Radical Halogenation of Alkanes: Chain Mechanism and Selectivity

Drop a bottle of methane into a flask of chlorine gas. In the dark, nothing happens. Shine ultraviolet light on it — or simply add a trace of heat or a radical initiator. The reaction then explodes into a self-sustaining chain. Methane and Cl₂ convert into chloromethane and HCl within milliseconds. This is one of the oldest reactions in organic chemistry. It is also one of the richest teaching cases in mechanistic reasoning. It forces you to think about bond homolysis and orbital geometry of radicals. It also demands attention to the Hammond Postulate. Finally, it requires subtle energetic bookkeeping. That bookkeeping separates a reagent giving "any C–H you like" from one that gives "only the best C–H." The reaction matters because it is the prototype for every radical chain process you will meet later. Polymerization, autoxidation, atmospheric ozone destruction, and industrial cracking all share its architecture. That architecture has three phases: initiation, propagation, and termination. But its deeper pedagogical value is in selectivity. Chlorination of isobutane gives a nearly statistical mixture of products. Bromination, by contrast, gives almost exclusively the tertiary bromide. Understanding why — not memorizing the numbers — requires connecting Bond Dissociation Energies and Radical Stability to transition-state theory. That connection runs via the Hammond postulate. It also requires recognizing something important about the rate-determining step. That step encodes how much "look-ahead" the transition state has toward product stability. The big picture is this: radical halogenation is a relay race. A chlorine or bromine radical abstracts a hydrogen atom, generating a carbon radical. That radical then attacks a halogen molecule. This regenerates a halogen radical and closes the loop. Every trip around the loop consumes one substrate C–H and one X–X bond. It also produces one C–X bond and one H–X bond. The overall thermodynamics are fixed by these four bond energies. But the kinetics — which C–H bond reacts, and how fast — are governed by the first propagation step alone. It is there that the story of selectivity is written. This article builds the chain mechanism from first principles. It uses full curved-arrow (fishhook) bookkeeping throughout. It derives the thermodynamics of each propagation step from bond dissociation energies. It then constructs the Hammond-postulate argument. That argument explains why bromination is dramatically more selective than chlorination — quantitatively, not just qualitatively. Three fully worked examples take you from methane through 2-methylbutane. The final example is a multi-step selectivity calculation. It resembles the kind found in graduate qualifying exams.

Key takeaways

Radical halogenation runs as a chain reaction: initiation (homolysis) generates radicals, two propagation steps regenerate the chain carrier, and termination steps (radical–radical combination) end it.
Each propagation step is elementary and bimolecular; curved (fishhook) arrows track single electrons, never pairs.
The rate-determining step is H-abstraction (X• + R–H → HX + R•); its activation energy sets the overall rate and the site selectivity.
The Hammond postulate says that for the endothermic, high-barrier H-abstraction by Br•, the transition state resembles the radical product — so radical stability (3° > 2° > 1°) is strongly felt in the TS energy.
Chlorination (nearly thermoneutral to exothermic abstraction, early TS) is fast but nearly unselective; bromination (endothermic abstraction, late TS) is slow but highly selective for tertiary/secondary C–H.
Selectivity is quantified by per-hydrogen relative rate factors (statistical correction), not just by counting hydrogens.
Key idea: selectivity tracks how "product-like" the rate-determining transition state is, and that resemblance is dictated by the endothermicity of the H-abstraction step for each halogen.

