Free radical halogenation of alkanes explains how relatively unreactive alkanes can form alkyl halides when exposed to halogens and light or heat. This cheat sheet helps students track the chain mechanism, predict major products, and compare chlorination with bromination. It is especially useful because product mixtures depend on both the number of available hydrogens and radical stability.
Grade 11 and 12 chemistry students need these ideas for organic reaction mechanisms and synthesis planning.
The reaction begins with initiation, where a halogen molecule forms radicals under or heat. Propagation steps replace a hydrogen on the alkane with a halogen while regenerating a radical, so the chain continues. Termination occurs when two radicals combine and stop the chain.
Product prediction often uses relative reactivity, radical stability, and the formula .
Key Facts
- The overall monosubstitution reaction is , where is usually or .
- Initiation breaks the halogen bond homolytically: .
- The first propagation step abstracts hydrogen: .
- The second propagation step forms the alkyl halide and regenerates a halogen radical: .
- Termination removes radicals from the reaction by combination, such as or .
- Radical stability usually follows , so more stable radicals are formed more easily.
- Approximate chlorination reactivity is , while bromination is much more selective at about .
- To estimate product ratios, use for each possible substitution site.
Vocabulary
- Free radical
- A free radical is a highly reactive species with an unpaired electron, often shown with a dot such as .
- Initiation
- Initiation is the first step in a radical chain reaction, where radicals are formed, such as .
- Propagation
- Propagation is a chain step that consumes one radical and forms another radical, allowing the reaction to continue.
- Termination
- Termination is a step where two radicals combine to form a stable molecule and stop that radical chain.
- Regioselectivity
- Regioselectivity is the preference for reaction at one position of a molecule over another, producing more of one constitutional isomer.
- Homolytic cleavage
- Homolytic cleavage is bond breaking where each atom receives one bonding electron, forming two radicals.
Common Mistakes to Avoid
- Forgetting the need for or heat is wrong because halogen molecules usually require energy to form radicals during initiation.
- Counting only radical stability and ignoring the number of hydrogens is wrong because product amount depends on both equivalent H atoms and relative reactivity.
- Treating bromination and chlorination as equally selective is wrong because bromination strongly favors the most stable radical, while chlorination gives broader mixtures.
- Writing propagation steps that do not regenerate a radical is wrong because a chain reaction must produce a new radical in each propagation cycle.
- Predicting only one product for an unsymmetrical alkane is wrong because different hydrogen environments can lead to different constitutional isomers.
Practice Questions
- 1 Write the initiation and two propagation steps for monochlorination of methane, .
- 2 Propane has primary hydrogens and secondary hydrogens. Using chlorination reactivity , estimate the product ratio of -chloropropane to -chloropropane.
- 3 For isobutane, , compare formation of the tertiary bromide and primary bromide using bromination reactivity and the available hydrogen counts.
- 4 Explain why bromination of an alkane is usually more selective than chlorination, using radical stability and activation energy ideas.
Understanding Free Radical Halogenation of Alkanes Reference
A radical is an atom or group with one unpaired electron. That unpaired electron makes it reactive because atoms tend to form electron pairs. In this mechanism, bonds split so that each atom takes one electron.
This differs from ionic bond breaking, where one atom takes both electrons. When drawing the mechanism, use single headed arrows to show movement of one electron at a time.
A full arrow is wrong here because it shows movement of an electron pair. Keeping this distinction clear helps prevent many mechanism errors.
The hydrogen removal step largely controls which carbon position reacts. Removing a hydrogen from a carbon that can form a more stable radical usually needs less energy. Alkyl groups around the radical center help spread out its electron deficiency through nearby bonds.
This effect makes tertiary radicals more stable than secondary radicals, followed by primary radicals and methyl radicals. Students should identify the type of every carbon before calculating products.
Count a hydrogen only if replacing it creates the radical type being considered. Hydrogens on the same carbon are often equivalent, though molecular symmetry can make hydrogens on different carbons equivalent too.
Chlorination and bromination give a useful lesson about reaction energy. Chlorine removes hydrogen relatively quickly, so it does not strongly favor the most stable radical. It commonly gives several products when an alkane has different kinds of hydrogen.
Bromine removes hydrogen less easily. Its hydrogen removal step is more sensitive to radical stability, so formation of a tertiary radical is strongly preferred over formation of a primary radical. This is why bromination often gives one major regioisomer.
Greater selectivity comes with lower speed. Chemists must balance these features when choosing a halogenation method.
Product prediction is not based on radical stability alone. A less favorable site may have many more hydrogens available. For each distinct substitution site, multiply the number of equivalent hydrogens by the relative reactivity for that type of hydrogen.
Compare the resulting values, then convert them into fractions or percentages if needed. For example, primary positions can still contribute a substantial product amount during chlorination because they are common in many alkanes. This kind of counting appears in exam problems, but it matters beyond class.
Mixtures of halogenated hydrocarbons are used as starting materials in synthesis, and unwanted isomers can make purification difficult. Real reactions may produce further substitution if halogen remains present, so the simple prediction works best when conditions favor replacement of only one hydrogen.