Organic Chemistry 1 • Radical Reactions

Radical Halogenation: Selectivity and Mechanism

Understanding how free-radical chain reactions transform alkanes into alkyl halides with predictable selectivity.

A molecular ring with laboratory flask and atom symbols for organic chemistry.
Inside this lesson 10 sections 5 questions 9 figures

SECTION 1

Historical Context & Motivation

Alkanes are among the least reactive organic molecules, earning them the historical name "paraffins" — from the Latin parum affinis, meaning "little affinity." Their strong, nonpolar C−H bonds resist most reagents that attack polar functional groups. Yet chemists in the nineteenth century observed that exposing methane or other simple hydrocarbons to chlorine gas under ultraviolet light produced hydrogen chloride and chlorinated products. This observation raised a fundamental question: how does a thermodynamically stable C−H bond break, and why do some C−H bonds react faster than others? The study of radical halogenation developed to answer these questions and became a cornerstone of free-radical chemistry.

1834 • Dumas Discovers Chlorination of Methane

Jean-Baptiste Dumas demonstrated that methane reacts with chlorine in sunlight to form chloromethane and HCl, providing early evidence that saturated hydrocarbons could undergo substitution reactions.

1900s • Rise of Free-Radical Theory

Moses Gomberg's work on the triphenylmethyl radical (1900) established that carbon-centered radicals could exist as discrete intermediates, laying the conceptual groundwork for radical chain mechanisms.

1935 • Rice & Herzfeld Propose Chain Mechanisms

Francis Rice and Karl Herzfeld formulated the initiation–propagation–termination framework for radical chain reactions, explaining the kinetics and high quantum yields observed in halogenation.

1947 • Hammond Postulate & Selectivity Rationalized

George Hammond's postulate linked transition-state structure to the thermodynamics of a step, explaining why bromine is far more selective than chlorine in radical halogenation.

1970s • Bond Dissociation Energies & Modern Selectivity Data

Precise BDE measurements enabled quantitative predictions of selectivity ratios, refining the per-hydrogen reactivity factors for chlorination and bromination at primary, secondary, and tertiary C−H sites.

The central question this lesson addresses is twofold: what is the step-by-step mechanism by which a halogen atom replaces a hydrogen on an alkane, and how can we predict which hydrogen will be replaced when multiple types of C−H bonds are present? The answers lie in bond dissociation energies, radical stability, and transition-state theory.

SECTION 2

Core Principles & Definitions

Radical halogenation converts an alkane R−H into an alkyl halide R−X (where X = Cl or Br) through a free-radical chain mechanism. The reaction requires an energy input — typically ultraviolet light (hν) or heat (Δ) — to generate the initial radical species. Understanding the mechanism and its selectivity depends on several interlocking concepts that form the intellectual backbone of radical chemistry.

1

Homolytic Bond Cleavage

A covalent bond breaks so that each fragment retains one electron, producing two radicals. This contrasts with heterolytic cleavage (ions). Represented by fishhook (single-barbed) arrows.
2

Bond Dissociation Energy (BDE)

The enthalpy required to homolytically break a specific bond in the gas phase. BDE values dictate the ΔH° of each propagation step and, ultimately, the selectivity of the halogen.
3

Radical Stability Order

Carbon radicals are stabilized by hyperconjugation: tertiary (3°) > secondary (2°) > primary (1°) > methyl. A more stable radical forms through a lower-energy transition state.
4

Chain Mechanism Phases

Every radical chain reaction has three phases: initiation (radical generation), propagation (radical-consuming and radical-producing steps), and termination (radical destruction by coupling).
5

Selectivity vs. Reactivity

A more reactive radical (Cl•) is less discriminating among C−H bond types; a less reactive radical (Br•) is highly selective. This inverse relationship is explained by the Hammond postulate.
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KEY TAKEAWAY

Think of radical selectivity like a hiring process. An employer desperate to fill a position immediately (chlorine — highly reactive) will hire almost anyone. A very selective employer with a low-urgency opening (bromine — less reactive) will carefully interview candidates and pick only the most qualified. In radical chemistry, the "most qualified" hydrogen is the one on the most substituted carbon, because removing it yields the most stable radical intermediate.