Where this lives today

Radical halogenation's mechanistic logic extends far past the flask. Industrially, chlorination of methane at scale (to make chloromethane, dichloromethane, chloroform, and carbon tetrachloride) is run as an uncontrolled radical chain precisely because chlorine's low selectivity is tolerable when a mixture of products is separated afterward by distillation; bromine's high selectivity, by contrast, is exploited in fine chemical synthesis whenever a single tertiary or benzylic C–H must be functionalized cleanly, such as in steroid and terpene functionalization where one specific methine hydrogen needs to be brominated among dozens of similar-looking C–H bonds. The same three-phase chain architecture — initiation, propagation, termination — underlies radical polymerization of vinyl monomers (the propagation step there is radical addition to a C=C π-bond rather than H-abstraction, but the topology is identical), atmospheric radical chemistry (chlorine- and bromine-atom-catalyzed ozone destruction in the stratosphere runs through formally analogous propagation cycles), and autoxidation of fats and oils, where peroxyl radicals abstract allylic hydrogens in a mechanistically parallel process that determines food and polymer shelf life. The Hammond-postulate reasoning connecting reaction thermicity to transition-state "lateness" recurs throughout mechanistic organic chemistry — in E1/E2 elimination transition states, in SN1 carbocation stability effects, and in Diels–Alder regiochemistry — making this halogenation case study a template for a much broader class of arguments a student will reuse constantly. In modern C–H functionalization chemistry, understanding why naive radical halogenation is so selectivity-limited for chlorine (and why it requires forcing, hot conditions for bromine to even proceed at a useful rate) is precisely what motivates the search for catalytic, milder alternatives — photoredox-catalyzed and metal-hydride-mediated hydrogen-atom-transfer (HAT) reactions that borrow the same fishhook mechanism but achieve site-selectivity through tuned catalyst sterics and electronics rather than relying solely on the blunt instrument of intrinsic bond-energy differences.

Ask the tutor: How would the selectivity analysis change if we used a radical initiator like AIBN instead of light for initiation — does the chain mechanism or the selectivity math differ at all?

The physical picture

Radical halogenation begins with bond homolysis. When Cl₂ absorbs a UV photon (or Br₂ is gently heated), the σ-bonding electron pair splits symmetrically — one electron to each chlorine — rather than heterolytically, because there is no polarizing counterion or solvent cage to stabilize an ion pair in a nonpolar hydrocarbon medium. This single event, initiation, produces two chlorine atoms, each a doublet radical with an electron in a singly-occupied 3p orbital. Once a single Cl• is born, the chain runs autocatalytically. Chlorine atom, an electrophilic, high-energy species craving an electron pair, attacks the weakest available C–H bond. The event is a hydrogen-atom transfer: the Cl• orbital overlaps with the C–H σ-bond, one electron from that bond pairs with the incoming radical to form H–Cl, and the other electron is left behind on carbon as a new radical center. The carbon radical is trigonal planar (or nearly so), sp²-ish with the odd electron in a p orbital perpendicular to the plane of the three remaining substituents — this geometry is exactly why radical stability parallels hyperconjugative and inductive stabilization patterns familiar from carbocations, though weaker in magnitude. The alkyl radical then meets an intact Cl₂ molecule. It donates its unpaired electron into the Cl–Cl σ* orbital while abstracting one chlorine atom, forming the new C–Cl bond and releasing a fresh Cl• to continue the cycle. Two propagation steps, each with one radical in and one radical out, keep the total spin and radical count conserved — this is the defining topological feature of a chain mechanism, distinguishing it from initiation and termination, which change the number of radicals present.

Figure

Overall chain architecture: initiation, propagation, termination

The curved-arrow (fishhook) formalism for the key propagation step deserves close attention. For H-abstraction, a single-barbed arrow travels from the midpoint of the C–H bond into the incoming radical's lone electron, and a second fishhook travels from the radical electron on X into the forming X–H bond electron count — actually the cleanest way to draw it uses two single-headed arrows: one from the C–H bonding pair (splitting it) toward the radical center on the incoming atom, forming the new bond, and one fishhook from the radical's own electron onto the same new bond, while the leftover electron from the C–H bond stays on carbon.

Figure

Fishhook mechanism for H-abstraction by Cl•

Note the geometry: the transition state for H-abstraction is nearly linear, X···H···C, because this maximizes overlap of the singly occupied orbital on X with the σ* character developing on the breaking C–H bond — a three-electron, three-center-like arrangement analogous to an SN2 backside approach, but with radicals instead of ions.

