Directive Effects in Aromatic Substitution

Master ortho/para and meta directors - electronic effects, activation/deactivation, and orientation control for JEE Advanced

The Hook: The Traffic Controllers of Benzene

Connect: Real Life → Chemistry

Imagine benzene ring as a circular road with 6 positions. When the first car (substituent) parks at one spot, where will the second car prefer to park?

  • Some “cars” attract others to park next to them or directly opposite (ortho/para directors)
  • Other “cars” force newcomers to park one spot away (meta directors)

The Chemistry Question: How does a -CH₃ group “know” to direct incoming electrophiles to ortho and para positions, while a -NO₂ group sends them to meta?

The answer lies in electron density distribution - and it’s one of the highest-yield topics for JEE Advanced!

Real-world impact: This controls synthesis of drugs like aspirin (ortho position), dyes, and explosives like TNT (2,4,6-trinitrotoluene).


The Core Concept

Why Directive Effects Matter

When benzene already has one substituent, the second substitution can occur at three different positions:

       R
       |
    1  2  ← ortho (o-)
   6      3  ← meta (m-)
    5  4  ← para (p-)

Three positions:

  1. Ortho (o-): Adjacent carbons (positions 2 and 6)
  2. Meta (m-): One carbon away (positions 3 and 5)
  3. Para (p-): Opposite carbon (position 4)
Two Key Questions

For any substituent already on benzene, we ask:

Question 1: Orientation - Where does the next group go?

  • Ortho/para positions? OR
  • Meta position?

Question 2: Reactivity - How fast does it react?

  • Activating: Makes ring MORE reactive than benzene
  • Deactivating: Makes ring LESS reactive than benzene

JEE Insight: These are independent properties!

  • Most o/p-directors are activating
  • BUT halogens are o/p-directors yet deactivating
  • ALL m-directors are deactivating

Related: Benzene Chemistry


Classification of Substituents

Group 1: Strongly Activating Ortho/Para Directors

Examples: -OH, -OR, -NH₂, -NHR, -NR₂

Electronic effect: +R (resonance donation) > -I (inductive withdrawal)

Mechanism: Donate electron density through resonance

Example: -OH group (Phenol)

Resonance structures:

     OH           OH           OH           OH
     |            |⁺           |⁺           |⁺
   /   \        /   \        /   \        /   \
              ⊝             ⊝             ⊝
   ortho      ortho        meta         para

Key points:

  • Lone pair on O/N enters the ring through resonance
  • Increases electron density at ortho and para positions
  • These positions are more nucleophilic → attract electrophiles

Reactivity: Phenol is ~1000× more reactive than benzene!

Why Not Meta?

Look at resonance structures:

In resonance forms, negative charge (high electron density) appears at:

  • Ortho positions (2, 6) ✓
  • Para position (4) ✓
  • Meta positions (3, 5) ✗ (no resonance structure puts charge here!)

Conclusion: Ortho and para have higher electron density → electrophile attacks there!

JEE Tip: Draw resonance structures to predict orientation!

Reactivity order (strongly activating):

$$\boxed{\text{-O}^- > \text{-OH} > \text{-OR} > \text{-NH}_2 > \text{-NHR} > \text{-NR}_2}$$

Why -O⁻ is strongest?

  • Full negative charge
  • Better resonance donation than -OH

Group 2: Moderately Activating Ortho/Para Directors

Examples: -NHCOCH₃ (acetamido), -OCOCH₃ (acetoxy)

Electronic effect: +R > -I (but weaker than Group 1)

Why weaker?

