UNIT 2 CHEMISTRY · PEARSON AAQ · E1.7.2
Substitution Reactions of Alkanes with Halogens
FREE CHEMISTRY REVISION · TOPIC 57
Substitution Reactions of Alkanes with Halogens
Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · E1.7.2
What you will learn
Write balanced equations for alkane substitution with chlorine or bromine under ultraviolet (UV) radiation, draw and name products, and evaluate why the method often gives mixtures. Pearson E1.7.2 includes further substitution and limitations in synthesis. Detailed reaction mechanisms and practical techniques are not required.
1. What substitution means
In substitution, an atom or group in a molecule is replaced by another. For the first halogenation of an alkane, one hydrogen is replaced by a halogen atom. The carbon skeleton remains intact. The displaced H combines with the other atom from X₂ to give hydrogen halide HX.
General first-substitution equation: R–H + X₂ → R–X + HX, with UV above the arrow. R represents the rest of the carbon-containing group; it is not a chemical element. Use Cl₂ or Br₂ for the standard examples. UV supplies energy to initiate the radical reaction; it is a condition, not an atom-containing reagent and not a catalyst.
| Feature | Alkane halogen substitution |
|---|---|
| Starting compound | Saturated alkane, containing C–C/C–H single bonds |
| Reagent | Cl₂ or Br₂ |
| Required named condition | UV radiation; room-temperature UV irradiation is a standard description |
| Organic change | One C–H becomes C–X per substitution |
| Other product | HCl for chlorine; HBr for bromine |
| Important limitation | Further substitution and different substitution positions can give mixtures |
Do not transfer the alkene addition pattern from E1.7.1. An alkane has no C=C bond across which the two halogen atoms can add. Ordinary dark, room-temperature mixing is not the stated UV-initiated process. Fluorine is exceptionally reactive and iodine substitution is generally unfavourable; do not assume every halogen is interchangeable.
2. First-substitution worked examples
Worked example 1 — methane with chlorine
CH₄ + Cl₂ → CH₃Cl + HCl (UV).
Organic product: chloromethane. Left: C=1, H=4, Cl=2. Right: C=1, H=3+1=4, Cl=1+1=2. Writing CH₄Cl would give carbon five bonds and would not describe substitution.
Worked example 2 — ethane with bromine
CH₃CH₃ + Br₂ → CH₃CH₂Br + HBr (UV).
Organic product: bromoethane, C₂H₅Br. The two carbon ends of ethane are equivalent, so replacing an H at either end gives the same constitutional product. There is one distinct monobromination product, but this does not guarantee the final mixture contains only that compound: further bromination remains possible.
Worked example 3 — propane with chlorine
Write two alternative first-substitution equations:
CH₃CH₂CH₃ + Cl₂ → CH₂ClCH₂CH₃ + HCl
CH₃CH₂CH₃ + Cl₂ → CH₃CHClCH₃ + HCl
Both require UV. Products are 1-chloropropane and 2-chloropropane. Both organic products are C₃H₇Cl. These are alternatives, not a claim that one mole of propane produces one mole of each. The middle and end H environments are different, so symmetry does not predict an equal product ratio.
3. Further substitution
Worked example 4 — replace a second H
Chloromethane still has three C–H bonds. It can react again:
CH₃Cl + Cl₂ → CH₂Cl₂ + HCl (UV).
The product is dichloromethane. The original chlorine remains; one more H is replaced. The first chlorination does not automatically stop when CH₃Cl appears.
Worked example 5 — the overall equation for four replacements
Add the four schematic substitution equations. Intermediate organic formulae cancel:
CH₄ + 4Cl₂ → CCl₄ + 4HCl (UV conditions).
Eight Cl atoms enter: four are in tetrachloromethane and four in HCl. All four methane H atoms appear in HCl. This balanced overall equation describes complete replacement; it does not establish a quantitative yield of pure CCl₄ in an uncontrolled mixture.
4. Why synthesis gives mixtures
| Cause | Consequence | What can and cannot be concluded |
|---|---|---|
| Different H positions | More than one monohalogenoalkane can form | Count distinct connectivities using symmetry; do not count reversed drawings twice |
| Remaining C–H bonds | Di-, tri- and more highly substituted products can form | A product with one halogen is still capable of substitution |
| Competing reactions at different sites | Product quantities need not be equal | Atom balance and the number of products do not determine ratios |
| Changing reactant proportions | Excess alkane relative to halogen can favour monosubstitution | It does not guarantee a pure single product |
A desired product can therefore be mixed with unreacted alkane, position isomers and more heavily halogenated compounds. This lowers the selectivity for the target and complicates obtaining a pure material. More reaction does not necessarily improve the yield of the desired mono-substituted compound: that compound can itself be consumed.
Worked evaluation 6 — choosing a selective route
A student wants pure 1-chloropropane from propane and chlorine under UV. Predict at least three concerns: 2-chloropropane can form; either monochloro product can undergo further substitution; unreacted propane can remain. Excess propane may reduce the relative extent of repeated substitution, but does not remove the different-position problem. Conclude that this method is not reliably selective for pure 1-chloropropane.
Worked example 7 — a stoichiometric upper limit
For the first-substitution equation, the alkane:Cl₂:monochloro product:HCl molar ratio is 1:1:1:1. If 0.0800 mol Cl₂ reacts only by first substitution with excess methane, the theoretical chloromethane amount is 0.0800 mol. This is a conditional upper-limit calculation. Actual useful yield may be lower because further substitution or other losses can occur.
Practice with answers
| Task | Answer |
|---|---|
| Name CHCl₃ | Trichloromethane; three methane H atoms have been replaced by Cl. |
| Give the two distinct monochlorination products of butane | 1-Chlorobutane and 2-chlorobutane, each C₄H₉Cl. C3/C4 drawings repeat these connectivities after renumbering. |
| Balance first bromination of methane | CH₄ + Br₂ → CH₃Br + HBr, UV. |
| Does one distinct mono-product prove a pure final mixture? | No. Methane/ethane illustrate how further substitution can still occur. |
Common misconceptions
- Both halogen atoms enter the alkane in the first step: one enters the organic product and the other forms HX.
- Substitution adds a fifth bond to carbon: an existing C–H bond is replaced.
- UV belongs in the atom balance: it is a reaction condition.
- Using exactly one equivalent of halogen guarantees monosubstitution of every molecule: reaction mixtures can still contain unreacted and more highly substituted material.
- Two positions mean a 50:50 mixture: product distributions depend on relative reactivity and conditions.
- Reversing a carbon chain creates a new isomer: compare connectivity after correct numbering.
Examination tips
- Put UV above or beside the reaction arrow.
- Show the hydrogen halide as well as the named organic product.
- Count C, H and halogen atoms on both sides.
- For mono-substitution products, replace one H at each distinct carbon environment and remove duplicate drawings.
- Explain both further substitution and positional product mixtures when evaluating limitations.
- Describe conditions and outcomes; detailed radical mechanism stages are outside the required scope.
Scientific cross-check: OpenStax, alkane halogenation. The diagrams, examples and questions here are original.
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