The shape of the lesson
By the end of the lesson, learners can
- Spot the functional groups: Build displayed, structural and skeletal formulae in the simulator and pick out each functional group before you predict any reaction.
- A toolkit of reagents and conditions: Click the arrows on the reaction map in the simulator: the halogenoalkane is the hub of most AS routes, and every arrow carries its reagents and conditions.
- Work it backwards: Use the work-it-backwards tab to ask what each target could have been made from, then follow the two worked routes.
- Groups first: name every functional group, then predict its reactions.
- Count carbons: one more carbon means a CN− step.
- Every step: reagent, conditions and type of reaction.
- Overall yield: multiply the step yields.
One picture every idea returns to
Organic synthesis means making a target molecule from a starting material by a sequence of known reactions. First identify the functional groups present, because each group has its own reactions and tests. Then work backwards from the target: compare the carbon skeleton and functional group of the target with the starting material. If the number of carbon atoms goes up by one, a cyanide step is needed (KCN on a halogenoalkane, or HCN with KCN on a carbonyl compound). Halogenoalkanes and alcohols are the key intermediates that connect most AS routes. For every step give the reagent, conditions and type of reaction (substitution, addition, elimination, oxidation, reduction, hydrolysis, condensation), and remember that the overall yield is the product of the step yields.
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, Plan the route, step by step.. Every AS reaction is a tool. 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: Spot the functional groups | Build displayed, structural and skeletal formulae in the simulator and pick out each functional group before you predict any reaction. Project tab 1 of the lab (Organic synthesis lab, spot the groups) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: A toolkit of reagents and conditions | Click the arrows on the reaction map in the simulator: the halogenoalkane is the hub of most AS routes, and every arrow carries its reagents and conditions. Project tab 2 of the lab (Organic synthesis lab, the reaction map) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: Work it backwards | Use the work-it-backwards tab to ask what each target could have been made from, then follow the two worked routes. Project tab 4 of the lab (Organic synthesis lab, work it backwards) and let a student drive it while the class predicts. | Mix-Pair-Share |
| 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, 19 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 Work it backwards, 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.
Mix-Pair-Share
Students mix around the room until you call pair; they pair with the nearest classmate, discuss the prompt, then mix again. Three prompts of rising difficulty, three different partners.
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 |
|---|---|
| Do not try to oxidise straight from an alkane or alkene to an acid at AS: go through the halogenoalkane or alcohol. | Give conditions as well as reagents: NaOH(aq) gives an alcohol but NaOH in ethanol gives an alkene; distilling gives an aldehyde but reflux gives a carboxylic acid. Use KCN in ethanol on a halogenoalkane, but HCN with a KCN catalyst on a carbonyl. And the overall yield is the product of the step yields, not their average or the lowest one. |
Organic synthesis and reaction pathways: worksheet
6 questions, 19 marks. Written in the style of the papers; no past paper question is reproduced.
Devise a three-step synthesis of ethyl ethanoate using ethene as the only organic starting material. Give the reagents and conditions for each step.
Bromoethane is converted into propanoic acid in two steps. (a) Give the reagents, conditions and intermediate. (b) 21.8 g of bromoethane (Mr 108.9) is used and the step yields are 75.0% and 80.0%. Calculate the mass of propanoic acid (Mr 74.0) formed.
Prop-2-en-1-ol is CH2=CHCH2OH. Name its two functional groups and give the organic product with (a) bromine water, (b) sodium, (c) acidified K2Cr2O7, distilling the product as it forms.
Ethanal is converted into 2-hydroxypropanoic acid. Give the two steps with reagents and conditions, and name the type of reaction in each.
A student makes propanal from 1-bromopropane: step 1 NaOH, heat; step 2 acidified K2Cr2O7, heat under reflux. Identify two mistakes in the conditions and the unwanted products they would cause.
A three-step synthesis has step yields of 90.0%, 80.0% and 75.0%. Starting with 0.150 mol of reactant, calculate the mass of product (Mr 88.0) obtained, assuming a 1 : 1 ratio in every step.
Answer key and marking notes
- Step 1: ethene + steam, H3PO4 catalyst, 300 °C, 6 MPa → ethanol (electrophilic addition).
- Step 2: oxidise part of the ethanol with excess acidified K2Cr2O7, heat under reflux → ethanoic acid.
- Step 3: heat ethanoic acid with the rest of the ethanol and a conc. H2SO4 catalyst → ethyl ethanoate + water (esterification).
- The ethanol from step 1 must be split: some is oxidised, the rest is the alcohol for step 3.
- (a) KCN in ethanol, heat under reflux, gives propanenitrile, CH3CH2CN (nucleophilic substitution).
- Then dilute HCl(aq), heat under reflux, hydrolyses the nitrile to CH3CH2COOH.