  • Lone pair on N/O is partially delocalized into C=O of the acyl group
  • Less available for donating to benzene ring
  • Still net activating, but weaker than -NH₂ or -OH

Example: Acetanilide

     NHCOCH₃
        |
    Benzene ring
  • Less reactive than aniline (-NH₂)
  • Still ortho/para directing
  • Useful in synthesis to “protect” amino group and control reactivity

Group 3: Weakly Activating Ortho/Para Directors

Examples: -CH₃, -C₂H₅, -R (alkyl groups)

Electronic effect: +I (inductive donation) only, NO resonance

Mechanism:

Alkyl groups are electron-donating through +I effect (hyperconjugation)

     CH₃
      |
   Benzene
  • CH₃ pushes electrons toward ring (σ donation)
  • Slightly increases electron density
  • Stabilizes positive charge in arenium intermediate at ortho/para

Reactivity:

  • Toluene is ~25× more reactive than benzene
  • Much weaker activation than -OH or -NH₂

Reactivity order (alkyl groups):

$$\boxed{\text{-C(CH}_3\text{)}_3 > \text{-CH(CH}_3\text{)}_2 > \text{-CH}_2\text{CH}_3 > \text{-CH}_3}$$

More alkyl groups → Stronger +I effect

Group 4: Weakly Deactivating Ortho/Para Directors

Examples: -F, -Cl, -Br, -I (halogens)

Electronic effect: -I (inductive withdrawal) > +R (resonance donation)

The Halogen Paradox - High-Yield for JEE!

This is THE most confusing and most tested concept!

Question: Why are halogens ortho/para directors despite being deactivating?

Answer: Two competing effects!

Effect 1: -I (Inductive Withdrawal)

  • Halogens are electronegative (F > Cl > Br > I)
  • Pull electrons through σ bond
  • Decreases overall electron density
  • Makes ring less reactive → Deactivating

Effect 2: +R (Resonance Donation)

  • Halogens have lone pairs
  • Can donate through resonance (like -OH, -NH₂)
  • Puts electron density at ortho/para

Net result:

  • Reactivity: -I wins → Deactivating (slower than benzene)
  • Orientation: +R wins → Ortho/para directing

Why does +R control orientation but not reactivity?

Reactivity is about ground state stabilization:

  • -I effect lowers ground state electron density
  • Ring is less nucleophilic → harder to react

Orientation is about transition state stabilization:

  • Once arenium forms, +R effect stabilizes ortho/para intermediates
  • Resonance puts lone pair into ring at ortho/para positions

JEE One-liner: “Halogens are reluctant directors - they slow you down (-I) but still show you the way (+R to o/p)!”

Related: Halogen Compounds

Interactive Demo: Visualize Directive Effects in Action

See how substituents control where electrophiles attack on benzene rings.

Reactivity order (halogens):

$$\boxed{\text{C}_6\text{H}_6 > \text{C}_6\text{H}_5\text{F} > \text{C}_6\text{H}_5\text{Cl} > \text{C}_6\text{H}_5\text{Br} > \text{C}_6\text{H}_5\text{I}}$$

All slower than benzene (deactivating)

Fluorine is least deactivating (strongest +R due to size match with 2p orbital)

Group 5: Moderately to Strongly Deactivating Meta Directors

Examples: -NO₂, -CN, -COOH, -CHO, -COR, -SO₃H

Electronic effect: -I and -R (both withdraw electrons!)

Mechanism: -NO₂ (Nitro Group)

Resonance structures:

     NO₂
      |
    /   \

Nitro group withdraws electrons by:

  1. -I effect: N⁺ pulls electrons through σ bond
  2. -R effect: Pulls electron density through π system

Where does electron density decrease most?

Draw resonance structures with positive charge in ring:

   NO₂           NO₂           NO₂
    |             |⁺            |⁺
  /   \         /   \         /   \
    ⊕           ⊕             ⊕
  ortho        para          meta

Actually, correct analysis:

NO₂ withdraws electrons → Ring becomes electron-deficient

For electrophilic attack, compare stability of arenium intermediates:

Ortho attack:

    NO₂             NO₂             NO₂
     |⁺              |⁺              |⁺
   H  E            H  E            H  E
     ⊕              (⊕ adjacent to NO₂)

Positive charge can be on carbon attached to NO₂ → VERY UNSTABLE (two +ve charges adjacent)

Meta attack:

    NO₂
     |
       H  E

Positive charge NEVER on carbon attached to NO₂ → More stable

Para attack:

    NO₂
     |

     ⊕  (can be on carbon attached to NO₂)
    H  E

Again, +ve charge can be adjacent to NO₂ → VERY UNSTABLE

Memory Trick: Meta Directors

“Electron Withdrawing? Go to Meta!”

