p}K_b = -\log K_b$$

Smaller pKb → Stronger base

pKa of Conjugate Acid

$$\text{R-NH}_3^+ + \text{H}_2\text{O} \rightleftarrows \text{R-NH}_2 + \text{H}_3\text{O}^+$$ $$K_a = \frac{[\text{R-NH}_2][\text{H}_3\text{O}^+]}{[\text{R-NH}_3^+]}$$

Relationship:

$$\boxed{K_a \times K_b = K_w = 10^{-14}}$$ $$\boxed{\text{p}K_a + \text{p}K_b = 14}$$

Stronger base → Higher pKa of conjugate acid

JEE Shortcut: Reading pKa Values

For amines, higher pKa of conjugate acid = stronger base

If given: “pKa of R-NH₃⁺ = 10.5”

  • Higher than NH₄⁺ (pKa ≈ 9.25)
  • R-NH₂ is more basic than NH₃

Memory trick: “PKa Propositional to basic strength”

When comparing bases, look at pKa of their conjugate acids: Higher pKa → Stronger base

Basicity in Gas Phase vs Aqueous Solution

Gas Phase Basicity

In gas phase (no solvent):

Order:

$$\boxed{3° > 2° > 1° > \text{NH}_3}$$

Example:

$$(CH_3)_3\text{N} > (\text{CH}_3)_2\text{NH} > \text{CH}_3\text{NH}_2 > \text{NH}_3$$

Reason: +I effect of alkyl groups

  • Alkyl groups are electron-donating (+I effect)
  • Increase electron density on nitrogen
  • More electron density → more available lone pair → stronger base
  • More alkyl groups → stronger +I effect → stronger base

Simple explanation: More methyl groups = More electron density on N = Easier to donate lone pair = Stronger base

Aqueous Solution Basicity

In aqueous solution:

Order:

$$\boxed{2° > 1° > 3° > \text{NH}_3}$$

Example (most important for JEE):

$$(\text{CH}_3)_2\text{NH} > \text{CH}_3\text{NH}_2 > (\text{CH}_3)_3\text{N} > \text{NH}_3$$

Why different from gas phase?

Two competing factors:

  1. Inductive effect (+I): 3° > 2° > 1° (electron donation)
  2. Solvation effect (H-bonding with H₂O): 1° > 2° > 3° (stabilization)

Winner in water: Solvation effect dominates for tertiary amines!

The Solvation Paradox

Why is tertiary amine LESS basic in water despite having strongest +I effect?

Key concept: Base strength depends on stability of conjugate acid, not just the base!

Primary amine (R-NH₃⁺):

  • Three N-H bonds
  • Can form 3 hydrogen bonds with water
  • Highly stabilized
  • More stable conjugate acid → stronger base

Secondary amine (R₂-NH₂⁺):

  • Two N-H bonds
  • Can form 2 hydrogen bonds with water
  • Moderately stabilized

Tertiary amine (R₃-NH⁺):

  • One N-H bond
  • Can form only 1 hydrogen bond with water
  • Least stabilized
  • Less stable conjugate acid → weaker base

JEE Insight:

Stronger +I effect (3°) + Poor solvation = Weaker base in water
Moderate +I effect (2°) + Good solvation = Stronger base in water

Diagram:

         H
         |
    H---O⁺---N---R     (3 H-bonds possible)
         |   |
         H   H
       (R-NH₃⁺ highly stabilized)

         H
         |
    H---O⁺---N---R     (2 H-bonds possible)
         |   |
         H   R
       (R₂NH₂⁺ moderately stabilized)

         H
         |
    H---O⁺---N---R     (1 H-bond possible)
             |
             R
       (R₃NH⁺ least stabilized)

Interactive Demo: Visualize Basicity Factors

See how solvation and electronic effects influence amine basicity.

Memory Trick: Aqueous Basicity Order

“Secondary Students score Higher Than Tertiary in Tests”

Secondary > Primary > Tertiary

Order: 2° > 1° > 3° > NH₃

Specific for aliphatic amines: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃

JEE loves this: Always assume aqueous solution unless stated otherwise!

pKb values (aqueous):

  • (CH₃)₂NH: pKb = 3.27 (strongest)
  • CH₃NH₂: pKb = 3.38
  • (CH₃)₃N: pKb = 4.22
  • NH₃: pKb = 4.75 (weakest)

Lower pKb = Stronger base ✓

Aromatic Amines: Aniline and Derivatives

Why is Aniline Less Basic than Ammonia?

