Oxidising an aldehyde (from 3.3.8) gives a carboxylic acid, and this page follows the family that grows from it. The unifying idea is the acyl group, R–C(=O)–: swap what is attached to it and you move between the acid, ester, acyl chloride, acid anhydride and amide. One mechanism, nucleophilic addition–elimination, links most of them.
Carboxylic acids
Carboxylic acids contain the –COOH group. The C=O and O–H make the group strongly polar, so carboxylic acids hydrogen-bond: to each other (high boiling points) and to water (the short-chain acids are very soluble).
They are weak acids (only partially dissociated in water), but they are acidic enough to show the usual reactions:
CH3COOH ⇌ CH3COO− + H+
- with carbonates and hydrogencarbonates they release CO2 (effervescence), the test that distinguishes a carboxylic acid from a phenol or an alcohol;
- with metals and bases they form carboxylate salts.
2CH3COOH + Na2CO3 → 2CH3COONa + H2O + CO2
Exam questions
Compounds V, W, X and Y are isomers with the molecular formula C5H10O2. Isomers V and W are carboxylic acids with formulas that can be written as C4H9COOH.
Give an equation for the reaction of C4H9COOH with sodium hydrogencarbonate.
Show answer
C4H9COOH + NaHCO3 → C4H9COONa + CO2 + H2O 1 mark
Isomer V has an asymmetric carbon atom. Deduce the structure of V.
Show answer
CH3CH2CH(CH3)COOH 1 mark
2-methylbutanoic acid: the CH carbon carries four different groups (CH3, C2H5, COOH and H), so it is the asymmetric centre. Later parts of this question deduce the other isomers from their 1H NMR spectra. That skill lives in NMR (3.3.15).
Source: AQA A-Level Chemistry past papers.
Esters & esterification
Esters have the group –COO– and are named alkyl alkanoate: the alkyl part comes from the alcohol, the alkanoate part from the acid. So ethanoic acid + ethanol gives ethyl ethanoate.
Ethyl ethanoate, CH3COOC2H5: ethyl = the C2H5 from ethanol; ethanoate = the CH3COO from ethanoic acid. Write the alcohol-derived group first, the acid-derived group second.
Warmed with an alcohol and a concentrated sulfuric acid catalyst, a carboxylic acid forms an ester. The reaction (esterification) is reversible, so the mixture reaches an equilibrium and the yield is never complete:
CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O
Making an ester in the lab
The ester preparation is a favourite practical question, and every step has a purpose you must be able to state. Vague answers about “purifying” do not score:
| Step | Why it is done |
|---|---|
| Reflux the acid + alcohol + conc H2SO4 (anti-bumping granules) | The catalyst speeds the reaction; reflux heats the flammable mixture without losing vapour. Heat with a water bath or electric heater, never a flame. |
| Distil off the product | The ester has the lowest boiling point (no hydrogen bonding between ester molecules). Collect the fraction over the ester’s boiling range. |
| Shake with aqueous sodium carbonate in a separating funnel | Neutralises / removes the acid impurities; the CO2 produced builds pressure, so release the pressure regularly. |
| Separate the layers | The ester is immiscible with water and less dense, so it forms the upper layer; run off and discard the aqueous layer. |
| Add anhydrous CaCl2 or MgSO4 | A drying agent: removes water. (Say “drying agent / removes water”, never “dehydrates”.) |
| Redistil, collecting over a narrow range | A sharp boiling point matching the data-book value confirms purity. |
14.8 g of butan-1-ol (Mr = 74.0) is warmed with 15.0 g of ethanoic acid (Mr = 60.0) and a concentrated sulfuric acid catalyst; 8.7 g of butyl ethanoate (Mr = 116.0) is collected. Find the percentage yield.
Step 1. Moles of each reactant:
n(butan-1-ol) = 14.874.0 = 0.200 mol n(ethanoic acid) = 15.060.0 = 0.250 mol
Step 2. Identify the limiting reagent. The equation is 1 : 1, so butan-1-ol (0.200 mol) runs out first, so it is limiting.
Step 3. Theoretical mass of ester from the limiting reagent:
0.200 × 116.0 = 23.2 g
Step 4. Percentage yield:
(8.723.2) × 100 = 37.5 %
Yields are low because esterification is an equilibrium, and more is lost in the separating funnel and distillations.
Their pleasant smells and low reactivity make esters useful as solvents, plasticisers, and in perfumes and food flavourings.
Exam questions
Ethyl ethanoate can be made by reacting ethanol with ethanoic acid in the presence of concentrated sulfuric acid.
