The shape of the lesson
By the end of the lesson, learners can
- Classes and preparation: Sort halogenoalkanes into primary, secondary and tertiary in the simulator by counting the carbons on the C–X carbon.
- Substitution and elimination: Run each reaction of bromoethane in the simulator.
- Which reacts fastest?: Warm 1-chlorobutane, 1-bromobutane and 1-iodobutane with aqueous silver nitrate in ethanol in the simulator.
- NaOH(aq) → alcohol; KCN/ethanol → nitrile (+1 C).
- NH3/ethanol, pressure → amine.
- NaOH/ethanol → alkene.
- Rate: C–I > C–Br > C–Cl (bond strength).
One picture every idea returns to
Halogenoalkanes contain a polar Cδ+–Xδ− bond, so the carbon is attacked by nucleophiles. They are classed as primary, secondary or tertiary by how many carbons are attached to the C–X carbon. With aqueous NaOH they give alcohols; with KCN in ethanol, nitriles (one extra carbon); with ammonia in ethanol under pressure, amines: all nucleophilic substitution. With NaOH in ethanol they undergo elimination to alkenes. The rate of hydrolysis is governed by bond strength, not polarity: C–I is the weakest bond, so iodoalkanes react fastest.
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 carbon with a leaving group.. Swap one hydrogen in an alkane for a halogen and the molecule becomes a chemical crossroads. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Classes and preparation | Sort halogenoalkanes into primary, secondary and tertiary in the simulator by counting the carbons on the C–X carbon. Project tab 1 of the lab (Halogen compounds lab, primary, secondary, tertiary) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Substitution and elimination | Run each reaction of bromoethane in the simulator. Project tab 4 of the lab (Halogen compounds lab, reactions) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: Which reacts fastest? | Warm 1-chlorobutane, 1-bromobutane and 1-iodobutane with aqueous silver nitrate in ethanol in the simulator. Project tab 3 of the lab (Halogen compounds lab, rate of hydrolysis) and let a student drive it while the class predicts. | Corners |
| 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, 15 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 Which reacts fastest?, 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.
Corners
Put one answer or viewpoint in each corner of the room. Students move to the corner they agree with, justify the choice to a partner there, then hear one voice from each corner before the reveal.
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 |
|---|---|
| Aqueous NaOH gives substitution (an alcohol); ethanolic NaOH gives elimination (an alkene). | For the rate of hydrolysis, bond strength wins over bond polarity: iodoalkanes react fastest. KCN is the reagent that lengthens the carbon chain. |
Halogenoalkanes: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
Give the reagents, conditions and product for converting 1-bromopropane into (a) propan-1-ol, (b) butanenitrile, (c) propylamine, (d) propene.
Give two ways of making 1-chloropropane from propan-1-ol, with equations.
Describe an experiment to compare the rates of hydrolysis of 1-chlorobutane, 1-bromobutane and 1-iodobutane, and explain the results.
Classify 1-bromobutane, 2-bromobutane and 2-bromo-2-methylpropane as primary, secondary or tertiary.
Explain why the reaction of a halogenoalkane with KCN is useful in synthesis.
Explain why the carbon atom of the C–Br bond is attacked by nucleophiles.
Answer key and marking notes
- (a) NaOH(aq), heat under reflux: propan-1-ol.
- (b) KCN in ethanol, heat under reflux: CH3CH2CH2CN.
- (c) excess NH3 in ethanol, heat in a sealed tube: CH3CH2CH2NH2.
- (d) NaOH in ethanol, heat: CH3CH=CH2.
- PCl5: CH3CH2CH2OH + PCl5 → CH3CH2CH2Cl + POCl3 + HCl.
- SOCl2: CH3CH2CH2OH + SOCl2 → CH3CH2CH2Cl + SO2 + HCl.
- Add each to aqueous silver nitrate in ethanol at the same temperature (about 50 °C) and time the appearance of a precipitate.
- Order: iodo (yellow) fastest, then bromo (cream), then chloro (white) slowest.
- The C–I bond is weakest and breaks most easily; bond strength matters more than polarity.