The core ideas

The overall (net) reaction, obtained by adding the two propagation steps, is

\text{R–H} + \text{X}_2 \longrightarrow \text{R–X} + \text{H–X}

where: R–H is the alkane, X₂ is Cl₂ or Br₂, R–X is the alkyl halide product, and H–X is the hydrogen halide byproduct. The enthalpy of this net reaction is obtained from bond dissociation energies (BDEs), since bonds broken minus bonds formed gives ΔH for a homolytic process:

\Delta H_{\text{rxn}} = \left[ D(\text{C–H}) + D(\text{X–X}) \right] - \left[ D(\text{C–X}) + D(\text{H–X}) \right]

where: $D$ denotes bond dissociation energy (kcal/mol, homolytic, 298 K gas phase); the bracketed sums are bonds broken (reactant side) and bonds formed (product side) respectively. Because the net reaction is the sum of two propagation steps, each with its own ΔH, we can decompose:

\Delta H_{\text{rxn}} = \Delta H_1 + \Delta H_2

\Delta H_1 = D(\text{C–H}) - D(\text{H–X}), \qquad \Delta H_2 = D(\text{X–X}) - D(\text{C–X})

where: ΔH₁ is the H-abstraction step (bond broken C–H, bond formed H–X) and ΔH₂ is the halogen-atom transfer step (bond broken X–X, bond formed C–X). Step 1 is rate-determining in both chlorination and bromination because it has the higher activation energy; step 2 is fast and strongly exothermic in both cases, so its transition state is reactant-like ("early") and contributes little to selectivity. The Hammond postulate converts these ΔH values into structural statements about the transition state:

\text{TS structure resembles reactants if } \Delta H \ll 0 \text{ (exothermic, "early" TS)}

\text{TS structure resembles products if } \Delta H \gg 0 \text{ (endothermic, "late" TS)}

where: the postulate is a statement about the geometry along the reaction coordinate: for a highly exothermic step the TS occurs early (close to reactants in energy and structure), while for an endothermic step the TS occurs late (close to products), because the TS sits at the energy maximum, and an energy maximum on a steep downhill curve must lie near the start.

Derivation ABond-energy bookkeeping for Cl2 vs Br2 abstracting a tertiary C–H

Goal. Compute ΔH₁ (H-abstraction) for Cl• and Br• attacking a generic tertiary C–H bond, and show how the sign and magnitude of ΔH₁ differ dramatically between the two halogens — this difference is the entire origin of the selectivity gap. Relevant BDEs (kcal/mol, gas-phase homolytic, 298 K, standard tabulated values):

D(3° C–H) ≈ 96.5
D(2° C–H) ≈ 98.6
D(1° C–H) ≈ 101.1
D(H–Cl) = 103.2
D(H–Br) = 87.5
D(Cl–Cl) = 58.0
D(Br–Br) = 46.1
D(3° C–Cl) ≈ 82.8 (typical), D(3° C–Br) ≈ 70 (typical)

Step 1 — write the H-abstraction step explicitly. For chlorine:

\text{Cl•} + \text{R}_3\text{C–H} \longrightarrow \text{R}_3\text{C•} + \text{H–Cl}

For bromine:

\text{Br•} + \text{R}_3\text{C–H} \longrightarrow \text{R}_3\text{C•} + \text{H–Br}

In both, the bond broken is the tertiary C–H (96.5 kcal/mol); the bond formed is H–X.

Step 2 — compute ΔH₁ for chlorine.

\Delta H_1(\text{Cl}) = D(\text{3° C–H}) - D(\text{H–Cl}) = 96.5 - 103.2 = -6.7\ \text{kcal/mol}

This is exothermic: the chlorine atom's abstraction of even the weakest tertiary C–H bond is downhill, because H–Cl is such a strong bond.

Step 3 — compute ΔH₁ for bromine.

\Delta H_1(\text{Br}) = D(\text{3° C–H}) - D(\text{H–Br}) = 96.5 - 87.5 = +9.0\ \text{kcal/mol}

This is strongly endothermic: H–Br is a much weaker bond (87.5 vs. 103.2 kcal/mol) than H–Cl, so bromine atom abstraction of a tertiary C–H costs energy even though it forms a stabilized tertiary radical.

