Free-Radical Reactions and Chain Mechanisms — Initiation, Propagation, Selectivity
A radical chain is one of the few mechanisms where a single event — one bond breaking homolytically — can convert thousands of molecules. That amplification is what makes chain reactions industrially enormous (polyethylene, autoxidation, combustion) and what makes their kinetics look strange: half-order concentrations, rate laws with a denominator, and product ratios that a beginner cannot predict from thermodynamics alone. This article works through the mechanism, the bond-energy arithmetic behind selectivity, and the steady-state kinetics that IIT-JAM, GATE and CSIR-NET actually test.
The three phases, and why each matters
Every chain mechanism separates into initiation (radicals are created), propagation (radicals are consumed and regenerated) and termination (radicals are destroyed). The defining feature of propagation is that the number of radicals does not change: a radical goes in, a different radical comes out.
| Phase | Radical count | Typical step | Controls |
|---|---|---|---|
| Initiation | increases (0 → 2) | Cl–Cl → 2 Cl• (heat or light) | How fast the chain starts; the overall rate's dependence on initiator |
| Propagation | unchanged | RH + Cl• → R• + HCl, then R• + Cl₂ → RCl + Cl• | The products, the selectivity, almost all the material turnover |
| Termination | decreases (2 → 0) | R• + R• → R–R; R• + Cl• → RCl | The steady-state radical concentration, hence the rate |
The kinetic chain length ν measures the amplification — how many propagation cycles one initiation event buys:
For a good industrial chlorination ν can be 10³–10⁶. A reaction with ν close to 1 is not a chain reaction at all; it is a stoichiometric radical reaction, and the distinction is a favourite one-mark trap.
Bond dissociation enthalpies do the predicting
Homolytic bond dissociation enthalpy (BDE) is the enthalpy needed to split a bond into two radicals in the gas phase. Since propagation steps are just bond-swaps, ΔH of any propagation step is simply BDE(bond broken) − BDE(bond made). Approximate values (textbooks differ by a few kJ/mol, so always use the set your paper supplies):
| Bond | BDE / kJ mol⁻¹ | Bond | BDE / kJ mol⁻¹ |
|---|---|---|---|
| CH₃–H (methane) | 439 | H–Cl | 431 |
| 1° C–H (ethane) | 423 | H–Br | 366 |
| 2° C–H (propane) | 412 | Cl–Cl | 243 |
| 3° C–H (2-methylpropane) | 404 | Br–Br | 193 |
| Benzylic C–H (toluene) | 375 | C–Cl (ethyl chloride) | 352 |
| Allylic C–H (propene) | 369 | C–Br (ethyl bromide) | 293 |
Read the C–H column downwards and you have the radical stability order directly: 3° > 2° > 1° > methyl, with allylic and benzylic radicals lower still because the unpaired electron is delocalised over a π system.
Worked example 1 — why chlorination is unselective and bromination is not
Chlorination of ethane, step by step.
Step 1 (abstraction): CH₃CH₂–H + Cl• → CH₃CH₂• + H–Cl
ΔH = 423 − 431 = −8 kJ mol⁻¹ (slightly exothermic)
Step 2 (halogen transfer): CH₃CH₂• + Cl–Cl → CH₃CH₂Cl + Cl•
ΔH = 243 − 352 = −109 kJ mol⁻¹
Sum of the two propagation steps = −8 + (−109) = −117 kJ mol⁻¹, which is the
overall enthalpy of CH₃CH₃ + Cl₂ → CH₃CH₂Cl + HCl.
Bromination of ethane.
Step 1: ΔH = 423 − 366 = +57 kJ mol⁻¹ (endothermic)
Step 2: ΔH = 193 − 293 = −100 kJ mol⁻¹
Sum = +57 − 100 = −43 kJ mol⁻¹.
The conclusion. Both overall reactions are exothermic, but the rate-determining abstraction is exothermic for Cl• and endothermic for Br•. By the Hammond postulate an exothermic step has an early, reactant-like transition state in which the C–H bond is barely broken, so the differences between 1°, 2° and 3° C–H bonds hardly show up. The endothermic bromine abstraction has a late, product-like transition state that already resembles the radical, so the full stability difference is expressed in the barrier. That single idea is the whole of halogenation selectivity.
Measured relative reactivities per hydrogen are roughly 1 : 3.8 : 5 (1° : 2° : 3°) for chlorine at room temperature and about 1 : 80 : 1600 for bromine. Different textbooks quote slightly different numbers and different temperatures, so quote the order of magnitude and the reason, not a memorised decimal.
Worked example 2 — predicting the actual product ratio
Monochlorination of 2-methylpropane, (CH₃)₃CH. There are 9 primary hydrogens and 1 tertiary hydrogen.
Statistical weight × reactivity:
primary: 9 × 1 = 9
tertiary: 1 × 5 = 5
Total = 14.
Fraction of 1-chloro-2-methylpropane = 9 ÷ 14 = 0.643 = 64.3%
Fraction of 2-chloro-2-methylpropane = 5 ÷ 14 = 0.357 = 35.7%
Now the same molecule with bromine (relative reactivity ≈ 1600 for 3°):
primary: 9 × 1 = 9; tertiary: 1 × 1600 = 1600; total 1609.
