The shape of the lesson
By the end of the lesson, learners can
- Spot every functional group: Highlight each functional group on the target molecules in the simulator until every group stands out at a glance.
- Chain the steps: Build the set routes in the simulator step by step, then compare the slow, reversible acid-and-alcohol ester route with the faster acyl chloride route used in step 6 of Fig. 1.
- Types, chirality and yield: Use the chirality and yield tab to see where a planar intermediate gives a racemic product, then rehearse the step types behind each route: electrophilic substitution, nucleophilic addition and nucleophilic substitution.
- Label every group before choosing a reagent.
- Work backwards from the target, one group at a time.
- Each step: reagent, conditions, reaction type.
- Overall yield: multiply the step yields.
One picture every idea returns to
Organic synthesis means getting from a starting material to a target using reactions you know. First identify every functional group in each molecule (arene, phenol, alcohol, halogenoalkane, alkene, aldehyde or ketone, carboxylic acid, ester, acyl chloride, amine, amide, nitrile) and predict how each reacts with the reagent given. Then work backwards from the target: what group must the previous compound have had? Each step needs a reagent, conditions and a reaction type (electrophilic substitution, nucleophilic addition, reduction, oxidation, hydrolysis, condensation). Nucleophilic addition to a planar C=O, such as HCN with ethanal, can create a chiral centre as a racemic mixture, but drugs are often needed as a single enantiomer. Yields multiply: the overall yield is the product of the step yields, so short routes are better.
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.. A Paper 4 synthesis question hands you a starting material and a target and asks for the route. 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 every functional group | Highlight each functional group on the target molecules in the simulator until every group stands out at a glance. Project tab 1 of the lab (Organic synthesis (A Level) 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: Chain the steps | Build the set routes in the simulator step by step, then compare the slow, reversible acid-and-alcohol ester route with the faster acyl chloride route used in step 6 of Fig. 1. Project tab 3 of the lab (Organic synthesis (A Level) lab, build a route) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: Types, chirality and yield | Use the chirality and yield tab to see where a planar intermediate gives a racemic product, then rehearse the step types behind each route: electrophilic substitution, nucleophilic addition and nucleophilic substitution. Project tab 4 of the lab (Organic synthesis (A Level) lab, chirality and yield) and let a student drive it while the class predicts. | Frayer Model |
| 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, 20 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 Types, chirality and yield, 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.
Frayer Model
Pairs fill a four-box grid for the key term: definition, characteristics, examples and non-examples. The non-examples box is where the misconceptions surface, so give it the most time.
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 add or average step yields: multiply them. | Give conditions as well as reagents (reflux, temperature, catalyst) and say dilute or concentrated. Diazotisation must be kept below 10 °C or the diazonium salt decomposes. Sn/HCl reduction needs NaOH(aq) afterwards to free the amine from its salt. Check whether a carbonyl product is chiral before claiming a racemic mixture, and remember that one reagent attacks every group that responds to it. |
Multi-step organic synthesis: worksheet
6 questions, 20 marks. Written in the style of the papers; no past paper question is reproduced.
Tyrosine is HOC6H4CH2CH(NH2)COOH. Name its three functional groups and predict what happens with (a) NaOH(aq), (b) Br2(aq), (c) dilute HCl.
Give the reagents and conditions for each step of the conversion benzene → nitrobenzene → phenylamine → benzenediazonium chloride → 4-(phenylazo)phenol.
Suggest a three-step route from methylbenzene to N-ethylbenzamide, C6H5CONHC2H5. Give reagents and conditions and the type of each reaction.
Propanone and ethanal each react with HCN and a KCN catalyst. Explain why only the product from ethanal is chiral, and why that product is formed as a racemic mixture.
15.6 g of benzene is converted into 4-(phenylazo)phenol (Mr 198.0) in three stages with yields of 85%, 70% and 60%. Calculate the overall yield and the mass of dye obtained.
Explain why pharmaceutical companies aim to make a drug as a single enantiomer, and give two ways of doing so.