- (b) n(bromoethane) = 21.8 ÷ 108.9 = 0.200 mol; overall yield = 0.750 × 0.800 = 0.600.
- n(acid) = 0.120 mol, mass = 0.120 × 74.0 = 8.88 g.
- Alkene (C=C) and primary alcohol (–CH2OH).
- (a) CH2BrCHBrCH2OH (bromine water decolourised); (b) CH2=CHCH2O−Na+ and H2.
- (c) the aldehyde CH2=CHCHO (propenal); the C=C is not attacked by dichromate.
- Step 1: HCN with a KCN catalyst gives 2-hydroxypropanenitrile, CH3CH(OH)CN.
- This is nucleophilic addition (CN− attacks the δ+ carbonyl carbon) and adds one carbon atom.
- Step 2: heat under reflux with dilute HCl(aq): hydrolysis of –CN to –COOH, giving CH3CH(OH)COOH.
- Step 1 must use NaOH(aq), heated under reflux; NaOH in ethanol would cause elimination and give propene.
- Step 2 under reflux oxidises the propanal further to propanoic acid.
- Instead, add the oxidant to warm propan-1-ol and distil the propanal off as it forms (b.p. 49 °C).
- Overall yield = 0.900 × 0.800 × 0.750 = 0.540 (54.0%).
- n(product) = 0.150 × 0.540 = 0.0810 mol; mass = 0.0810 × 88.0 = 7.13 g.
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.
Compound Q is 4-hydroxybut-2-enal, shown in the figure. Which statement about Q is correct?
- Ait decolourises bromine water, gives an orange precipitate with 2,4-DNPH and turns warm acidified K2Cr2O7 from orange to green
- Bit decolourises bromine water but gives no precipitate with 2,4-DNPH, because the C=C bond stops the carbonyl group reacting
- Cit fizzes with aqueous sodium carbonate, because its -OH group is acidic enough to release CO2
- Dit gives a pale yellow precipitate with alkaline iodine, because it contains a C=O group
Bromoethane is converted into propanoic acid in two steps. Which reagents and conditions are used?
- Astep 1: KCN in ethanol, heat under reflux; step 2: dilute HCl(aq), heat under reflux
- Bstep 1: HCN with a KCN catalyst; step 2: NaOH in ethanol, heat
- Cstep 1: NaOH(aq), heat under reflux; step 2: excess acidified K2Cr2O7, heat under reflux
- Dstep 1: KCN in water, warm; step 2: NaBH4
Ethyl ethanoate is made from ethene in three steps: (1) steam with an H3PO4 catalyst; (2) oxidation of part of the ethanol with excess acidified K2Cr2O7 under reflux; (3) heating the product of step 2 with the rest of the ethanol and concentrated H2SO4. What types of reaction are steps 1, 2 and 3?
- Aelectrophilic addition, oxidation, condensation (esterification)
- Bnucleophilic addition, reduction, hydrolysis
- Celectrophilic addition, reduction, condensation (esterification)
- Dfree-radical substitution, oxidation, hydrolysis
The figure shows a two-step route from propan-2-ol to 2-hydroxy-2-methylpropanenitrile. What is X, and what type of reaction is step 2?
- AX is propanone; step 2 is nucleophilic addition
- BX is propanal; step 2 is nucleophilic addition
- CX is propanone; step 2 is electrophilic addition
- DX is propanoic acid; step 2 is nucleophilic substitution
Which reaction gives an organic product with more carbon atoms than the organic reactant?
- A1-bromopropane heated under reflux with KCN in ethanol
- B1-bromopropane heated with excess NH3 in ethanol in a sealed tube
- Cpropanal reduced with NaBH4
- D1-bromopropane heated with NaOH in ethanol
7.00 g of ethene (Mr 28.0) is converted into ethanoic acid (Mr 60.0) in two steps with yields of 80.0% (hydration) and 70.0% (oxidation). What mass of ethanoic acid is obtained?
- A8.40 g
- B15.0 g
- C12.0 g
- D10.5 g
Which four-step route converts propene into 2-hydroxy-2-methylpropanoic acid, (CH3)2C(OH)COOH?
- Asteam with H3PO4; acidified K2Cr2O7, heat; HCN with a KCN catalyst; dilute HCl(aq), reflux
- BHBr; KCN in ethanol, reflux; dilute HCl(aq), reflux; acidified K2Cr2O7, heat
- Csteam with H3PO4; acidified K2Cr2O7, heat; HCN with a KCN catalyst; NaBH4
- DHBr; NaOH(aq), reflux; acidified K2Cr2O7, distil; KCN in ethanol, reflux
A student tries to make propan-2-ol by heating 2-bromopropane with NaOH dissolved in ethanol. What is the main organic product, and how should the method be changed?