Aniline (C₆H₅-NH₂): pKb = 9.42

Ammonia (NH₃): pKb = 4.75

Aniline is about 10⁶ times weaker base than ammonia!

Reason: Resonance delocalization

The lone pair on nitrogen delocalizes into the benzene ring through resonance.

Resonance structures:

   NH₂              NH₂⁺             NH₂⁺             NH₂⁺
    |                ||               ||               ||
    ⟨⟩     ←→       ⟨⟩⁻     ←→      ⟨⟩⁻     ←→      ⟨⟩⁻
 (neutral)      (ortho⁻)         (para⁻)         (meta)

Consequences:

  1. Lone pair is less available for protonation
  2. Nitrogen-carbon bond has partial double bond character
  3. Electron density on nitrogen decreases
  4. Weaker base

In anilinium ion (C₆H₅-NH₃⁺):

  • No resonance possible (no lone pair)
  • Less stable than expected
  • Makes aniline weak base
Comparing Benzylamine and Aniline

Q: Why is benzylamine more basic than aniline?

Benzylamine: C₆H₅-CH₂-NH₂ (pKb = 4.70) Aniline: C₆H₅-NH₂ (pKb = 9.42)

Benzylamine is ~30,000 times more basic!

Reason:

In benzylamine:

  • Nitrogen is on CH₂, not directly on benzene ring
  • No resonance between N lone pair and benzene ring
  • Behaves like aliphatic amine
  • Full lone pair availability

In aniline:

  • Nitrogen directly on benzene ring
  • Lone pair delocalizes into ring
  • Reduced lone pair availability

JEE Trap: Don’t confuse these two!

  • Benzylamine → Aliphatic character (strong base)
  • Aniline → Aromatic character (weak base)

Related: Classification of amines

Effect of Substituents on Basicity

Electron-Donating Groups (EDG)

Increase basicity of aniline

Examples: -CH₃, -OCH₃, -NH₂ (at ortho/para positions)

Mechanism: +I or +R effect

  • Increase electron density on nitrogen
  • Lone pair more available
  • Stronger base

Basicity order:

$$\text{p-CH}_3\text{O-C}_6\text{H}_4\text{-NH}_2 > \text{p-CH}_3\text{-C}_6\text{H}_4\text{-NH}_2 > \text{C}_6\text{H}_5\text{-NH}_2$$

p-Methoxyaniline (p-anisidine) > p-Toluidine > Aniline

Why?

  • -OCH₃ has +R effect (donates electrons via resonance)
  • -CH₃ has +I effect (donates electrons via σ-bond)
  • Both increase electron density on NH₂
  • More electron density → more basic

Electron-Withdrawing Groups (EWG)

Decrease basicity of aniline

Examples: -NO₂, -CN, -COOH, -X (halogens), -CHO

Mechanism: -I or -R effect

  • Decrease electron density on nitrogen
  • Lone pair less available
  • Weaker base

Basicity order:

$$\text{C}_6\text{H}_5\text{-NH}_2 > \text{p-Cl-C}_6\text{H}_4\text{-NH}_2 > \text{p-NO}_2\text{-C}_6\text{H}_4\text{-NH}_2$$

Aniline > p-Chloroaniline > p-Nitroaniline

Why?

  • -NO₂ has strong -R and -I effects
  • Withdraws electrons from ring and nitrogen
  • Destabilizes lone pair
  • Much weaker base

p-Nitroaniline: pKb ≈ 13 (almost neutral!)

Position of Substituent Matters!

Effect of -NO₂ group at different positions:

ortho-Nitroaniline: pKb ≈ 14.3 meta-Nitroaniline: pKb ≈ 11.5 para-Nitroaniline: pKb ≈ 13.0

Order: meta > para > ortho

Why meta is most basic among nitroanilines?

ortho/para-NO₂:

  • Direct resonance with NH₂ through ring
  • Strong electron withdrawal via -R effect
  • Maximum decrease in electron density on N

meta-NO₂:

  • No direct resonance with NH₂
  • Only -I effect (through σ-bonds)
  • Less electron withdrawal
  • Relatively more basic

JEE Note: For -R groups (like -NO₂), meta isomer is MORE basic than ortho/para!

