Free-Radical Initiation

Introduction

Free-radical polymerisation begins by generating reactive radicals. Heat, light or redox chemistry can decompose an initiator; however, not every radical produced successfully starts a polymer chain. Radical efficiency links decomposition to the actual initiation rate. That distinction matters when predicting how many chains will grow.

Core explanation

A common radical initiator contains a weak bond that undergoes homolysis, forming two radical fragments per initiator molecule. Each fragment can add to a vinyl monomer, such as styrene, producing a new carbon-centred radical at the end of the first unit. If initiator concentration is [I] and its first-order decomposition constant is k d, initiator molecules disappear at rate k d[I]. Two primary radicals are formed per decomposition. Let f, between zero and one, be the fraction that actually initiate growing chains. Then the initiation rate is R i = 2 f k d[I]. The factor two counts radical fragments, while f accounts for cage recombination, side reactions and radicals that fail to add monomer. Efficient initiation requires the fragment radical to survive long enough to encounter monomer and add in a productive orientation. Increasing initiator concentration generally raises the number of growing radicals but can lower average chain length because more chains share the monomer and termination becomes more frequent. The initiator is consumed; it is not a catalyst in the strict sense of being regenerated. The fragments often remain as end groups of polymer chains, which can sometimes be detected analytically. Initiation conditions must also avoid uncontrolled exotherm and oxygen inhibition, since oxygen readily traps many carbon-centred radicals.

Step-by-step reasoning

Write the initiator homolysis yielding two radicals. Show one radical adding to monomer to make a chain radical. Calculate decomposition rate k d[I], multiply by two fragments and then by efficiency f. Interpret the result in concentration per time and distinguish it from the rate of monomer consumption during propagation.

Visual explanation

Draw one initiator molecule splitting into two dots. Some dots recombine inside a solvent cage; the remaining dots attach to monomer circles and turn into growing chain-end dots. The fraction reaching chains is f.

Real-world analogy

Lighting two matches does not mean two fires will spread: a match may go out before igniting fuel. Initiator decomposition produces primary radicals, but only the successful fraction begins propagating polymer chains.

Real-world example

Peroxide and azo initiators are chosen for their decomposition rates at a planned polymerisation temperature. A formulation may need oxygen removal because dissolved oxygen captures radicals and delays the appearance of sustained chain growth.

Why?

The relation R i = 2 f k d[I] separates radical production from successful chain initiation. Radical efficiency matters because primary radicals can recombine or undergo competing chemistry before they add monomer.

Common misconception

Do not equate initiator disappearance directly with the number of new chains. Each decomposition creates two candidate radicals, but f may be less than one. Also, an initiator is consumed rather than repeatedly regenerated like a catalyst.

Worked example

Question: If k d[I] = 0.010 mol L⁻¹ s⁻¹ and f = 0.60, calculate R i. Reasoning: Two radicals form per decomposed molecule, and 60% of those begin chains. Multiply 2 × 0.60 × 0.010. Answer: R i = 0.012 mol L⁻¹ s⁻¹ of successfully initiated chain radicals.

Quick check

1. What does the factor two represent in R i = 2 f k d[I]? Answer: Two primary radicals formed from each ideal initiator decomposition event.

Exam focus

Give units for k d, [I] and R i, and explain the meaning of f. A typical exam calculation uses the factor two and efficiency separately; omitting either changes the predicted initiation rate.

Advanced insight

Primary radicals may be born as a close solvent-cage pair and recombine before diffusing away. This cage effect is one microscopic reason f can be below one, even when every initiator molecule decomposes as expected.

Summary

An initiator decomposes to radicals, some of which add monomer and start chains. If decomposition is first order and two radicals form per event, R i = 2 f k d[I]. Efficiency accounts for unsuccessful radicals, and initiator concentration affects both the number and typical length of chains.

Practice questions

1. What is radical efficiency f? Answer: The fraction of primary radicals that successfully initiate polymer chains.

2. Calculate R i when f = 0.5 and k d[I] = 0.020 mol L⁻¹ s⁻¹. Answer: R i = 2 × 0.5 × 0.020 = 0.020 mol L⁻¹ s⁻¹.

3. Why might f be less than one? Answer: Radicals can recombine in a solvent cage or undergo side reactions before adding monomer.

4. Is an initiator regenerated after starting a chain? Answer: Generally no. Its radical fragment is consumed in initiation and may become a polymer end group.