Source: Baldwin, CHEM 2510, Ohio State University
Tags: radical halogenation, chain mechanism, initiation, propagation, termination, chlorination, bromination, fluorination, selectivity, Hammond's postulate, transition state, reaction coordinate diagram, halogen reactivity, CHEM 2510
Difficulty: Intermediate Prerequisites: Bond dissociation energy, radical stability order (Sections 3.1–3.2), enthalpy of reaction basics.
Radical halogenation is one of the first full reaction mechanisms you meet in organic chemistry. It shows how radicals, once formed, can sustain a chain reaction that converts an alkane into an alkyl halide. The mechanism is three steps (initiation, propagation, termination), and the product distribution depends on which hydrogen gets abstracted. That, in turn, depends on the halogen you use: chlorine is relatively unselective, bromine is highly selective. Understanding why comes down to transition-state theory and Hammond's postulate. If you are coming in cold, make sure you are solid on BDE values and radical stability from Sections 3.1–3.2 first.
Radical halogenation replaces a C–H bond with a C–X bond through a three-stage chain mechanism. Chlorination is fast but gives mixtures; bromination is slower but targets the most substituted C–H bond almost exclusively. The selectivity difference traces back to whether the transition state for hydrogen abstraction is early (chlorine) or late (bromine), explained by Hammond's postulate.
Radical chain mechanism
A self-sustaining sequence of radical reactions consisting of initiation (radical generation), propagation (radicals react and regenerate), and termination (radicals pair up and the chain stops). Think of it as: a relay race where each step creates the radical needed for the next step, until two radicals accidentally collide and end the race.
Initiation
The first stage of a radical chain reaction, in which radicals are generated from a non-radical species. Typically, a halogen molecule (X₂) absorbs light (hν) or heat (Δ) and undergoes homolytic cleavage to form two halogen radicals. In simple terms, this is the spark that starts the chain.
Propagation
The stage where radicals react with stable molecules to form new radicals and products. In halogenation, there are two propagation steps that cycle: (1) X· abstracts H from the alkane to give HX and a carbon radical; (2) the carbon radical reacts with X₂ to give the alkyl halide and regenerate X·. Think of it as: the engine of the reaction, a loop that keeps turning over.
Termination
Any step in which two radicals combine to form a stable (non-radical) product, ending the chain. Examples include X· + X· → X₂, R· + R· → R–R, or R· + X· → R–X. In simple terms, two radicals bump into each other and cancel out.
Hammond's postulate
States that the transition state of a reaction step resembles the species (reactant or product) that is closer to it in energy. For an exothermic step, the transition state is early and resembles the reactants. For an endothermic step, the transition state is late and resembles the products. Think of it as: the transition state looks like whichever side of the energy hill it is nearest to.
Selectivity (in halogenation)
The preference of a halogen radical to abstract one type of hydrogen over another. High selectivity means the reaction targets the most substituted position; low selectivity means it attacks more randomly, roughly in proportion to the number of each type of hydrogen.
CH₄ + Cl₂ → CH₃Cl + HCl (in CCl₄, with Δ or hν)
The reaction is exothermic. You can verify this from BDE values:
Bonds broken: C–H (105 kcal/mol) + Cl–Cl (58 kcal/mol) = 163 kcal/mol
Bonds formed: C–Cl (85 kcal/mol) + H–Cl (103 kcal/mol) = 188 kcal/mol
ΔH° ≈ 163 – 188 = –25 kcal/mol (exothermic)
1. Initiation
Cl–Cl absorbs hν (or heat) and undergoes homolytic cleavage → 2 Cl·
This is the only step that creates radicals from scratch.
2. Propagation (two steps, looping)
Step A: Cl· + H–CH₃ → H–Cl + ·CH₃ (hydrogen abstraction)
Step B: ·CH₃ + Cl–Cl → H₃C–Cl + Cl· (radical reacts with Cl₂, regenerating Cl·)
The Cl· produced in step B feeds back into step A. This is why it is called a chain.
