Free Radicals

1. Free Radicals and Unpaired Electrons

A free radical is any species with an odd number of valence electrons, so at least one electron is unpaired. A carbon radical has seven valence electrons: three bonds (6 e⁻) + one unpaired electron (1 e⁻). Compare a neutral carbon (8 e⁻), a carbocation (6 e⁻), and a carbanion (8 e⁻ with a lone pair). Radicals are neutral; the odd electron carries no formal charge. The unpaired electron is shown as a single dot, the hallmark of radical ("fishhook") chemistry.

2. Radical Geometry and the Orbital

A carbon radical is essentially trigonal planar (≈ sp²-hybridized): three σ bonds lie ~120° apart, and the unpaired electron occupies the p orbital above and below that plane. (Methyl is planar; larger radicals may be slightly shallow-pyramidal, but planar is the working model.) Because the odd electron is in a p orbital, attack can occur from either face, which is why reactions at radical centers erase stereochemistry (racemization) — used in topic 38.

3. Radical Stability: Substitution, Hyperconjugation, Resonance

Stability follows methyl < primary < secondary < tertiary (CH₃• < RCH₂• < R₂CH• < R₃C•). Two effects drive this: inductive donation by alkyl groups, and — more important — hyperconjugation, in which adjacent C–H/C–C σ bonds donate electron density into the half-filled p orbital. More substituents = more hyperconjugation = more stable.

Resonance stabilization is a larger, separate effect. An allylic radical (CH₂=CH–CH₂•) is on a carbon next to a C=C; its odd electron delocalizes over two carbons (CH₂=CH–CH₂• ↔ •CH₂–CH=CH₂). A benzylic radical (C₆H₅–CH₂•) spreads the odd electron into the ring. Both beat a tertiary radical. A vinylic radical (CH₂=CH•) holds its odd electron on an sp² carbon of the double bond — higher s-character, held tighter, and not delocalizable — so vinylic/aryl radicals are less stable than even primary. Full order:

benzylic ≈ allylic > 3° > 2° > 1° > methyl > vinylic/aryl

How it works

1. A bond to carbon breaks homolytically, each atom keeping one electron (topic 37).
2. The carbon is left with seven valence electrons — a radical — in ~trigonal planar geometry, odd electron in a p orbital.
3. Adjacent σ bonds (hyperconjugation) and any π system (resonance) donate density into that p orbital.
4. The more the odd electron spreads out, the more stable the radical — which decides which product dominates (topics 38-40).

Common confusions

Memory aids

"Radicals are RUSHY: Resonance wins, Unpaired electron = dot, Substitution (3° > 2° > 1° > methyl), Hyperconjugation, You'll never find a stable vinyl." Order: "Methyl < Primary < Secondary < Tertiary < Allylic/Benzylic," with vinyl/aryl last.

Quick review

Topic Recap

A free radical is a neutral, odd-electron species; carbon radicals have seven valence electrons and are drawn with a dot. They are ~trigonal planar with the odd electron in a p orbital. Stability follows methyl < primary < secondary < tertiary via hyperconjugation and induction, with allylic/benzylic radicals further stabilized by resonance and vinylic/aryl radicals destabilized. Radical stability predicts which radicals form and thus which products dominate in the reactions that follow.

Knowledge Check

1. How many valence electrons surround the carbon of a methyl radical, and is it charged?
2. What orbital holds the unpaired electron, and what geometry results?
3. Rank by stability: methyl, allylic, tertiary, vinylic, secondary.
4. Why is a tertiary radical more stable than a primary radical?
5. Why is an allylic radical more stable than a tertiary radical?

Answers and Rationales

1. Seven (three bonds = 6, plus one unpaired), and it is neutral. Rationale: radicals are odd-electron but uncharged; the dot is the seventh electron.
2. A p orbital, giving ~trigonal planar geometry. Rationale: ~sp² carbon with three σ bonds and the odd electron in the perpendicular p orbital.
3. Allylic > tertiary > secondary > methyl > vinylic. Rationale: resonance tops hyperconjugation; among alkyl radicals more substitution wins; vinylic cannot delocalize.
4. Hyperconjugation: more adjacent C–H/C–C σ bonds donate into the half-filled p orbital, plus more inductive donation. Rationale: more substituents = more stabilization.
5. Resonance: the odd electron delocalizes over two carbons (CH₂=CH–CH₂• ↔ •CH₂–CH=CH₂). Rationale: delocalization over a π system beats hyperconjugation alone.