The Hook: The Mystery Molecule That Defied Logic
In 1825, Michael Faraday discovered a mysterious compound in illuminating gas - molecular formula C₆H₆. Scientists were puzzled: such a high degree of unsaturation should make it extremely reactive, yet this compound was remarkably stable!
Even more mysterious: It didn’t behave like alkenes at all!
- Didn’t decolorize Br₂ readily
- Didn’t give addition reactions easily
- Preferred substitution over addition
The JEE question: How can a molecule with four degrees of unsaturation (equivalent to four π bonds!) be so stable? The answer revolutionized organic chemistry: Aromaticity!
Fun fact: Benzene is the smell in petrol/gasoline, and it’s found in coal tar, crude oil, and even coffee!
The Core Concept
What is Benzene?
Benzene (C₆H₆) is the parent aromatic hydrocarbon with a unique cyclic structure that exhibits exceptional stability.
Molecular Formula: C₆H₆
Structure: Hexagonal planar ring with delocalized π electrons
H
|
H—C C—H
║ ║
H—C C—H
|
H
But this structure doesn’t tell the full story!
The Puzzle of Benzene Structure
Degree of Unsaturation:
$$\text{DBE} = \frac{2C + 2 - H}{2} = \frac{2(6) + 2 - 6}{2} = 4$$Four degrees of unsaturation suggest:
- Four double bonds OR
- Three double bonds + one ring OR
- Two double bonds + two rings OR
- One triple bond + one double bond, etc.
August Kekulé proposed alternating single and double bonds in a hexagon:
C═C
╱ ╲
C C
║ ║
C C
╲ ╱
C═C
Problems with this structure:
Should have different C-C bond lengths
- C=C: 134 pm (double bond)
- C-C: 154 pm (single bond)
- Reality: All C-C bonds are 139 pm (intermediate!)
Should give two isomers of 1,2-dibromobenzene
- One with Br on C=C (short distance)
- One with Br on C-C (long distance)
- Reality: Only one isomer exists!
Should readily undergo addition (like alkenes)
- Reality: Prefers substitution!
Conclusion: Simple alternating structure is inadequate!
Modern Picture: Resonance and Delocalization
Resonance Structures:
Benzene is represented as a resonance hybrid of two Kekulé structures:
C═C C—C
╱ ╲ ╱ ╲
C C ↔ C C
║ ║ ║ ║
C C C C
╲ ╱ ╲ ╱
C═C C═C
Better representation: Circle inside hexagon (showing delocalized π electrons)
⬡
Reality:
- All six C-C bonds are equivalent
- Each C-C bond is partial double bond (bond order = 1.5)
- Bond length: 139 pm (between 134 and 154 pm)
- Six π electrons delocalized over entire ring
Resonance Energy = Extra stability due to delocalization
For benzene: 152 kJ/mol
What this means:
- Benzene is 152 kJ/mol more stable than hypothetical cyclohexatriene
- This extra stability explains why benzene resists addition reactions
- Adding across double bond would destroy aromaticity (lose 152 kJ/mol of stability!)
JEE Concept: This is why benzene undergoes substitution (retains aromaticity) rather than addition (loses aromaticity)
Related: Chemical Bonding
Interactive Demo: Visualize Benzene’s Aromatic Structure
Explore the delocalized π-electron system that makes benzene so stable.
Aromaticity: Hückel’s Rule
Criteria for Aromaticity
For a compound to be aromatic, it must satisfy ALL four conditions:
1. Cyclic Structure
- Compound must have a closed ring of atoms
2. Planar Structure
- All atoms in the ring must lie in the same plane
- Required for π orbital overlap
3. Complete Conjugation (Continuous π System)
- Every atom in the ring must have a p-orbital
- Usually means sp² or sp hybridized carbons
- p-orbitals overlap to form continuous π system
4. Hückel’s Rule: (4n + 2) π electrons
$$\boxed{\text{Number of } \pi \text{ electrons} = 4n + 2, \text{ where } n = 0, 1, 2, 3...}$$Allowed values: 2, 6, 10, 14, 18… π electrons
Aromatic: 6 π electrons is most common (n=1)
Benzene Satisfies All Criteria
- ✓ Cyclic: Six-membered ring
- ✓ Planar: All carbons sp² hybridized, hexagon is flat
- ✓ Conjugated: Each carbon has one p-orbital, continuous overlap
- ✓ Hückel’s Rule: 6 π electrons (4×1 + 2 = 6, where n=1)
Conclusion: Benzene is aromatic!
