Free Radicals: Structure, Stability & Bond Dissociation Energy

Section	Topic
1	What is a Free Radical? Structure and Electronic Nature
2	Homolytic vs Heterolytic Bond Cleavage
3	Radical Stability: The Stability Order and Why It Exists
4	Bond Dissociation Energy (BDE) and Its Role in Radical Reactions

1. What Is a Free Radical? Structure and Electronic Nature

A free radical is any species that contains an unpaired electron. In organic chemistry, carbon-centered radicals are the most important — these are species where one carbon atom bears an unpaired electron rather than a full electron pair.

Free radicals are fundamentally different from the ionic intermediates (carbocations and carbanions) covered in substitution and elimination chemistry.

Intermediate	Electrons at Carbon	Charge	Geometry
Carbocation	6 electrons (empty p-orbital)	Positive (+)	sp², trigonal planar
Carbanion	8 electrons (lone pair)	Negative (−)	sp³, tetrahedral
Carbon radical	7 electrons (one unpaired)	Neutral (0)	sp³ or sp² (near-planar)

Carbon radicals are drawn with a dot (•) representing the unpaired electron: CH₃• (methyl radical), (CH₃)₃C• (tert-butyl radical). The dot is not just a drawing convention — it signals that the species will react rapidly with almost anything able to supply an electron or a bond partner.

How Are Radicals Formed?

Radicals are generated by homolytic bond cleavage, which requires an input of energy. The most common energy sources in laboratory radical reactions are:

Ultraviolet (UV) or visible light — photochemical initiation, abbreviated hν
Heat — thermal initiation, typically above 100°C for halogen molecules
Radical initiators such as peroxides (ROOR) that fragment homolytically at moderate temperatures

Why this matters: Radical reactions require an external energy source to get started — they do not simply occur when two reagents are mixed at room temperature in the dark. This is why methane and Cl₂ can be stored together with no reaction until light or heat is applied.

How Does Radical Chemistry Differ From Ionic Chemistry?

Feature	Ionic Chemistry (SN1, SN2, E1, E2, Electrophilic Addition)	Radical Chemistry
Bond breaking	Heterolytic — both electrons go to one atom	Homolytic — one electron to each atom
Intermediates	Carbocations, carbanions, halonium ions	Radicals (neutral, unpaired electron)
Driving force	Charge stabilization, nucleophile/electrophile interaction	Radical stability, BDE of bonds broken/formed
Initiation	No initiation needed — reagents react directly	Requires light, heat, or initiator
Chain reaction?	No	Yes
Curved arrows	Two-headed arrows (pairs of electrons)	Fishhook (one-headed) arrows (single electrons)

2. What Is the Difference Between Homolytic and Heterolytic Bond Cleavage?

When a covalent bond breaks, the two electrons that made up that bond must go somewhere. There are exactly two possibilities, and they lead to completely different types of chemistry.

Type of Cleavage	What Happens
Heterolytic cleavage	Both electrons go to the more electronegative atom. One atom becomes electron-rich (anion or neutral nucleophile); the other becomes electron-poor (cation or electrophile).
Homolytic cleavage	One electron goes to each atom. Both atoms end up with an unpaired electron and are neutral (or carry radical character).

Visualizing Homolytic Cleavage

The key halogen bond that initiates radical halogenation is the X–X bond in Cl₂ or Br₂. When this bond undergoes homolytic cleavage:

Homolytic Cleavage of Cl₂

Cl–Cl → Cl• + Cl• (homolytic cleavage; each Cl gets one electron)

Each chlorine atom now has seven valence electrons and one unpaired electron. This chlorine radical is highly reactive and will immediately seek to form a bond to stabilize itself, initiating the chain reaction.

Contrast this with what happens in ionic reactions, such as HCl dissolving in water:

Heterolytic Cleavage of H–Cl in Water

H–Cl + H₂O → H₃O⁺ + Cl⁻ (heterolytic cleavage; both electrons go to Cl)

Critical notation rule: In radical mechanisms, always use fishhook (half-headed) arrows to show the movement of single electrons. Using full double-headed arrows in a radical mechanism is a significant error. One fishhook arrow represents one electron moving.

When Does Homolytic Cleavage Occur?