Comprehensive Basicity Order

Aliphatic Amines (in water)

$$\boxed{2° > 1° > 3° > \text{NH}_3}$$

Specific example:

$$(\text{CH}_3)_2\text{NH} > \text{C}_2\text{H}_5\text{NH}_2 > (\text{CH}_3)_3\text{N} > \text{NH}_3$$

pKb values:

  • (CH₃)₂NH: 3.27
  • C₂H₅NH₂: 3.35
  • (CH₃)₃N: 4.22
  • NH₃: 4.75

Aromatic vs Aliphatic

$$\boxed{\text{Aliphatic amines} >> \text{Aromatic amines}}$$

Example:

$$\text{CH}_3\text{NH}_2 >> \text{C}_6\text{H}_5\text{NH}_2$$

Reason: Resonance in aromatic amines reduces lone pair availability

Substituted Anilines

Electron-donating substituents:

$$\text{p-CH}_3\text{O-C}_6\text{H}_4\text{-NH}_2 > \text{p-CH}_3\text{-C}_6\text{H}_4\text{-NH}_2 > \text{C}_6\text{H}_5\text{-NH}_2$$

Electron-withdrawing substituents:

$$\text{C}_6\text{H}_5\text{-NH}_2 > \text{p-Cl-C}_6\text{H}_4\text{-NH}_2 > \text{p-NO}_2\text{-C}_6\text{H}_4\text{-NH}_2$$

Combined order (decreasing basicity):

$$\text{p-Anisidine} > \text{p-Toluidine} > \text{Aniline} > \text{p-Chloroaniline} > \text{p-Nitroaniline}$$

The Ultimate JEE Order

For typical JEE questions:

$$\boxed{(\text{CH}_3)_2\text{NH} > \text{CH}_3\text{NH}_2 > \text{C}_6\text{H}_5\text{CH}_2\text{NH}_2 > \text{NH}_3 > \text{C}_6\text{H}_5\text{NH}_2}$$

In words: Dimethylamine > Methylamine > Benzylamine > Ammonia > Aniline

Key points:

  1. 2° aliphatic is strongest
  2. All aliphatic amines > NH₃ > aromatic amines
  3. Benzylamine behaves as aliphatic (strong base)
  4. Aniline is weakest due to resonance
Memory Trick: Complete Basicity Order

“Di-Met Mary Ben Before Amy Ann”

  • Dimethylamine (strongest)
  • Methylamine
  • Benzylamine
  • (Am)monia (NH₃)
  • Aniline (weakest)

JEE Strategy:

  1. If comparing aliphatic amines only → Remember: 2° > 1° > 3°
  2. If comparing with aromatic → All aliphatic > NH₃ > aromatic
  3. If comparing substituted anilines → EDG increase, EWG decrease basicity
  4. Always assume aqueous solution (unless gas phase mentioned)

Quick test: “Which is more basic, (C₂H₅)₂NH or C₆H₅NH₂?” Answer: (C₂H₅)₂NH (aliphatic 2° » aromatic)

Special Cases and Exceptions

Guanidine: The Exceptionally Strong Base

Structure:

      NH₂
       |
   H₂N-C=NH

pKb ≈ 0.4 (extremely strong base, almost like NaOH!)

Why so basic?

Conjugate acid (guanidinium ion) is highly stabilized:

      NH₂⁺              NH₂               NH₂
       |                ||                ||
   H₂N-C-NH    ←→   H₂N-C-NH₂⁺   ←→   H₂N⁺-C-NH₂

Three equivalent resonance structures:

  • Positive charge delocalized over three nitrogen atoms
  • Highly stable conjugate acid
  • Makes guanidine very strong base

JEE Relevance: Found in arginine (amino acid)

Pyrrole: The “Amine” That’s Not Basic

Structure:

    NH
   /  \
  |    |
  \____/

Pyrrole is NOT basic! (pKb > 14)

Why?

  • Lone pair on N is part of aromatic sextet (6 π electrons)
  • Delocalized in the ring (aromatic stability)
  • Not available for protonation
  • Protonation would destroy aromaticity

JEE Note: Don’t confuse with aniline!

