Lesson 7: Haloalkanes & Reaction Schemes
Choose reagents and conditions to convert haloalkanes by substitution or elimination, then build multi-step pathways across organic functional groups.
🎯 Ngā Whāinga Akoranga | Learning Intentions
Classification of haloalkanes as primary, secondary or tertiary. The three things a haloalkane does: substitution with aqueous KOH to an alcohol, substitution with ammonia to a primary amine, and elimination with KOH in ethanol to an alkene. Multi-step pathways linking alkanes, haloalkanes, alkenes, alcohols, carboxylic acids and amines.
Classify haloalkanes, predict which product the same haloalkane gives with each of three reagents, identify the major product of an elimination from an asymmetric haloalkane, and construct multi-step synthesis flowcharts with reagents and conditions at every arrow.
🎥 Media Anchor & Pedagogical Scaffold
Organic Reaction Schemes & Reagents Summary
Video Clip: Completing reaction schemes — NCEA Level 2 Organic Chemistry | Delene Holm (Runtime: 9m 08s).
🧠 1. Before Viewing (Activate & Predict)
How do synthetic organic chemists map out multi-step reaction pathways to create complex medicines from basic hydrocarbons?
👁️ 2. During Viewing (Watch With a Job)
- Steps in a scheme: using the video as a guide, what information must you identify at each arrow in a reaction scheme? (reactant, reagent, product — what else?)
- Work through: the video shows how to complete a 3-step reaction scheme. Describe in words the strategy: do you start at the beginning, end, or wherever you have the most information?
- Apply: draw a 2-step scheme converting ethene → ethanol → ethanoic acid. Label each arrow with the reagents and conditions required.
🗣️ 3. After Viewing & Kaiako Move (Process & Apply)
Kaiako Move: Provide the 'KOH Trap' memory mnemonic: KOH in Alcohol = Alkene (elimination); KOH in Aqua = Alcohol (substitution).
Immediate Task: Complete Section 7 of your Organic Portfolio: Master Reaction Pathways Map & Reagent Reconciler.
⚡ Whakaoho | Do Now: Organic Recall Challenge (10 mins)
You cannot get there in one step. Start with ethane. Finish with ethanoic acid. There is no single reagent that does it.
Two minutes: list every reaction type you have met so far — substitution, addition, elimination, oxidation. Then sketch any route from ethane to ethanoic acid, even a clumsy one. Do not worry about reagents yet; get the sequence of functional groups right first.
📖 Activity 1: Core Reaction Mechanism & Structure Analysis (25 mins)
⚠️ Paper lesson. This lesson names reagents — PCl5, SOCl2, KOH(alc), H2SO4(conc), acidified dichromate — as steps on a scheme, and adds ammonia and aqueous KOH. None of it is a bench instruction. SOCl2 and PCl5 react violently with water and are not school-laboratory reagents; the practical work in this unit is lessons 3, 4 and 8, each with its own hazard panel.
First, the node you have not met yet (12 min). Every conversion in this unit so far has started or ended at an alkene, an alcohol, a carboxylic acid or an amine. The haloalkane is the junction that connects them, and it is the one node the earlier lessons only pointed at.
Classify it. A haloalkane is primary, secondary or tertiary on the same test you used for alcohols in Lesson 4 — count the carbons attached to the carbon carrying the halogen. Classify these four: 1-bromobutane, 2-bromobutane, 2-bromo-2-methylpropane, 1-bromo-2-methylpropane.
Then react it. Take one haloalkane — 2-bromobutane — and put it with each of three reagents in turn. Write the organic product of each, and name it:
- warmed with aqueous KOH
- heated with excess ammonia in ethanol, in a sealed tube
- heated with KOH dissolved in ethanol
The third one has two possible products. Draw both, say which is major, and justify it by comparing the two structures — not by naming a rule. Then answer the question that makes the whole scheme work: the first and third reagents are both KOH. What is the solvent actually doing?
Build the map (15 min). In pairs, draw a single diagram with alkane, haloalkane, alkene, alcohol, carboxylic acid and amine as nodes. Draw an arrow for every conversion you have been taught in this unit and label it with reagent AND condition. Every node now has at least one arrow in. Four of the six also have arrows out — but carboxylic acid and amine are dead ends on this map, and that is a fact about the standard, not a gap in your work. The only reaction the standard lists for either of them is an acid–base one, and that gives you a salt rather than a different family of compound. So if you cannot find a way out of those two, stop looking: there isn't one. Leave arrows you cannot justify off the map — a wrong arrow costs more than a missing one.
Solve three (10 min). Solve these three: (a) ethene → ethanoic acid; (b) 1-bromopropane → propan-2-ol; (c) ethanol → ethylamine. For each, give the shortest valid route with reagents and conditions at every step. Then answer the extension question: for one of them, explain why a shorter-looking route does not actually work.
Kaiako answer key — Build the map. The complete map this unit supports: twelve arrows across six nodes. Anything beyond these is outside the standard's reaction list.