Examples: Aromatic vs Non-Aromatic vs Anti-Aromatic
AROMATIC (Extra stable, 4n+2 π electrons)
- Benzene (C₆H₆): 6 π electrons ✓
- Naphthalene (C₁₀H₈): 10 π electrons ✓
- Pyridine (C₅H₅N): 6 π electrons ✓
- Furan (C₄H₄O): 6 π electrons ✓
ANTI-AROMATIC (Highly unstable, 4n π electrons)
- Cyclobutadiene: 4 π electrons ✗
- Cyclooctatetraene (if planar): 8 π electrons ✗
Note: Cyclooctatetraene is actually non-planar (tub-shaped) to avoid anti-aromaticity!
NON-AROMATIC (Normal stability, doesn’t follow rules)
- Cyclohexene: Not fully conjugated ✗
- Cyclooctatetraene: Not planar ✗
- Cyclohexane: No π electrons ✗
JEE Tip: If 4n π electrons and planar → Anti-aromatic (very unstable)
“Hückel’s Magic Numbers: 2, 6, 10, 14…”
For quick checking:
- n=0: 2 π electrons (rare, but possible)
- n=1: 6 π electrons ← Most common (benzene!)
- n=2: 10 π electrons (naphthalene)
- n=3: 14 π electrons (anthracene)
Anti-aromatic numbers to AVOID: 4, 8, 12, 16… (4n with n=1,2,3…)
JEE Shortcut: Count π electrons
- If 6 or 10 → Likely aromatic (if planar and cyclic)
- If 4 or 8 → Anti-aromatic (highly unstable)
Reactions of Benzene: Electrophilic Aromatic Substitution
Key Point: Benzene undergoes substitution (not addition) to preserve aromaticity!
General Mechanism:
$$\boxed{\text{C}_6\text{H}_6 + \text{E}^+ \xrightarrow{\text{catalyst}} \text{C}_6\text{H}_5\text{-E} + \text{H}^+}$$General Mechanism (Two Steps)
Step 1: Electrophilic Attack → Carbocation Intermediate
Benzene + E⁺
↓
Arenium ion (σ complex/Wheland intermediate)
H E
\/
/ \
- Electrophile attacks π electrons
- Carbocation intermediate forms (loses aromaticity temporarily!)
- This is the slow, rate-determining step
Step 2: Deprotonation → Restoration of Aromaticity
Arenium ion → C₆H₅-E + H⁺
- Lose H⁺ to restore aromaticity
- Fast step
- Aromaticity regained!
Why substitution, not addition?
$$\Delta G_{\text{sub}} < \Delta G_{\text{add}}$$- Substitution: Regains aromatic stability (−152 kJ/mol)
- Addition: Loses aromatic stability permanently
Benzene “prefers to stay aromatic”
Even though the carbocation intermediate is less stable (lost aromaticity), the system immediately eliminates H⁺ to restore the aromatic ring.
This is the defining characteristic of aromatic compounds!
Specific Reactions of Benzene
Reaction 1: Halogenation
Chlorination
$$\boxed{\text{C}_6\text{H}_6 + \text{Cl}_2 \xrightarrow{\text{FeCl}_3} \text{C}_6\text{H}_5\text{Cl} + \text{HCl}}$$Mechanism:
Step 1: Generation of electrophile (Cl⁺)
$$\text{Cl}_2 + \text{FeCl}_3 \rightarrow \text{Cl}^+ + [\text{FeCl}_4]^-$$FeCl₃ is a Lewis acid (electron pair acceptor) - polarizes Cl₂
Step 2: Electrophilic attack
$$\text{C}_6\text{H}_6 + \text{Cl}^+ \rightarrow \text{C}_6\text{H}_6\text{Cl}^+ \text{ (arenium ion)}$$Step 3: Deprotonation
$$\text{C}_6\text{H}_6\text{Cl}^+ \rightarrow \text{C}_6\text{H}_5\text{Cl} + \text{H}^+$$Bromination: Similar mechanism with Br₂/FeBr₃ or Br₂/AlBr₃
Q: Why doesn’t benzene react with Cl₂ directly (without FeCl₃)?
