Initiation
A concise revision guide to the initiation step in free-radical substitution, including UV light, homolytic fission, halogen radicals and how unpaired electrons are represented.
- 4.7i
- 4.18i
- 4.18ii
What these spec points say
- 4.7i understand that bond breaking can be: homolytic, to produce free radicals
- 4.18i understand the mechanism of the free radical substitution reaction between an alkane and a halogen: using free radicals, which are species with an unpaired electron, represented by a single dot
- 4.18ii understand the mechanism of the free radical substitution reaction between an alkane and a halogen: showing the initiation step of the mechanism, with curly half-arrows for free radical formation
Radical Chain Reaction: Methane and Chlorine
Choose a step to follow every electron: UV light makes the radicals, the chain repeats, and two radicals meeting end it.
© Dr. Mohammed Al-Fatah – onlinelearningsystem.net
GCSE Recap: Bond Energy and Breaking Bonds
Start with the GCSE ideas about the energy needed to break bonds.
Watch: ultraviolet light meets chlorine
The first step of free-radical substitution: ultraviolet light and a chlorine molecule.
What Happens During Initiation?
Initiation is the first step in the free-radical substitution mechanism. It produces the free radicals needed to start the chain reaction.
In this step, a halogen molecule absorbs energy from ultraviolet, UV, light. The energy breaks the halogen-halogen bond by homolytic fission.
For chlorine, the initiation equation is:
Initiation equation: Cl2 → 2Cl•
Key idea: Initiation does not directly make the halogenoalkane product. It produces the reactive halogen radicals that allow the next stage, propagation, to begin.
Quick Check: One Flash of Light
Use the data from one flash of UV light to decide what initiation does.
Why UV Light Breaks the Halogen Bond
UV light provides enough energy to break the Cl-Cl bond in chlorine or the Br-Br bond in bromine.
In chlorination, the Cl-Cl bond is broken in preference to other bonds because it is the weakest bond present in the starting mixture. UV light does not have sufficient energy to break the much stronger C-H bond in the alkane under these conditions.
| Bond considered | What UV light does | Importance in initiation |
|---|---|---|
| Cl-Cl or Br-Br | Breaks the halogen-halogen bond by homolytic fission. | Forms halogen free radicals, such as Cl• or Br•. |
| C-H | Does not break this bond during initiation. | The C-H bond is stronger, so it is not the bond selected by UV light at this stage. |
Exam focus: Do not write that UV light breaks a C-H bond in the initiation step. The initiation step breaks the halogen-halogen bond.
The halogen–halogen bond is the weakest bond in the mixture, so it is the one UV light breaks.
Quick Check: Explain Why Sunlight Is Needed
Use bond enthalpies to explain why light is needed and which bond breaks.
Homolytic Fission
The halogen-halogen bond breaks by homolytic fission. This means that each atom takes one electron from the shared covalent bond.
Because each atom receives one electron, the products are not ions. Instead, the products are free radicals with unpaired electrons.
Homolytic fission: bond breaking in which each atom takes one electron from the shared covalent bond.
Each atom receives one electron from the shared pair, producing two free radicals rather than ions.
Free Radicals Formed in Initiation
A free radical is a reactive species with an unpaired electron. This unpaired electron is represented using a dot, such as Cl• or Br•.
Free radicals have no overall charge. The dot shows the unpaired electron, not a positive or negative charge.
| Halogen molecule | Initiation equation | Radical formed |
|---|---|---|
| Chlorine, Cl2 | Cl2 → 2Cl• | Chlorine radicals, Cl• |
| Bromine, Br2 | Br2 → 2Br• | Bromine radicals, Br• |
Remember: The dot must stay with the radical symbol. It represents the unpaired electron that makes the species highly reactive.
Writing the Initiation Step Correctly
When writing initiation, keep the equation simple and specific. The reactant is the halogen molecule, and the products are two halogen radicals.
For chlorination, the correct initiation step is:
Cl2 → 2Cl•
For bromination, the equivalent initiation step is:
Br2 → 2Br•
Common mistake: Do not add charges to the radicals. Cl• and Br• are neutral radicals, not Cl– or Br– ions.
A chlorine radical is neutral and has an unpaired electron; do not confuse it with the chloride ion or the chlorine molecule.
Quick Check: Write the Step for Iodine
Complete the description and equation for the initiation step with iodine.
Quick Check: Pick the Accurate Statement
Finish the page by choosing the accurate statement about radicals, ions and bond breaking in each round.
Initiation FAQs
These questions target the main points students often confuse when first learning the initiation step.
What is initiation in free radical substitution?
Initiation is the first stage of the mechanism. UV light breaks a halogen-halogen bond by homolytic fission, forming halogen free radicals.
What condition is needed for initiation?
Ultraviolet light is needed. It provides enough energy to break the Cl-Cl or Br-Br bond and produce radicals.
What is homolytic fission?
Homolytic fission is bond breaking where each atom takes one electron from the shared covalent bond. This forms free radicals.
Why does UV light break Cl-Cl rather than C-H?
The Cl-Cl bond is weaker than the C-H bond, so UV light breaks the chlorine-chlorine bond preferentially during initiation.
How are free radicals shown in equations?
Free radicals are shown using a dot to represent an unpaired electron, for example Cl• or Br•.
Copyright notice: This OLS revision content, including the explanations, layout, diagrams, tables and embedded learning structure, is authored for Online Learning System by Dr. Mohammed Al-Fatah. It may not be copied, reproduced, redistributed or adapted without written permission.
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