Why is the halogenation of alkanes called a chain reaction?

Because each propagation step regenerates a reactive radical that goes on to start another propagation cycle. A single chlorine atom produced in initiation can convert thousands of alkane molecules before two radicals finally meet and terminate the chain. The radical is consumed and re-created repeatedly, so the reaction self-sustains until reactants run out or termination removes the radicals.

Why does fluorine react more violently than bromine with alkanes?

The reactivity order of halogens towards alkanes is F2 > Cl2 > Br2 > I2. Fluorine is so reactive that its halogenation is explosive and uncontrollable, while iodination is essentially non-spontaneous and reversible. The high reactivity of fluorine traces to the very exothermic, low-barrier hydrogen-abstraction step, whereas the bromine atom abstracts hydrogen far more reluctantly.

Why is bromine more selective than chlorine in radical substitution?

The hydrogen-abstraction step for bromine has a higher, later transition state, so the energy difference between forming a 3 degree, 2 degree or 1 degree radical is strongly reflected in the rates. Bromine therefore strongly prefers the position that gives the most stable radical (3 degree > 2 degree > 1 degree). Chlorine abstracts almost indiscriminately and gives a closer-to-statistical mixture, so it is far less selective.

Why does monohalogenation of an alkane give a mixture of products?

An alkane usually has several non-equivalent hydrogens (primary, secondary, tertiary). A radical can abstract any of them, so substitution occurs at more than one position, giving isomeric monohalides. The product ratio depends both on how many hydrogens of each type are present and on the relative ease of abstracting them. Further substitution also gives di- and poly-halogenated products.

Why does NBS brominate at the allylic position rather than adding to the double bond?

N-Bromosuccinimide supplies a very low, steady concentration of bromine, which favours the radical substitution pathway over electrophilic addition to the alkene. The allylic C-H bond is abstracted easily because the resulting allylic radical is resonance-stabilised, so substitution occurs next to the double bond while the double bond itself is preserved.

Why does benzene not undergo free radical substitution like alkanes?

The aromatic ring is stabilised by delocalisation of its pi electrons, and abstracting a ring hydrogen as a radical would disrupt that aromatic system at a high energy cost. Instead, benzene reacts through electrophilic substitution, in which an electrophile attacks the pi cloud and aromaticity is restored after loss of a proton. Radical substitution on the ring is therefore not the favoured route.

Free Radical Substitution — NEET Chemistry

What free radical substitution is

A substitution reaction replaces one atom or group in a molecule with another. When the species doing the work are neutral free radicals rather than ions, the process is a free radical substitution. The defining NEET example is the halogenation of alkanes: NIOS defines halogenation as "the chemical reactions in which a hydrogen atom of an alkane is replaced by a halogen atom" (Section 24.1.4).

Methane reacting with chlorine in diffused sunlight is the canonical case. The hydrogens are removed one at a time, so the reaction does not stop at a single product unless conditions are controlled; successive substitution gives the full series:

$$\ce{CH4 + Cl2 ->[\text{diffused sunlight}] CH3Cl + HCl}$$ $$\ce{CH3Cl + Cl2 -> CH2Cl2 + HCl}$$ $$\ce{CH2Cl2 + Cl2 -> CHCl3 + HCl}$$ $$\ce{CHCl3 + Cl2 -> CCl4 + HCl}$$

The four products are chloromethane, dichloromethane, trichloromethane (chloroform) and tetrachloromethane (carbon tetrachloride). The reaction needs ultraviolet light or heat to begin, and that requirement is the first clue that radicals, not ions, are involved.

Homolytic fission and the radical

To understand the mechanism we first need the right kind of bond breaking. NIOS (Section 23.3.1, Nomenclature and General Principles) distinguishes two ways a covalent bond can break. In homolytic fission the bond breaks "with equal sharing of bonding electrons" so each fragment keeps one electron, producing neutral free radicals. In heterolytic fission the electrons are shared unequally, producing ions (a carbocation and a carbanion).

Free radicals are "neutral but reactive species having an unpaired electron," and NIOS notes they "can also initiate a chemical reaction." Homolysis is promoted by heat or light:

$$\ce{Cl2 ->[h\nu] 2Cl^.}$$ $$\ce{H3C-CH3 ->[\Delta][\text{light}] 2 \overset{.}{C}H3}$$

Figure 1 · Homolytic vs heterolytic fission

Halogenation runs entirely through the left-hand pathway: light splits a halogen molecule homolytically, and every reactive species in the mechanism carries an unpaired electron.