Rule: If substituent withdraws electrons by -I AND -R, it’s meta directing

Common meta directors - All have electron-withdrawing atoms:

  • -NO₂ (N⁺=O)
  • -CN (C≡N is electron-poor)
  • -COOH, -CHO, -COR (C=O pulls electrons)
  • -SO₃H (S with multiple bonds to O)

Pattern recognition: Look for atoms with:

  • Multiple bonds to O or N
  • Positive charge (like N⁺ in NO₂)
  • Carbonyl groups (C=O)

JEE Shortcut: “If you see C=O, think meta!”

Exception: -OCOR is o/p (because O’s lone pair donates through +R)

Reactivity order (deactivating, meta directors):

$$\boxed{\text{-NR}_3^+ > \text{-NO}_2 > \text{-CN} > \text{-SO}_3\text{H} > \text{-COOH} > \text{-CHO} > \text{-COR}}$$

-NR₃⁺ (ammonium) is most deactivating (full positive charge!)


Master Summary Table

GroupExamplesTypeEffectReactivity vs Benzene
Strongly Activating o/p-O⁻, -OH, -NH₂+R » -IOrtho/Para10³-10⁶× faster
Moderately Activating o/p-NHCOCH₃, -OCOCH₃+R > -IOrtho/Para10-100× faster
Weakly Activating o/p-CH₃, -C₂H₅, -R+I onlyOrtho/Para2-25× faster
Weakly Deactivating o/p-F, -Cl, -Br, -I-I > +ROrtho/Para0.001-0.1× slower
Strongly Deactivating m-NO₂, -CN, -SO₃H-I, -RMeta10⁻⁶-10⁻⁸× slower
Moderately Deactivating m-COOH, -CHO, -COR-I, -RMeta10⁻²-10⁻⁴× slower
The Golden Rules - Memorize These!

Rule 1: Ortho/Para vs Meta

  • Electron-donating groups (+I or +R) → Ortho/Para
  • Electron-withdrawing groups (-I and -R) → Meta
  • Exception: Halogens (-I > +R) → Ortho/Para (orientation controlled by +R)

Rule 2: Activation vs Deactivation

  • If +R or +I net effect → Activating
  • If -I or -R net effect → Deactivating
  • Halogens: -I wins → Deactivating

Rule 3: All Meta Directors are Deactivating

  • If meta directing → automatically deactivating
  • NEVER: “activating meta director”

Rule 4: Most o/p Directors are Activating

  • Exception: Halogens (o/p but deactivating)

JEE Mantra:Ortho-Para directors are electron-Donating (except halogens)” “Meta directors are electron-Withdrawing (always!)”


Mechanism Analysis: Why Ortho/Para or Meta?

Case Study 1: Toluene (C₆H₅-CH₃)

Electrophilic attack at ortho position:

Step 1: Electrophile (E⁺) attacks ortho carbon

     CH₃
      |
    ⊕ E
   /H  \

Resonance structures of arenium intermediate:

  CH₃         CH₃         CH₃
   |           |⁺          |
  ⊕-E        ⊕-E         ⊕-E
  H                      H

Key resonance structure: Positive charge is on carbon bearing CH₃

CH₃ group stabilizes this positive charge through:

  • +I effect (donates electrons)
  • Hyperconjugation (C-H bonds donate into empty p-orbital)

Result: Ortho intermediate is STABILIZED → favored

Similarly for para attack.

Meta attack:

     CH₃
      |

    H E

Resonance structures: Positive charge NEVER on carbon bearing CH₃

→ Can’t benefit from CH₃ stabilization

Meta intermediate is LESS STABLE → disfavored

Conclusion: CH₃ directs ortho/para because it stabilizes carbocation at those positions

Case Study 2: Nitrobenzene (C₆H₅-NO₂)

Electrophilic attack at ortho position:

    NO₂
     |
    ⊕ E
   /H  \

Resonance structures:

   NO₂          NO₂          NO₂
    |⁺           |⁺           |⁺
   ⊕-E          ⊕-E         ⊕-E
   H                        H
   (VERY UNSTABLE!)