Wrong answer: “Benzene is unreactive”
Correct answer:
- Cl₂ is not electrophilic enough to attack benzene’s π electrons
- FeCl₃ polarizes Cl-Cl bond: Cl-Cl → Cl^(δ+)-Cl^(δ-)
- Generates Cl⁺ (strong electrophile)
- Without catalyst, no reaction at room temperature
Contrast with alkenes:
- Alkenes have exposed π electrons (above/below plane)
- React with Cl₂ directly without catalyst
- Benzene’s π electrons are delocalized → less reactive
JEE One-liner: “Benzene needs catalyst, alkenes don’t”
Reaction 2: Nitration
Formation of Nitrobenzene
$$\boxed{\text{C}_6\text{H}_6 + \text{HNO}_3 \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{C}_6\text{H}_5\text{NO}_2 + \text{H}_2\text{O}}$$Mechanism:
Step 1: Generation of nitronium ion (NO₂⁺)
$$\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^-$$Detailed:
$$\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{NO}_3^+ + \text{HSO}_4^-$$ $$\text{H}_2\text{NO}_3^+ \rightarrow \text{NO}_2^+ + \text{H}_2\text{O}$$Step 2: Electrophilic attack by NO₂⁺
$$\text{C}_6\text{H}_6 + \text{NO}_2^+ \rightarrow \text{C}_6\text{H}_6(\text{NO}_2)^+ \text{ (arenium ion)}$$Step 3: Deprotonation
$$\text{C}_6\text{H}_6(\text{NO}_2)^+ \rightarrow \text{C}_6\text{H}_5\text{NO}_2 + \text{H}^+$$Product: Nitrobenzene (pale yellow liquid, almond-like odor)
Uses:
- Reduction → Aniline (important dye intermediate)
- Precursor for explosives (TNT)
Related: Nitrogen Compounds
Reaction 3: Sulfonation
Formation of Benzenesulfonic Acid
$$\boxed{\text{C}_6\text{H}_6 + \text{H}_2\text{SO}_4 \xrightarrow{\text{fuming H}_2\text{SO}_4} \text{C}_6\text{H}_5\text{SO}_3\text{H} + \text{H}_2\text{O}}$$Fuming H₂SO₄ = Concentrated H₂SO₄ + SO₃ (oleum)
Mechanism:
Step 1: Electrophile = SO₃ (or HSO₃⁺)
$$\text{SO}_3 \text{ acts as electrophile}$$Step 2: Electrophilic attack
$$\text{C}_6\text{H}_6 + \text{SO}_3 \rightarrow \text{C}_6\text{H}_6(\text{SO}_3)^+ $$Step 3: Deprotonation
$$\text{C}_6\text{H}_6(\text{SO}_3)^+ + \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_5\text{SO}_3\text{H} + \text{H}^+$$Special property: Sulfonation is reversible!
$$\text{C}_6\text{H}_5\text{SO}_3\text{H} + \text{H}_2\text{O} \xrightarrow{\text{heat, steam}} \text{C}_6\text{H}_6 + \text{H}_2\text{SO}_4$$Use: Sulfonation followed by desulfonation can be used to block a position temporarily in multi-step synthesis!
Reaction 4: Friedel-Crafts Alkylation
Adding alkyl group (R-)
$$\boxed{\text{C}_6\text{H}_6 + \text{R-Cl} \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5\text{-R} + \text{HCl}}$$Example:
$$\text{C}_6\text{H}_6 + \text{CH}_3\text{Cl} \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5\text{-CH}_3 + \text{HCl}$$(Benzene → Toluene)
Mechanism:
Step 1: Generation of carbocation
$$\text{R-Cl} + \text{AlCl}_3 \rightarrow \text{R}^+ + [\text{AlCl}_4]^-$$Step 2: Electrophilic attack
$$\text{C}_6\text{H}_6 + \text{R}^+ \rightarrow \text{C}_6\text{H}_6\text{R}^+$$Step 3: Deprotonation
$$\text{C}_6\text{H}_6\text{R}^+ \rightarrow \text{C}_6\text{H}_5\text{R} + \text{H}^+$$Common mistake: Expecting primary carbocation to stay as-is
Example:
$$\text{C}_6\text{H}_6 + \text{CH}_3\text{CH}_2\text{Cl} \xrightarrow{\text{AlCl}_3} ?$$Expected: C₆H₅-CH₂-CH₃ (ethylbenzene)
Reality: Mixture!
