A representative example — structures drawn live.
The stability order
A carbon radical is a carbon with only seven valence electrons and one unpaired electron — electron-deficient, much like a carbocation. Its stability increases with the number of attached alkyl groups:
- tertiary (3°) > secondary (2°) > primary (1°) > methyl
This is the same ordering you learned for carbocations, and for the same underlying reason: alkyl groups stabilize an electron-poor center. If you already know carbocation stability, radical stability comes almost for free.
Why alkyl groups help: hyperconjugation and induction
Two effects stabilize a more-substituted radical:
- Hyperconjugation. Adjacent C–H (and C–C) bonds can overlap with the half-filled orbital on the radical carbon, spreading the unpaired electron over more of the molecule. More neighboring bonds means more hyperconjugation, and 3° radicals have the most.
- Inductive donation. Alkyl groups are weakly electron-donating and help supply electron density to the electron-poor radical center.
Together these lower the energy of the radical, so the barrier to forming it is smaller.
Resonance: allylic and benzylic radicals
The strongest stabilization comes from resonance. When a radical sits next to a π system, the unpaired electron delocalizes across it:
- Allylic radicals (next to a C=C) are spread over two carbons through resonance.
- Benzylic radicals (next to a benzene ring) delocalize into the ring across several resonance structures.
This delocalization makes allylic and benzylic radicals even more stable than a typical 3° radical, which is why allylic and benzylic C–H bonds are the weakest and most easily abstracted.
The Hammond postulate and selectivity
Radical stability governs product distribution because the hydrogen-abstraction step in halogenation builds the carbon radical. The Hammond postulate says that a transition state resembles whichever species — reactant or product — it is closer to in energy.
- For bromination, abstraction by Br• is endothermic. Its transition state is late (product-like) and looks a lot like the carbon radical. So the full stability difference between a 3° and a 1° radical is felt in the transition state, and bromination is highly selective for the most stable radical.
- For chlorination, abstraction by Cl• is exothermic. Its transition state is early (reactant-like), so radical stability barely registers and chlorination is unselective, giving mixtures.
This is why the choice of halogen matters so much: the more stable radical is always formed faster, but only when the transition state resembles the radical (as in bromination) does that translate into strong selectivity.
Predicting the major product
To predict the outcome of a radical reaction, identify which C–H, when removed, gives the most stable radical. Rank the candidate positions using resonance (allylic/benzylic) > 3° > 2° > 1°. That position gives the major product with a selective reagent like Br2 or NBS. With an unselective reagent like Cl2, you weight by the number of each type of hydrogen as well, since selectivity is low. Tying it together: radical stability sets both the rate of formation and, through the Hammond postulate, the selectivity that determines product ratios.
Draw this on the whiteboard
Open the OChem Board whiteboard — benzene rings, curved arrows, wedge/dash bonds and a clickable periodic table built in. No account needed.