3. Termination
Any two radicals combine: Cl· + Cl· → Cl₂, or ·CH₃ + Cl· → CH₃Cl, or ·CH₃ + ·CH₃ → CH₃CH₃.
Not all termination products are useful. Some (like ethane from two methyl radicals) are side products.
The two propagation steps each have their own transition state:
Transition state 1 (H-abstraction): Cl· approaching H–CH₃. In the chlorination of methane, this step is exothermic, so the TS is early and resembles the reactants (looks like CH₄).
Transition state 2 (radical + Cl₂): ·CH₃ approaching Cl–Cl. Also has its own activation energy.
The overall energy profile shows two humps (one per propagation step) with the products lower in energy than the reactants.
This is the conceptual bridge between energy and product distribution.
Chlorination – The H-abstraction step is exothermic (ΔH is negative). By Hammond's postulate, the transition state is early, resembling the starting alkane. Because the TS does not closely resemble the radical product, differences in radical stability do not strongly influence the activation energy. Chlorine is therefore relatively unselective.
Bromination – The H-abstraction step is endothermic (ΔH ≈ +6 kcal/mol for a 1° C–H). The transition state is late, resembling the product radical. Because the TS looks like the radical, the stability of that radical directly influences the activation energy. Bromine is therefore highly selective, overwhelmingly favouring the most stable (most substituted) radical.
Halogen | ΔH (H-abstraction) | Selectivity | Notes |
|---|---|---|---|
F₂ | –106 kcal/mol | None (essentially zero) | F· is less stable than ·CH₃; too exothermic to control |
Cl₂ | –25 kcal/mol | Low | Early TS; some preference for 3° > 2° > 1° |
Br₂ | +6 kcal/mol | High | Late TS; strongly favours most stable radical |
I₂ | +15 kcal/mol | N/A | Too endothermic; reaction does not proceed |
F· is less stable than ·CH₃, so fluorination is explosively exothermic and completely uncontrollable in practice.
I· is more stable than ·CH₃, so iodination is too endothermic to be useful.
Chlorine has a measurable, if modest, preference for more substituted hydrogens:
Selectivity ratio for Cl: 3° : 2° : 1° ≈ 5 : 4 : 1
To predict product yields, multiply the number of each type of hydrogen by the selectivity factor, then convert to percentages.
Example: chlorination of 2-methylbutane gives a mixture of products. The expected ratio based purely on hydrogen count does not match observation, because chlorine is about 4× more selective for a 2° H over a 1° H and about 5× more selective for a 3° H over a 1° H.
Selectivity ratio for Br: 3° : 2° : 1° ≈ 1600 : 800 : 1
Bromination occurs almost exclusively at the most substituted carbon. This makes it synthetically useful when you want a single product.
Fluorination is extremely exothermic and not selective. The F· radical is less stable than even a methyl radical, so it reacts too quickly for selectivity to matter. Not practically controllable.
Bromination is mildly endothermic for H-abstraction, producing a late transition state that closely resembles the product radical. It always favours the most stable radical, so it is the halogen of choice when selectivity matters.
Enthalpy of reaction from BDE:
ΔH° = Σ BDE(bonds broken) – Σ BDE(bonds formed)
Chlorination selectivity calculation:
Predicted yield (%) for a given position = (number of H's at that position × selectivity factor) / (total weighted H count) × 100
Selectivity factors (approximate):
Cl: 3° = 5, 2° = 4, 1° = 1
Br: 3° = 1600, 2° = 800, 1° = 1
Radical halogenation is one of the few reliable ways to functionalise an unactivated alkane, which is otherwise nearly inert. Industrial chlorination of methane produces chloromethane, dichloromethane, chloroform, and carbon tetrachloride, all of which are common laboratory solvents. Selective bromination is used in synthesis to install a bromine handle at a specific position, which can then be converted to other functional groups via substitution or elimination.
Students often forget that propagation is a loop, not a one-shot event. The radical produced in step B feeds directly into step A. Drawing the loop helps you see why small amounts of initiating radicals can convert large quantities of starting material.