- Some ethylbenzene
- Also get C₆H₅-CH(CH₃)₂ (isopropylbenzene)
Why?
$$\text{CH}_3\text{CH}_2^+ \rightarrow \text{CH}_3\text{CH}_2^+ \text{ (1°)} $$Can rearrange via hydride shift:
$$\text{CH}_3\text{-CH}_2^+ \xrightarrow{\text{H-shift}} \text{CH}_3\text{-CH}^+\text{-H} \text{ (wait, this doesn't work)}$$Actually for ethyl:
$$\text{CH}_3\text{CH}_2^+ \text{ is primary, stays as-is}$$Better example: n-propyl chloride
$$\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + \text{AlCl}_3 \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2^+ \text{ (1°)}$$Rearranges to:
$$\text{CH}_3\text{CH}^+\text{CH}_3 \text{ (2°, more stable!)}$$Product: Mainly isopropylbenzene, not n-propylbenzene!
JEE Rule: “Friedel-Crafts with 1° halides → Expect rearrangements!”
To avoid: Use acyl chlorides (Friedel-Crafts acylation) - no rearrangement!
Limitations of Friedel-Crafts Alkylation:
- Polyalkylation - Product is more reactive than starting material
- Carbocation rearrangement - Especially with 1° and 2° halides
- Doesn’t work with deactivated rings (nitrobenzene, etc.)
Reaction 5: Friedel-Crafts Acylation
Adding acyl group (R-CO-)
$$\boxed{\text{C}_6\text{H}_6 + \text{R-CO-Cl} \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5\text{-CO-R} + \text{HCl}}$$Example:
$$\text{C}_6\text{H}_6 + \text{CH}_3\text{COCl} \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5\text{-CO-CH}_3$$(Benzene → Acetophenone)
Mechanism:
Step 1: Generation of acylium ion
$$\text{R-CO-Cl} + \text{AlCl}_3 \rightarrow [\text{R-C≡O}]^+ + [\text{AlCl}_4]^-$$Acylium ion: R-C≡O⁺ (resonance stabilized, doesn’t rearrange!)
Step 2: Electrophilic attack
$$\text{C}_6\text{H}_6 + [\text{R-C≡O}]^+ \rightarrow \text{C}_6\text{H}_6(\text{COR})^+$$Step 3: Deprotonation
$$\text{C}_6\text{H}_6(\text{COR})^+ \rightarrow \text{C}_6\text{H}_5\text{-CO-R} + \text{H}^+$$Advantages over alkylation:
- No rearrangement - Acylium ion is resonance stabilized
- No polyacylation - Product (ketone) is deactivated, won’t react further
- Can reduce to alkyl - Reduce C₆H₅-CO-R → C₆H₅-CH₂-R (Clemmensen/WK)
Want to add alkyl group to benzene?
Direct alkylation issues:
- Polysubstitution
- Rearrangement (for 1°, 2° carbocations)
Better approach: Acylation followed by reduction!
$$\text{C}_6\text{H}_6 \xrightarrow[\text{AlCl}_3]{\text{RCOCl}} \text{C}_6\text{H}_5\text{-CO-R} \xrightarrow{\text{Zn-Hg/HCl}} \text{C}_6\text{H}_5\text{-CH}_2\text{-R}$$Benefits:
- No polysubstitution (ketone deactivates ring)
- No rearrangement (acylium ion is stable)
- Clean, single product
JEE Mantra: “For clean alkylation, go via acylation!”
Related: Carbonyl Compounds
Orientation in Disubstituted Benzenes
Quick preview - Detailed coverage in Directive Effects
When benzene already has one substituent, where does the second group attach?
Three positions possible:
R
|
1 2 ortho (o-)
6 3 meta (m-)
5 4 para (p-)
- Ortho (o-): Adjacent to existing group (positions 2, 6)
- Meta (m-): One carbon away (positions 3, 5)
- Para (p-): Opposite side (position 4)
Substituents are classified as:
Ortho-para directors (OH, NH₂, CH₃, halogens)
- Direct incoming group to ortho and para positions
- Most are activating (make ring more reactive)
- Exception: Halogens are deactivating but still o/p-directing
Meta directors (NO₂, SO₃H, COOH, CN)
- Direct incoming group to meta position
- All are deactivating (make ring less reactive)
“Electron Donating = Ortho/Para” “Electron Withdrawing = Meta”
Exception: Halogens (withdraw by induction, donate by resonance)
- Net effect: Deactivating
- But still ortho/para directors (resonance wins for orientation!)