Step 4 — apply the Hammond postulate to each case. For chlorination (ΔH₁ = −6.7 kcal/mol, exothermic, low barrier ≈ 1 kcal/mol experimentally): the TS is early, resembling the starting alkane. The C–H bond is only slightly stretched and R retains little radical character at the TS; consequently the energy difference between abstracting a 1°, 2°, or 3° C–H is small — the TS barely "feels" how stable the eventual radical will be. For bromination (ΔH₁ = +9.0 kcal/mol, endothermic, barrier ≈ 13–18 kcal/mol experimentally): the TS is late, resembling the radical product. The C–H bond is nearly fully broken and substantial radical character has developed on carbon at the TS; consequently the TS energy closely tracks the radical's stability, so 3° vs 2° vs 1° differences (roughly 96.5, 98.6, 101.1 kcal/mol — spanning about 4.6 kcal/mol) are strongly amplified into rate differences.

Step 5 — connect ΔΔH‡ to rate ratios via Arrhenius/Eyring reasoning. If activation energies track the radical stability differences (as Hammond dictates for bromination), a difference of even 2–3 kcal/mol in Ea translates, via

\frac{k_1}{k_2} = \exp\!\left(-\frac{E_{a,1}-E_{a,2}}{RT}\right)

into rate ratios of 10–100 at room temperature, explaining why bromination is so selective while chlorination, with a much smaller ΔEa across C–H types, is nearly statistical.

Go deeper · Derivation B · 2 minFrom per-hydrogen rate constants to product ratios: the selectivity formula

Derivation BFrom per-hydrogen rate constants to product ratios: the selectivity formula

Goal. Derive the working formula used to predict product distributions in radical halogenation from tabulated relative reactivities per hydrogen, and show how it must be corrected for the number of equivalent hydrogens of each type.

Step 1 — identify what governs product ratio. Because propagation step 1 (H-abstraction) is rate-determining, the ratio of two products R–X and R′–X formed from the same starting material is set entirely by the relative rates of abstracting the corresponding C–H bonds, not by their relative bond energies alone (activation energy, not ΔH, controls kinetics — though the two are correlated via Hammond for the late TS of bromination).

Step 2 — write rate expressions for each distinguishable H-abstraction pathway. If a substrate has $n_1$ primary, $n_2$ secondary, and $n_3$ tertiary hydrogens, each type reacting with rate constant $k_1, k_2, k_3$ per hydrogen, the total rate of forming each product class is

\text{rate}_i = n_i k_i [\text{X•}][\text{R–H}]

where: $i \in \{1,2,3\}$ labels the H-type, $n_i$ is the number of equivalent hydrogens of that type, and $k_i$ is the per-hydrogen rate constant (this distinction — per-hydrogen vs per-molecule — is the step most often botched).

Step 3 — form the product ratio.

\frac{\text{product}_3}{\text{product}_1} = \frac{n_3 k_3}{n_1 k_1}

where: the ratio of relative rate factors $k_3/k_1$ is what is tabulated experimentally (often normalized to $k_1 \equiv 1$ for primary), and $n_3/n_1$ is the purely statistical (counting) factor.

Step 4 — separate statistical and electronic contributions explicitly.

\frac{\text{product}_3}{\text{product}_1} = \underbrace{\frac{n_3}{n_1}}_{\text{statistical}} \times \underbrace{\frac{k_3}{k_1}}_{\text{electronic (Hammond-governed)}}

This factorization is the single most useful equation for predicting halogenation product ratios: always multiply "how many H's of each kind" by "how reactive is each kind," never confuse the two.

Step 5 — connect $k_3/k_1$ back to the transition-state argument. For chlorination at 25 °C, typical per-hydrogen relative rates are roughly 1(1°) : 4(2°) : 5(3°) — a narrow range consistent with the early, reactant-like TS computed above. For bromination at 125 °C (typically run hot because the reaction is otherwise too slow, itself a consequence of the higher Ea from the endothermic first step), typical per-hydrogen relative rates are roughly 1(1°) : 82(2°) : 1600(3°) — a huge range consistent with the late, radical-like TS. These numbers are not arbitrary; they are direct experimental manifestations of the ΔH₁ values computed in the derivation above, filtered through the Hammond postulate.

Worked examples

Example 1 (foundational): Monochlorination of methane — full mechanism with arrows and thermodynamics. Methane has only one type of hydrogen, so there is no selectivity question, but working the full mechanism through cleanly is essential groundwork. Initiation.

