Radical Halogenation: Mechanism, Regiochemistry & Selectivity

Q: What are the three stages of radical halogenation?

Radical halogenation proceeds through initiation (homolytic cleavage of X₂ by light or heat to generate two X• radicals), propagation (two cycling steps: X• abstracts H from the alkane to give R• + HX, then R• reacts with X₂ to give R–X + X•, regenerating the chain carrier), and termination (two radicals combine to form a stable closed-shell molecule, ending the chain).

Q: Why is bromination more selective than chlorination?

Bromination's H-abstraction step is endothermic (ΔH ≈ +5 to +18 kcal/mol), giving a late transition state that resembles the product radical. Radical stability therefore strongly controls the TS energy, resulting in ~1600:1 selectivity for 3° over 1° H. Chlorination's H-abstraction is exothermic (ΔH ≈ −2 to −5 kcal/mol), giving an early TS where radical stability barely registers — only ~5:1 selectivity for 3° over 1° H.

Q: Why can't you use F₂ or I₂ for radical halogenation?

F₂ is far too reactive — the H-abstraction step is so exothermic that the reaction is explosive and uncontrollable. I₂ is too unreactive — the H-abstraction step is endothermic overall (ΔH > 0 for the full reaction), making iodination thermodynamically unfavorable. Only Cl₂ and Br₂ give practically useful radical halogenation.

Q: What is NBS and when do you use it?

NBS (N-Bromosuccinimide) is a reagent used for selective allylic bromination. It provides a very low steady-state concentration of Br₂, which favors the slower, selective H-abstraction at the allylic position over addition across the C=C double bond. Use NBS when you need to brominate specifically at the allylic position without reacting with the alkene.

Section	Topic
1	Radical Halogenation: Overview and Conditions
2	Mechanism: Initiation, Propagation, Termination
3	Regiochemistry: Which C–H Bond Gets Halogenated?
4	Selectivity: Chlorination vs Bromination
5	The Hammond Postulate: Why Bromination Is More Selective
6	Allylic and Benzylic Radicals: Resonance Stabilization

1. What Is Radical Halogenation and When Does It Occur?

Radical halogenation is a free radical chain reaction in which one hydrogen atom of an alkane is replaced by a halogen atom (Cl or Br) to give an alkyl halide. It is one of the very few reactions that alkanes — otherwise notoriously unreactive — undergo under standard laboratory conditions.

R–H + X₂ → R–X + H–X (X = Cl or Br; requires hν or heat)

Feature	Detail
Reaction type	Free radical substitution (homolytic substitution)
Reagents	Cl₂ or Br₂ with an alkane substrate
Energy source	UV light (hν) or heat (Δ)
Product	Alkyl halide (R–X) + hydrogen halide (H–X) as byproduct
Mechanism	Chain reaction: initiation → propagation → termination
Selectivity	Moderate for Cl₂; high for Br₂

Why Only Cl₂ and Br₂?

Halogen	Practical?	Reason
F₂	No	Too reactive — H-abstraction is explosively exothermic and uncontrollable
Cl₂	Yes	Reactive, moderate selectivity — good for large-scale synthesis where mixtures are tolerable
Br₂	Yes	Slower, high selectivity — preferred when a specific product is needed
I₂	No	Too slow — H-abstraction is endothermic overall; reaction does not proceed usefully

2. What Are the Three Stages of the Radical Halogenation Mechanism?

The radical halogenation mechanism is a chain reaction. A reactive radical produced in one step is regenerated in the next, allowing the reaction to cycle continuously with only a tiny amount of initiation energy.

Initiation — Starting the Chain

Cl–Cl + hν → Cl• + Cl•

Homolytic cleavage of the halogen molecule by UV light or heat. This is the only step requiring external energy input. Each photon absorbed breaks one X–X bond, generating two halogen radicals. Only a small number of radicals need to be generated because the propagation steps sustain the chain through thousands of cycles.

ΔH = +58 kcal/mol for Cl–Cl (endothermic; energy from light)

Exam trap: The initiation step produces radicals from stable molecules — it never produces product directly. Do not draw the alkyl halide product forming in the initiation step.

Propagation — The Productive Cycle

The propagation stage consists of exactly two steps that cycle continuously. The radical consumed in Step 1 is regenerated in Step 2, creating a self-sustaining chain.

