The aromatic ring, one Kekulé structure, and the two forms that average into it:
All three describe the same real molecule — drawn live.
1. Benzene Is a Regular Hexagon With Three Alternating Double Bonds
Draw a regular hexagon, then double three bonds in an alternating pattern — double, single, double, single — giving a Kekulé structure.
2. Every Ring Carbon Is sp² Hybridized, Planar, and Carries One Hydrogen
Each carbon is sp², bonded at 120° in one flat plane to two neighbors and one H; replace that H to add a substituent.
3. The Two Kekulé Structures Are Resonance Forms, Not Two Molecules
Shift the double bonds by one position and you get a second Kekulé form; the real molecule is the resonance hybrid of both, with six equal ~1.39 Å bonds.
4. The Inner-Circle Notation Shows Delocalization — but Kekulé Wins for Mechanisms
The inner circle is clean shorthand for the delocalized π system, but you can't push arrows from it — use the circle to show a ring, Kekulé to react it.
5. Name Positions on a Disubstituted Ring as Ortho, Meta, and Para
Two groups are ortho (1,2, adjacent), meta (1,3, one carbon apart), or para (1,4, directly across) — as in the three xylenes.
6. Most "New" Aromatics Are Just Benzene With a Group Swapped In
Attach one group to a vertex for its trivial name — phenol, aniline, benzoic acid, styrene, nitrobenzene — or fuse/substitute the ring for naphthalene and pyridine.
7. Benzene Is Special Because It Is Aromatic
Benzene's six π electrons satisfy Hückel's 4n+2 rule, giving ~36 kcal/mol of extra stability — so it reacts by substitution, not addition (see aromaticity and Hückel's rule).
8. Summary
Hexagon + three alternating double bonds = Kekulé · two forms resonate into one delocalized ring · six equal sp² CH carbons · circle to show, Kekulé to react · ortho/meta/para for disubstitution · it all traces back to aromaticity.
Quiz yourself
Tap a question to reveal the answer — free, no login.
Because each sp² carbon can form only one π (double) bond, and a hexagon has an even number of carbons, the double bonds must fall on every other bond — single, double, single, double — so that all six carbons are used exactly once. Two adjacent double bonds would give one carbon two π bonds, which is impossible here.
No. They are resonance structures of a single molecule. The real benzene is the average (resonance hybrid) of both, which is why all six C–C bonds are identical in length (~1.39 Å) rather than alternating short and long.
Use the circle for quick structural sketches where you only need to show that the ring is aromatic and delocalized. Switch to explicit Kekulé double bonds whenever you have to push curved arrows in a mechanism, because the circle is ambiguous about where the electrons are.
Para (1,4): the two groups sit directly across the ring from each other, with two carbons between them on either side. Ortho would be 1,2 (adjacent) and meta would be 1,3 (one carbon apart).
Draw this on the whiteboard
Open the OChem Board whiteboard — benzene rings, wedge/dash bonds, and a clickable periodic table built in. No account needed.