Bond fission and free-radical substitution

What you’ll learn

• Distinguish homolytic from heterolytic bond breaking.
• Identify initiation, propagation and termination steps.
• Explain why radical substitution produces mixtures.

Understand the idea

Homolytic bond fission gives one bonding electron to each atom and forms radicals with unpaired electrons. A half-headed curly arrow represents the movement of one electron. Heterolytic fission gives both bonding electrons to one atom, forming ions, and is represented with an ordinary full-headed curly arrow for the electron pair.

In the conceptual photochemical substitution of an alkane, light provides energy for halogen–halogen bond homolysis. Initiation creates radicals. A propagation step consumes a radical but forms another, allowing a chain to continue. A termination step combines two radicals without producing another radical. These labels describe the mechanism, not separate bulk batches of reaction.

Substitution replaces a hydrogen atom by a halogen atom. Further substitution and attack at different carbon positions can generate several products, limiting selectivity. Mechanism equations should conserve every atom and explicitly show unpaired electrons; counting only the stable overall products hides the chain carriers.

Key formulae

$$\mathrm{Br_2}\xrightarrow{h\nu}2\mathrm{Br}^{\bullet}$$

$$\mathrm{Br}^{\bullet}+\mathrm{CH_4}\rightarrow\mathrm{HBr}+\mathrm{CH_3}^{\bullet}$$

$$\mathrm{CH_3}^{\bullet}+\mathrm{Br_2}\rightarrow\mathrm{CH_3Br}+\mathrm{Br}^{\bullet}$$

$$2\mathrm{CH_3}^{\bullet}\rightarrow\mathrm{C_2H_6}$$

Worked example

Classify $\mathrm{CH_3^{\bullet}+Br_2\rightarrow CH_3Br+Br^{\bullet}}$ and explain how it sustains a chain reaction.

1. A methyl radical is consumed on the left.
2. A bromine radical is produced on the right.
3. Because one radical replaces another, the product radical can react with another alkane molecule and continue the cycle.

Answer: This is a propagation step.