Step 1 (H-abstraction): R–H + X• → R• + H–X

The halogen radical (Cl• or Br•) abstracts a hydrogen atom from the alkane. One electron from the C–H bond goes to form H–X; the other stays on carbon, generating an alkyl radical. This step controls selectivity — it is the rate-determining step for determining which C–H bond reacts.

Chlorination: ΔH ≈ −2 to −5 kcal/mol (slightly exothermic for most C–H bonds)
Bromination: ΔH ≈ +5 to +18 kcal/mol (endothermic for most C–H bonds)

Step 2 (halogen abstraction): R• + X–X → R–X + X•

The alkyl radical reacts with a molecule of X₂, taking one halogen atom to form the product R–X while regenerating the halogen radical X•. This is where the vast majority of product forms. This step is exothermic for both Cl₂ and Br₂.

Termination — Ending the Chain

Termination steps occur when two radicals collide and combine, destroying two radical species and producing a stable closed-shell molecule:

Cl• + Cl• → Cl₂ (regenerates starting material)
R• + Cl• → R–Cl (gives product, but minor pathway)
R• + R• → R–R (dimerization side product)

Exam trap: Although some product IS formed in the R• + X• termination step, never draw the final product as coming exclusively from termination. The vast majority of product forms through propagation. On an exam, always show product formation through the propagation cycle.

3. Regiochemistry: Which C–H Bond Gets Halogenated?

For any alkane with more than one type of C–H bond, two factors together control which position is halogenated:

Radical stability — more-substituted radicals are more stable and form more easily (lower activation energy for H-abstraction).
Statistical factor — more hydrogens at a given position means more opportunities for the radical to abstract from that position.

The predicted product distribution combines both factors:

Predicted product % ≈ (number of H's at position) × (relative reactivity per H)

Worked Example — Bromination of 2-Methylpropane (Isobutane)

2-methylpropane has: 1 tertiary H (at the central carbon) and 9 primary H's (three CH₃ groups).

Relative reactivity per H: 3° H is ~1600× more reactive than 1° H toward Br•.
Predicted yield from 3° position: (1 × 1600) / (1 × 1600 + 9 × 1) ≈ 99%.
Observed: >99% tertiary product. Bromination is essentially completely selective for the more stable radical.

The key insight: For chlorination, the statistical factor (number of H's) can compete with the reactivity difference (~5:1 per H for 3° vs 1°), giving significant amounts of all possible products. For bromination, the reactivity difference is so large (~1600:1 per H for 3° vs 1°) that the statistical factor is essentially irrelevant — the major product is almost entirely from the most stable radical.

4. How Do Chlorination and Bromination Differ in Selectivity?

Selectivity refers to how strongly a radical halogenation reaction prefers one type of C–H bond over another. This is one of the most commonly tested topics in radical chemistry.

Feature	Chlorination (Cl₂)	Bromination (Br₂)	Fluorination (F₂)	Iodination (I₂)
Selectivity	Low (poor)	High (excellent)	None (explosive)	N/A
Reactivity	High (fast)	Low (slow)	Explosive	Too slow
H-abstraction ΔH	Exothermic (~−2 to −5)	Endothermic (+5 to +18)	Strongly exothermic	Endothermic overall
Transition state	Early (reactant-like)	Late (product-like)	Very early	—
3° vs 1° per H	~5:1	~1600:1	~1:1	—
2° vs 1° per H	~4.5:1	~82:1	~1:1	—
Practical use?	Yes, but mixtures likely	Yes, high purity	No	No

Why Is Chlorination Less Selective?

Chlorination is highly exothermic in the H-abstraction step. When a reaction step is strongly exothermic, the activation energy is low and the transition state comes early — before much C–H bond breaking has occurred.

At an early transition state, very little radical character has developed at carbon. The stability difference between 1°, 2°, and 3° radicals barely registers in the transition state energy, so chlorination shows poor selectivity (~5:1 per H for 3° vs 1°).

Why Is Bromination More Selective?

Bromination is endothermic in the H-abstraction step. When a step is endothermic, the transition state comes late — after substantial C–H bond breaking has occurred. The carbon center already has significant radical character at the transition state.