Detailed mechanisms and examples in next topic: Directive Effects
Common Mistakes to Avoid
Wrong: “Benzene has 3 π bonds, so 3 × 2 = 6 π electrons”
Correct thinking process:
- Count total π electrons in the system
- Don’t count π bonds separately
- Each C contributes 1 p-electron
- 6 carbons × 1 electron = 6 π electrons total
- Check if 4n+2 (n=1): 4(1)+2 = 6 ✓
JEE Trap: Some students count double bonds instead of electrons
Wrong: Benzene + Br₂ → C₆H₆Br₂ (addition product)
Correct: Benzene + Br₂/FeBr₃ → C₆H₅Br (substitution product)
Why?
- Addition would destroy aromaticity (lose 152 kJ/mol stability!)
- Substitution preserves aromaticity
- System “prefers” to regain aromatic stabilization
JEE Rule: Aromatic compounds → Substitution, not addition
Wrong: Benzene reacts with Cl₂ at room temperature
Correct: Need FeCl₃ or AlCl₃ catalyst to generate Cl⁺
Why benzene needs catalyst but alkenes don’t:
| Property | Alkenes | Benzene |
|---|---|---|
| π electrons | Localized, exposed | Delocalized, stable |
| Reactivity | High | Low (aromatic stability) |
| Electrophile | Cl₂ sufficient | Need Cl⁺ (stronger) |
JEE Tip: Always write catalyst in benzene reactions!
Practice Problems
Level 1: Foundation (NCERT)
Q: Why does benzene show exceptional stability compared to hypothetical cyclohexatriene?
Solution:
Key concept: Resonance energy
Comparison:
Hypothetical cyclohexatriene:
- Three localized C=C double bonds
- Three C-C single bonds
- Alternating bond lengths
Actual benzene:
- Six π electrons delocalized over entire ring
- All C-C bonds equivalent (139 pm)
- Resonance stabilization = 152 kJ/mol
Answer: Delocalization of π electrons over the entire ring provides extra stability (resonance energy of 152 kJ/mol), making benzene much more stable than expected for a compound with three double bonds.
Practical consequence: Resists addition reactions (would lose aromaticity)
Q: Is cyclooctatetraene (C₈H₈) aromatic?
Solution:
Check all four criteria:
- Cyclic? Yes ✓
- Planar? NO ✗ (tub-shaped to avoid anti-aromaticity)
- Conjugated? Would be if planar
- Hückel’s rule? 8 π electrons = 4n (n=2) ✗
Count π electrons: 4 C=C double bonds = 8 π electrons
Check Hückel: 4n+2 = ?
- If n=1: 4(1)+2 = 6 ✗
- 8 = 4(2) = 4n → Anti-aromatic if planar!
Reality: To avoid anti-aromaticity, cyclooctatetraene adopts non-planar tub shape
Answer: NOT aromatic (it’s non-aromatic because not planar)
JEE Insight: 8 π electrons would be anti-aromatic if planar, so molecule distorts to avoid this!
Level 2: JEE Main
Q: Why does benzene undergo electrophilic substitution rather than electrophilic addition?
Solution:
Energy considerations:
Addition pathway:
$$\text{C}_6\text{H}_6 + \text{Br}_2 \rightarrow \text{C}_6\text{H}_6\text{Br}_2$$- Loses aromaticity permanently
- Loses 152 kJ/mol resonance energy
- Product has no special stability
Substitution pathway:
$$\text{C}_6\text{H}_6 + \text{Br}_2 \xrightarrow{\text{FeBr}_3} \text{C}_6\text{H}_5\text{Br} + \text{HBr}$$- Temporarily loses aromaticity (arenium intermediate)
- Regains aromaticity in product
- Net: Retains aromatic stabilization
Energy diagram:
Energy
| ‡ (Arenium ion)
| /\
| / \___ C₆H₅Br + HBr (aromatic - stable)
| /
|/___________ C₆H₆ (aromatic - stable)
|
Answer: Substitution pathway allows benzene to regain its aromatic stability, making it thermodynamically and kinetically more favorable than addition which permanently destroys aromaticity.
JEE Key Point: Loss of 152 kJ/mol is too costly!
Q: What is the major product when benzene reacts with CH₃CH₂COCl in presence of AlCl₃?