Answer key and marking notes
- Phenol, primary amine, carboxylic acid (it also has a chiral carbon).
- (a) Both the COOH and the phenol OH are deprotonated, giving a salt.
- (b) The phenol-activated ring is substituted by bromine: decolourised, white precipitate.
- (c) The NH2 group is protonated to –NH3+, giving a salt.
- Conc. HNO3 and conc. H2SO4, 25–60 °C.
- Sn and conc. HCl, heat under reflux, then NaOH(aq).
- NaNO2 and dilute HCl (HNO2), below 10 °C.
- Phenol in NaOH(aq), below 10 °C.
- Hot alkaline KMnO4, reflux, then dilute acid: benzoic acid (oxidation).
- SOCl2 (or PCl5): benzoyl chloride (substitution).
- Ethylamine, room temperature: N-ethylbenzamide (addition-elimination, condensation).
- Ethanal gives CH3CH(OH)CN: the central carbon has four different groups (CH3, H, OH, CN), so it is chiral. Propanone gives (CH3)2C(OH)CN, with two identical CH3 groups, so it is achiral.
- The C=O carbon is trigonal planar.
- CN− attacks equally from above and below the plane, so equal amounts of the two enantiomers form.
- Overall yield = 0.85 × 0.70 × 0.60 = 0.357 = 35.7%.
- n(benzene) = 15.6 ÷ 78.0 = 0.200 mol, so the maximum is 0.200 × 198.0 = 39.6 g.
- Mass obtained = 39.6 × 0.357 = 14.1 g.
- Usually only one enantiomer binds to the chiral receptor or enzyme; the other may be inactive or cause harmful side effects.
- A single enantiomer means a smaller dose, fewer side effects and no wasted half of the product.
- Methods: a chiral catalyst or enzyme, a naturally chiral starting material, or separating (resolving) the racemic mixture.
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.
The route shown makes phenyl benzoate from methylbenzene. Which reagents are X and Y?
- AX: hot alkaline KMnO4, then dilute acid; Y: SOCl2
- BX: acidified K2Cr2O7, distil; Y: HCl(aq)
- CX: hot alkaline KMnO4, then dilute acid; Y: Cl2 with AlCl3
- DX: Cl2 in UV light, then NaOH(aq); Y: SOCl2
Aspirin is CH3COOC6H4COOH. Which functional groups does it contain?
- Aan ester, a carboxylic acid and an arene ring
- Ba ketone, an ether and an arene ring
- Ca phenol, a carboxylic acid and a ketone
- Dtwo carboxylic acid groups and an arene ring
A four-step synthesis has step yields of 80%, 75%, 90% and 60%. What is the overall yield?
- A32.4%
- B76.3%
- C60.0%
- D36.0%
Propanenitrile, CH3CH2CN, is to be converted into propylamine, CH3CH2CH2NH2. Which reagent is used?
- Ahydrogen with a nickel catalyst (or LiAlH4)
- Bdilute HCl(aq), heated under reflux
- CNaBH4 in water
- Dammonia in ethanol, heated in a sealed tube
Propanone is converted into 2-hydroxy-2-methylpropanoic acid in two steps: HCN with KCN, then heating under reflux with dilute HCl(aq). What types of reaction are these?
- Anucleophilic addition, then hydrolysis
- Belectrophilic addition, then hydrolysis
- Cnucleophilic substitution, then oxidation
- Dnucleophilic addition, then reduction
Which of these syntheses gives an organic product that is a racemic mixture?
- Aethanal with HCN and a KCN catalyst
- Bpropanone with HCN and a KCN catalyst
- Cmethylbenzene with hot alkaline KMnO4, then dilute acid
- Dbenzoyl chloride with ethanol at room temperature
9.20 g of methylbenzene (Mr 92.0) is converted into ethyl benzoate, C6H5COOC2H5 (Mr 150.0), in three steps with the yields shown. What mass of ethyl benzoate is obtained?