The stability of the developing radical now strongly affects the transition state energy, so bromination is highly sensitive to whether the radical is 1°, 2°, or 3° — giving ~1600:1 selectivity per H for 3° vs 1°.

5. How Does the Hammond Postulate Explain Bromination's Selectivity?

The Hammond Postulate explains the selectivity difference in a single principle: the transition state of any reaction step resembles whichever species is closest to it in energy.

Feature	Chlorination	Bromination
H-abstraction ΔH	Exothermic (~−2 to −5 kcal/mol)	Endothermic (+5 to +18 kcal/mol)
Transition state position	Early (reactant-like)	Late (product-like)
Radical character at TS	Minimal — C–H barely broken	Substantial — C–H mostly broken
Effect of radical stability on TS	Barely registers	Strongly controls TS energy
Selectivity	Low — mixture of products	High — essentially one product

The core argument: The energy difference between a 3° and 1° radical is roughly the same for both reactions (~3–4 kcal/mol). What changes is how much of that difference shows up at the transition state. Early TS (Cl•) → almost none shows up. Late TS (Br•) → nearly all of it shows up. This is the entire mechanistic basis of the selectivity difference.

For a deeper treatment with full energy diagrams and applications across SN1, Markovnikov's rule, and radical chemistry, see the companion guide: The Hammond Postulate — Transition States, Energy Diagrams, and Reaction Selectivity.

6. What Makes Allylic and Benzylic Radicals Especially Stable?

Beyond the 1°/2°/3° alkyl radical series, radicals formed adjacent to a π system are substantially more stable due to resonance delocalization of the unpaired electron. This makes allylic and benzylic C–H bonds much more reactive toward radical H-abstraction.

Allylic Radicals — Delocalization Into C=C

An allylic radical forms when a C–H bond adjacent to a C=C double bond undergoes homolytic cleavage. The unpaired electron delocalizes into the adjacent π system, spreading across two carbon atoms.

The allylic C–H BDE is only ~88 kcal/mol, significantly lower than a typical primary C–H BDE of ~100 kcal/mol. This means allylic hydrogens are far easier to abstract than ordinary primary hydrogens — even by the less reactive Br•.

Critical exam point: When the allylic radical reacts with Br₂ in Propagation Step 2, Br• can bond to either end of the delocalized radical, potentially giving a mixture of allylic products. Always check both ends of an allylic radical when predicting products — draw out both resonance structures of the allylic radical before predicting.

NBS — Selective Allylic Bromination

N-Bromosuccinimide (NBS) is a specialized reagent for selectively brominating at the allylic position. NBS provides a very low steady-state concentration of Br₂, which:

Favors the slower, selective H-abstraction at the allylic position (NBS conditions favor Br•, which is selective)
Avoids addition across the C=C double bond (addition requires higher Br₂ concentration)

Conditions: NBS, CCl₄, hν or AIBN (radical initiator).

Benzylic Radicals — Delocalization Into the Aromatic Ring

A benzylic radical forms when a C–H bond directly attached to a benzene ring undergoes homolytic cleavage. The unpaired electron delocalizes into the aromatic π system, spreading across multiple ring positions through several resonance contributors.

Benzylic radicals are even more stable than simple allylic radicals because the aromatic ring provides multiple resonance contributors. The benzylic C–H BDE (~88–90 kcal/mol) is similar to the allylic value and much lower than a primary C–H. Benzylic positions are highly reactive toward radical halogenation even with the less reactive Br•.

Complete Radical Stability Ranking

Methyl < 1° < 2° < 3° < Allylic ≈ Benzylic

Radical Type	Example	Approx. BDE of C–H (kcal/mol)
Methyl	CH₃•	~105
Primary	CH₃CH₂•	~100
Secondary	(CH₃)₂CH•	~98–99
Tertiary	(CH₃)₃C•	~95–96
Allylic	CH₂=CHCH₂•	~88
Benzylic	C₆H₅CH₂•	~88–90