- 1-bromobutane: primary; 2-bromobutane: secondary.
- 2-bromo-2-methylpropane: tertiary.
- It adds one carbon atom to the chain, forming a nitrile.
- The nitrile can then be converted into other compounds, e.g. hydrolysed to a carboxylic acid or reduced to an amine.
- Bromine is more electronegative than carbon, so the bond is polar, Cδ+–Brδ−.
- The δ+ carbon attracts nucleophiles, which donate a lone pair.
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.
How is 2-bromo-2-methylpropane, (CH3)3CBr, classified?
- Atertiary, because the carbon bonded to Br is attached to three other carbon atoms
- Bprimary, because the molecule contains only one bromine atom
- Csecondary, because the bromine is on carbon 2 of the chain
- Dquaternary, because the central carbon makes four bonds
Bromoethane is heated under reflux with aqueous sodium hydroxide. What is the organic product?
- Aethanol, C2H5OH, by nucleophilic substitution (hydrolysis)
- Bethene, C2H4, by elimination of HBr
- Cethylamine, C2H5NH2, by nucleophilic substitution
- Dethane, C2H6, by reduction
Bromoethane is heated under reflux with potassium cyanide dissolved in ethanol. What is the organic product?
- Apropanenitrile, C2H5CN, which has one more carbon atom than bromoethane
- Bethanenitrile, CH3CN, which has the same number of carbon atoms as bromoethane
- Cethanol, C2H5OH, because cyanide ions act as a base
- Dethene, C2H4, by elimination of HBr
1-chlorobutane, 1-bromobutane and 1-iodobutane are each added to aqueous silver nitrate in ethanol at 55 °C. The chart shows the C-X bond energies. Which tube forms a precipitate first, and why?
- A1-iodobutane gives yellow AgI first, because the C-I bond is the weakest and breaks most easily
- B1-chlorobutane gives white AgCl first, because the C-Cl bond is the most polar
- C1-bromobutane gives cream AgBr first, because bromine is in the middle of the group
- Dall three appear together, because silver ions react with every halide ion at the same rate
Which reagent converts ethanol into chloroethane at room temperature?
- Aphosphorus(V) chloride, PCl5
- Bchlorine gas in ultraviolet light
- Caqueous sodium chloride
- Ddilute aqueous sodium hydroxide
2-bromopropane is heated with sodium hydroxide. Under one set of conditions the main product is propan-2-ol; under another it is propene. Which pairing is correct?
- ANaOH(aq) under reflux gives propan-2-ol; NaOH in ethanol gives propene
- BNaOH in ethanol gives propan-2-ol; NaOH(aq) under reflux gives propene
- Cboth give propan-2-ol; propene needs concentrated H2SO4
- Ddilute NaOH gives propene; concentrated NaOH gives propan-2-ol
1-bromopropane is heated with ammonia in ethanol in a sealed tube. What are the conditions for, and the main organic product of, this reaction?
- Aexcess ammonia, heated under pressure; propylamine, CH3CH2CH2NH2
- Bexcess 1-bromopropane, room temperature; propylamine
- Caqueous ammonia under reflux; propan-1-ol
- Dexcess ammonia, heated under pressure; propene, by elimination
CFCs such as CCl2F2 were used as refrigerants because they are unreactive and non-toxic. Why do they damage the ozone layer?
- Athey survive long enough to reach the stratosphere, where UV breaks C-Cl bonds to give Cl• radicals that catalyse the breakdown of O3
- Bthey react directly with ozone in the lower atmosphere, because C-F bonds are very weak
- CUV light breaks the C-F bonds, releasing F• radicals, because C-F is the weakest bond
- Dthey dissolve in rain to give Cl− ions, which react with O3 at ground level
Quiz key
The classification depends on how many carbon atoms are bonded to the C-X carbon. In (CH3)3CBr that carbon carries three CH3 groups, so it is tertiary. Primary, secondary and tertiary do not count halogen atoms: a molecule with one Br can be any of the three. The number 2 in the name is a position, not the classification; 2-bromopropane is secondary, but here carbon 2 also carries a methyl branch. Every carbon makes four bonds, so that cannot be the basis; there is no quaternary halogenoalkane, because one of the four bonds is always to the halogen.