\text{Cl–Cl} \xrightarrow{h\nu} 2\,\text{Cl}\bullet

Two fishhook arrows diverge from the midpoint of the Cl–Cl bond, one to each chlorine, giving two chlorine radicals. Bond broken: Cl–Cl (58.0 kcal/mol); this energy is supplied by the photon (Cl₂ absorbs around 330 nm, corresponding to ≈ 87 kcal/mol per mole of photons — more than enough). Propagation 1.

\text{Cl}\bullet + \text{CH}_4 \longrightarrow \text{CH}_3\bullet + \text{HCl}

\Delta H_1 = D(\text{CH}_3\text{–H}) - D(\text{H–Cl}) = 104.9 - 103.2 = +1.7\ \text{kcal/mol}

Slightly endothermic — the C–H bond in methane (104.9 kcal/mol) is unusually strong (no hyperconjugative stabilization of the resulting methyl radical, and no adjacent alkyl groups), so even chlorine atom finds this step mildly uphill, though far less so than any bromine abstraction. Propagation 2.

\text{CH}_3\bullet + \text{Cl–Cl} \longrightarrow \text{CH}_3\text{Cl} + \text{Cl}\bullet

\Delta H_2 = D(\text{Cl–Cl}) - D(\text{CH}_3\text{–Cl}) = 58.0 - 83.7 = -25.7\ \text{kcal/mol}

Strongly exothermic — the methyl radical's singly occupied orbital attacks the σ of Cl–Cl, forming a strong C–Cl bond and releasing a fresh Cl• fishhook. Net reaction.*

\Delta H_{\text{rxn}} = \Delta H_1 + \Delta H_2 = 1.7 + (-25.7) = -24.0\ \text{kcal/mol}

Check against the direct BDE formula: $[D(\text{C–H}) + D(\text{Cl–Cl})] - [D(\text{C–Cl}) + D(\text{H–Cl})] = [104.9+58.0]-[83.7+103.2] = 162.9-186.9=-24.0$ kcal/mol — consistent, confirming the additivity of the two propagation steps. Termination (any of three combinations, all radical–radical, both barrierless):

\text{Cl}\bullet + \text{Cl}\bullet \to \text{Cl}_2, \qquad \text{CH}_3\bullet + \text{Cl}\bullet \to \text{CH}_3\text{Cl}, \qquad \text{CH}_3\bullet + \text{CH}_3\bullet \to \text{CH}_3\text{CH}_3

The last of these is why trace ethane is always found as a byproduct in radical chlorination of methane — a classic mechanistic fingerprint confirming radical, not ionic, chemistry.

Example 2 (applied): Monochlorination vs monobromination of 2-methylbutane — full statistical and electronic analysis. 2-Methylbutane, (CH₃)₂CHCH₂CH₃, has four distinct hydrogen environments:

C1 and its mirror methyl (the two methyls on the isopropyl-like carbon): 6 equivalent primary H's — call this "1°(a)"
C3 (the CH₂): 2 secondary H's
C2 (the CH): 1 tertiary H
C4 (the terminal CH₃ of the ethyl-like chain): 3 primary H's — call this "1°(b)"

For simplicity (as is standard in this classic problem) treat all 9 primary H's as one class with $n_1=9$ , secondary $n_2=2$ , tertiary $n_3=1$ . Chlorination, per-H relative rates 1(1°) : 4(2°) : 5(3°), 25 °C.

\text{ratio (1° : 2° : 3°)} = (9\times1) : (2\times4) : (1\times5) = 9 : 8 : 5

Total parts = $9+8+5=22$ .

\%\text{primary} = \frac{9}{22}\times100 = 40.9\%, \quad \%\text{secondary} = \frac{8}{22}\times100=36.4\%, \quad \%\text{tertiary}=\frac{5}{22}\times100=22.7\%

This is a nearly statistical mixture with three products all formed in significant amounts — chlorination is a poor synthetic tool for selective functionalization, precisely because its TS is early (Hammond) and barely discriminates C–H types. Bromination, per-H relative rates 1(1°) : 82(2°) : 1600(3°), 125 °C.

