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Aromaticity Checker (Hückel’s 4n + 2 Rule)
In short: Draw a molecule or paste its SMILES. The checker tests every ring for the four Hückel conditions (cyclic, fully conjugated, planar, 4n + 2 π electrons), counts the π electrons each atom gives and labels the ring aromatic, antiaromatic or non-aromatic with the reason. It is free and works in your web browser.
Draw a ring or paste its SMILES. The checker finds every ring and fused ring system, works through the four Hückel tests and counts the π electrons atom by atom, explaining which lone pairs are in the π system and which are not.
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Hückel’s rule: the four tests
- Cyclic. The π system must form a closed ring.
- Fully conjugated. Every ring atom needs a p orbital: an atom in a ring double bond, a carbocation (empty p), a carbanion or heteroatom whose lone pair can sit in a p orbital. One sp³ atom (such as the CH₂ in cyclopentadiene) breaks the ring.
- Planar. The p orbitals must line up, so the ring must be able to lie flat.
- 4n + 2 π electrons (n = 0, 1, 2 …): 2, 6, 10, 14, 18 → aromatic. 4n π electrons (4, 8, 12) in a flat, conjugated ring → antiaromatic.
π electrons = 4n + 2 → aromatic π electrons = 4n → antiaromatic (if flat) fails tests 1–3 → non-aromatic
How many π electrons does each ring atom give?
| Ring atom | π electrons | Example |
|---|---|---|
| C or N in a ring double bond | 1 | benzene C, pyridine N |
| N–H, N–R, O or S with a lone pair and no double bond | 2 (one lone pair) | pyrrole N, furan O, thiophene S |
| Carbanion C⁻ | 2 | cyclopentadienyl anion |
| Carbocation C⁺, or B with three bonds | 0 (empty p orbital) | tropylium, cyclopropenyl cation |
| C with an exocyclic C=O | 0 | tropone, 2-pyridone |
| sp³ carbon (CH₂, CR₂) | no p orbital: ring is non-aromatic | cyclopentadiene, cyclohexene |
A heteroatom that is already in a double bond (pyridine N) uses that bond for its 1 π electron; its lone pair lies in the ring plane and is not counted. A heteroatom with two lone pairs (furan O) puts only one of them into the π system.
Worked examples
Benzene
6 sp² carbons, three C=C bonds: 6 × 1 = 6 π electrons.
6 = 4n + 2 with n = 1.
Aromatic
Cyclobutadiene
Flat, conjugated, 4 × 1 = 4 π electrons.
4 = 4n with n = 1.
Antiaromatic
Furan
4 carbons (4) + one O lone pair (2) = 6. The second O lone pair lies in the plane.
Aromatic (n = 1)
Tropylium cation, C₇H₇⁺
6 carbons in three C=C (6) + C⁺ with an empty p orbital (0) = 6.
Aromatic (n = 1)
Cyclopropenyl cation
One C=C (2) + C⁺ (0) = 2 π electrons.
2 = 4n + 2 with n = 0.
Aromatic, the smallest one
Naphthalene (JEE style)
10 sp² carbons, 10 π electrons over the fused system.
10 = 4n + 2 with n = 2.
Aromatic
Common mistakes
- Counting the pyridine lone pair. It is in an sp² orbital in the ring plane, so pyridine has 6 π electrons, not 8.
- Counting both lone pairs of furan or thiophene. Only one lone pair is in the p orbital.
- Calling cyclooctatetraene antiaromatic. It is tub-shaped, so it is non-aromatic.
- Ignoring an sp³ ring atom. Cyclopentadiene and cycloheptatriene are non-aromatic; their ions can be aromatic (C₅H₅⁻ and C₇H₇⁺) or antiaromatic (C₅H₅⁺).
- Forgetting that ions change the count. Each negative charge on a ring carbon adds 2 π electrons; each positive charge adds 0.
- Applying 4n + 2 to each ring of a fused system separately. In azulene the 5- and 7-membered rings share a bond; count the 10 π electrons of the whole system.
The checker applies the textbook Hückel rules. Very large or strained rings, and rings with unusual bridges, may need a real 3D structure to decide whether they are flat.
Draw aromatic rings for your notes. Sketch rings, lone pairs and resonance arrows in MolDraw, a free online chemical structure editor, and export a PNG or SVG.
Open MolDraw editorFrequently asked questions
What is Hückel’s rule?
A ring is aromatic if it is cyclic, planar, fully conjugated (every ring atom has a p orbital) and has 4n + 2 π electrons, where n = 0, 1, 2, 3 … So 2, 6, 10, 14 and 18 π electrons are aromatic numbers.
What is the difference between aromatic, antiaromatic and non-aromatic?
Aromatic rings meet all four conditions with 4n + 2 π electrons and are extra stable. Antiaromatic rings are cyclic, planar and fully conjugated but have 4n π electrons, which destabilises them. Non-aromatic rings fail at least one condition, for example they contain an sp³ atom or are not flat.
Why is pyrrole aromatic?
Pyrrole has four ring carbons in two C=C bonds (4 π electrons) and an N–H nitrogen whose lone pair sits in a p orbital (2 π electrons). Total 6 = 4n + 2 with n = 1, so pyrrole is aromatic.
Why does the lone pair of pyridine not count?
The pyridine nitrogen is already in a C=N double bond, which gives 1 π electron. Its lone pair lies in an sp² orbital in the plane of the ring, at right angles to the π system, so it is not counted. Pyridine has 6 π electrons.
Is cyclooctatetraene antiaromatic?
No, it is non-aromatic. With 8 π electrons it would be antiaromatic if it were flat, but the ring bends into a tub shape so the four C=C bonds behave as separate alkenes.
Is the cyclopentadienyl anion aromatic?
Yes. Cyclopentadiene itself is non-aromatic because its CH2 carbon is sp³. Removing H+ gives a carbanion whose lone pair sits in a p orbital, so the ring has 6 π electrons and is aromatic. This is why cyclopentadiene is unusually acidic (pKa about 16).
How do you count π electrons in fused rings like naphthalene?
Count the π electrons of the whole fused system: naphthalene has 10, anthracene and phenanthrene 14, and azulene 10. All fit 4n + 2. Hückel’s rule is exact only for single rings, so for fused systems this is a good rule of thumb.
Related chemistry tools
Read the full guide: Aromaticity and Hückel’s 4n + 2 rule explained.
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