Aromaticity = extra stability shown by certain cyclic compounds due to delocalization of π-electrons.
4 Conditions for Aromaticity
A compound must be:
- Cyclic (ring structure)
- Planar (all atoms in one plane)
- Fully conjugated (continuous p-orbitals)
- Have (4n + 2) π electrons (Hückel Rule)
Hückel Rule
4n+2
Values of π electrons:
- n = 0 → 2 π e⁻
- n = 1 → 6 π e⁻
- n = 2 → 10 π e⁻
- n = 3 → 14 π e⁻
These are aromatic. (Chemistry LibreTexts)
Antiaromatic Compounds
Must be:
- Cyclic
- Planar
- Fully conjugated
- Have 4n π electrons
Examples:
- Cyclobutadiene (4 π e⁻)
- Cyclopentadienyl cation (4 π e⁻)
Very unstable.
Antiaromatic Rule
4n
Non-Aromatic Compounds
If any one condition fails:
- Not planar OR
- Not cyclic OR
- Not fully conjugated
Examples:
- Cyclooctatetraene (non-planar)
- Cyclohexene
Non-aromatic compounds are neither specially stable nor specially unstable.
JEE/NEET Electron Counting Tricks
π-electrons contributed by:
| Species | π electrons |
|---|---|
| Double bond | 2 |
| Negative charge in p-orbital | 2 |
| Positive charge | 0 |
| Radical | 1 |
| Pyrrole-type lone pair | 2 |
| Pyridine-type lone pair | 0 |
Important Aromatic Species
✅ Benzene → 6 π e⁻
✅ Cyclopropenyl cation (C₃H₃⁺) → 2 π e⁻
✅ Cyclopentadienyl anion (C₅H₅⁻) → 6 π e⁻
✅ Tropylium ion (C₇H₇⁺) → 6 π e⁻
✅ Naphthalene → 10 π e⁻
✅ Pyridine, Pyrrole, Furan, Thiophene → Aromatic
Most Important JEE/NEET Shortcut
CAP + H Rule
Check in this order:
C = Cyclic
A = All atoms conjugated
P = Planar
H = Hückel (4n+2)
If all four satisfy → Aromatic
If CAP satisfies but H gives 4n → Antiaromatic
If CAP itself fails → Non-aromatic
Frequently Asked Examples
| Compound | Nature |
|---|---|
| Benzene | Aromatic |
| Cyclobutadiene | Antiaromatic |
| Cyclooctatetraene | Non-aromatic |
| Cyclopentadienyl anion | Aromatic |
| Cyclopentadienyl cation | Antiaromatic |
| Tropylium ion | Aromatic |
For JEE/NEET, remember this table:
| Compound | Lone pairs on heteroatom | Lone pairs involved in aromatic resonance | Lone pairs NOT involved |
|---|---|---|---|
| Pyridine | 1 | 0 | 1 |
| Pyrrole | 1 | 1 | 0 |
| Furan | 2 | 1 | 1 |
| Thiophene | 2 | 1 | 1 |
Why?
Pyridine
- N is sp² hybridized.
- Its lone pair lies in an sp² orbital (in the plane of the ring).
- It does not participate in resonance.
- Aromatic π-electrons = 6 from three double bonds only.
Pyrrole
- N forms three σ bonds.
- Its lone pair occupies a p-orbital perpendicular to the ring.
- It participates in resonance.
- Aromatic π-electrons = 4 (from two double bonds) + 2 (from lone pair) = 6.
Furan
- O has two lone pairs.
- One lone pair is in a p-orbital and participates in resonance.
- The other is in an sp² orbital and does not participate.
- Aromatic π-electrons = 4 + 2 = 6.
Thiophene
- S has two lone pairs.
- One lone pair participates in the aromatic π-system.
- One lone pair remains localized.
- Aromatic π-electrons = 4 + 2 = 6.
Quick Memory Trick
Pyridine → 0 LP involved
Pyrrole → 1 LP involved
Furan → 1 of 2 LPs involved
Thiophene → 1 of 2 LPs involved
So the answer is:
Pyridine = 0, Pyrrole = 1, Furan = 1, Thiophene = 1 lone pair involved in resonance.
Sarvan sir- Chemistry For All 