Quick Recap — Radical Halogenation

Concept	Key Fact
Reaction type	Free radical substitution chain reaction
Conditions	Cl₂ or Br₂ + alkane + hν or heat
Initiation	X₂ → 2X•; requires energy; produces radicals NOT product
Propagation Step 1	X• + R–H → R• + H–X (H-abstraction; controls selectivity)
Propagation Step 2	R• + X₂ → R–X + X• (makes product; regenerates chain carrier)
Termination	Radical + Radical → stable molecule; never draw main product from here
Chlorination	Fast, low selectivity — exothermic H-abstraction, early TS
Bromination	Slow, high selectivity — endothermic H-abstraction, late TS
Hammond Postulate	Endothermic step → late TS → radical stability matters → high selectivity
Allylic/Benzylic C–H	BDE ~88 kcal/mol; very reactive; NBS for selective allylic bromination

Do's and Don'ts for Exam Problems

Use fishhook (half-headed) arrows for all radical steps
Show product forming through the propagation cycle
Check both ends of an allylic radical when predicting bromination products
Apply statistical correction: multiply reactivity per H by number of H's
Link Br₂ selectivity to endothermic H-abstraction and late TS (Hammond)
Never draw product forming in the initiation step
Never draw the main product as coming from termination
Don't use full double-headed curved arrows in radical mechanisms
Don't use F₂ (explosive) or I₂ (endothermic overall)
Don't forget Br can bond to either end of a delocalized allylic radical

Common Questions

What are the three stages of radical halogenation?

Radical halogenation proceeds through: initiation (homolytic cleavage of X₂ by light or heat generating two X• radicals), propagation (two cycling steps: X• abstracts H from the alkane to give R• + HX, then R• reacts with X₂ to give R–X + X•, regenerating the chain carrier), and termination (two radicals combine to form a stable closed-shell molecule, ending the chain).

Why is bromination more selective than chlorination?

Bromination's H-abstraction step is endothermic (ΔH ≈ +5 to +18 kcal/mol), giving a late transition state that resembles the product radical. Radical stability therefore strongly controls the TS energy, resulting in ~1600:1 selectivity for 3° over 1° H. Chlorination's H-abstraction is exothermic (ΔH ≈ −2 to −5 kcal/mol), giving an early TS where radical stability barely registers — only ~5:1 selectivity for 3° over 1° H.

Why can't you use F₂ or I₂ for radical halogenation?

F₂ is far too reactive — the H-abstraction step is so exothermic that the reaction is explosive and uncontrollable. I₂ is too unreactive — the H-abstraction step is endothermic overall (ΔH > 0 for the full reaction), making iodination thermodynamically unfavorable. Only Cl₂ and Br₂ give practically useful radical halogenation.

Where does product form in the radical halogenation mechanism?

The vast majority of product (R–X) forms in Propagation Step 2, where the alkyl radical R• reacts with X₂ to give R–X + X•. Some product also forms in the termination step R• + X• → R–X, but this is a minor pathway. On exams, always show product forming through the propagation cycle — never draw all product as coming from termination.

Why are allylic and benzylic radicals especially stable?

Allylic radicals (adjacent to C=C) and benzylic radicals (adjacent to benzene ring) are stabilized by resonance delocalization of the unpaired electron into the adjacent π system. The allylic C–H BDE is only ~88 kcal/mol vs ~100 kcal/mol for a primary C–H, making allylic and benzylic positions far more reactive toward radical H-abstraction. Benzylic radicals have multiple resonance contributors through the aromatic ring, making them even more stable.

What is NBS and when do you use it?

NBS (N-Bromosuccinimide) is a reagent for selective allylic bromination. It provides a very low steady-state concentration of Br₂, which favors the slower, selective H-abstraction at the allylic position over addition across the C=C double bond. Use NBS (with CCl₄ solvent and hν or AIBN initiator) when you need to brominate specifically at the allylic position without reacting with the alkene.

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.
Kharasch, M. S.; Hered, W.; Mayo, F. R. J. Org. Chem. 1941, 6 (6), 818–829. DOI: 10.1021/jo01206a005 (Original radical chain mechanism paper)
Hammond, G. S. J. Am. Chem. Soc. 1955, 77 (2), 334–338. DOI: 10.1021/ja01607a027 (Hammond Postulate)
Chemistry LibreTexts: Free Radical Halogenation of Alkanes. chem.libretexts.org
Master Organic Chemistry: Initiation, Propagation, Termination. masterorganicchemistry.com