Homolytic cleavage is favored under three conditions:

Low bond polarity — nonpolar or weakly polar bonds like Cl–Cl, Br–Br, or C–C. There is no reason for both electrons to go to one atom if the atoms have similar electronegativities.
External energy supplied — light or heat. Homolytic cleavage is always endothermic (bond breaking requires energy); the energy from a photon or heat drives the initiation step.
Stable radicals are formed — the more stable the radicals produced, the lower the activation energy for cleavage.

3. What Determines Radical Stability?

Just as carbocation stability is fundamental to SN1, E1, and electrophilic addition, radical stability is fundamental to radical halogenation. The more stable the radical intermediate, the more easily it forms, and the lower the activation energy for the step that produces it.

The Radical Stability Order

Tertiary (3°) > Secondary (2°) > Primary (1°) > Methyl

Radical Type	Example
Methyl	CH₃•
Primary (1°)	CH₃CH₂•
Secondary (2°)	(CH₃)₂CH•
Tertiary (3°)	(CH₃)₃C•
Allylic (next to C=C)	CH₂=CHCH₂•
Benzylic (next to ring)	C₆H₅CH₂•

This is the same substitution trend as carbocation stability, but for a different reason. Carbocations are stabilized because alkyl groups donate electron density into an empty p-orbital (hyperconjugation and inductive effects). Radicals are stabilized by the same interactions, but the orbital being stabilized contains one electron rather than being empty.

Why Are More-Substituted Radicals More Stable?

Reason 1: Hyperconjugation

Alkyl groups adjacent to the radical center donate electron density through hyperconjugation — partial overlap of adjacent C–H and C–C σ bonds with the half-filled orbital on the radical carbon. Each additional alkyl group provides more of this stabilizing overlap. A tertiary radical has three alkyl groups doing this; a primary radical has only one.

Reason 2: Inductive Electron Donation

Alkyl groups are weakly electron-donating through the σ bond framework (inductive effect). This slight donation of electron density toward the radical center provides additional stabilization. More alkyl groups means greater inductive stabilization.

Magnitude matters for selectivity: The stability difference between radical types is smaller than for carbocations — a tertiary radical is only about 3–4 kcal/mol more stable than a primary radical (consistent with the BDE differences in Section 4). This relatively small gap matters enormously for halogenation selectivity, discussed further below.

What Is the Geometry of a Carbon Radical?

Carbon radicals have a geometry close to but not quite planar. The radical carbon is approximately sp² hybridized, similar to a carbocation, with the unpaired electron in a p-type orbital perpendicular to the three substituents.

However, the radical center inverts rapidly (the "umbrella flip"), so radicals do not lead to retention of configuration at the radical center. If a radical forms at a stereocenter, the product is typically a racemic mixture.

4. What Is Bond Dissociation Energy and How Do You Use It?

Bond Dissociation Energy (BDE) is the energy required to break a bond homolytically in the gas phase, producing two radicals. It is always positive (endothermic) because bond breaking always requires an input of energy.

A–B → A• + B• ΔH = BDE (always positive)

Key BDE Values for Radical Halogenation

Bond	BDE (kcal/mol)
Cl–Cl	58
Br–Br	46
F–F	38
I–I	36
H–Cl	103
H–Br	87
1° C–H (R–CH₃)	~100
2° C–H (R₂CH₂)	~98–99
3° C–H (R₃CH)	~95–96
Allylic C–H	~88

How Do You Calculate Reaction Enthalpy From BDE?

For any radical reaction step, calculate the enthalpy change (ΔH) using BDE values:

ΔH = ΣBDE(bonds broken) − ΣBDE(bonds formed)

Bonds broken are endothermic (+); bonds formed are exothermic (−, subtracted). The sign of ΔH tells you whether the step is favorable (exothermic, ΔH < 0) or unfavorable (endothermic, ΔH > 0).

Worked Example — Propagation Step 1 of Chlorination of Methane

CH₄ + Cl• → CH₃• + HCl

Bond broken: C–H in methane. BDE = +105 kcal/mol (endothermic, energy in).
Bond formed: H–Cl. BDE = −103 kcal/mol (exothermic, energy released).
ΔH = +105 − 103 = −2 kcal/mol. Slightly exothermic overall. The step is thermodynamically favorable.