- Apropene, by elimination; use NaOH(aq) and heat under reflux instead
- Bpropan-1-ol, by nucleophilic substitution; use a lower temperature instead
- Cpropan-2-ol, by nucleophilic substitution; the method is already correct
- Dpropanone, by oxidation; add acidified K2Cr2O7 to reduce it back
Quiz key
Each group reacts independently. The C=C adds bromine, so bromine water is decolourised. The aldehyde C=O reacts with 2,4-DNPH to give an orange precipitate. Both the primary alcohol and the aldehyde can be oxidised, so acidified dichromate(VI) turns green. The C=C does not stop the carbonyl group reacting, so the 2,4-DNPH test is still positive. Alcohols are far too weakly acidic to react with carbonates; only a -COOH group fizzes with Na2CO3. The tri-iodomethane test needs a CH3CO- or CH3CH(OH)- group, and Q has no methyl group at all.
The cyanide ion substitutes for bromine: CH3CH2Br + CN− → CH3CH2CN + Br− (KCN in ethanol, reflux). Hydrolysis with dilute acid under reflux then turns -CN into -COOH, giving CH3CH2COOH. HCN with KCN adds to aldehydes and ketones, not halogenoalkanes, and ethanolic NaOH causes elimination. NaOH(aq) followed by oxidation gives ethanoic acid, which has only two carbon atoms. KCN must be dissolved in ethanol (in water, OH− substitution gives the alcohol), and NaBH4 is a reducing agent that cannot make a carboxylic acid.
Step 1 is hydration: the electrophile H+ attacks the electron-rich C=C, so it is electrophilic addition. Step 2 turns CH3CH2OH into CH3COOH, gaining oxygen and losing hydrogen: oxidation. Step 3 joins the acid and alcohol with loss of water: a condensation reaction (esterification). Alkenes undergo electrophilic, not nucleophilic, addition. Dichromate(VI) is an oxidising agent, so step 2 cannot be a reduction. Hydrolysis breaks an ester apart using water; step 3 is the reverse. Free-radical substitution needs an alkane and UV light.
Oxidation of the secondary alcohol propan-2-ol, CH3CH(OH)CH3, gives the ketone propanone, CH3COCH3. In step 2 the nucleophile CN− attacks the δ+ carbonyl carbon and the product is protonated by HCN: nucleophilic addition, giving CH3C(OH)(CN)CH3 with one extra carbon. Propanal comes from the primary alcohol propan-1-ol, not propan-2-ol. The C=O carbon is electron-poor, so it is attacked by a nucleophile, not an electrophile. A secondary alcohol cannot be oxidised to a carboxylic acid, and HCN adds to C=O rather than substituting into an acid.
Cyanide forms a new C-C bond: CH3CH2CH2Br + CN− → CH3CH2CH2CN, so butanenitrile (4 C) forms from a 3-carbon reactant. This is the standard chain-lengthening step. Ammonia substitutes an NH2 group to give propylamine, still with 3 carbons. NaBH4 reduces propanal to propan-1-ol, again 3 carbons. Ethanolic NaOH eliminates HBr to give propene, with 3 carbons.
n(ethene) = 7.00 ÷ 28.0 = 0.250 mol. Overall yield = 0.800 × 0.700 = 0.560, so n(ethanoic acid) = 0.250 × 0.560 = 0.140 mol, and mass = 0.140 × 60.0 = 8.40 g. 15.0 g is the theoretical mass at 100% yield. 12.0 g applies only the 80.0% yield of step 1, and 10.5 g applies only the 70.0% yield of step 2; both steps lose product, so both yields must be multiplied together.
The target is the hydrolysed hydroxynitrile from propanone. Hydration of propene gives mainly propan-2-ol (H adds to the CH2 end); oxidation gives propanone; HCN with KCN adds across C=O to give (CH3)2C(OH)CN; acid hydrolysis turns -CN into -COOH. The HBr route gives 2-bromopropane, then 2-methylpropanenitrile and 2-methylpropanoic acid, which has no -OH group, and dichromate cannot add one. NaBH4 reduces carbonyl groups; it does not hydrolyse a nitrile, so the -COOH never forms. KCN in ethanol substitutes into halogenoalkanes but does not add to a ketone (HCN is needed), and that route also lacks the hydrolysis step.
In ethanol, OH− acts mainly as a base: it removes an H from a carbon next to C-Br and HBr is eliminated, giving propene, CH3CH=CH2. In water, OH− acts mainly as a nucleophile, giving propan-2-ol by substitution, so NaOH(aq) under reflux is needed. Substitution keeps the -OH on the carbon that carried the Br, so it can never give propan-1-ol. The ethanolic conditions are the mistake, so the method is not correct as it stands. NaOH is not an oxidising agent, so no propanone forms, and K2Cr2O7 is an oxidising agent, not a reducing agent.
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.