SECTION 3

The Radical Chain Mechanism — Visual Overview

The following diagram illustrates the complete radical chain mechanism for the chlorination of methane, the simplest case. Each phase — initiation, propagation, and termination — is shown with the relevant bond-breaking and bond-forming events. Note the use of fishhook arrows (single-barbed) to indicate the movement of single electrons, as opposed to the double-barbed curved arrows used in ionic mechanisms.

INITIATIONCl–Cl → Cl• + Cl•UV light (hν) or heat (Δ) homolyzes the Cl−Cl bond (BDE = 243 kJ/mol)This step generates the chain-carrying radicals.PROPAGATIONStep 1 — Hydrogen AbstractionCl• + CH₄ → HCl + •CH₃ΔH° = BDE(C−H) − BDE(H−Cl) = 435 − 431 = +4 kJ/mol (slightly endothermic)Step 2 — Halogen Abstraction•CH₃ + Cl₂ → CH₃Cl + Cl•ΔH° = BDE(Cl−Cl) − BDE(C−Cl) = 243 − 351 = −108 kJ/mol (exothermic)CyclerepeatsTERMINATIONAny two radicals combine, destroying the chain:Cl• + Cl• → Cl₂•CH₃ + Cl• → CH₃Cl•CH₃ + •CH₃ → CH₃CH₃ (ethane — a coupling by-product)Termination steps are rare because radical concentrations are very low.
Complete radical chain mechanism for the chlorination of methane. The initiation phase generates Cl• radicals. The two propagation steps cycle continuously, each consuming one radical and producing another. Termination events destroy radicals in pairs. The overall ΔH° equals the sum of both propagation steps: +4 + (−108) = −104 kJ/mol.

Several features of this mechanism deserve emphasis. First, notice that the chlorine radical consumed in propagation step 1 is regenerated in propagation step 2 — this is the defining characteristic of a chain reaction. A single initiation event can trigger thousands of propagation cycles before a termination event occurs. Second, the overall reaction enthalpy is obtained by summing the ΔH° values of the two propagation steps, not the initiation step (which is thermodynamically "paid back" during propagation). Third, the rate-determining step for selectivity purposes is propagation step 1 — the hydrogen-abstraction step — because this is where the halogen radical discriminates among different C−H bonds.

SECTION 4

Thermodynamic Analysis & the Hammond Postulate

The selectivity of radical halogenation is governed by the enthalpy of the hydrogen-abstraction step (propagation step 1). To calculate ΔH° for this step, we apply the relationship between bond dissociation energies: ΔH° = Σ(BDEs of bonds broken) − Σ(BDEs of bonds formed). Since only one bond breaks and one bond forms in this step, the equation simplifies considerably.

ENTHALPY OF HYDROGEN ABSTRACTION

ΔH°=BDE(R−H)−BDE(H−X)ΔH° = BDE(R−H) − BDE(H−X)

R−H = the C−H bond being broken; H−X = the H−Cl or H−Br bond being formed. When ΔH° is negative, the step is exothermic; when positive, endothermic.

BondBDE (kJ/mol)Radical Type Formed
CH₃−H (methyl)439Methyl radical (•CH₃)
RCH₂−H (1°)423Primary radical
R₂CH−H (2°)413Secondary radical
R₃C−H (3°)400Tertiary radical
H−Cl431—
H−Br366—

Key Bond Dissociation Energies for Radical Halogenation

For chlorination, the hydrogen-abstraction step is slightly endothermic for a 1° C−H bond: ΔH° = 423 − 431 = −8 kJ/mol (actually mildly exothermic) and more exothermic for 3° C−H bonds: ΔH° = 400 − 431 = −31 kJ/mol. The energy differences between these ΔH° values are small (about 23 kJ/mol spread across all types), so the transition-state energies are closely spaced, and chlorine shows only modest selectivity.

For bromination, the hydrogen-abstraction step is significantly endothermic: ΔH° = 423 − 366 = +57 kJ/mol (1° C−H) versus +34 kJ/mol (3° C−H). The Hammond postulate tells us that for an endothermic step, the transition state resembles the products (the carbon radical). Since 3° radicals are substantially more stable than 1° radicals, the transition state leading to a 3° radical is significantly lower in energy. This large energy gap between transition states gives bromine its extraordinary selectivity for tertiary C−H bonds.

