The shape of the lesson
By the end of the lesson, learners can
- Making and naming esters: Pick an acid and an alcohol in the simulator and watch the ester assemble.
- Acid or alkali: hydrolysis: Reflux the same ester with dilute acid and with sodium hydroxide in the simulator, and compare how far each reaction goes.
- Fats, oils and soap: Build a triglyceride in the simulator, compare a saturated fat with an unsaturated oil, then boil it with sodium hydroxide.
- Make: RCOOH + R′OH, conc. H2SO4 catalyst, warm, reversible.
- Name: alcohol part first, then -oate.
- Acid hydrolysis: dilute H+, reflux, equilibrium.
- NaOH hydrolysis: complete, gives RCOO−Na+ + R′OH; fats give soap.
One picture every idea returns to
An ester, RCOOR′, forms when a carboxylic acid reacts with an alcohol, heated with a few drops of concentrated H2SO4 as a catalyst; water is the other product and the reaction is reversible. Esters are named alkyl alkanoate: the alcohol part first, then the acid part, so CH3COOC2H5 is ethyl ethanoate. Acid hydrolysis (dilute acid, heat under reflux) is the reverse of esterification and reaches equilibrium. Alkaline hydrolysis with NaOH(aq) under reflux goes to completion, giving the carboxylate salt and the alcohol, because the carboxylate ion cannot react with the alcohol. Fats and oils are triesters of glycerol (propane-1,2,3-triol); boiling them with NaOH is saponification and gives soap. Esters are used as flavourings, perfumes and solvents.
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, Join an acid and an alcohol, then split them.. A carboxylic acid and an alcohol join to make a sweet-smelling ester and water, slowly and reversibly. 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: Making and naming esters | Pick an acid and an alcohol in the simulator and watch the ester assemble. Project tab 3 of the lab (Carboxylic acids lab, making an ester) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Acid or alkali: hydrolysis | Reflux the same ester with dilute acid and with sodium hydroxide in the simulator, and compare how far each reaction goes. Project tab 4 of the lab (Carboxylic acids lab, hydrolysing an ester) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: Fats, oils and soap | Build a triglyceride in the simulator, compare a saturated fat with an unsaturated oil, then boil it with sodium hydroxide. Project tab 5 of the lab (Carboxylic acids lab, fats, oils and soap) 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, 16 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 Fats, oils and soap, 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 |
|---|---|
| Name the alcohol part first: CH 3 COOC 2 H 5 is ethyl ethanoate, and HCOOCH 3 is methyl methanoate. | Concentrated H 2 SO 4 is a catalyst in esterification, but in alkaline hydrolysis the NaOH is a reactant that is used up. Alkaline hydrolysis gives the salt (RCOO − Na +), not the free acid, until you acidify. One fat molecule uses three NaOH and gives three soap molecules. |
Esters: worksheet
6 questions, 16 marks. Written in the style of the papers; no past paper question is reproduced.
Ethanoic acid is warmed with propan-1-ol and a few drops of concentrated sulfuric acid. Write the equation, name the ester and state the role of the sulfuric acid.
Give the names of the carboxylic acid and the alcohol needed to make (a) methyl butanoate, (b) ethyl methanoate, HCOOC2H5.
Compare the hydrolysis of methyl ethanoate by dilute sulfuric acid and by aqueous sodium hydroxide. Give the products of each and explain why only one goes to completion.
6.00 g of ethanoic acid (Mr 60.0) is heated with excess ethanol and concentrated sulfuric acid. 5.72 g of ethyl ethanoate (Mr 88.0) is obtained. Calculate the percentage yield and suggest why it is below 100%.
17.8 g of glyceryl tristearate (Mr 890) is boiled with excess NaOH(aq). Write a word equation and calculate the mass of NaOH used and the mass of sodium stearate (Mr 306) formed.
State two uses of esters and explain why unsaturated oils are liquids at room temperature while saturated fats are solids.
Answer key and marking notes
- CH3COOH + CH3CH2CH2OH ⇌ CH3COOCH2CH2CH3 + H2O.
- The ester is propyl ethanoate.
- Concentrated H2SO4 is a catalyst.
- (a) Butanoic acid and methanol.
- (b) Methanoic acid and ethanol.
- Dilute acid, reflux: ethanoic acid and methanol, in an equilibrium mixture (reversible).
- NaOH(aq), reflux: sodium ethanoate and methanol.
- The ethanoate ion cannot react with methanol, so there is no reverse reaction and alkaline hydrolysis goes to completion.
- n(acid) = 6.00 ÷ 60.0 = 0.100 mol; maximum ester = 0.100 × 88.0 = 8.80 g.
- Yield = 5.72 ÷ 8.80 × 100 = 65.0%.
- The reaction is reversible and reaches equilibrium, and some ester is lost during separation and purification.
- Glyceryl tristearate + sodium hydroxide → glycerol + sodium stearate (1 : 3 : 1 : 3).
- n(fat) = 17.8 ÷ 890 = 0.0200 mol, so n(NaOH) = 0.0600 mol = 0.0600 × 40.0 = 2.40 g.
- n(soap) = 0.0600 mol; mass = 0.0600 × 306 = 18.4 g.
- Uses: flavourings and perfumes (fruity smells); solvents, e.g. ethyl ethanoate in glues and nail varnish remover.
- C=C double bonds put kinks in the chains, so oil molecules cannot pack closely; intermolecular forces are weaker, so melting points are lower.
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.
What is the name of the ester shown?
- Amethyl propanoate
- Bpropyl methanoate
- Cethyl ethanoate
- Dmethyl ethanoate
Which statement about the reaction of ethanoic acid with ethanol to make ethyl ethanoate is correct?