C2H5OH + CH3COOH ⇌ CH3COOC2H5 + H2O
Method
- A mixture of ethanol, ethanoic acid, and concentrated sulfuric acid, with anti-bumping granules, is heated under reflux for 10 minutes.
- The apparatus is rearranged for distillation.
- The mixture is heated to collect the liquid that distils between 70 and 85 °C
- The distillate is placed in a separating funnel. Aqueous sodium carbonate is added, and a stopper is placed in the funnel. The mixture is shaken, releasing pressure as necessary.
- The lower aqueous layer is removed and the upper organic layer is placed in a small conical flask.
- Anhydrous calcium chloride is added to the sample in the conical flask. The flask is shaken well and left for a few minutes.
- The liquid from the flask is redistilled and the distillate is collected between 74 and 79 °C
State the role of concentrated sulfuric acid in this reaction.
Show answer
Catalyst. 1 mark
Allow “reduces Ea”. Ignore “speeds up reaction”, “provides alternative path”, “proton donor” and “dehydrating agent” on their own.
The reaction mixture is flammable. Suggest how the reaction mixture should be heated in step 1.
Show answer
Electric heater / heating mantle or (hot) water bath. 1 mark
“Not with a Bunsen / naked flame” on its own is ignored: you must say what is used. Any indication of direct heat from a Bunsen loses the mark.
State why sodium carbonate is added to the distillate in step 4. Explain why there is a build-up of pressure in the separating funnel.
Show answer
M1: to neutralise / react with / remove the acid. 1 mark
M2: carbon dioxide / gas is produced. 1 mark
Either acid (ethanoic and/or sulfuric) is allowed for M1; naming the wrong acid or the wrong gas scores nothing. “Effervescence / bubbles” is allowed for M2.
Give a reason why two layers form in the separating funnel. Suggest why ethyl ethanoate forms the upper layer.
Show answer
M1: ethyl ethanoate is immiscible with / insoluble in water. 1 mark
M2: ethyl ethanoate is less dense / has a lower density (than water). 1 mark
“Lighter” is not accepted for M2, and “different / low density” is ignored. Compare it to water.
State why anhydrous calcium chloride is added in step 6.
Show answer
To remove / absorb water: a drying agent. 1 mark
“Dehydrates” is ignored, and “to dry the reactants” or “to remove soluble impurities” is not accepted. It dries the product.
A student uses the method to prepare some ethyl ethanoate. The student adds 10.0 cm3 of ethanol (Mr = 46.0) to 5.25 g of ethanoic acid (Mr = 60.0) and obtains 5.47 g of ethyl ethanoate (Mr = 88.0). For ethanol, density = 0.790 g cm−3. Determine the limiting reagent. Calculate the percentage yield of ethyl ethanoate.
Show answer
M1: mass of ethanol = 10 × 0.790 (= 7.90 g) 1 mark
M2: n(ethanol) = 7.9046.0 (= 0.172 mol) and n(ethanoic acid) = 5.2560.0 (= 0.0875 mol) 1 mark
M3: (limiting reagent is) ethanoic acid. 1 mark
M4: max mass of ethyl ethanoate = 88.0 × 0.0875 (= 7.70 g) 1 mark
M5: % yield = (5.477.70) × 100 = 71.0 % 1 mark
70.6 to 71.1 (to a minimum of 2 s.f.) is accepted, with error carried forward at each stage. M2 needs the numbers or sums for both substances; M4 is independent of M3.
Suggest a reason why the percentage yield is not 100%.
Show answer
The reaction is an equilibrium / reversible. 1 mark
Losses during distillation / isolation / transfer, incomplete reaction, or side reactions are also allowed. “Water is also produced” is ignored.
The original question also asks you to spot two mistakes in a printed diagram of the reflux set-up (bung in the condenser; water connected the wrong way). Reflux apparatus is drawn on the required practicals page.
Ester F can be prepared from propan-2-ol and ethanoic acid. Give an equation for this reaction. Name ester F.
Show answer
M1: CH3CH(OH)CH3 + CH3COOH ⇌ CH3COOCH(CH3)2 + H2O 1 mark
M2: propan-2-yl ethanoate. 1 mark
Other valid names are allowed: 1-methylethyl ethanoate, isopropyl ethanoate, 2-propyl ethanoate. The ethanoate half must come from the acid.
Source: AQA A-Level Chemistry past papers.