- A7.56 g
- B10.5 g
- C12.0 g
- D9.45 g
4-(hydroxymethyl)phenol, HOC6H4CH2OH, is shaken with excess cold NaOH(aq). What is the organic product?
- ANa+ −OC6H4CH2OH
- BNa+ −OC6H4CH2O− Na+
- CHOC6H4CH2O− Na+
- Dno reaction, because neither OH group is acidic enough
Quiz key
Hot alkaline KMnO4 oxidises the side chain of an alkylarene to COO−; dilute acid then gives benzoic acid. SOCl2 (or PCl5) converts the acid into benzoyl chloride, which reacts with phenol in alkali to give the ester; phenol does not esterify usefully with the carboxylic acid itself. Acidified dichromate does not oxidise a CH3 group on a ring, and HCl(aq) does not convert an acid into an acyl chloride. Cl2 with AlCl3 substitutes the ring, it does not make COCl. Cl2 in UV light then NaOH(aq) gives phenylmethanol, C6H5CH2OH, not benzoic acid.
CH3COO–Ar is an ester (the ethanoate ester of a phenol) and –COOH on the ring is a carboxylic acid; the C6H4 is an arene ring. The C=O in CH3COO– is part of the ester, not a ketone, and the single-bonded O is not an ether because it is next to C=O. There is no free phenol OH: it has been esterified (hydrolysis would release one). Only one of the two C(=O)O groups has an O–H, so there are not two acid groups.
Overall yield = 0.80 × 0.75 × 0.90 × 0.60 = 0.324, so 32.4%. 76.3% is the mean of the four yields, but losses compound, they do not average. 60.0% assumes the worst step sets the yield, ignoring losses in the other steps. 36.0% multiplies only three of the yields, leaving out the 90% step.
Reduction: CH3CH2CN + 4[H] → CH3CH2CH2NH2, using H2/Ni or LiAlH4 in dry ether. Acid hydrolysis turns a nitrile into a carboxylic acid (propanoic acid), not an amine. NaBH4 reduces aldehydes and ketones but is too weak to reduce a nitrile. Ammonia in ethanol makes amines from halogenoalkanes, not from nitriles.
CN− (the nucleophile, from the KCN catalyst) attacks the δ+ carbon of C=O and the alkoxide is then protonated by HCN: nucleophilic addition, giving (CH3)2C(OH)CN. Heating with dilute acid hydrolyses CN to COOH. The C=O is attacked by a nucleophile, not an electrophile, because the carbon is δ+. Nothing is replaced in step 1, so it is not substitution, and step 2 adds water rather than oxygen. Reduction would turn CN into CH2NH2, not COOH.
Ethanal gives CH3CH(OH)CN, whose central carbon carries four different groups (CH3, H, OH, CN). CN− attacks the planar C=O equally from either face, so equal amounts of the two enantiomers form: a racemic mixture with no optical activity. Propanone gives (CH3)2C(OH)CN, which has two identical CH3 groups and is not chiral. Benzoic acid and ethyl benzoate contain no chiral carbon at all.
n(methylbenzene) = 9.20 / 92.0 = 0.100 mol, so the maximum is 0.100 × 150.0 = 15.0 g. Overall yield = 0.70 × 0.90 × 0.80 = 0.504, so the mass = 15.0 × 0.504 = 7.56 g. 10.5 g applies only the first step yield. 12.0 g uses the mean yield of 80%. 9.45 g stops after the first two steps and forgets the 80% esterification.
The ring OH is a phenol: the O lone pair is delocalised into the ring, stabilising the phenoxide ion, so phenol is a weak acid that reacts with NaOH(aq). The CH2OH group is a primary alcohol, which is a weaker acid than water and does not react with NaOH(aq); only sodium metal would deprotonate it. So only the phenol becomes O− Na+. Deprotonating both OH groups, or only the alcohol, reverses their acid strengths. Phenol is acidic enough, so there is a reaction.
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.