  • Aniline: Weak base (pKb = 9.42)
  • Pyrrole: Not a base (pKb > 14)

Aniline vs Diphenylamine vs Triphenylamine

Order:

$$\text{Aniline} > \text{Diphenylamine} > \text{Triphenylamine}$$

Reason:

  • More phenyl groups → More resonance delocalization
  • Greater decrease in lone pair availability
  • Weaker base

pKb values:

  • Aniline: 9.42
  • Diphenylamine: 13.1
  • Triphenylamine: ~14 (essentially non-basic)

Common Mistakes to Avoid

Mistake #1: Using Gas Phase Order in Aqueous Solution

Wrong: “(CH₃)₃N is most basic in water because it has most +I effect”

Correct: “(CH₃)₂NH is most basic in water due to optimal balance of +I effect and solvation”

Remember:

  • Gas phase: 3° > 2° > 1°
  • Aqueous: 2° > 1° > 3°

JEE assumes aqueous solution unless specified!

Mistake #2: Confusing Benzylamine with Aniline

Wrong: “Both have C₆H₅ group, so both are weak bases”

Correct:

  • Benzylamine (C₆H₅-CH₂-NH₂) → Strong base (aliphatic)
  • Aniline (C₆H₅-NH₂) → Weak base (aromatic)

Key: Is N directly on the ring?

  • Yes (aniline) → Weak base
  • No (benzylamine) → Strong base
Mistake #3: Wrong Effect of Substituents

Wrong: “All substituents on aniline decrease basicity”

Correct:

  • EDG (like -OCH₃, -CH₃) → Increase basicity
  • EWG (like -NO₂, -CN) → Decrease basicity

Also wrong: “Position doesn’t matter”

Correct: For -R groups (like -NO₂):

  • meta isomer is MORE basic than ortho/para
  • Ortho/para have direct resonance with NH₂
Mistake #4: Ignoring Resonance

Wrong: “Aniline should be more basic than ammonia because phenyl is electron-donating”

Correct: Aniline is LESS basic because:

  • Lone pair delocalizes into benzene ring (resonance)
  • Reduced availability for protonation
  • Resonance effect > inductive effect

JEE tip: When aromatic ring is attached, always consider resonance!

Practice Problems

Level 1: Foundation (NCERT)

Problem 1: Basic Order

Q: Arrange in increasing order of basicity: C₆H₅NH₂, (C₂H₅)₂NH, C₂H₅NH₂

Solution:

Analysis:

  • C₆H₅NH₂: Aromatic amine (weak base, resonance)
  • C₂H₅NH₂: Aliphatic primary amine
  • (C₂H₅)₂NH: Aliphatic secondary amine

In aqueous solution: Secondary > Primary > Aromatic

Order (increasing basicity):

$$\boxed{\text{C}_6\text{H}_5\text{NH}_2 < \text{C}_2\text{H}_5\text{NH}_2 < (\text{C}_2\text{H}_5)_2\text{NH}}$$

pKb values (for verification):

  • C₆H₅NH₂: 9.42
  • C₂H₅NH₂: 3.35
  • (C₂H₅)₂NH: 3.27

Lower pKb = Stronger base ✓

Problem 2: Substituent Effect

Q: Why is aniline more basic than p-nitroaniline?

Solution:

Aniline (C₆H₅-NH₂):

  • Lone pair on N partially delocalized into ring
  • pKb = 9.42

p-Nitroaniline (NO₂-C₆H₄-NH₂):

  • -NO₂ is strong electron-withdrawing group (-R and -I effects)
  • Withdraws electrons from ring AND from NH₂
  • Further decreases electron density on nitrogen
  • Lone pair less available for protonation
  • pKb ≈ 13 (much weaker base)

Conclusion: Aniline » p-Nitroaniline

JEE Insight: EWG on aromatic amines decrease basicity even further

Level 2: JEE Main

Problem 3: Comprehensive Comparison

Q: Arrange in decreasing order of basicity: (A) C₆H₅CH₂NH₂ (B) C₆H₅NH₂ (C) C₆H₅N(CH₃)₂ (D) CH₃NH₂

Solution:

Classification:

  • (A) Benzylamine: Aliphatic 1° (N on CH₂, not ring)
  • (B) Aniline: Aromatic 1°
  • (C) N,N-Dimethylaniline: Aromatic 3°
  • (D) Methylamine: Aliphatic 1°

Key principles:

  1. Aliphatic > Aromatic (resonance effect)
  2. Among aliphatic: 2° > 1° > 3° (in water)
  3. Among aromatic: EDG increase basicity

Analysis:

(D) CH₃NH₂: Aliphatic 1° (strong base)

