Learn · Organic Chemistry

Frost Circles

The polygon-in-a-circle trick for π molecular-orbital energy levels

Quick answer A Frost circle inscribes a ring's polygon in a circle with one vertex pointing down, so each vertex marks a π MO energy — below center is bonding, at center nonbonding, above center antibonding. Fill electrons from the bottom up: 4n+2 electrons that exactly fill all bonding MOs give a stable aromatic ring, while 4n electrons leave two unpaired in nonbonding orbitals and give an unstable antiaromatic ring.

1. A Frost circle turns any cyclic conjugated ring into its π MO energy diagram.

Inscribe the ring's polygon in a circle with one vertex pointing straight down; every point where a vertex touches the circle is one π molecular orbital.

Benzene — a hexagon inscribed point-down gives 6 π MOs

2. The circle's center is the nonbonding line that sorts the orbitals.

Vertices below center are bonding (stabilizing), a vertex exactly at center is nonbonding, and vertices above center are antibonding (destabilizing).

Cyclobutadiene — a square gives one bottom, two center, one top MO

3. Filling 4n+2 π electrons into only bonding MOs gives a closed shell — aromatic.

Benzene's 6 π electrons exactly fill its three bonding orbitals with no unpaired electrons, so the ring is aromatic and unusually stable.

Benzene — 6 π, three bonding MOs full → aromatic
Naphthalene — a 10 π fused system, still aromatic

4. Filling 4n π electrons leaves two unpaired in the nonbonding pair — antiaromatic.

Cyclobutadiene's 4 π electrons fill the bottom MO then split singly into the two degenerate nonbonding orbitals, an open shell that is antiaromatic and unstable.

Cyclobutadiene — 4 π, two unpaired → antiaromatic
Cyclooctatetraene — 8 π would be antiaromatic, so it puckers non-planar instead

5. The same picture explains why charged rings become aromatic.

Adding or removing electrons shifts the count to 4n+2: cyclopentadiene's ring gains an electron to form the aromatic 6 π cyclopentadienyl anion, and cycloheptatriene loses one to form the aromatic 6 π tropylium cation.

Cyclopentadiene — the CH₂ loses H⁺ → aromatic 6 π anion
Cycloheptatriene — loses hydride → aromatic 6 π tropylium cation

6. Summary

Polygon point-down in a circle · vertices = π MOs · below center bonding, center nonbonding, above antibonding · fill bottom-up · 4n+2 closed shell = aromatic · 4n open shell = antiaromatic.

Quiz yourself

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One vertex points straight down; this puts a single lowest bonding MO at the bottom of the circle.

Exactly at the circle's center — vertices below it are bonding, vertices above it are antibonding.

Its square gives two degenerate nonbonding MOs; after filling the bottom orbital, the last two electrons go in singly and unpaired — an unstable open shell.

Gaining one electron gives 6 π electrons (4n+2) that exactly fill all three bonding MOs — a closed aromatic shell.

Draw this on the whiteboard

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