HAMMOND POSTULATE (QUALITATIVE)

Endothermicstep→product−likeTS→selectivityreflectsradicalstabilityEndothermic step → product-like TS → selectivity reflects radical stability

TS = transition state. For exothermic (or thermoneutral) steps, the TS resembles reactants, and radical stability matters less. This is why Cl• (exothermic abstraction) is less selective than Br• (endothermic abstraction).

ℹ

Why Not Fluorine or Iodine?

Fluorination is explosively exothermic (H−F BDE = 570 kJ/mol) and virtually non-selective — F• abstracts any hydrogen with reckless speed. Iodination is endothermic overall (H−I BDE = 297 kJ/mol), making the propagation cycle thermodynamically unfavorable; the reaction simply does not proceed. Thus, only chlorination and bromination are synthetically useful radical halogenation reactions.

SECTION 5

Selectivity Factors & Product Prediction

Predicting the product distribution in radical halogenation requires two pieces of information: the number of each type of hydrogen and the relative reactivity (selectivity factor) of each hydrogen type toward the halogen radical. By convention, the reactivity of a primary C−H bond is set to 1.0, and the reactivities of secondary and tertiary C−H bonds are expressed relative to that baseline. The commonly accepted selectivity factors at 25 °C are as follows.

C−H TypeChlorination (relative rate)Bromination (relative rate)
Primary (1°)1.01
Secondary (2°)3.982
Tertiary (3°)5.21600

Relative Reactivity per Hydrogen at ~25 °C

The product ratio is calculated by multiplying the number of hydrogens of each type by its relative reactivity factor. The general formula is:

PRODUCT RATIO CALCULATION

% product at site A = (nₐ × rₐ) / Σ(nᵢ × rᵢ) × 100%

nₐ = number of equivalent H atoms at site A; rₐ = relative reactivity of that H type; Σ(nᵢ × rᵢ) sums over all distinct hydrogen types in the molecule.

Reaction Coordinate: Hydrogen Abstraction StepCHLORINATION (Cl•)Reaction Coordinate →Energy →R−H + Cl•TS (1°)TS (2°)TS (3°)R• + HClSmall ΔΔG‡ → low selectivity≈ smallBROMINATION (Br•)Reaction Coordinate →Energy →R−H + Br•TS (1°)TS (2°)TS (3°)R• + HBrLarge ΔΔG‡ → HIGH selectivity≈ large1° C−H2° C−H3° C−H
Reaction coordinate diagrams for the hydrogen-abstraction step. Chlorination (left) is mildly exothermic with a reactant-like transition state — the energy gaps between 1°, 2°, and 3° TS are small, producing low selectivity. Bromination (right) is endothermic with a product-like transition state — the TS reflects radical stability, creating large energy gaps and high selectivity.

The strikingly different selectivity factors for chlorination and bromination emerge directly from these energy diagrams. In chlorination, the three transition-state energies for 1°, 2°, and 3° C−H abstraction are clustered closely together — hence selectivity factors of only 1.0 : 3.9 : 5.2. In bromination, the same transition states are spread far apart — giving selectivity factors of 1 : 82 : 1600. The practical consequence is that bromination is the reagent of choice when you want a single constitutional isomer from a substrate that has a tertiary C−H bond.

SECTION 6

Worked Example: Monochlorination of 2-Methylbutane

Let us predict the product distribution for the radical monochlorination of 2-methylbutane (isopentane, C₅H₁₂) at 25 °C. This molecule has four distinct types of C−H bonds, making it an excellent test of the selectivity-factor method.

