Radical Stability and BDE Calculator

Calculate relative radical stability and C-H bond dissociation energies for organic molecules.

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What Bond Dissociation Energy Measures

BDE is the energy needed to break a bond homolytically — each fragment keeping one electron, producing two radicals. This is different from heterolytic cleavage, where one fragment takes both electrons and ions form instead.

A–B → A· + ·B     ΔH = BDE

A lower BDE means a weaker bond, and it means the resulting radical is more stable. The two statements are the same thing: if the radical produced is stabilised, less energy is required to make it. BDE is therefore a direct experimental measure of radical stability.

C–H typeApprox. BDE (kJ/mol)Relative to methylWhy
Methyl (CH4)439referenceNo stabilisation
Primary423−16One alkyl group hyperconjugates
Secondary413−26Two alkyl groups
Tertiary400−39Three alkyl groups
Allylic368−71Resonance delocalisation
Benzylic375−64Resonance into the ring
Vinylic465+26sp² carbon holds electrons tightly

Two distinct effects appear here. Alkyl substitution stabilises through hyperconjugation — adjacent C–H bonding electrons overlapping with the half-filled orbital — which gives the modest 13 kJ/mol steps from primary to tertiary. Resonance is far more powerful: an allylic radical delocalises over two carbons, dropping the BDE by 71 kJ/mol in one step.

Why This Controls Selectivity

Radical halogenation abstracts hydrogen from the weakest available C–H bond, so BDE predicts which product dominates. But selectivity also depends on the halogen. Bromination is highly selective — roughly 1600:1 favouring tertiary over primary — while chlorination is only about 5:1.

The reason is the Hammond postulate. Bromine abstraction is endothermic, giving a late transition state that closely resembles the radical product, so radical stability differences translate strongly into rate differences. Chlorine abstraction is exothermic with an early transition state resembling the reactants, so those differences barely register.

Worked Examples

Example 1: Tertiary C-H in isobutane
BDE=400 kJ/mol
Result: 39 kJ/mol weaker than methyl
Tertiary radicals form preferentially
Example 2: Allylic C-H (propene)
BDE=368 kJ/mol
Result: 71 kJ/mol weaker - resonance stabilized
Allylic radicals delocalized over 3 carbons
Example 3: Benzylic versus primary
Toluene: benzylic C–H is 375 kJ/mol, methyl C–H in ethane is 423
Result: 48 kJ/mol weaker
The benzylic radical delocalises into the aromatic ring. This is why toluene undergoes side-chain halogenation readily while benzene itself does not.
Example 4: Selectivity in bromination
Isobutane brominated: tertiary 400, primary 423 kJ/mol
Result: Roughly 1600:1 favouring tertiary
Only 23 kJ/mol difference, yet the ratio is enormous because the late transition state amplifies it. Chlorination of the same substrate gives about 5:1.
Example 5: Vinylic is stronger, not weaker
Vinylic C–H is 465 kJ/mol, above methyl
Result: Vinyl radicals are unstable
The sp² carbon has more s character, holding electrons closer to the nucleus. Vinylic hydrogens are essentially never abstracted in radical reactions.

Common Mistakes

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Confusing homolytic with heterolytic cleavage

BDE refers to homolytic cleavage producing radicals. Heterolytic cleavage produces ions and has a completely different energy, usually much higher in the gas phase.

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Assuming a weak bond means a fast reaction

BDE is thermodynamic. Reaction rate depends on the activation barrier, which correlates with BDE only when the transition state resembles the products.

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Treating tabulated BDE values as universal

Values are specific to the molecule. The C–H BDE in ethane differs from that in toluene, and substituents shift them by tens of kJ/mol.

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Expecting chlorination to be as selective as bromination

Chlorination is only about 5:1 selective for tertiary over primary because its early transition state barely distinguishes radical stabilities.

Frequently Asked Questions

Why is allylic radical stable?
Radical delocalized over C=C-C system by resonance. Two equivalent resonance structures share the radical, stabilizing it by ~71 kJ/mol vs methyl. Same applies to benzylic radicals on benzene ring.
BDE and reaction selectivity?
Free radical halogenation: Br2 is selective (large BDE differences matter). Cl2 is less selective (lower activation energy, exothermic for all C-H). Predict major product from most stable radical.
Why are allylic and benzylic radicals so stable?
Resonance delocalises the unpaired electron over several atoms. This lowers the BDE by 60–70 kJ/mol, far more than the 13 kJ/mol steps from alkyl substitution.
What is the difference between homolytic and heterolytic cleavage?
Homolytic splits the bonding pair evenly, giving two radicals — this is what BDE measures. Heterolytic gives both electrons to one fragment, producing ions.
Why is bromination more selective than chlorination?
Bromine abstraction is endothermic with a late, product-like transition state, so radical stability differences strongly affect rate. Chlorine abstraction is exothermic with an early transition state that barely distinguishes them.
Why is a vinylic C–H bond stronger than an alkyl one?
The sp² carbon has 33% s character versus 25% for sp³, holding the bonding electrons closer to the nucleus and strengthening the bond.
Does a low BDE always mean a fast reaction?
No. BDE is thermodynamic, describing bond strength. Rate depends on the activation barrier, which tracks BDE closely only when the transition state resembles the radical product.

Formula Explorer connections

Interpretation: This relationship tracks energy transfer, state-function change or the balance between enthalpy and entropy in a chemical process. Assumption: Keep energy units compatible, use kelvin for absolute temperature, and match standard states and reaction stoichiometry. Thermodynamic favorability does not determine reaction speed.

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