Key resonance structure: Positive charge on carbon attached to NO₂

This puts two positive charges adjacent (⊕-C-N⁺) → VERY UNSTABLE!

Similarly for para attack - also destabilizes

Meta attack:

    NO₂
     |

    H E

Resonance structures: Positive charge NEVER adjacent to NO₂

→ Avoids the unfavorable ⊕-⊕ repulsion

Meta intermediate is MORE STABLE (relatively) → favored

Conclusion: NO₂ directs meta because ortho/para intermediates are destabilized by charge repulsion


Competitive Directive Effects

What happens when benzene has two different substituents?

Priority Rules

Rule 1: If both direct to same position → No conflict

Example: -OH and -CH₃ (both o/p directors)

   OH
    |
        → Both direct ortho/para
    CH₃

Rule 2: If they direct to different positions → Stronger director wins

Activating groups overrule deactivating groups

Strength order (strongest first):

$$\boxed{\text{-NH}_2, \text{-OH} > \text{-OR} > \text{-NHCOR} > \text{-R} > \text{-Hal} > \text{Meta directors}}$$

Example:

    OH          OH
     |           |
         vs
    NO₂         NO₂
                 |
              New E

-OH is strongly activating ortho/para director -NO₂ is deactivating meta director

Result: -OH wins! New electrophile goes ortho/para to -OH (ignores NO₂)

Rule 3: Sterically hindered positions are disfavored

Between ortho and para:

  • Para is usually major (less steric crowding)
  • Ortho products form but in lower yield

JEE Tip: In competitive situations:

  1. Identify the strongest activating group
  2. That group controls orientation
  3. If o/p director, expect para as major product

Practice Example: Directive Competition

Problem: Where does electrophile attack?

Q: In p-nitrotoluene, where will the next electrophilic substitution occur?

    CH₃
     |

    NO₂

Solution:

Substituent 1: -CH₃ (weakly activating, o/p director) Substituent 2: -NO₂ (strongly deactivating, m director)

Strength: CH₃ (activating) > NO₂ (deactivating)

Winner: CH₃ controls orientation

CH₃ directs ortho/para:

Positions ortho to CH₃:

  • Position 2 (between CH₃ and NO₂) ← Available
  • Position 6 (between CH₃ and NO₂) ← Available

Position para to CH₃:

  • Position 4 ← Already occupied by NO₂!

Answer: Positions 2 and 6 (ortho to CH₃)

Product:

    CH₃
     |
   E    E

    NO₂

2,4-disubstituted and 2,6-disubstituted (where NO₂ is at 4-position originally)

JEE Insight: Activating group wins, even if weakly activating!


Applications in Synthesis

Strategic Use of Directive Effects

Problem: Want to synthesize m-bromonitrobenzene

Approach 1: Nitration then bromination

C₆H₆ → C₆H₅NO₂ → m-BrC₆H₄NO₂
      (HNO₃)    (Br₂/FeBr₃)

Why this works:

  • NO₂ is meta director
  • Br goes to meta position
  • ✓ Correct product!

Approach 2 (WRONG): Bromination then nitration

C₆H₆ → C₆H₅Br → ?
      (Br₂)    (HNO₃)

Problem:

  • Br is ortho/para director
  • NO₂ goes to ortho/para
  • ✗ Wrong product! (get o- and p-bromonitrobenzene)
Strategic Synthesis Rule

“First substituent controls second substituent’s position”

For meta products:

  • Add meta director first (like -NO₂)
  • Then add second group

For ortho/para products:

  • Add ortho/para director first (like -CH₃, -Cl)
  • Then add second group

JEE Strategy: Work backwards!

  1. Identify target structure
  2. Determine what relationship groups have (o, m, or p)
  3. Decide which group to add first

Example: Want p-chlorotoluene?