Monochlorination of 2-Methylbutane: Product Distribution

  1. 1

    Step 1 — Draw the Structure and Classify All Hydrogens

    2-Methylbutane is CH₃CH(CH₃)CH₂CH₃. Identify each unique hydrogen type. Position a: the two equivalent CH₃ groups attached to C2 → 6 primary (1°) H's. Position b: the single H on C2 (the tertiary carbon) → 1 tertiary (3°) H. Position c: the CH₂ group at C3 → 2 secondary (2°) H's. Position d: the terminal CH₃ at C4 → 3 primary (1°) H's.
    6 × 1° H (a), 1 × 3° H (b), 2 × 2° H (c), 3 × 1° H (d) = 12 total H atoms
  2. 2

    Step 2 — Apply the Selectivity Factors for Chlorination

    Using the relative reactivity values: 1° = 1.0, 2° = 3.9, 3° = 5.2. Calculate the weighted reactivity for each site. Site a: 6 × 1.0 = 6.0. Site b: 1 × 5.2 = 5.2. Site c: 2 × 3.9 = 7.8. Site d: 3 × 1.0 = 3.0.
    Weighted reactivities: a = 6.0, b = 5.2, c = 7.8, d = 3.0; Total = 22.0
  3. 3

    Step 3 — Calculate the Percentage of Each Product

    Divide each site's weighted reactivity by the total (22.0) and multiply by 100%. Site a (1-chloro-2-methylbutane, on the gem-dimethyl CH₃ groups): (6.0/22.0) × 100% = 27.3%. Site b (2-chloro-2-methylbutane, the 3° product): (5.2/22.0) × 100% = 23.6%. Site c (3-chloro-2-methylbutane, the 2° product): (7.8/22.0) × 100% = 35.5%. Site d (1-chloro-3-methylbutane, on the terminal CH₃): (3.0/22.0) × 100% = 13.6%.
    Product distribution: 27% (a), 24% (b), 35% (c), 14% (d) — a complex mixture
  4. 4

    Step 4 — Interpret the Result

    No single product dominates. Despite the tertiary C−H being the most reactive per hydrogen, there is only one such hydrogen, so the 3° product constitutes only about 24% of the mixture. The 2° product (site c) is actually the largest fraction because 2 hydrogens × 3.9 = 7.8 gives the highest weighted reactivity. This illustrates the critical point: selectivity depends on both the number of hydrogens AND their relative reactivity.
    Chlorination gives a mixture of four constitutional isomers — poor for synthesis.
  5. 5

    Step 5 — Compare with Bromination (for perspective)

    Using bromination selectivity factors (1° = 1, 2° = 82, 3° = 1600): Site a: 6 × 1 = 6. Site b: 1 × 1600 = 1600. Site c: 2 × 82 = 164. Site d: 3 × 1 = 3. Total = 1773. Site b: (1600/1773) × 100% = 90.2%. The tertiary product overwhelmingly dominates.
    Bromination gives ~90% 2-bromo-2-methylbutane — excellent selectivity for the 3° product

SECTION 7

Chlorination vs. Bromination — Strengths and Limitations

Chlorination and bromination each have distinct advantages and disadvantages in synthetic planning. The choice between them depends on the substrate structure and the desired outcome. The following comparison highlights the practical trade-offs a chemist must consider when designing a radical halogenation procedure.

PropertyChlorination (Cl₂ / hν)Bromination (Br₂ / hν)
SelectivityLow (1° : 2° : 3° ≈ 1 : 3.9 : 5.2)Very high (1° : 2° : 3° ≈ 1 : 82 : 1600)
ReactivityHigh — reacts with virtually any alkaneModerate — may not react well with 1° C−H only substrates
ΔH° of H-abstraction (3° C−H)−31 kJ/mol (exothermic)+34 kJ/mol (endothermic)
TS character (Hammond)Reactant-like — does not reflect radical stability wellProduct-like — strongly reflects radical stability
Synthetic utilityUseful only for methane or symmetrical alkanes (single product)Excellent for substrates with 3° C−H; gives predominant single product
Polyhalogenation riskHigher — product R−Cl is still reactive toward Cl•Lower — high selectivity limits over-reaction

Comparison of Chlorination and Bromination in Radical Halogenation

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KEY TAKEAWAY

In the broader context of organic synthesis, radical halogenation is most valuable when bromination can be used to selectively install a bromine at a tertiary carbon. This C−Br bond can then serve as a handle for further transformations — elimination to form an alkene (E2), nucleophilic substitution (SN1), or coupling reactions. Chlorination, by contrast, rarely provides a clean enough product distribution to be the first choice for complex molecule synthesis.

