The shape of the lesson
By the end of the lesson, learners can
- Initiation, propagation, termination: Shine UV light on the methane and chlorine mixture in the simulator and follow the radicals step by step.
- A mixture of products: Radicals attack any C–H bond they meet, including those in molecules that have already been substituted.
- Initiation: homolytic fission in UV light.
- Propagation: one radical in, one radical out.
- Termination: two radicals in, none out.
- Mixture of products from further substitution.
One picture every idea returns to
Alkanes react with chlorine or bromine in ultraviolet light by free radical substitution. Initiation: UV light breaks Cl–Cl by homolytic fission into two chlorine radicals. Propagation: a radical reacts with a molecule to give a product and a new radical, so the chain continues. Termination: two radicals combine and the chain stops. Because radicals are so reactive, further substitution occurs and a mixture of products forms (CH3Cl, CH2Cl2, CHCl3, CCl4, and some ethane).
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, A chain reaction in sunlight.. Alkanes ignore most reagents, but shine ultraviolet light on methane and chlorine and they react explosively. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Initiation, propagation, termination | Shine UV light on the methane and chlorine mixture in the simulator and follow the radicals step by step. Project tab 2 of the lab (Hydrocarbons lab, free radical substitution) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: A mixture of products | Radicals attack any C–H bond they meet, including those in molecules that have already been substituted. | Cooperative Concept Mapping |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 13 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in A mixture of products, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Cooperative Concept Mapping
Teams receive the key terms on cards and arrange them on a large sheet, drawing and labelling the links. Compare maps between teams before showing the model map.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Radicals form by homolytic fission and carry an unpaired electron, shown by a dot. | Every propagation step has one radical on each side; every termination step has two radicals in and none out. A step that makes HCl and a methyl radical is propagation, not termination. |
Free radical substitution: worksheet
6 questions, 13 marks. Written in the style of the papers; no past paper question is reproduced.
Describe the mechanism for the reaction of methane with chlorine to form chloromethane, naming each stage.
State the type of bond fission in the initiation step and the condition needed.
Explain why the product mixture contains a little ethane.
Write the two propagation steps for the formation of chloroethane from ethane.
Suggest how the conditions could be changed to make tetrachloromethane the main product.
Draw the structures of the two monobromo products formed when propane reacts with bromine in UV light.
Answer key and marking notes
- Initiation: Cl2 → 2Cl• (UV light, homolytic fission).
- Propagation: Cl• + CH4 → •CH3 + HCl.
- Propagation: •CH3 + Cl2 → CH3Cl + Cl•.
- Termination: e.g. •CH3 + Cl• → CH3Cl.
Homolytic fission of Cl–Cl, in ultraviolet light.
- Two methyl radicals can combine in a termination step.
- •CH3 + •CH3 → C2H6.
- Cl• + C2H6 → •C2H5 + HCl.
- •C2H5 + Cl2 → C2H5Cl + Cl•.
- Use a large excess of chlorine.
- Then radicals are much more likely to meet chlorinated molecules and substitute every H atom.
- 1-bromopropane, CH3CH2CH2Br.
- 2-bromopropane, CH3CHBrCH3.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Methane reacts with chlorine in ultraviolet light. What happens in the initiation step?
- AUV light breaks the Cl-Cl bond by homolytic fission, giving two chlorine radicals, Cl2 → 2Cl•
- BUV light breaks the Cl-Cl bond by heterolytic fission, giving Cl+ and Cl−
- CUV light breaks a C-H bond in methane, giving CH3• and H•
- Dtwo chlorine radicals join to form a chlorine molecule
The steps in the chlorination of methane are shown. Why is the sequence of steps 2 and 3 called a chain reaction?
- Astep 3 regenerates a Cl• radical, which can start step 2 again with another CH4 molecule
- Bstep 3 forms a C-Cl bond, which joins the carbon atoms into a chain
- Ceach propagation step needs a fresh photon of UV light
- Dstep 2 removes all the radicals, so step 3 cannot happen without more UV light
Which statement about termination steps in free-radical substitution is correct?
- Atwo radicals combine to form a molecule, so the number of radicals falls
- Ba radical reacts with a molecule to produce two new radicals
- Ca molecule absorbs UV light and splits into two radicals
- Dtermination steps only produce hydrogen chloride
Methane is mixed with a large excess of chlorine and exposed to UV light for a long time. Which product is present in the greatest amount at the end?
- ACCl4
- BCH3Cl
- CC2H6
- DH2
What is a free radical?
- Aa species with an unpaired electron
- Ba negatively charged ion with a lone pair of electrons
- Ca positively charged ion with an empty orbital
- Dany atom that has a lone pair of electrons
When ethane reacts with chlorine in UV light, the main organic product is C2H5Cl, but a small amount of butane, C4H10, is also detected. How is the butane formed?