Target: CH₃-C₆H₄-Cl (para)

Route 1: CH₃ first, then Cl

  • Toluene + Cl₂/FeCl₃ → o/p-chlorotoluene ✓ (can separate)

Route 2: Cl first, then CH₃

  • Would need Friedel-Crafts alkylation on chlorobenzene
  • Problem: Cl deactivates ring → F-C doesn’t work well ✗

Best route: Route 1

Related: Organic Synthesis Strategy

Blocking and Protecting Groups

Problem: Want to make o-bromophenol, but -OH is very activating

Issue: -OH is so activating that multiple brominations occur!

$$\text{C}_6\text{H}_5\text{OH} + \text{Br}_2 \xrightarrow{\text{excess}} \text{2,4,6-tribromophenol}$$

Solution: Protect -OH by making it less activating!

Strategy:

Step 1: Convert to less activating group

$$\text{C}_6\text{H}_5\text{OH} + \text{(CH}_3\text{CO)}_2\text{O} \rightarrow \text{C}_6\text{H}_5\text{-OCOCH}_3$$

(phenyl acetate)

Step 2: Monohalogenation

$$\text{C}_6\text{H}_5\text{-OCOCH}_3 + \text{Br}_2 \rightarrow \text{o/p-Br-C}_6\text{H}_4\text{-OCOCH}_3$$

Step 3: Deprotection (hydrolyze ester)

$$\text{o-Br-C}_6\text{H}_4\text{-OCOCH}_3 + \text{H}_2\text{O/OH}^- \rightarrow \text{o-Br-C}_6\text{H}_4\text{-OH}$$

Result: Controlled monobromination!


Common Mistakes to Avoid

Mistake #1: Confusing Halogen Behavior

Wrong: “Halogens withdraw electrons, so they’re meta directors”

Correct: “Halogens are ortho/para directors (despite being deactivating)”

Why?

  • -I effect controls reactivity (deactivating)
  • +R effect controls orientation (ortho/para)

JEE Trap Question: “Is chlorobenzene more or less reactive than benzene?”

  • Answer: LESS reactive (deactivated by -I)
  • But still directs o/p!

Memory: “Halogens are unique - deactivating ortho/para directors”

Mistake #2: Assuming All o/p Directors Activate

Wrong: “If it’s ortho/para directing, it must be activating”

Correct: Most o/p directors activate, BUT halogens don’t!

Classification:

  • Activating o/p: -OH, -NH₂, -OR, -R
  • Deactivating o/p: -F, -Cl, -Br, -I

JEE Tip: Make a special note for halogens - they’re the exception!

Mistake #3: Wrong Synthesis Order

Wrong: Brominating first to make m-bromonitrobenzene

Correct: Nitrate first (NO₂ is meta director), then brominate

Rule:

  • For meta relationship → Use meta director first
  • For ortho/para → Use o/p director first

JEE Question Type: “Suggest a synthesis route for compound X” → Always determine which substituent to add first!

Mistake #4: Forgetting Steric Effects

Wrong: “Equal amounts of ortho and para products”

Correct: Para is usually major due to less steric hindrance

Example: Nitration of toluene

CH₃ → ortho (42%) + para (58%) + meta (trace)

Para predominates because:

  • Two ortho positions vs one para
  • But ortho is sterically hindered
  • Net result: para is major

JEE Tip: If asked for “major product” in o/p substitution → Answer para!


Practice Problems

Level 1: Foundation (NCERT)

Problem 1: Classification

Q: Classify the following as activating/deactivating and ortho-para/meta directing: (a) -COOH (b) -OCH₃ (c) -Br

Solution:

(a) -COOH (carboxyl group)

  • Has C=O (electron-withdrawing)
  • -I effect from electronegative O
  • -R effect from C=O
  • Classification: Deactivating, meta director

(b) -OCH₃ (methoxy group)

  • Oxygen has lone pair
  • +R effect (donates through resonance)
  • -I effect (O is electronegative)
  • Net: +R > -I
  • Classification: Activating, ortho/para director

(c) -Br (bromo group)

  • Has lone pairs
  • +R effect (weak, due to poor 2p-4p overlap)
  • -I effect (Br is electronegative)
  • Net: -I > +R
  • Classification: Deactivating, ortho/para director

Remember: Halogens are the exception - deactivating but still o/p!