Hydrolysis, fats & soap
Hydrolysis splits an ester back apart. Which products you get depends on the conditions:
- Acid hydrolysis (dilute acid, heat under reflux) is the reverse of esterification, so it is reversible and reaches an equilibrium, giving the carboxylic acid + alcohol.
- Alkaline hydrolysis (aqueous NaOH, heat under reflux) goes to completion, because the carboxylate salt formed does not react back, giving the carboxylate salt + alcohol.
CH3COOC2H5 + H2O ⇌ CH3COOH + C2H5OH
CH3COOC2H5 + NaOH → CH3COONa + C2H5OH
Fats, oils, soap & biodiesel
Vegetable oils and animal fats are triesters (triglycerides) of propane-1,2,3-triol (glycerol) with three long-chain carboxylic (fatty) acids. Their ester chemistry is the ester chemistry above, three times over:
- Alkaline hydrolysis of a fat or oil gives soap (the sodium or potassium salts of the long-chain carboxylic acids) plus glycerol. (This is saponification, the reaction soap-making has used for millennia.)
- Biodiesel is a mixture of methyl esters of long-chain carboxylic acids, made by reacting vegetable oils with methanol in the presence of a catalyst. The methanol displaces the glycerol from the ester links (a transesterification).
Write the equations for making soap and biodiesel from a triglyceride whose three R groups are identical.
Step 1. Soap (alkaline hydrolysis). Three NaOH cut the three ester links; the salt is the soap:
(RCOO)3C3H5 + 3NaOH → 3RCOO−Na+ + CH2(OH)CH(OH)CH2OH
Step 2. Biodiesel (transesterification). Three methanol swap in; no water is involved:
(RCOO)3C3H5 + 3CH3OH → 3RCOOCH3 + CH2(OH)CH(OH)CH2OH
Both release glycerol. Count your products: three salt/ester molecules per triglyceride, one glycerol.
Exam questions
Coconut oil contains a triester with three identical R groups. This triester reacts with potassium hydroxide.
Complete the equation by drawing the structure of the other product of this reaction in the box. Name the type of compound shown by the formula RCOOK. Give one use for this type of compound.
Show answer
CH2OHCH(OH)CH2OH 1 mark
(Potassium) carboxylate salt. 1 mark
Soap. 1 mark
“Fatty acid salt” / “salt of a carboxylic acid” is allowed for the type; “detergent / surfactant” is allowed for the use.
The triester in coconut oil has a relative molecular mass, Mr = 638.0. In the equation shown at the start of this question, R represents an alkyl group that can be written as CH3(CH2)n. Deduce the value of n in CH3(CH2)n. Show your working.
Show answer
M1: 638 = 173 + 3(15 + 14n). The Mr of the ester fragment (the backbone plus three COO groups) is 173 1 mark
M2: 638 − 173 − 3(15) = 420, then divide by 3 × 14 = 42 1 mark
M3: n = 10 (n must be an integer). 1 mark
A 1.450 g sample of coconut oil is heated with 0.421 g of KOH in aqueous ethanol until all of the triester is hydrolysed. The mixture is cooled. The remaining KOH is neutralised by exactly 15.65 cm3 of 0.100 mol dm−3 HCl. Calculate the percentage by mass of the triester (Mr = 638.0) in the coconut oil.
Show answer
M1: n(HCl) = 0.100 × 0.01565 = 1.565 × 10−3 mol 1 mark
M2: initial n(KOH) = 0.42156.1 = 7.50 × 10−3 mol 1 mark
M3: n(KOH) used = M2 − M1 = 5.939 × 10−3 mol 1 mark
M4: n(ester) = M3 divided by 3 = 1.980 × 10−3 mol 1 mark
M5: mass of ester = 1.980 × 10−3 × 638 = 1.263 g 1 mark
M6: % by mass = (1.2631.450) × 100 = 87.1 % 1 mark
87.0 to 87.1 is allowed (2 s.f. accepted); an answer above 100% cannot score M6. This is a back titration: the HCl counts the KOH left over, and each triester molecule used three KOH.
Suggest why aqueous ethanol is a suitable solvent when heating the coconut oil with KOH. Give a safety precaution used when heating the mixture. Justify your choice.
Show answer
M1: it dissolves both the oil and the KOH (a mutual solvent), so the reactants are miscible. 1 mark
M2: use a water bath (or electric heater / mantle / sand bath) for heating. 1 mark
M3: ethanol is flammable. This prevents the risk of fire. 1 mark
The precaution must be linked to heating, and the justification must match the precaution given.
The diagram below shows an incomplete mechanism for the reaction of an ester with aqueous sodium hydroxide.