Worked Example — Propagation Step 1 of Bromination of Methane

CH₄ + Br• → CH₃• + HBr

Bond broken: C–H in methane. BDE = +105 kcal/mol.
Bond formed: H–Br. BDE = −87 kcal/mol.
ΔH = +105 − 87 = +18 kcal/mol. Endothermic. This step is thermodynamically unfavorable for methane bromination.

Why this is the mechanistic root of bromination's selectivity: The endothermicity of Step 1 in bromination is the reason bromination is far more selective than chlorination. An endothermic step has a later transition state that resembles the product (the radical) — by the Hammond Postulate, radical stability therefore matters much more for bromination than for chlorination. See our Hammond Postulate guide for the full energy-diagram argument.

Quick Recap

Radicals are neutral species with one unpaired electron — formed by homolytic bond cleavage
Use fishhook (half-headed) arrows for radical mechanisms, never full curved arrows
Radical stability: 3° > 2° > 1° > methyl; allylic ≈ benzylic >> 3° (resonance)
BDE = energy to break a bond homolytically; lower BDE = weaker bond = easier to break
Use ΔH = ΣBDE(broken) − ΣBDE(formed) to assess each propagation step
Bromination H-abstraction is endothermic; chlorination H-abstraction is exothermic — this drives selectivity
Don't confuse homolytic (radicals) and heterolytic (ions) cleavage
Don't confuse radical stability with carbocation stability — same order, different underlying reason

Common Questions

What is a free radical in organic chemistry?

A free radical is a neutral species containing one unpaired electron, most commonly on a carbon atom. It forms via homolytic bond cleavage and is drawn with a dot (•), such as CH₃• for the methyl radical. Radicals are highly reactive and distinct from carbocations and carbanions.

What is the difference between homolytic and heterolytic bond cleavage?

In heterolytic cleavage, both bonding electrons go to one atom, producing a cation and an anion — this is how ionic mechanisms like SN1 and SN2 work. In homolytic cleavage, one electron goes to each atom, producing two neutral radicals. Homolytic cleavage requires an external energy source like heat or light.

Why are tertiary radicals more stable than primary radicals?

Tertiary radicals are more stable due to hyperconjugation and inductive electron donation from three adjacent alkyl groups, compared to only one for a primary radical. This is the same underlying logic as carbocation stability, but the stabilized orbital holds one electron instead of being empty. The difference is about 3–4 kcal/mol between tertiary and primary.

What is Bond Dissociation Energy (BDE)?

Bond Dissociation Energy is the energy required to homolytically break a bond in the gas phase, producing two radicals. It is always a positive (endothermic) value. Lower BDE means a weaker bond that breaks more easily — for example, a 3° C–H bond (~95–96 kcal/mol) breaks more easily than a 1° C–H bond (~100 kcal/mol).

How do you calculate the enthalpy of a radical reaction step using BDE?

Use the formula ΔH = ΣBDE(bonds broken) − ΣBDE(bonds formed). Bonds broken require energy input (positive), while bonds formed release energy (negative, subtracted). A negative ΔH means the step is exothermic and favorable; a positive ΔH means it is endothermic and unfavorable.

Why is bromination more selective than chlorination?

The H-abstraction step is endothermic for bromination (+18 kcal/mol for methane) but exothermic for chlorination (−2 kcal/mol). By the Hammond Postulate, an endothermic step has a late transition state resembling the radical product, so radical stability strongly controls the rate. This makes bromination highly selective for more stable (more substituted) radicals, while chlorination is much less selective.

References & Further Reading

Clayden, J.; Greeves, N.; Warren, S. Organic Chemistry, 2nd ed.; Oxford University Press, 2012. Chapter 39.
McMurry, J. Organic Chemistry, 9th ed.; Cengage Learning, 2016. Chapter 6.
Wade, L. G. Organic Chemistry, 9th ed.; Pearson, 2017. Chapter 5.
Hammond, G. S. A Correlation of Reaction Rates. J. Am. Chem. Soc. 1955, 77 (2), 334–338. DOI: 10.1021/ja01607a027
Chemistry LibreTexts: Radical Stability and BDE. chem.libretexts.org
Master Organic Chemistry: Free Radical Stability. masterorganicchemistry.com