Problem 2: Predicting Products

Q: What are the main products when toluene reacts with Br₂/FeBr₃?

Solution:

Toluene structure: C₆H₅-CH₃

CH₃ is: Weakly activating, ortho/para director

Possible positions:

    CH₃
     |
  2     6  (ortho positions)
3           (meta)
    4       (para)

Bromination at:

  • Ortho (2, 6) ← Directed here
  • Para (4) ← Directed here
  • Meta (3, 5) ← Minor

Products:

  1. o-bromotoluene (2-bromotoluene)
  2. p-bromotoluene (4-bromotoluene) ← Major (less steric hindrance)
  3. m-bromotoluene (trace amounts)

Approximate ratio: ortho:meta:para = 40:2:58

Answer: Mixture of ortho and para bromides, with para as major product

Level 2: JEE Main

Problem 3: Synthesis Route

Q: How will you synthesize m-nitrochlorobenzene from benzene?

Solution:

Target: m-ClC₆H₄NO₂

Analysis:

  • Cl and NO₂ are meta to each other
  • Need a meta director to be added first

Directive properties:

  • Cl: ortho/para director
  • NO₂: meta director

Correct sequence: Add NO₂ first!

Route:

Step 1: Nitration

$$\text{C}_6\text{H}_6 + \text{HNO}_3/\text{H}_2\text{SO}_4 \rightarrow \text{C}_6\text{H}_5\text{NO}_2$$

Step 2: Chlorination

$$\text{C}_6\text{H}_5\text{NO}_2 + \text{Cl}_2/\text{FeCl}_3 \rightarrow \text{m-ClC}_6\text{H}_4\text{NO}_2$$

Why this order?

  • NO₂ is meta director
  • Directs incoming Cl to meta position
  • ✓ Correct product!

Wrong route: Cl first, then NO₂

  • Would give o/p-nitrochlorobenzene ✗

Answer:

  1. C₆H₆ + HNO₃/H₂SO₄ → C₆H₅NO₂
  2. C₆H₅NO₂ + Cl₂/FeCl₃ → m-ClC₆H₄NO₂
Problem 4: Reactivity Comparison

Q: Arrange the following in order of increasing reactivity towards electrophilic substitution: (a) Benzene (b) Nitrobenzene (c) Toluene (d) Phenol

Solution:

Analyze each substituent:

(a) Benzene: Reference compound (reactivity = 1)

(b) Nitrobenzene (C₆H₅-NO₂):

  • NO₂ is strongly deactivating
  • Reactivity ≈ 10⁻⁸ (extremely slow)

(c) Toluene (C₆H₅-CH₃):

  • CH₃ is weakly activating
  • Reactivity ≈ 25 (faster than benzene)

(d) Phenol (C₆H₅-OH):

  • OH is strongly activating
  • Reactivity ≈ 10³ (thousand times faster than benzene)

Order of increasing reactivity:

$$\boxed{\text{Nitrobenzene} < \text{Benzene} < \text{Toluene} < \text{Phenol}}$$

(b) < (a) < (c) < (d)

JEE Tip:

  • Activating groups → reactivity > benzene
  • Deactivating groups → reactivity < benzene
  • Phenol is super reactive (can react with Br₂ without catalyst!)

Related: Phenols

Level 3: JEE Advanced

Problem 5: Competitive Direction

Q: When p-cresol (4-methylphenol) undergoes bromination, what are the major products?

    OH
     |

    CH₃

Solution:

Substituents:

  • -OH: Strongly activating, ortho/para director
  • -CH₃: Weakly activating, ortho/para director

Both are o/p directors!

Which one controls?

Strength: -OH » -CH₃

-OH is ~1000× more activating than -CH₃

Winner: -OH controls the reaction

Available positions ortho to -OH:

    OH
     |
  2     6  ← ortho to OH (both available!)

