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.
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).
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.
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.
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.
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
Open the OChem Board whiteboard — benzene rings, wedge/dash bonds, and a clickable periodic table built in. No account needed.