\text{ratio (1° : 2° : 3°)} = (9\times1) : (2\times82) : (1\times1600) = 9 : 164 : 1600

Total parts = $9+164+1600=1773$ .

\%\text{tertiary} = \frac{1600}{1773}\times100 = 90.2\%, \quad \%\text{secondary}=\frac{164}{1773}\times100=9.3\%, \quad \%\text{primary}=\frac{9}{1773}\times100=0.5\%

Bromination is essentially a clean, single-product reaction here: 2-bromo-2-methylbutane forms in over 90% selectivity despite there being nine primary hydrogens competing against just one tertiary hydrogen. This dramatic reversal — outnumbered 9-to-1 yet winning 90-to-1 — is the signature result that every student should be able to reconstruct from the Hammond argument alone, without memorizing the final percentages. Limiting-case check. If we artificially set $k_3/k_1 \to 1$ (imagining a hypothetical, perfectly unselective halogen with a very early TS for every C–H type), the formula collapses to pure hydrogen-counting: ratio $9:2:1$ , giving 75% primary, 16.7% secondary, 8.3% tertiary — exactly what you'd expect from statistics alone with no electronic preference, confirming that the $k_i$ factors are the sole carriers of "chemistry" in the model and that setting them all equal correctly recovers the trivial statistical limit.

Example 3 (exam/competition level): Predicting and rationalizing the major monobromination product of a more complex substrate, with quantitative selectivity and an explicit rate-constant back-calculation. Consider 2,3-dimethylbutane-derived reasoning extended to 3-methylpentane, CH₃CH₂CH(CH₃)CH₂CH₃, brominated at 125 °C with per-H relative rate factors 1(1°) : 82(2°) : 1600(3°). Step 1 — classify all hydrogens.

C1 and C5 (two equivalent terminal CH₃ groups): 6 primary H's, $n_1 = 6$
C2 and C4 (two equivalent CH₂ groups): 4 secondary H's, $n_2 = 4$
C3 (the CH bearing the methyl branch): 1 tertiary H, $n_3 = 1$
the branch methyl itself (on C3): 3 more primary H's, chemically distinct from the C1/C5 methyls but same type — add to primary count: $n_1 = 6+3 = 9$

Step 2 — apply the weighted-rate formula.

\text{ratio (1° : 2° : 3°)} = (9\times1):(4\times82):(1\times1600) = 9:328:1600

Total = $9+328+1600 = 1937$ .

\%\text{tertiary}= \frac{1600}{1937}\times100 = 82.6\%,\quad \%\text{secondary}=\frac{328}{1937}\times100=16.9\%,\quad\%\text{primary}=\frac{9}{1937}\times100=0.46\%

Step 3 — reverse-engineer: given an experimental product ratio, extract the hidden rate constant. Suppose an experiment on a related tertiary/primary pair (n₃ = 1, n₁ = 9, as in Example 2) at a different temperature gives an observed product ratio of 95.0% tertiary to 5.0% primary (instead of the 90.2%/0.5% figure computed earlier — a deliberately different, non-matching value to illustrate the back-calculation, since at a higher temperature selectivity typically drops somewhat due to the smaller relative importance of Ea differences as $RT$ grows, per the Arrhenius exponential). We want $k_3/k_1$ at this new temperature. From the ratio formula:

\frac{\%\text{tert}}{\%\text{prim}} = \frac{n_3 k_3}{n_1 k_1} \implies \frac{95.0}{5.0} = \frac{(1)(k_3)}{(9)(k_1)}

19.0 = \frac{k_3}{9k_1} \implies \frac{k_3}{k_1} = 19.0\times 9 = 171

So at this (implied, higher) temperature the per-hydrogen selectivity factor has fallen from 1600 to about 171 — a physically sensible trend, since raising $T$ shrinks the Boltzmann weighting $\exp(-\Delta E_a/RT)$ that amplifies small activation-energy gaps, exactly as the Arrhenius relation in the derivation above predicts. This kind of back-calculation — going from an observed product ratio to an implied rate-constant ratio, and then checking that the trend with temperature makes physical sense — is exactly the reasoning tested in advanced problem sets and qualifying exams.