    CH₃ (at position 4)

Position para to -OH:

  • Position 4 ← Already occupied by CH₃

Expected products:

Bromination occurs at positions 2 and 6 (ortho to -OH)

With excess Br₂:

    OH
     |
  Br    Br

    CH₃

2,6-dibromo-4-methylphenol

Actually, with Br₂/water (no catalyst needed!):

All three available positions (2, 3, 6) can be brominated!

    OH
     |
  Br  Br Br

    CH₃

Wait, let me reconsider positions:

    OH (position 1)
     |
  2     6
3          5
    4 (CH₃)

Ortho to OH: positions 2, 6 ✓ Meta to OH: positions 3, 5 Para to OH: position 4 (occupied)

For controlled monobromination:

  • Products: 2-bromo-4-methylphenol + 6-bromo-4-methylphenol (same by symmetry!)

For excess Br₂:

  • Product: 2,6-dibromo-4-methylphenol

Answer:

  • With 1 equiv Br₂: 2-bromo-4-methylphenol (positions 2 and 6 are equivalent)
  • With excess Br₂: 2,6-dibromo-4-methylphenol

JEE Insight: Stronger activator controls, even when both direct to same positions!

Problem 6: Multi-Step Mechanism

Q: Explain using resonance why -OH directs ortho/para while -NO₂ directs meta in electrophilic aromatic substitution.

Solution:

This requires drawing arenium intermediate resonance structures!

For Phenol (-OH group):

Ortho attack:

   OH          OH          OH          OH
    |⁺          |⁺          |           |
  H-E         H-E         H-E⁺        H-E
   ⊕                                   ⊕

Key structure: Third resonance form

Oxygen’s lone pair donates into ring, putting positive charge on oxygen

    O⁺H
   H-E

This is VERY STABLE because:

  • Full octet on all atoms (except O has +1)
  • Positive charge on more electronegative atom (O)

Para attack: Similarly stable (lone pair donation possible)

Meta attack:

    OH          OH          OH
     |           |           |
       ⊕       H-E          ⊕
     H-E

No resonance structure with O⁺ - lone pair can’t reach meta position!

Conclusion: Ortho/para intermediates more stable → products favored


For Nitrobenzene (-NO₂ group):

Ortho attack:

   NO₂         NO₂         NO₂
    |⁺          |⁺          |⁺
  H-E         H-E⊕        H-E

Key unstable structure:

Positive charge on carbon directly attached to N⁺ in NO₂

This is VERY UNSTABLE because:

  • Two positive charges on adjacent atoms
  • Strong electrostatic repulsion

Para attack: Same problem - +ve charge can be adjacent to NO₂

Meta attack:

    NO₂
     |
     H-E

No resonance form puts positive charge adjacent to NO₂

Relatively more stable (avoids ⊕-⊕ repulsion)

Conclusion: Meta intermediate least unstable → meta product favored

Answer:

  • -OH stabilizes ortho/para intermediates through resonance donation (+R)
  • -NO₂ destabilizes ortho/para intermediates through charge repulsion
  • Meta attack avoids both extremes → favored for -NO₂

JEE Advanced expects: Clear resonance structures showing electron movement and charge distribution!


Quick Revision Table

Complete Classification

SubstituentTypeDirective EffectRelative Reactivity
-O⁻, -NH₂, -OH+R » -Io/p10³-10⁶
-OR, -NHCOR+R > -Io/p10-10²
-R (alkyl)+I onlyo/p2-25
-F, -Cl, -Br, -I-I > +Ro/p10⁻¹-10⁻³
-CHO, -COR, -COOH-I, -Rm10⁻²-10⁻⁴
-CN, -SO₃H-I, -Rm10⁻⁴-10⁻⁶
-NO₂, -NR₃⁺-I, -Rm10⁻⁶-10⁻⁸

Decision Tree for Synthesis

Target: Disubstituted benzene
├─ Groups are meta to each other?
│  └─ Add meta director first
│     (NO₂, CN, COOH, etc.)
└─ Groups are ortho or para?
   └─ Add ortho/para director first
      (OH, NH₂, R, halogens, etc.)