The three steps of the chain

NIOS states that "the free radical mechanism involves the following three steps": chain initiation, chain propagation and chain termination. They must be written in this order and with the radicals balanced.

Step 1 — Initiation

Light supplies the energy that homolyses the halogen molecule into two halogen atoms. This is the only step that creates radicals from a non-radical, and it is slow because it depends on absorption of a photon:

$$\ce{Cl2 ->[h\nu] 2Cl^.}$$

Step 2 — Propagation

NIOS describes propagation as the stage where "the free radicals give rise to the formation of more free radicals." Two complementary reactions repeat in tandem. First the chlorine atom abstracts a hydrogen from methane to give a methyl radical and hydrogen chloride; then the methyl radical attacks a fresh chlorine molecule to give chloromethane and regenerate a chlorine atom:

$$\ce{\overset{.}{C}l + CH4 -> \overset{.}{C}H3 + HCl}$$ $$\ce{\overset{.}{C}H3 + Cl2 -> CH3Cl + \overset{.}{C}l}$$

The second reaction hands back a chlorine atom, which feeds the first reaction again. This regeneration is the engine of the chain.

Step 3 — Termination

In termination, NIOS notes, "free radicals combine with one another and the further reaction stops." Any two radicals that meet pair their electrons and form a stable bond, removing both from the cycle:

$$\ce{\overset{.}{C}H3 + \overset{.}{C}l -> CH3Cl}$$ $$\ce{\overset{.}{C}l + \overset{.}{C}l -> Cl2}$$ $$\ce{\overset{.}{C}H3 + \overset{.}{C}H3 -> CH3-CH3}$$

The last of these gives ethane, a tell-tale minor by-product that confirms methyl radicals were genuinely present in the mixture.

NEET Trap

Initiation makes radicals; propagation conserves them

A common slip is to count the H-abstraction step as initiation. Initiation is only the photochemical splitting of the halogen. Propagation steps neither create nor destroy the radical count overall — they consume one radical and regenerate another. Termination is the only step that reduces the radical count.

Tag each step by what it does to the radical tally: initiation 0→2, propagation no net change, termination 2→0.

The chain nature of the reaction

The reason halogenation is called a chain reaction is that one initiation event triggers many product-forming cycles. NIOS captures this directly: the reaction "continuously takes place till it is stopped or the reactants completely react to form the products." A single chlorine atom can run the two propagation reactions thousands of times before it is finally consumed in a termination collision.

Figure 2 · The propagation cycle

The chlorine atom is regenerated at the end of every loop, so a tiny amount of initiation drives a large amount of product. The chain only breaks when two radicals collide (termination).

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Build the foundation

New to how mechanisms are classified? Start with Types of Reaction Mechanisms to see where radical chains sit alongside ionic pathways.

Halogen reactivity and selectivity

Not all halogens behave the same. NIOS gives the order of reactivity of the halogens towards alkanes directly: F₂ > Cl₂ > Br₂ > I₂. The Chapter 25 account adds the practical limits: "Direct iodination is not possible with iodine as the reaction is reversible," and "Direct fluorination is also not possible because due to the high reactivity of the fluorine, the reaction cannot be controlled." Fluorination is explosively fast; iodination effectively does not go.

Halogen	Behaviour with alkanes	Practical status
`F2`	Extremely reactive, highly exothermic	Violent / uncontrollable
`Cl2`	Reactive, needs UV or heat	Useful, but unselective
`Br2`	Slower, needs UV or heat	Useful and highly selective
`I2`	Endothermic abstraction; reaction reverses	Does not proceed directly

Selectivity is a separate idea from reactivity. The hydrogen abstracted in propagation can be primary (1°), secondary (2°) or tertiary (3°), and the radical formed differs in stability. The order of radical stability mirrors that of carbocations: 3° > 2° > 1° > methyl. NIOS develops this stability ordering for carbon-centred species through hyperconjugation in Chapter 25, noting that a tertiary species has nine C–H bonds available for hyperconjugation against six for secondary, so it is more stable.

Figure 3 · Radical stability

A more stable radical forms more easily, so the position that gives the most stable radical is preferred — most strongly when the halogen is bromine.

Because the bromine atom abstracts hydrogen reluctantly, its transition state strongly reflects the stability of the radical being formed, so bromination is highly selective for the 3° position. Chlorine abstracts almost any hydrogen readily, so chlorination is far less selective and gives a product spread closer to the statistical ratio of available hydrogens. The memory line is: bromine is choosy, chlorine is greedy.