SECTION 8

Connection to Advanced Radical Chemistry

The principles learned in radical halogenation form the foundation for understanding far more sophisticated radical reactions encountered in advanced organic chemistry. The same chain-mechanism framework, selectivity principles, and thermodynamic reasoning apply to a range of transformations that exploit radical intermediates for strategic bond formation.

Concept in This LessonAdvanced Extension
Radical chain mechanism (initiation / propagation / termination)Radical polymerization — chains grow by repeated radical additions to alkenes (e.g., polyethylene, polystyrene)
Selectivity via BDE and Hammond postulateBarton reaction — remote C−H functionalization using alkoxy radicals, guided by geometric selectivity and BDE considerations
Radical stability (3° > 2° > 1°)Allylic and benzylic bromination (NBS reactions) — stabilized radicals direct selectivity at allylic/benzylic positions
Anti-Markovnikov addition excluded hereRadical addition of HBr to alkenes — peroxide-initiated anti-Markovnikov hydrobromination follows the same chain mechanism
Termination by radical couplingPersistent radical effect and TEMPO-mediated oxidation — controlled radical reactions that exploit selective termination

From Radical Halogenation to Advanced Radical Chemistry

Perhaps the most exciting modern extension is the renaissance of radical-mediated C−H functionalization in synthetic methodology. Chemists now use photoredox catalysis, hydrogen-atom transfer (HAT) catalysts, and metal-mediated radical generation to achieve selective C−H bond transformations under mild conditions — reactions that would have seemed miraculous to the pioneers of radical halogenation. Yet at their core, these modern methods rely on the same interplay of bond strengths, radical stability, and transition-state theory that govern the simple halogenation of an alkane.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain why the initiation step of radical halogenation is not included when calculating the overall ΔH° of the reaction. What role does initiation play thermodynamically, and why are only the propagation steps summed?
Show answer
The initiation step (homolysis of X₂) is endothermic and requires energy input, but it occurs only once (or rarely) to start the chain. The propagation steps are the net chemical transformation — they cycle repeatedly, and their sum gives the stoichiometric equation: R−H + X₂ → R−X + HX. Since the Cl• consumed in propagation step 1 is regenerated in step 2, the halogen radical is catalytic in each cycle. The initiation bond that was broken is effectively "repaid" when the chain terminates by radical recombination. Therefore, the overall thermodynamics are determined solely by the enthalpy changes in the propagation cycle: ΔH°(overall) = ΔH°(step 1) + ΔH°(step 2) = [BDE(R−H) − BDE(H−X)] + [BDE(X−X) − BDE(R−X)].

Question 2: BASIC CALCULATION

Calculate ΔH° for the hydrogen-abstraction step (propagation step 1) when Br• abstracts a secondary hydrogen. Use BDE(2° C−H) = 413 kJ/mol and BDE(H−Br) = 366 kJ/mol. Is this step exothermic or endothermic?
Show answer
ΔH° = BDE(bond broken) − BDE(bond formed) = BDE(2° C−H) − BDE(H−Br) = 413 − 366 = +47 kJ/mol. This step is endothermic by 47 kJ/mol. According to the Hammond postulate, the transition state for this endothermic step resembles the product (the secondary carbon radical plus HBr). Because radical stability is reflected in the transition state, bromine shows high selectivity — the transition state leading to a more stable radical (e.g., 3° at +34 kJ/mol) has a measurably lower activation energy than the one leading to a less stable radical (1° at +57 kJ/mol).

Question 3: INTERMEDIATE

Propane (CH₃CH₂CH₃) undergoes monochlorination. It has 6 primary hydrogens and 2 secondary hydrogens. Using chlorination selectivity factors (1° = 1.0, 2° = 3.9), predict the percentage of 1-chloropropane and 2-chloropropane in the product mixture.
Show answer
Weighted reactivity for the 1° site: 6 × 1.0 = 6.0. Weighted reactivity for the 2° site: 2 × 3.9 = 7.8. Total weighted reactivity = 6.0 + 7.8 = 13.8. Percentage of 1-chloropropane (from 1° H abstraction): (6.0 / 13.8) × 100% = 43.5%. Percentage of 2-chloropropane (from 2° H abstraction): (7.8 / 13.8) × 100% = 56.5%. Despite having three times as many primary hydrogens, the secondary product slightly dominates because each 2° H is 3.9 times more reactive. This result underscores chlorine's modest selectivity — both products form in substantial amounts.