Remember: Stronger director wins if conflicting!

Connection to Other Topics

Prerequisites:

Related Topics:

Applications:


Teacher’s Summary

Key Takeaways

1. Two Independent Properties of Substituents:

Orientation (where next group goes):

  • Ortho/Para directors: -OH, -OR, -NH₂, -R, -Hal
  • Meta directors: -NO₂, -CN, -COOH, -CHO, -COR

Reactivity (how fast):

  • Activating: Makes ring MORE reactive
  • Deactivating: Makes ring LESS reactive

2. The Golden Rules - MUST MEMORIZE!

RuleDetails
Rule 1Electron-donating groups → ortho/para
Rule 2Electron-withdrawing groups → meta
Rule 3All meta directors are deactivating
Rule 4Most o/p directors are activating
ExceptionHalogens: o/p directors but deactivating!

3. The Halogen Paradox (Highest-Yield for JEE!)

Halogens are deactivating ortho/para directors

Why deactivating? -I effect > +R effect (overall electron withdrawal) Why o/p? +R effect controls orientation (resonance stabilizes o/p intermediates)

4. Mechanism Understanding:

Ortho/Para directors:

  • Stabilize arenium intermediate at o/p positions
  • Through resonance (+R) or induction (+I)
  • Example: -OH puts lone pair into ring

Meta directors:

  • Destabilize arenium intermediate at o/p positions
  • Withdraw electrons through -I and -R
  • Meta position “least bad” option
  • Example: -NO₂ creates ⊕-⊕ repulsion at o/p

5. Synthesis Strategy:

For meta products:

Add meta director → Add second group
Example: C₆H₆ → C₆H₅NO₂ → m-ClC₆H₄NO₂

For ortho/para products:

Add o/p director → Add second group
Example: C₆H₆ → C₆H₅CH₃ → o/p-ClC₆H₄CH₃

Competitive situations:

  • Stronger activator wins
  • Order: -NH₂, -OH > -OR > -R > -Hal » Meta directors

6. JEE Strategy:

High-Yield Question Types:

  1. Classify substituents (activating/deactivating, o/p/m)
  2. Predict major product of substitution
  3. Design synthesis route (order matters!)
  4. Explain using resonance structures
  5. Compare reactivities

Common Traps:

  • Forgetting halogens are deactivating (but still o/p!)
  • Wrong order in synthesis (always think: which substituent first?)
  • Ignoring competitive effects (stronger group wins)
  • Not considering steric effects (para usually major over ortho)

7. Must-Know Patterns:

Look for these structural features:

If you see…It’s probably…
Lone pair on atom attached to ringo/p director
C=O attached to ringm director
Positive charge/multiple bonds to Om director
Alkyl groupo/p activator
Halogeno/p deactivator (exception!)

8. Quick Mental Model:

“Electrons attract, lack repels”

  • Groups that donate electrons → ortho/para (electrons attract electrophile nearby)
  • Groups that withdraw electrons → meta (lack repels electrophile far away)
  • Exception: Halogens donate weakly through resonance → still o/p despite withdrawing overall

“In synthesis, strong beats weak, smart beats brute force”

  • Stronger activator controls orientation
  • Plan synthesis with directive effects in mind
  • Order of addition is crucial!

Final JEE Mantra:

“Ortho-Para directors Donate electrons (except halogens)” “Meta directors Withdraw electrons (always, no exceptions)” “In synthesis: First substituent controls second’s position”

Master directive effects and you’ve mastered aromatic chemistry! This is the key to solving complex multi-step synthesis problems and predicting product distributions in JEE Advanced.


Congratulations! You’ve completed the comprehensive hydrocarbon series. You now understand:

  • Alkanes - Saturated systems and conformations
  • Alkenes - Addition reactions and Markovnikov’s rule
  • Alkynes - Acidic character and triple bond chemistry
  • Benzene - Aromaticity and electrophilic substitution
  • Directive Effects - Controlling reactivity and orientation

Next, explore halogen compounds to see how these directive principles apply in substitution and elimination reactions!