Tertiary product = 1600 ÷ 1609 = 0.994 = 99.4%.
Bromination is synthetically useful precisely because it gives one product; chlorination gives a mixture you have to separate.
Worked example 3 — how much does the barrier difference matter?
Selectivity is a ratio of rate constants, and rate constants follow the Arrhenius equation.
Two abstraction pathways differ by ΔEa = 20 kJ mol⁻¹ at 298 K. Take R = 8.314 J K⁻¹ mol⁻¹.
RT = 8.314 × 298 = 2477.6 J mol⁻¹
ΔEa/RT = 20000 ÷ 2477.6 = 8.072
k₁/k₂ = e8.072 = 3.2 × 10³
So a 20 kJ mol⁻¹ difference — small on a bond-energy scale — is a three-thousand-fold selectivity. Raise the temperature to 500 K and RT = 8.314 × 500 = 4157 J mol⁻¹, ΔEa/RT = 4.812, ratio = e4.812 = 123. Selectivity always falls as temperature rises, which is why selective radical brominations are run cool.
Steady-state kinetics: where the half-orders come from
Radical concentrations are tiny and nearly constant during the reaction, so we set d[radical]/dt = 0. For an initiator I decomposing with rate constant kd and efficiency f, and bimolecular termination with rate constant kt:
⇒ [R•] = ( f kd[I] / kt )1/2
Substituting that into the propagation rate gives the classic free-radical polymerisation rate law:
The half-order in initiator is the fingerprint of a chain reaction with bimolecular termination. If an examiner gives you kinetic data showing rate ∝ [I]0.5, the answer they want is "radical chain, second-order termination". Note the practical consequence: doubling the initiator only multiplies the rate by √2 ≈ 1.41, while it halves the kinetic chain length and so lowers the molecular mass — you cannot speed up a polymerisation for free.
The hydrogen–bromine reaction gives the other famous example. Its experimental rate law is
The [Br₂]1/2 comes from the equilibrium dissociation of Br₂ into atoms; the denominator comes from the inhibition step H• + HBr → H₂ + Br•, in which the product destroys a chain carrier. Early in the reaction [HBr] ≈ 0, the denominator is 1 and the order is 3/2 overall.
Inhibitors, autoxidation and why food goes rancid
Anything that traps a chain carrier and gives a radical too stable to propagate is an inhibitor. Molecular oxygen is a ground-state triplet diradical and reacts with carbon radicals at close to the diffusion limit to give peroxyl radicals, ROO•. That is the autoxidation chain: RH + ROO• → R• + ROOH, then R• + O₂ → ROO•. Phenolic antioxidants such as BHT donate their O–H hydrogen (a weak bond, and the resulting phenoxyl radical is delocalised and sterically shielded), breaking the chain. The same chemistry explains why radical reactions must be degassed, why peroxides accumulate in old ether bottles, and why polymer stabilisers exist.
Mistakes that cost marks
- Calling the initiation step rate-determining for the products. Initiation fixes how many chains start; propagation fixes what is formed. Product ratios come from the competing propagation barriers.
- Using ΔH of the overall reaction to explain selectivity. Both chlorination and bromination are exothermic overall. Only the abstraction step's sign differs, and that is the step that decides.
- Forgetting the statistical factor. A tertiary C–H is far more reactive, but there is only one of it. Always multiply reactivity by the number of equivalent hydrogens.
- Predicting rearrangements. Radicals do not undergo the 1,2-hydride and 1,2-alkyl shifts that carbocations do. Do not import carbocation habits.
- Writing arrows wrongly. Radical mechanisms use single-barbed (fish-hook) arrows for one-electron movement. A double-barbed arrow in a radical mechanism is an automatic deduction.
- Ignoring the [I]1/2. Writing a first-order dependence on initiator shows the steady-state treatment was never applied.
Where this appears in competitive papers
| Exam | Typical use |
|---|---|
| IIT-JAM Chemistry | Halogenation selectivity, allylic/benzylic bromination, radical stability ordering |
| CUET-PG Chemistry | Identifying initiation/propagation/termination steps; simple product prediction |
| GATE Chemistry (CY) | Steady-state derivations, chain length, polymerisation rate laws, autoxidation |
| CSIR-NET (Chemical Sciences) | Full kinetic analysis, inhibition terms, Hammond-postulate reasoning, radical clocks |
Check the numbers yourself. Every selectivity argument on this page reduces to an Arrhenius ratio. Enter two activation energies (or a rate constant at two temperatures) and the calculator returns Ea, the pre-exponential factor and the rate ratio, so you can see exactly how fast selectivity collapses as temperature rises.
Open the Arrhenius Equation Calculator →Preparing for IIT-JAM, GATE, CSIR-NET or CUET-PG? ABC Chemistry runs dedicated competitive-exam batches at its coaching centre and online for students anywhere in India — course details at abcchemistry.in.