- alkane → haloalkane — Cl2 or Br2, UV light; monosubstitution only (L3)
- alkene → alkane — H2, Pt catalyst (L3)
- alkene → haloalkane — HX, or X2 for the dihaloalkane; Markovnikov decides the major product from an asymmetric alkene (L3)
- alkene → alcohol — H2O/H+, conc. H2SO4; Markovnikov again (L3)
- alkene → alcohol (a diol) — cold dilute MnO4−; a second, different route to an –OH, and students who find both have read L3 properly (L3)
- alkene → polymer — addition polymerisation (L6)
- alcohol → haloalkane — HX, PCl3, PCl5 or SOCl2 (L4)
- alcohol → alkene — conc. H2SO4, heat; Saytzeff decides the major product (L4)
- alcohol → carboxylic acid — MnO4−/H+ or Cr2O72−/H+, heat under reflux; primary alcohols (L4)
- haloalkane → alcohol — KOH(aq), warm (L7)
- haloalkane → amine — excess NH3 in ethanol, sealed tube, heat (L7)
- haloalkane → alkene — KOH in ethanol, heat; Saytzeff (L7)
What to mark. A condition missing from an arrow is the commonest lost mark, and the KOH pair is where it costs most — water substitutes, ethanol eliminates. Terminal nodes: carboxylic acid and amine have arrows in and none out. A student who draws an arrow out of either has invented chemistry this standard does not contain, and esterification is the usual invention. Their acid–base reactions from L5 are worth adding as a labelled side-branch to a salt, not as a seventh node.
Kaiako answer key — Classify it. 1-bromobutane primary (one carbon attached to the C–Br carbon); 2-bromobutane secondary (two); 2-bromo-2-methylpropane tertiary (three); 1-bromo-2-methylpropane primary — this is the one students get wrong, because the branch is on the next carbon along, not on the carbon holding the bromine.
Kaiako answer key — Then react it. From 2-bromobutane, CH3CHBrCH2CH3:
• aqueous KOH, warm → butan-2-ol + KBr. Substitution: the OH− takes the place of the bromine.
• excess NH3 in ethanol, sealed tube, heat → butan-2-amine. Written as CH3CHBrCH2CH3 + 2 NH3 → CH3CH(NH2)CH2CH3 + NH4Br. The ammonia is in excess because the amine that forms can itself react further; excess ammonia keeps the primary amine as the main product.
• KOH in ethanol, heat → elimination. Two alkenes are possible: but-2-ene (major) and but-1-ene (minor), plus KBr and H2O. The justification wanted is structural: the hydrogen is removed from the neighbouring carbon carrying fewer hydrogens, which leaves the double bond with more carbon groups on it. This is the same Saytzeff decision students made for dehydration in Lesson 4 — the point worth making out loud is that it is one rule, met twice.
The solvent question. Same reagent, two different reactions, and the solvent is the switch: KOH in water substitutes and gives the alcohol; KOH in ethanol eliminates and gives the alkene. Students who can state that and use it in a scheme have the single most examinable fact in this lesson.
Kaiako answer key — Solve three. (a) ethene + H2O with conc. H2SO4 → ethanol; then acidified dichromate (or MnO4−/H+), heat under reflux → ethanoic acid. (b) 1-bromopropane + KOH in ethanol, heat → propene (elimination); then H2O/H+ → propan-2-ol by Markovnikov addition. (c) ethanol + PCl5 (or SOCl2, or HCl) → chloroethane; then excess NH3 in ethanol, sealed tube, heat → ethylamine. The extension question is (b): the shorter-looking route — KOH(aq) straight onto 1-bromopropane — is a substitution and gives propan-1-ol, the wrong isomer. The elimination-then-addition detour is what moves the functional group to carbon 2.
📝 Activity 2: Level 2 Chemistry Portfolio Task & Merit/Excellence Scaffolding (20 mins)
Organic Portfolio — Section 7. Submit: (1) four haloalkanes classified, with the reason for each; (2) the three reactions of 2-bromobutane, with the elimination's major product justified structurally; (3) your completed conversion map, every arrow labelled with reagent and condition; (4) three multi-step syntheses written out in full; (5) one paragraph explaining a route you rejected and why.
🏫 Kaiako Planning & Pedagogy Notes
Where this sits in the standard. This lesson carries three of the standard's reactions that appear nowhere else in the unit: substitution of haloalkanes with ammonia and with aqueous potassium hydroxide, and elimination of a hydrogen halide from a haloalkane including its major and minor products. Classification of haloalkanes as primary, secondary or tertiary is in the property list.
Exit ticket (3 min, on a slip, collected). 2-bromobutane with KOH in water, and 2-bromobutane with KOH in ethanol. Name both organic products and say what makes the difference.
What to watch for while they work. In "Build the map", an arrow with no condition on it is a wrong arrow — KOH appears twice on most maps and only the solvent distinguishes them. Check every map for that pair specifically before students move on to the syntheses.
Differentiation. Support: Give a partly completed map with three of the six nodes and two arrows already drawn, and let students fill the rest. The blank-diagram version defeats students who understand every individual reaction. Extension: Ask for a conversion that cannot be done with the reactions in this unit and have them say what is missing, or ask them to find two valid routes to the same product and argue for one.
The error to head off. Students treat KOH as one reagent with one outcome. Water gives substitution and the alcohol; ethanol gives elimination and the alkene. Nothing else in this standard turns on a solvent, so it is worth naming as the exception it is.