Free-Radical Substitution – Halogenation of Alkanes
ChemistryOrganic ChemistryAges 17–18
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Sign in to playUV light splits Cl₂ or Br₂ into radicals, which react with methane or ethane in a chain: initiation, propagation and termination, each drawn with fish-hook arrows and its enthalpy change. Change the halogen : alkane ratio and the UV intensity and watch the product mixture build up, from CH₃Cl to CCl₄, with traces of ethane from termination.
Lesson: Free-radical substitution of alkanes
What it shows
Alkanes react with chlorine or bromine only in UV light. The light breaks the halogen–halogen bond by homolytic fission, giving two radicals, each with an unpaired electron (initiation). A halogen radical removes an H atom from the alkane, forming HX and an alkyl radical, which takes a halogen atom from X₂ and regenerates the halogen radical (propagation), so one photon can make many product molecules. The chain stops when two radicals meet (termination), which also explains traces of ethane. Further substitution gives a mixture of products, controlled by the ratio of reactants.
How to use
Choose the Halogen and the Alkane, set the Halogen : alkane ratio and the UV light intensity, then press Start. Watch the vessel for flashes, the mechanism panel for the step that just happened, and the bar chart for the products. Use One step to follow single events with their fish-hook arrows. Compare a large excess of alkane with a large excess of halogen, and switch the UV light off.
Parameters you can change
- Halogen Chlorine (Cl₂), Bromine (Br₂)
- Alkane Methane (CH₄), Ethane (C₂H₆)
- Halogen : alkane ratio (molecules) 0.25–4
- UV light intensity 0–100 %
Questions to explore
- Why does a mixture of methane and chlorine not react in the dark?
- How does the halogen : alkane ratio change which product is formed most?
- Why does finding ethane among the products support a radical chain mechanism?