Solution:
This is Friedel-Crafts Acylation
Reagent: CH₃CH₂COCl (propanoyl chloride)
Step 1: Form acylium ion
$$\text{CH}_3\text{CH}_2\text{COCl} + \text{AlCl}_3 \rightarrow [\text{CH}_3\text{CH}_2\text{C≡O}]^+ + [\text{AlCl}_4]^-$$Step 2: Electrophilic substitution
$$\text{C}_6\text{H}_6 + [\text{CH}_3\text{CH}_2\text{C≡O}]^+ \rightarrow \text{C}_6\text{H}_5\text{-CO-CH}_2\text{CH}_3 + \text{H}^+$$Product: Propiophenone (C₆H₅-CO-CH₂-CH₃)
Key points:
- No rearrangement (acylium ion is stable)
- Only monosubstitution (product is deactivated)
- Clean, predictable reaction
Answer: C₆H₅-CO-CH₂-CH₃ (propiophenone)
Level 3: JEE Advanced
Q: Arrange the following in order of increasing aromatic character: (a) Benzene (b) Cyclopentadienyl anion (C₅H₅⁻) (c) Cycloheptatrienyl cation (C₇H₇⁺, tropylium) (d) Pyrrole (C₄H₄NH)
Solution:
Check each for aromaticity:
(a) Benzene (C₆H₆):
- 6 π electrons (4×1+2) ✓
- Planar ✓
- Cyclic, conjugated ✓
- Aromatic
(b) Cyclopentadienyl anion (C₅H₅⁻):
[ ]⁻
- 5 carbons, each contributes 1 π electron = 5
- Plus negative charge (2 electrons in p-orbital) = 6 total
- 6 π electrons (4×1+2) ✓
- Planar ✓
- Aromatic (remarkably stable for anion!)
(c) Tropylium cation (C₇H₇⁺):
[ ]⁺
- 7 carbons, but one has empty p-orbital (positive charge)
- 3 double bonds = 6 π electrons
- 6 π electrons ✓
- Planar ✓
- Aromatic (remarkably stable for cation!)
(d) Pyrrole (C₄H₄NH):
NH
||
/ \
- 4 carbons: 2 double bonds = 4 π electrons
- N has lone pair in p-orbital = 2 π electrons
- Total: 4 + 2 = 6 π electrons ✓
- Planar ✓
- Aromatic
All are aromatic! But question asks for “increasing aromatic character”
Consider resonance energy and stability:
Resonance energies (approximate):
- Benzene: 152 kJ/mol (most stable, perfect symmetry)
- Tropylium: ~150 kJ/mol
- Cyclopentadienyl: ~105 kJ/mol
- Pyrrole: ~90 kJ/mol (heteroatom disrupts symmetry)
Order: Pyrrole < Cyclopentadienyl anion < Tropylium < Benzene
Answer: (d) < (b) < (c) < (a)
JEE Insight:
- All satisfy Hückel’s rule (6 π electrons)
- But resonance energy varies
- Benzene is “gold standard” for aromaticity
- Ions and heteroaromatic compounds have lower resonance energy
Related: Heterocyclic Compounds
Q: How will you convert benzene to n-propylbenzene using Friedel-Crafts reaction?
Solution:
Target: C₆H₅-CH₂-CH₂-CH₃
Problem: Direct alkylation with n-propyl chloride gives rearrangement!
$$\text{C}_6\text{H}_6 + \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5\text{CH(CH}_3\text{)}_2$$(isopropylbenzene, major)
Why? 1° carbocation rearranges to 2°:
$$\text{CH}_3\text{-CH}_2\text{-CH}_2^+ \rightarrow \text{CH}_3\text{-CH}^+\text{-CH}_3$$Solution: Use Friedel-Crafts Acylation + Reduction
Step 1: Acylation (no rearrangement!)
$$\text{C}_6\text{H}_6 + \text{CH}_3\text{CH}_2\text{COCl} \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5\text{-CO-CH}_2\text{CH}_3$$(Propiophenone)
Step 2: Reduction (Clemmensen or Wolff-Kishner)
$$\text{C}_6\text{H}_5\text{-CO-CH}_2\text{CH}_3 \xrightarrow{\text{Zn-Hg/HCl}} \text{C}_6\text{H}_5\text{-CH}_2\text{-CH}_2\text{CH}_3$$Final product: n-Propylbenzene (no rearrangement!)
Answer:
- C₆H₆ + CH₃CH₂COCl/AlCl₃ → C₆H₅-CO-CH₂-CH₃
- Clemmensen reduction → C₆H₅-CH₂-CH₂-CH₃
JEE Strategy: “Can’t trust Friedel-Crafts alkylation with 1° halides? Go via acylation!”