Limiting cases, scope & common traps

Situation	What happens	Why
Cl₂, room temperature, simple alkane	Nearly statistical mixture of all possible monochlorides	Early (reactant-like) TS for exothermic H-abstraction barely feels radical stability (Hammond)
Br₂, 125 °C (typical conditions), simple alkane	Overwhelming preference for most-substituted C–H	Late (product-like) TS for endothermic H-abstraction strongly feels radical stability
No light, no heat, no initiator	No reaction (or negligible rate)	No mechanism to homolyze X–X; initiation step has a large barrier without photon/thermal input
Substrate with only 1° and no 3°/2° H's (e.g., neopentane)	Chlorination gives single product regardless of halogen	No competition exists; selectivity question is moot
F₂ instead of Cl₂/Br₂	Explosively unselective, often destructive (C–C cleavage)	H–F bond so strong (136 kcal/mol) that ΔH₁ is hugely exothermic; TS is even earlier than Cl₂'s, essentially zero barrier, reaction is barely controllable
I₂ instead of Cl₂/Br₂	Reaction does not proceed usefully	Propagation step 2 (R• + I₂ → RI + I•) is endothermic because C–I is weak; the chain cannot close efficiently
High [X₂] vs low [X₂]	Slight effect on chain length, not on H-abstraction selectivity	Selectivity is set by relative $k_i[\text{X•}]$ terms in step 1; concentration of X₂ affects step 2 kinetics and overall rate, not the ratio of H-abstraction pathways

The chemist behind it

The framework of radical chain reactions traces to the early twentieth-century recognition, largely through the work of Moses Gomberg (who first isolated a stable radical, triphenylmethyl, in 1900) and later the systematic kinetic studies of Morris S. Kharasch, that many organic reactions proceed not through polar (ionic) pathways but through free-radical intermediates. Kharasch's studies of the "peroxide effect" in HBr addition to alkenes (anti-Markovnikov addition under radical conditions) in the 1930s were pivotal in convincing the community that radical chains were real, kinetically distinct pathways — not mere curiosities. The quantitative selectivity framework used throughout this article — relative rate factors per hydrogen, and their rationalization via bond energies and the Hammond postulate — owes its modern form to the American physical organic chemist George S. Hammond, whose 1955 paper articulated the postulate as a general principle relating reaction thermicity to transition-state structure, unifying observations across radical, ionic, and pericyclic reactions alike.

Origins · 1 minGeorge Simms Hammond and how this was discovered

George Simms Hammond

1921–2005 · American physical organic chemist, Iowa State/Caltech/Allied Chemical

Hammond's 1955 Journal of the American Chemical Society paper, "A Correlation of Reaction Rates," proposed that the transition state's geometry and energy resemble whichever adjacent species (reactant or product) is closer to it in energy. Though framed generally, it found its most celebrated early application precisely in explaining the Cl₂/Br₂ selectivity contrast in alkane halogenation — a result so clean and quantitatively verifiable by bond-energy bookkeeping that it became a textbook staple almost immediately. Hammond later worked extensively on photochemistry and energy transfer, but the postulate bearing his name remains one of the most load-bearing heuristics in all of physical organic chemistry.

History

- 1900 — Gomberg isolates triphenylmethyl radical, establishing that stable free radicals exist.

- 1930s — Kharasch and Mayo identify the peroxide effect, cementing radical-chain mechanisms as real kinetic pathways distinct from ionic ones.

- 1942 — Hass, McBee, and Weber report quantitative relative reactivities of primary/secondary/tertiary C–H bonds toward Cl₂, giving the first numerical selectivity data.

- 1955 — Hammond publishes the postulate correlating reaction rate with transition-state structure, providing the theoretical explanation for the Cl₂ vs. Br₂ contrast.

- 1960s–70s — Detailed kinetic (Arrhenius) studies quantify per-hydrogen rate constants for Cl•, Br•, and other radicals across many substrates, refining the numbers used in modern textbooks.

Why does radical halogenation need light or heat to start, but not to continue?