(A) C₆H₅CH₂NH₂: Aliphatic 1° (strong base, similar to D)

(C) C₆H₅N(CH₃)₂: Aromatic 3°, but has two +I CH₃ groups on N

  • More basic than simple aniline

(B) C₆H₅NH₂: Aromatic 1° (weakest)

Order (decreasing basicity):

$$\boxed{\text{D} > \text{A} > \text{C} > \text{B}}$$

Or: CH₃NH₂ > C₆H₅CH₂NH₂ > C₆H₅N(CH₃)₂ > C₆H₅NH₂

Problem 4: pKb Interpretation

Q: The pKb values of three amines are given below:

  • Amine X: pKb = 3.3
  • Amine Y: pKb = 9.4
  • Amine Z: pKb = 4.7

Arrange them in increasing order of basicity.

Solution:

Key relationship: Lower pKb = Stronger base

Given pKb values:

  • X: 3.3 (lowest → strongest base)
  • Y: 9.4 (highest → weakest base)
  • Z: 4.7 (middle)

Order (increasing basicity):

$$\boxed{\text{Y} < \text{Z} < \text{X}}$$

Likely identities:

  • X (pKb 3.3): Aliphatic amine (e.g., CH₃NH₂ or (CH₃)₂NH)
  • Z (pKb 4.7): Ammonia (NH₃)
  • Y (pKb 9.4): Aromatic amine (e.g., C₆H₅NH₂)

JEE Tip: Can also use pKa of conjugate acid:

  • pKa + pKb = 14
  • Higher pKa of conjugate acid → Stronger base

Level 3: JEE Advanced

Problem 5: Substituent Position Effect

Q: Arrange in decreasing order of basicity: (A) m-Nitroaniline (B) p-Nitroaniline (C) o-Nitroaniline (D) Aniline

Solution:

Principle: -NO₂ is electron-withdrawing (decreases basicity)

Position matters:

(D) Aniline: No substituent (reference)

(A) m-Nitroaniline:

  • -NO₂ at meta position
  • Only -I effect (through σ-bonds)
  • Cannot directly resonate with NH₂
  • Least decrease in basicity

(B) p-Nitroaniline:

  • -NO₂ at para position
  • Both -I and -R effects
  • Direct resonance with NH₂ possible
  • Significant decrease in basicity

(C) o-Nitroaniline:

  • -NO₂ at ortho position
  • Both -I and -R effects
  • Direct resonance with NH₂
  • Plus steric hindrance (intramolecular H-bonding)
  • Maximum decrease in basicity

Resonance in p-nitroaniline:

  NH₂⁺              NH₂
   ||                |
   ⟨⟩      ←→       ⟨⟩=O
   ||
  NO₂              NO₂⁻

Electron density withdrawn from N → weaker base

Order (decreasing basicity):

$$\boxed{\text{D} > \text{A} > \text{B} > \text{C}}$$

Or: Aniline > m-Nitroaniline > p-Nitroaniline > o-Nitroaniline

pKb values:

  • Aniline: 9.42
  • m-Nitroaniline: 11.5
  • p-Nitroaniline: 13.0
  • o-Nitroaniline: 14.3
Problem 6: Complex Reasoning

Q: Explain why (C₂H₅)₂NH is more basic than (C₂H₅)₃N in aqueous solution, but the reverse is true in gas phase.

Solution:

In Gas Phase:

Basicity order: (C₂H₅)₃N > (C₂H₅)₂NH

Reason: Only inductive effect matters

  • Tertiary has three electron-donating ethyl groups
  • Maximum +I effect → most electron density on N
  • Lone pair most available → strongest base

No solvation to complicate matters

In Aqueous Solution:

Basicity order: (C₂H₅)₂NH > (C₂H₅)₃N

Reason: Solvation of conjugate acid is crucial

(C₂H₅)₂NH₂⁺ (from secondary amine):

  • Has two N-H bonds
  • Can form two H-bonds with water
  • Well-stabilized conjugate acid
  • Equilibrium favors product
  • Stronger base

(C₂H₅)₃NH⁺ (from tertiary amine):

  • Has only one N-H bond
  • Can form only one H-bond with water
  • Poorly stabilized conjugate acid
  • Equilibrium doesn’t favor product as much
  • Weaker base

Key Insight:

Gas phase: Inductive effect dominates → 3° > 2°
Aqueous:   Solvation effect dominates → 2° > 3°

JEE Conclusion: Base strength depends on medium!