NEET Trap

Reactivity and selectivity move in opposite directions

Chlorine is more reactive than bromine, yet bromine is more selective. Students often assume the more reactive halogen must also be the more selective one. The opposite is true: the less reactive, more reluctant radical (Br·) is the picky one.

High reactivity → low selectivity (Cl₂). Low reactivity → high selectivity (Br₂).

Orientation and product mixtures

Because most alkanes carry several non-equivalent hydrogens, free radical halogenation rarely gives a single clean product. NIOS itself emphasises that even successive substitution of methane produces the whole sequence from chloromethane to carbon tetrachloride. With larger alkanes there is an additional layer: positional isomers.

Worked reasoning

Propane is monochlorinated. What governs the ratio of 1-chloropropane to 2-chloropropane?

Propane has six primary hydrogens (on the two end carbons) and two secondary hydrogens (on the middle carbon). Two factors compete: the number of each kind of hydrogen, and the relative ease of abstracting them (secondary is easier than primary because the secondary radical is more stable). Chlorine, being unselective, gives a substantial amount of both products; bromine, being selective, would strongly favour the 2-bromo product from the secondary position.

For NEET, the practical takeaway is to expect a mixture, identify the major product by the most stable radical (especially with Br₂), and remember that polysubstitution becomes important when the halogen is in excess.

Allylic substitution with NBS

A special and heavily examined case is substitution at the allylic position — the saturated carbon directly attached to a C=C double bond. The reagent of choice is N-bromosuccinimide (NBS), which supplies a low, steady concentration of bromine. The low Br₂ concentration is what tips the balance away from electrophilic addition across the double bond and towards radical substitution at the allylic C–H.

The reason allylic substitution is favoured is that the resulting allylic radical is resonance-stabilised: the unpaired electron is delocalised over two carbons through the adjacent double bond. Schematically:

$$\ce{CH2=CH-\overset{.}{C}H2 <-> \overset{.}{C}H2-CH=CH2}$$

This delocalisation lowers the energy of the radical, so abstraction of an allylic hydrogen is unusually easy. A representative transformation, with the double bond preserved and the new bromine at the allylic carbon:

$$\ce{CH2=CH-CH3 ->[\text{NBS}][h\nu] CH2=CH-CH2Br}$$

The same resonance logic explains the benzylic position (the carbon attached to a benzene ring), which is also readily substituted by radicals because the benzylic radical is delocalised into the ring.

Why aromatic rings resist it

Benzene does not undergo free radical substitution on the ring. NIOS is explicit that "aromatic hydrocarbons generally undergo electrophilic substitution reactions," and that "benzene undergoes electrophilic substitution reactions." Halogenation of benzene proceeds in the presence of a Lewis acid catalyst by an ionic, electrophilic route — not by light-driven radicals.

The reason is the stability of the aromatic system. The six π electrons are delocalised over the ring, and abstracting a ring hydrogen as a radical, or adding a halogen atom to the ring, would disrupt this aromatic stabilisation at a steep energy cost. Electrophilic substitution, by contrast, lets the ring lose a proton at the end and restore aromaticity. The contrast is summarised below.

Feature	Alkane (radical substitution)	Benzene (electrophilic substitution)
Reactive species	Neutral radicals (X·, R·)	Electrophile (e.g. `X+`)
Conditions	UV light or heat	Lewis acid catalyst
Bond fission	Homolytic	Heterolytic
What is preserved	Carbon skeleton; H replaced	Aromaticity restored after substitution

For contrast with the ionic counterpart of this whole family, the electrophilic substitution mechanism is worth reading directly after this page.

Quick Recap

Free radical substitution in one screen

Halogenation of alkanes replaces an H with a halogen and runs through a free radical chain (NIOS 24.1.4).
Three steps: initiation (light homolyses X₂ → 2X·), propagation (radical consumed and regenerated), termination (two radicals combine).
It is a chain reaction: one initiation drives many product cycles until reactants run out or radicals meet.
Halogen reactivity: F₂ > Cl₂ > Br₂ > I₂; F₂ uncontrollable, I₂ does not go.
Radical stability 3° > 2° > 1° > methyl; Br₂ is highly selective, Cl₂ is unselective.
Larger alkanes give product mixtures; the major product comes from the most stable radical.
NBS brominates the resonance-stabilised allylic (and benzylic) position; benzene rings react by electrophilic, not radical, substitution.

Free Radical Substitution