- Ait is reversible and catalysed by a few drops of concentrated sulfuric acid
- Bit goes to completion once concentrated sulfuric acid is added
- Cit needs aqueous sodium hydroxide as a catalyst
- Dit releases hydrogen gas as the second product
Ethyl ethanoate is heated under reflux with aqueous sodium hydroxide. What are the products?
- Asodium ethanoate and ethanol
- Bethanoic acid and ethanol
- Cethanoic acid and sodium ethoxide
- Dsodium ethanoate and ethene
Why does the alkaline hydrolysis of an ester go to completion, while acid hydrolysis reaches equilibrium?
- Athe carboxylate ion, RCOO−, formed in alkali does not react with the alcohol, so the reverse reaction cannot occur
- Bhydroxide ions are a more efficient catalyst than H+ ions
- Cthe alcohol evaporates from the alkaline mixture during reflux
- Dthe ester is more soluble in alkali than in acid
Which is a common use of simple esters such as ethyl ethanoate and ethyl butanoate?
- Aas solvents and as flavourings or perfumes, because they are volatile with fruity smells
- Bas strong acids for removing limescale
- Cas soaps, because they have an ionic head
- Das reducing agents in organic synthesis
8.90 g of the fat shown is boiled with excess aqueous sodium hydroxide. What mass of sodium stearate, C17H35COONa (Mr 306), forms?
- A9.18 g
- B3.06 g
- C8.52 g
- D1.20 g
7.40 g of propanoic acid (Mr 74.0) is heated with excess methanol and a little concentrated H2SO4. 5.72 g of methyl propanoate (Mr 88.0) is collected. What is the percentage yield?
- A65.0%
- B77.3%
- C129%
- D35.0%
Ethanoic acid is esterified with ethanol in which the oxygen atom is the isotope 18O, CH3CH218OH. Where is the 18O found after the reaction?
- Ain the ester, as the single-bonded oxygen linking the two parts
- Bin the water molecule formed
- Cin the C=O group of the ester
- Dhalf in the ester and half in the water
Quiz key
The acid part is CH3CH2CO-, three carbons including the C=O carbon, so it is propanoate. The alcohol part attached through the O is CH3, methyl, and it is named first: methyl propanoate, made from propanoic acid and methanol. Propyl methanoate reverses the two parts, the classic naming error. Ethyl ethanoate has the same molecular formula, C4H8O2, but its two halves have two carbons each, so it is a different isomer. Methyl ethanoate forgets that the C=O carbon belongs to the acid part.
CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O. Water can hydrolyse the ester again, so the reaction reaches equilibrium; the concentrated H2SO4 is a catalyst (and absorbs a little water). A catalyst does not change the position of equilibrium, so it does not make the reaction go to completion. NaOH would react with the acid to form a salt, and it hydrolyses esters, so it is not used. The small molecule lost is water, not hydrogen.
CH3COOC2H5 + NaOH → CH3COONa + C2H5OH. Any ethanoic acid formed is immediately neutralised by the alkali, so the product is its sodium salt; the free acid is obtained only after acidifying, which is why ethanoic acid and ethanol is wrong here (that is the acid hydrolysis result). The ethyl group leaves as ethanol, not as sodium ethoxide, which cannot exist in water in significant amounts. No elimination occurs, so ethene is not formed.
Esterification needs the free carboxylic acid. In alkali the acid is converted to RCOO−, which has no tendency to react with an alcohol, so the reverse reaction is removed and hydrolysis continues until all the ester is used up. NaOH is not a catalyst: it is a reactant, consumed in a 1 : 1 ratio. Reflux returns any alcohol vapour to the flask, so nothing escapes. Solubility can affect the rate, but it cannot change whether a reaction is reversible.
Short-chain esters are volatile and smell fruity, so they are used as food flavourings and in perfumes; ethyl ethanoate is also a solvent in glues and nail varnish remover. Esters are neutral, not acidic, so they cannot remove limescale. Soaps are the sodium salts of long-chain carboxylic acids made by hydrolysing fats; an ester has no ionic head. Esters are not reducing agents; they can themselves be reduced, for example by LiAlH4.
n(fat) = 8.90 / 890 = 0.0100 mol. Fat + 3NaOH → glycerol + 3C17H35COONa, so n(soap) = 0.0300 mol and the mass is 0.0300 × 306 = 9.18 g. 3.06 g uses a 1 : 1 ratio and forgets that there are three ester links. 8.52 g is the mass of 0.0300 mol of stearic acid (Mr 284), the free acid, not its sodium salt. 1.20 g is the mass of NaOH used (0.0300 × 40.0), not the soap.
n(acid) = 7.40 / 74.0 = 0.100 mol, so the maximum ester is 0.100 mol × 88.0 = 8.80 g. Yield = 5.72 / 8.80 × 100 = 65.0%, typical for a reversible reaction plus losses during purification. 77.3% divides the mass of ester by the mass of acid, ignoring the different Mr values. 129% inverts that ratio; a yield above 100% is impossible. 35.0% is the shortfall (100 − 65.0), the percentage of the maximum that was not collected, not the yield.
In esterification the acid loses its OH group and the alcohol loses only the H from its O-H. The alcohol's oxygen therefore stays bonded to its ethyl group and becomes the link oxygen: CH3CO18OCH2CH3. The water is made from the acid's OH and the alcohol's H, so it contains ordinary 16O; saying the label ends up in the water is the classic error. The C=O oxygen comes from the acid. The label is not split, because the reaction follows one specific route.
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.