A frequent error is assuming chlorination gives only one product from a complex alkane. Chlorine is not very selective, so you will typically get a mixture. Only bromination gives near-exclusive selectivity for the most substituted position.
Some students mix up "early transition state" and "late transition state." Remember: exothermic step = early TS (looks like reactants); endothermic step = late TS (looks like products).
Students sometimes think termination is a useful, productive step. Most termination events are not helpful: they consume radicals and can form unwanted by-products.
⚠️ You will very likely be asked to draw the full radical chain mechanism (initiation, both propagation steps, and termination). Practise this until it is automatic.
⚠️ Be prepared to calculate ΔH° for a halogenation reaction using BDE values.
⚠️ Hammond's postulate shows up repeatedly in organic chemistry. Understand it here and you will recognise it in E1, SN1, and electrophilic addition later.
⚠️ Expect a problem asking you to predict the major product of chlorination vs. bromination of a branched alkane, using selectivity factors and hydrogen counts.
⚠️ Know why fluorination is impractical and why iodination does not proceed.
True or false: The propagation step of radical chlorination is endothermic.
Fill in the blank: According to Hammond's postulate, the transition state for an endothermic step resembles the __________.
True or false: Bromination of an alkane with both 2° and 3° C–H bonds will produce roughly equal amounts of 2° and 3° alkyl bromides.
Fill in the blank: In a radical chain mechanism, the stage that generates radicals from non-radical precursors is called __________.
True or false: Iodination of alkanes by a radical mechanism is too endothermic to proceed.
Answers: 1. False (exothermic overall, though individual steps vary by halogen). 2. Products. 3. False (bromination is highly selective and will overwhelmingly favour the 3° product). 4. Initiation. 5. True.
Q: Write the two propagation steps for the radical chlorination of ethane (CH₃CH₃) and identify the organic product.
A: Step 1: Cl· + CH₃CH₃ → HCl + ·CH₂CH₃. Step 2: ·CH₂CH₃ + Cl₂ → CH₃CH₂Cl + Cl·. The organic product is chloroethane.
Q: Using BDE values (C–H = 105, Cl–Cl = 58, C–Cl = 85, H–Cl = 103 kcal/mol), calculate ΔH° for the chlorination of methane.
A: ΔH° = (105 + 58) – (85 + 103) = 163 – 188 = –25 kcal/mol. The reaction is exothermic.
Q: 2-Methylpropane has nine 1° hydrogens and one 3° hydrogen. Using chlorination selectivity factors (3° = 5, 1° = 1), predict the approximate percentage of tert-butyl chloride in the product mixture.
A: Weighted contribution of 3°: 1 × 5 = 5. Weighted contribution of 1°: 9 × 1 = 9. Total = 14. Percentage of 3° product: 5/14 × 100 ≈ 36%. The remaining ~64% is isobutyl chloride (1° product).
Q: Explain, using Hammond's postulate, why bromine is more selective than chlorine in radical halogenation.
A: The hydrogen-abstraction step is endothermic for bromine but exothermic for chlorine. By Hammond's postulate, the endothermic step has a late transition state that resembles the product radical. Differences in radical stability therefore have a large effect on the activation energy for bromination, causing it to favour the most stable radical. Chlorination has an early transition state that resembles the reactants, so radical stability differences have less influence on selectivity.
Hammond's postulate reappears in virtually every mechanistic discussion in the course, including SN1, E1, and electrophilic addition. The selectivity concepts here also connect to regioselectivity in later chapters (e.g., Markovnikov vs. anti-Markovnikov addition). The thermodynamic analysis using BDE is the same framework you will use when evaluating any reaction's feasibility from bond energies.
radical halogenation, radical chain mechanism, initiation, propagation, termination, free radical substitution, chlorination, bromination, fluorination, iodination, selectivity, Hammond's postulate, early transition state, late transition state, reaction coordinate diagram, BDE calculation, enthalpy of reaction, 3° selectivity, product distribution, CCl₄ solvent, hν, homolytic cleavage, CHEM 2510, organic chemistry chapter 3