Always assume aqueous solution for JEE unless stated otherwise.

Quick Revision Box

ComparisonOrder (Decreasing Basicity)Reason
Aliphatic (aq)2° > 1° > 3° > NH₃Solvation + Inductive
Aliphatic (gas)3° > 2° > 1° > NH₃Inductive only
AromaticEDG-aniline > Aniline > EWG-anilineSubstituent effect
OverallAliphatic » NH₃ » AromaticResonance in aromatic
Specific(CH₃)₂NH > CH₃NH₂ > C₆H₅CH₂NH₂ > NH₃ > C₆H₅NH₂Combined effects
Substitutedp-OCH₃ > p-CH₃ > H > p-Cl > p-NO₂EDG vs EWG

Key Values to Remember:

  • (CH₃)₂NH: pKb = 3.27 (strongest aliphatic)
  • NH₃: pKb = 4.75 (reference)
  • C₆H₅NH₂: pKb = 9.42 (aromatic)
  • p-NO₂-C₆H₄-NH₂: pKb ≈ 13 (very weak)

Decision Tree: Comparing Basicity

Compare two amines?
│
├─ Both aliphatic?
│  ├─ In gas phase → 3° > 2° > 1°
│  └─ In water → 2° > 1° > 3°
│
├─ Both aromatic?
│  ├─ Check substituents
│  │  ├─ EDG (+I, +R) → More basic
│  │  └─ EWG (-I, -R) → Less basic
│  └─ Check position (for -R groups)
│     └─ meta > para > ortho
│
├─ One aliphatic, one aromatic?
│  └─ Aliphatic >> Aromatic
│
└─ Check special cases
   ├─ Benzylamine → Aliphatic character
   ├─ Pyrrole → Non-basic
   └─ Guanidine → Extremely basic

Connection to Other Topics

Prerequisites:

Related Topics:

Applications:

Summary

Key Takeaways

1. Basicity Fundamentals:

What makes amines basic?

  • Lone pair on nitrogen can accept H⁺
  • Strength depends on lone pair availability and conjugate acid stability

Measurement:

  • Kb, pKb, or pKa of conjugate acid
  • Higher Kb (lower pKb) = Stronger base
  • Higher pKa of conjugate acid = Stronger base

2. Aliphatic Amines (MOST IMPORTANT FOR JEE):

In aqueous solution (default):

$$\boxed{2° > 1° > 3° > \text{NH}_3}$$

Example: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃

Reason: Optimal balance of +I effect and H-bonding solvation

In gas phase:

$$3° > 2° > 1° > \text{NH}_3$$

Reason: Only +I effect matters (no solvation)

3. Aromatic vs Aliphatic:

$$\boxed{\text{Aliphatic} >> \text{NH}_3 >> \text{Aromatic}}$$

Why aniline is weak:

  • Lone pair delocalized into benzene ring (resonance)
  • Reduced availability for protonation
  • ~10⁶ times weaker than NH₃

Why benzylamine is strong:

  • N on CH₂, not directly on ring
  • No resonance with ring
  • Behaves as aliphatic amine

4. Substituent Effects on Aniline:

EDG (Electron-Donating):

  • -OCH₃, -CH₃, -NH₂
  • Increase electron density
  • Increase basicity

EWG (Electron-Withdrawing):

  • -NO₂, -CN, -Cl, -CHO
  • Decrease electron density
  • Decrease basicity

Position matters: For -R groups, meta > para > ortho

5. The Ultimate JEE Order:

$$\boxed{(\text{CH}_3)_2\text{NH} > \text{CH}_3\text{NH}_2 > \text{C}_6\text{H}_5\text{CH}_2\text{NH}_2 > \text{NH}_3 > \text{C}_6\text{H}_5\text{NH}_2}$$

Memory: “Di-Met Mary Ben Before Amy Ann”

6. JEE Strategy:

✓ Always assume aqueous solution (unless specified) ✓ For aliphatic: Remember 2° > 1° > 3° ✓ For aromatic: All aliphatic » aromatic ✓ For substituents: EDG increase, EWG decrease ✓ Don’t confuse benzylamine (strong) with aniline (weak)

“Basicity of amines is a beautiful interplay of electronic effects and solvation - master the aqueous order and you’ll ace JEE questions!”

Next, explore diazonium salts to see how aromatic amines become versatile synthetic intermediates!