Initiation requires homolyzing a strong X–X σ-bond (58 kcal/mol for Cl₂, 46 kcal/mol for Br₂) with no assistance from any other bond-forming event to offset the cost — it is purely endothermic, so it needs an external energy source (a UV photon or thermal energy) to occur at a useful rate. Once a handful of radicals exist, however, each propagation step is either mildly exothermic/endothermic on its own (step 1) or strongly exothermic (step 2), and crucially each propagation step regenerates a radical, so the chain is self-sustaining: one initiation event can drive thousands of propagation cycles before termination catches up. This is why even a trace of initiator or a brief flash of light suffices to convert a whole flask of reagents.

Does the Hammond postulate apply to the whole reaction or to individual steps?

Strictly to individual elementary steps. The overall halogenation reaction (alkane + X₂ → alkyl halide + HX) is exothermic for both Cl₂ and Br₂, but that overall thermicity is irrelevant to selectivity because it is not the overall reaction that has a single transition state — each elementary step has its own transition state and its own thermicity. Selectivity in radical halogenation is governed exclusively by the thermicity of the first propagation step (H-abstraction), since that is the rate-determining step; the second propagation step, though also having a Hammond-describable TS, is fast in both cases and does not discriminate between which alkyl radical is present, so it does not affect product ratios.

Why is fluorine not used for selective radical halogenation, and why is iodine also unsuitable?

Fluorination is not practically useful because the H–F bond is so strong (136 kcal/mol) that H-abstraction by F• is enormously exothermic for any C–H bond, giving an extremely early, nearly barrierless transition state with almost no discrimination between C–H types — worse selectivity than chlorine, combined with so much heat release per event that reactions can become uncontrollably fast or even detonate, frequently cleaving C–C bonds too. Iodination fails for the opposite reason: the second propagation step, R• + I₂ → RI + I•, is endothermic (C–I bonds are weak, ~52 kcal/mol, while I–I is also weak, ~36 kcal/mol, but the H-abstraction step by I• from a C–H bond is so endothermic that the chain effectively stalls at step 1), so the chain cannot propagate efficiently and iodoalkanes are not obtained in useful yield by this direct route.

How do you correctly account for "equivalent" hydrogens when computing selectivity ratios?

Group hydrogens by chemical equivalence — same connectivity and same resulting radical upon abstraction — and count how many such hydrogens exist per molecule, including any symmetry-related duplicates (for example, the two methyl groups flanking a central CH in an isopropyl-type fragment together contribute six, not three, primary hydrogens). Multiply that count by the appropriate per-hydrogen relative rate constant, sum over all classes, and take ratios; never use the total number of hydrogens of one class only in the numerator without applying the same weighting consistently across every class, since forgetting the count factor for even one class silently converts a correct calculation into the classic "more H's must win" fallacy.

Why is bromination typically carried out at higher temperature than chlorination?

Because the rate-determining H-abstraction step by Br• is endothermic (ΔH₁ ≈ +9 to +16 kcal/mol depending on C–H type) and correspondingly has a substantial activation energy (roughly 13–18 kcal/mol), the reaction is intrinsically slow at room temperature. Raising the temperature supplies the thermal energy needed to cross this barrier at a practical rate via the Arrhenius factor $\exp(-E_a/RT)$ . Chlorination's much smaller activation energy (often only a few kcal/mol, since ΔH₁ is near zero or even negative) means it proceeds readily even at or below room temperature, sometimes requiring cooling to prevent runaway reaction rather than heating to initiate it.

Does the selectivity trend (3° > 2° > 1°) ever reverse for a different radical abstractor?

The ordering itself (tertiary radicals more stable, hence tertiary C–H bonds weaker and more easily abstracted in a late TS) is a structural fact about hyperconjugative stabilization of the radical center and does not reverse for ordinary alkyl/aryl radicals. What does change dramatically across different abstracting radicals is the magnitude of the selectivity — governed by how endothermic the abstraction step is for that particular X•, exactly as the Hammond postulate predicts. Very reactive, early-TS abstractors (F•, alkoxyl radicals in some contexts) show weak site selectivity; more finely balanced, late-TS abstractors (Br•, and many catalytic HAT reagents used in modern C–H functionalization) show strong site selectivity, but the underlying stability order of the radicals themselves stays the same.