Related: Alkanes for reduction mechanisms
Quick Revision Table
| Reaction | Reagent | Electrophile | Product |
|---|---|---|---|
| Halogenation | Cl₂/FeCl₃ or Br₂/FeBr₃ | Cl⁺ or Br⁺ | C₆H₅-X |
| Nitration | HNO₃/H₂SO₄ | NO₂⁺ | C₆H₅-NO₂ |
| Sulfonation | H₂SO₄ (fuming) | SO₃ or HSO₃⁺ | C₆H₅-SO₃H |
| F-C Alkylation | R-Cl/AlCl₃ | R⁺ | C₆H₅-R (+ rearrangement risk) |
| F-C Acylation | RCOCl/AlCl₃ | RC≡O⁺ | C₆H₅-CO-R (no rearrangement) |
Aromaticity Criteria:
- Cyclic ✓
- Planar ✓
- Conjugated ✓
- (4n+2) π electrons ✓
Connection to Other Topics
Prerequisites:
- Alkenes - Understanding π bonds and electrophilic addition
- Chemical Bonding - Resonance and delocalization
- Reaction Mechanisms - Electrophilic mechanisms
Next Topics:
- Directive Effects - Ortho/para vs meta orientation
- Halogen Compounds - Aromatic halides
- Phenols - Aromatic alcohols
- Aromatic Amines - Aniline derivatives
Applications:
- Dyes and Pigments - Azo dyes from benzene
- Polymers - Polystyrene from styrene
- Pharmaceuticals - Many drugs have benzene rings
Teacher’s Summary
1. Benzene (C₆H₆) is the archetypal aromatic compound
Structure:
- Hexagonal planar ring
- All C-C bonds equivalent (139 pm, bond order 1.5)
- 6 π electrons delocalized over ring
- Resonance energy: 152 kJ/mol
2. Aromaticity Criteria (Hückel’s Rule):
Must have ALL four:
- ✓ Cyclic structure
- ✓ Planar geometry
- ✓ Complete conjugation (continuous p-orbitals)
- ✓ (4n+2) π electrons where n = 0, 1, 2, 3…
Common aromatic number: 6 π electrons (n=1)
3. Characteristic Reaction: Electrophilic Aromatic Substitution
Why substitution, not addition?
- Substitution preserves aromaticity (retains 152 kJ/mol stability)
- Addition destroys aromaticity (permanent loss)
General mechanism:
- Electrophile attacks → Arenium ion (loses aromaticity temporarily)
- Lose H⁺ → Restore aromaticity (fast)
4. Five Important Reactions:
| Reaction | Key Reagent | Product | Remember |
|---|---|---|---|
| Halogenation | X₂/FeX₃ | C₆H₅-X | Need catalyst (vs alkenes) |
| Nitration | HNO₃/H₂SO₄ | C₆H₅-NO₂ | NO₂⁺ is electrophile |
| Sulfonation | Fuming H₂SO₄ | C₆H₅-SO₃H | Reversible! |
| F-C Alkylation | RCl/AlCl₃ | C₆H₅-R | Rearrangement risk, polysubstitution |
| F-C Acylation | RCOCl/AlCl₃ | C₆H₅-CO-R | No rearrangement, clean |
5. JEE Strategy Points:
Aromaticity questions:
- Always check all four criteria
- Count π electrons carefully (4n+2 = aromatic, 4n = anti-aromatic if planar)
Mechanism questions:
- Benzene needs stronger electrophiles (Cl⁺ not Cl₂)
- Arenium intermediate is key
- Deprotonation restores aromaticity
Synthesis questions:
- Avoid F-C alkylation with 1° halides (rearrangement!)
- Better: F-C acylation → reduction
- Sulfonation useful for blocking positions
6. Common JEE Traps:
- Forgetting catalyst in halogenation (benzene ≠ alkenes!)
- Expecting addition products (benzene → substitution!)
- Counting π bonds instead of π electrons for Hückel’s rule
- Ignoring carbocation rearrangements in F-C alkylation
“Benzene’s aromatic stability is worth 152 kJ/mol - it will fight to keep it!”
The concept of aromaticity extends far beyond benzene - it’s the foundation for understanding all aromatic chemistry. Master these principles, and you’ll excel in aromatic compound questions!
Next, study directive effects to learn how substituents control orientation in disubstituted benzenes!