- Atwo C2H5• radicals combine in a termination step
- Bethane molecules join together in a propagation step
- CCl• removes a carbon atom from ethane and adds it to another ethane
- Dbutane is an impurity that comes from heterolytic fission of C2H5Cl
Pentane, CH3CH2CH2CH2CH3, reacts with chlorine in UV light. How many structural isomers of C5H11Cl can form?
- A3
- B5
- C12
- D2
One product of the chlorination of methane contains 83.5% chlorine by mass. Which product is it? (Ar H 1.0, C 12.0, Cl 35.5)
- ACH2Cl2
- BCH3Cl
- CCHCl3
- DCCl4
Quiz key
The Cl-Cl bond (242 kJ mol−1) is much weaker than C-H (about 410 kJ mol−1), so UV light breaks it. Each Cl atom keeps one of the two bonding electrons (homolytic fission), giving two Cl• radicals, each with an unpaired electron. Heterolytic fission, giving ions, does not occur here; that is the classic wrong idea, and radicals are not ions. UV does not have enough energy to break C-H bonds readily. Two radicals joining to form Cl2 is a termination step, the opposite of initiation.
In each propagation step one radical reacts and one radical is formed. Step 3 produces a new Cl•, which attacks another methane molecule in step 2, so one initiation can lead to thousands of cycles: a chain. 'Chain' refers to this cycle of radicals, not a chain of carbon atoms; CH3Cl has only one carbon. UV light is only needed for initiation, to split Cl2; propagation steps continue in the dark once radicals exist. Step 2 does not remove radicals: Cl• is replaced by CH3•.
In termination, two radicals pair their unpaired electrons to form a covalent bond, for example Cl• + Cl• → Cl2, CH3• + Cl• → CH3Cl or CH3• + CH3• → C2H6. Radicals are removed, so the chain stops. Termination does not produce radicals; that is the classic error, and a step that increases the number of radicals would be a branching or initiation step. Splitting a molecule with UV is initiation. HCl is made in a propagation step (Cl• + CH4), never in termination, because no termination step involves an H atom radical here.
Cl• can remove H from CH3Cl as well as from CH4, so further substitution gives CH2Cl2, then CHCl3 and finally CCl4. With a large excess of chlorine and a long time, the hydrogens are replaced one by one until CCl4 dominates. CH3Cl is the main product only when methane is in excess. Ethane forms only in small amounts, from the termination step CH3• + CH3•. Hydrogen gas is not formed; the H atoms leave as HCl.
A free radical has an unpaired electron (shown as a dot), which makes it very reactive: it pairs up by taking an atom from a molecule or by combining with another radical. Cl• and CH3• are uncharged. A negative ion with a lone pair, such as OH−, is a nucleophile, not a radical. A positive ion with an empty orbital, such as NO2+, is an electrophile. Many stable molecules contain atoms with lone pairs (for example the O in water), but all their electrons are paired.
Propagation: Cl• + C2H6 → C2H5• + HCl. When two ethyl radicals meet, they pair their electrons in a termination step: C2H5• + C2H5• → C4H10. This is the ethane equivalent of methyl radicals forming ethane. Two stable molecules do not join in propagation; propagation always involves a radical reacting with a molecule. Cl• abstracts H atoms, not carbon atoms. No heterolytic fission occurs in this mechanism; radicals form by homolytic fission.
Pentane is symmetrical, so C1 and C5 are equivalent, and C2 and C4 are equivalent. Substituting an H on C1, C2 or C3 gives 1-chloropentane, 2-chloropentane and 3-chloropentane: 3 isomers. (In the same way propane gives 2: 1-chloropropane and 2-chloropropane.) 5 counts each carbon separately and ignores the symmetry; 4-chloropentane is just 2-chloropentane numbered from the wrong end. 12 counts every hydrogen atom, but H atoms on the same or equivalent carbons give the same product. 2 misses the central carbon, C3. The random attack of Cl• on any C-H bond is why free-radical substitution gives a mixture that must be separated.
For CH2Cl2, Mr = 12.0 + 2.0 + 71.0 = 85.0, and %Cl = 71.0 / 85.0 × 100 = 83.5%. CH3Cl (Mr 50.5) has 70.3% Cl, the first substitution product. CHCl3 (Mr 119.5) has 89.1% Cl. CCl4 (Mr 154.0) has 92.2% Cl. The percentage rises with each further substitution because heavy Cl atoms replace light H atoms. Bromine reacts in the same way, giving CH3Br, CH2Br2 and so on.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.