In water, OH− acts as a nucleophile: it attacks the δ+ carbon and replaces Br, so C2H5Br + OH− → C2H5OH + Br−. This is nucleophilic substitution, also called hydrolysis. Ethene forms only when NaOH is dissolved in ethanol, where OH− acts as a base and removes HBr; thinking that NaOH(aq) gives the alkene is a classic error. Ethylamine needs ammonia, not hydroxide, as the nucleophile. Nothing here is a reducing agent, so ethane is not formed.
The cyanide ion is the nucleophile; its carbon bonds to the δ+ carbon and Br− leaves: C2H5Br + CN− → C2H5CN + Br−. The product has three carbons, so this reaction is used to lengthen a carbon chain by one. Ethanenitrile has only two carbons: the error is forgetting that the nitrile carbon is new; KCN does lengthen the chain. Ethanol would need OH− in water. Elimination to ethene needs NaOH in ethanol; CN− is used here as a nucleophile.
Water in the mixture hydrolyses each halogenoalkane, releasing X−, which immediately gives a silver halide precipitate. The rate is controlled by breaking the C-X bond. C-I (240 kJ mol−1) is the weakest, so yellow AgI appears first, then cream AgBr (C-Br 280), and white AgCl last (C-Cl 340). C-Cl is indeed the most polar bond, which would attract a nucleophile most strongly, but the experiment shows bond strength outweighs polarity; choosing C-Cl is the classic error. Bromine's middle position gives a middle rate, not the fastest. Ag+ reacts instantly with any free halide ion, but the halide ions are released at very different rates.
PCl5 reacts with alcohols at room temperature: C2H5OH + PCl5 → C2H5Cl + POCl3 + HCl (steamy fumes of HCl). PCl3 (with heating) and SOCl2 do the same job. Chlorine in UV light is the free-radical substitution used to make halogenoalkanes from alkanes, not alcohols; it would attack C-H bonds and give a mixture. Chloride ions in NaCl(aq) are not acidic and cannot remove the OH; a halide needs concentrated acid (for example KBr + concentrated H2SO4) to work on an alcohol. NaOH(aq) converts halogenoalkanes into alcohols, the reverse direction.
In water OH− acts mainly as a nucleophile and substitutes Br, giving propan-2-ol (CH3CH(OH)CH3). In hot ethanol OH− acts mainly as a base: it removes H from a carbon next to the C-Br carbon and Br− leaves, an elimination giving CH3CH=CH2 + H2O + Br−. Swapping the conditions is the classic error. Concentrated H2SO4 dehydrates alcohols to alkenes, but ethanolic NaOH is the standard route from a halogenoalkane. The deciding factor is the solvent, not the concentration of NaOH.
CH3CH2CH2Br + 2NH3 → CH3CH2CH2NH2 + NH4Br. The lone pair on N attacks the δ+ carbon. The mixture is heated in a sealed tube because ammonia is a volatile gas, and a large excess of ammonia makes it much more likely that each halogenoalkane meets NH3 rather than the amine product, so further substitution to secondary and tertiary amines is reduced. Excess halogenoalkane does the opposite and favours multiple substitution, and the reaction is too slow at room temperature. Ammonia is the nucleophile, so the product is an amine, not an alcohol. Ammonia is a weak base and gives substitution, not elimination, here.
Because CFCs are so unreactive they are not broken down in the lower atmosphere and slowly diffuse to the stratosphere. There, UV breaks the weakest bond, C-Cl, giving Cl•. The radical destroys ozone in a chain: Cl• + O3 → ClO• + O2, then ClO• + O → Cl• + O2, so one chlorine radical is regenerated and removes thousands of O3 molecules. CFCs do not react in the lower atmosphere, which is exactly why they persist. C-F is the strongest carbon-halogen bond (about 485 kJ mol−1), so it is not broken. The damage is caused by radicals in the stratosphere, not by chloride ions at ground level.
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.