Question 4: APPLIED

A chemist wants to prepare 2-bromoadamantane from adamantane (C₁₀H₁₆). Adamantane has 12 secondary C−H bonds and 4 tertiary C−H bonds. Using bromination selectivity factors (2° = 82, 3° = 1600), predict whether this is a synthetically practical route to the tertiary bromide 1-bromoadamantane.
Show answer
Weighted reactivity for the 2° site: 12 × 82 = 984. Weighted reactivity for the 3° site: 4 × 1600 = 6400. Total = 984 + 6400 = 7384. Percentage of 1-bromoadamantane (3° product): (6400 / 7384) × 100% = 86.7%. Percentage of 2-bromoadamantane (2° product): (984 / 7384) × 100% = 13.3%. The 3° product overwhelmingly dominates at ~87%, making radical bromination an excellent route to 1-bromoadamantane (the bridgehead tertiary bromide). However, the question asks about 2-bromoadamantane (the secondary product) — at only ~13%, radical bromination is NOT a practical route to the 2° product. This example illustrates that bromine's powerful selectivity for 3° positions can be a disadvantage if you actually want the 2° isomer.

Question 5: CRITICAL THINKING

Neopentane, (CH₃)₄C, has only primary C−H bonds (12 equivalent 1° H's) but no secondary or tertiary hydrogens. Predict and explain: (a) Will monobromination of neopentane give a single product? (b) Why is radical bromination of neopentane expected to be extremely slow compared to bromination of 2-methylbutane? Relate your answer to the Hammond postulate and BDE values.
Show answer
(a) Yes, monobromination of neopentane gives a single product — neopentyl bromide, (CH₃)₃CCH₂Br — because all 12 hydrogens are equivalent primary hydrogens. There is only one type of C−H bond to abstract. (b) The hydrogen-abstraction step for a 1° C−H by Br• has ΔH° = BDE(1° C−H) − BDE(H−Br) = 423 − 366 = +57 kJ/mol. By the Hammond postulate, the transition state for this highly endothermic step closely resembles the product radical, which is a primary radical — the least stable type. The activation energy (Eₐ) is therefore very high, close to or exceeding ΔH° = +57 kJ/mol. In contrast, bromination at a 3° C−H bond in 2-methylbutane has ΔH° = +34 kJ/mol, and the transition state leads to a much more stable tertiary radical, lowering Eₐ considerably. The rate constant for 1° H abstraction by Br• is roughly 1600 times smaller per hydrogen than for 3° abstraction, so neopentane reacts extraordinarily slowly. This demonstrates that selectivity and reactivity are inversely related: bromine's high selectivity means it is very reluctant to abstract a hydrogen from a position that would generate an unstable radical.

SUMMARY

Lesson Summary

Summary

Radical halogenation converts alkanes (R−H) into alkyl halides (R−X) through a free-radical chain mechanism consisting of three phases: initiation (homolysis of X₂ by UV light or heat), propagation (hydrogen abstraction followed by halogen abstraction in a self-sustaining cycle), and termination (coupling of any two radicals). The thermodynamics of each propagation step are calculated using bond dissociation energies (BDE): ΔH° = BDE(broken) − BDE(formed).

Selectivity in radical halogenation depends on two factors: the number of each type of hydrogen and its relative reactivity toward the halogen radical. Chlorine is reactive but unselective (1° : 2° : 3° = 1 : 3.9 : 5.2) because its exothermic H-abstraction step has a reactant-like transition state. Bromine is less reactive but highly selective (1 : 82 : 1600) because its endothermic H-abstraction step has a product-like transition state that strongly reflects radical stability (3° > 2° > 1°), as predicted by the Hammond postulate. Product ratios are predicted using the formula: % product = (n × r) / Σ(nᵢ × rᵢ) × 100%.

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