Radical Reaction Mechanisms

Introduction

Radicals are atoms or molecules with an unpaired electron. They react quickly and indiscriminately, so their concentrations stay tiny, yet they control the rates of combustion, polymerisation, oxidation of fats and the ageing of plastics. Because radicals are short-lived intermediates, their kinetics is analysed with a radical balance : at steady state, radicals are made as fast as they are destroyed. This page applies that balance to radical chain polymerisation and shows how inhibitors and antioxidants work at the level of rate laws.

Core explanation

Consider free-radical addition polymerisation of a monomer M, started by an initiator I that decomposes into two radicals.

Initiation. I → 2R• with rate constant k d. Not every radical starts a chain; some recombine inside the solvent "cage". The fraction that succeed is the initiator efficiency , f, typically 0.3–0.8. The rate of chain starts is R i = 2fk d[I]. Each R• then adds to monomer to form a growing chain radical M•.

Propagation. M• + M → M• (one unit longer) with rate constant k p. Monomer is consumed almost entirely in this step, so the rate of polymerisation is R p = k p[M][M•].

Termination. Two chain radicals meet and combine or disproportionate: 2M• → dead polymer, with rate R t = 2k t[M•]².

Radical balance. Setting d[M•]/dt = 0 gives R i = R t, so 2fk d[I] = 2k t[M•]², and

[M•] = (fk d[I]/k t)^(1/2)

Substituting into R p:

R p = k p (fk d/k t)^(1/2) [M][I]^(1/2)

The rate is first order in monomer and half order in initiator. Doubling the initiator concentration raises the rate by only √2 ≈ 1.41, because more radicals also means faster second-order termination.

Chain length. The kinetic chain length ν = R p/R i is proportional to [M]/[I]^(1/2). More initiator gives faster polymerisation but shorter chains and so lower molar mass. This trade-off governs industrial recipes.

Inhibitors and retarders. An inhibitor Z reacts with radicals so quickly (R• + Z → inactive products) that almost no propagation occurs while Z remains. Polymerisation shows an induction period whose length is roughly the amount of inhibitor divided by the rate of radical generation. Once Z is consumed, the normal rate resumes. A retarder reacts more slowly with radicals, competing with propagation; it lowers the rate throughout without a sharp induction period. Dissolved oxygen, quinones and hindered phenols act in these ways.

Formulae

R i = 2fk d[I]; R p = k p[M][M•]; R t = 2k t[M•]².

Steady state: [M•] = (fk d[I]/k t)^(1/2).

R p = k p (fk d/k t)^(1/2) [M][I]^(1/2); ν = R p ÷ R i.

Induction time ≈ n(Z) ÷ (rate of radical generation), for an inhibitor that removes one radical per molecule.

Step-by-step reasoning

To predict the effect of changing initiator concentration:

1. Write R i in terms of [I]. 2. Balance R i against termination to find [M•]. 3. Substitute [M•] into R p. 4. Compare the new and old rates using the power of [I]. 5. Use ν = R p/R i to see how chain length responds.

Visual explanation

Plot conversion against time for three samples: no inhibitor (a straight line from the origin at early times), an inhibitor (a flat line followed by the same slope after a delay) and a retarder (a single line with a smaller slope). The shapes distinguish the two kinds of additive at a glance.

Real-world analogy

Think of radicals as sparks landing on dry grass. An inhibitor is like a patch of wet ground that smothers every spark until it dries out; nothing burns for a while, then burning proceeds normally. A retarder is like damp grass that burns, but more slowly throughout.

Real-world example

Commercial monomers such as styrene are shipped with a small amount of phenolic inhibitor to prevent spontaneous polymerisation during storage. Similarly, antioxidants such as vitamin E and synthetic hindered phenols slow the radical chain oxidation that turns cooking oils rancid.

Why?

Why does more initiator give shorter chains? Increasing [I] raises [M•] as [I]^(1/2) but raises termination as [M•]², that is, in proportion to [I]. Each chain therefore meets a partner and dies sooner, so fewer monomer units are added per chain.

Common misconception

"An inhibitor is a negative catalyst that permanently slows the reaction." A true inhibitor is consumed as it destroys radicals. It delays the reaction; it does not change the rate constants of the steps that follow.

Worked example

Question: A polymerisation obeys R p ∝ [M][I]^(1/2). The initiator concentration is increased from 0.010 to 0.040 mol dm⁻³ with [M] unchanged. By what factor do R p and the kinetic chain length change?

Reasoning: [I] rises fourfold, so R p rises by 4^(1/2) = 2. R i rises fourfold, so ν = R p/R i changes by 2/4 = 0.5.

Answer: The rate doubles and the chain length halves.

Quick check

1. What kinetic feature distinguishes an inhibitor from a retarder in a conversion–time plot? Answer: An inhibitor gives an induction period with almost no reaction; a retarder lowers the rate throughout without a sharp delay.

Exam focus

Derive R p ∝ [M][I]^(1/2) from the radical balance and explain the half order. Link initiator concentration to rate and molar mass in opposite directions. Interpret induction periods quantitatively when told the radical generation rate.

Advanced insight

At high conversion the reaction mixture becomes viscous and chain radicals diffuse slowly, so k t falls sharply while k p changes little. The radical concentration rises and polymerisation accelerates, the gel effect . This is a case where the simple steady-state rate law fails because a rate constant depends on the changing medium.

Summary

Radical mechanisms are analysed by balancing radical formation against termination. For chain polymerisation this gives R p ∝ [M][I]^(1/2) and chain length ∝ [M]/[I]^(1/2). Inhibitors consume radicals and cause induction periods; retarders compete with propagation and lower the rate throughout.

Practice questions

1. State the radical balance condition for a chain with no branching. Answer: The rate of radical formation by initiation equals the rate of radical destruction by termination. 2. Why is initiator efficiency f usually less than 1? Answer: Some radical pairs recombine or react inside the solvent cage before they can escape and add to monomer. 3. An inhibitor removes one radical per molecule. Radicals are generated at 1.0 × 10⁻⁸ mol dm⁻³ s⁻¹ and the inhibitor concentration is 2.0 × 10⁻⁵ mol dm⁻³. Estimate the induction period. Answer: 2.0 × 10⁻⁵ ÷ 1.0 × 10⁻⁸ = 2.0 × 10³ s, about 33 minutes. 4. How would halving [M] at fixed [I] affect R p and the kinetic chain length? Answer: Both halve, since each is proportional to [M].