Add three curly arrows to complete the mechanism in the diagram.
Show answer
M1: arrow from the C=O bond to the O. 1 mark
M2: arrow from the correct C–O bond to the O. 1 mark
M3: arrow from the O–H bond to the O. 1 mark
You are not expected to recall this mechanism: the question supplies it, and the skill is placing arrows precisely: each starts at a bond midpoint and lands on the oxygen gaining the electrons.
Name the type of reaction shown in the diagram.
Show answer
(Alkaline / base) hydrolysis. 1 mark
Deduce the role of the CH3O− ion in step 3 shown in the diagram.
Show answer
Base. 1 mark
“Proton acceptor” is allowed: in step 3 it takes the H+ from the carboxylic acid.
A triester in vegetable oil reacts with sodium hydroxide in a similar way. Give a use for a product of this reaction.
Show answer
Soap. 1 mark
“Soap only”: the mark scheme accepts nothing else here.
Source: AQA A-Level Chemistry past papers.
Acyl chlorides & acid anhydrides
Replace the –OH of a carboxylic acid with –Cl and you get an acyl chloride (e.g. ethanoyl chloride, CH3COCl); join two acid molecules with loss of water and you get an acid anhydride (e.g. ethanoic anhydride, (CH3CO)2O). Both are acylating agents: they hand the acyl group CH3CO– to a nucleophile, far faster than the acid itself would. Acyl chlorides are the more reactive of the two. The third derivative to recognise is the amide, RCONH2, the nitrogen-containing product below.
The four nucleophiles: one pattern
Each of the four common nucleophiles attacks the acyl group and displaces the leaving group. The product is set by the nucleophile:
| Nucleophile | Organic product | Equation |
|---|---|---|
| Water | carboxylic acid | CH3COCl + H2O → CH3COOH + HCl |
| Alcohol (e.g. ethanol) | ester | CH3COCl + C2H5OH → CH3COOC2H5 + HCl |
| Ammonia | amide | CH3COCl + 2NH3 → CH3CONH2 + NH4Cl |
| Primary amine (e.g. CH3NH2) | N-substituted amide | CH3COCl + 2CH3NH2 → CH3CONHCH3 + CH3NH3Cl |
| Aromatic amine: phenylamine, C6H5NH2 | N-phenyl-substituted amide | CH3COCl + 2C6H5NH2 → CH3CONHC6H5 + C6H5NH3Cl |
Phenylamine behaves exactly like any other primary amine here. Its nitrogen lone pair attacks the carbonyl carbon and the product is the N-substituted amide N-phenylethanamide. It is worth knowing by name because it is the aromatic amine AQA use, and because the amide it forms is a solid that can be purified by recrystallisation and identified by its melting point.
An acid anhydride does the same four reactions, just more gently. Its leaving group is the ethanoate ion, so the by-product is ethanoic acid instead of HCl.
One molecule of the nucleophile forms the product; a second mops up the HCl released (as NH4+Cl− or the alkylammonium salt). Balance the equation with 2NH3 / 2 amine, not one.
The mechanism: nucleophilic addition–elimination
Every one of those reactions goes by the same two-stage mechanism. The carbonyl carbon is Cδ+ (electron-poor), so:
- Addition: the nucleophile’s lone pair attacks Cδ+; the C=O π bond breaks onto the oxygen, giving a tetrahedral intermediate with the negative charge on that oxygen.
- Elimination: the C=O reforms and pushes out the leaving group (Cl− from an acyl chloride, or the carboxylate from an anhydride); the nucleophile then loses H+ to give the neutral product.
Interactive: one mechanism, four nucleophiles
Ethanoyl chloride reacts with methylamine. Name the mechanism and the organic product, and write the balanced equation.
Step 1. The mechanism is nucleophilic addition–elimination (the amine’s nitrogen lone pair attacks Cδ+).
Step 2. Build the product. The nucleophile replaces the Cl: CH3CO– joins –NHCH3, giving CH3CONHCH3, an amide with a methyl on the nitrogen, so it is named N-methylethanamide.
Step 3. Balance with the second equivalent of amine mopping up the HCl:
CH3COCl + 2CH3NH2 → CH3CONHCH3 + CH3NH3Cl
Observations: how acyl chlorides give themselves away
Acyl chlorides react violently with cold water, giving off misty fumes of HCl. Add aqueous silver nitrate and the chloride released gives an immediate white precipitate of AgCl. A chloroalkane, by contrast, barely reacts at room temperature. Acid anhydrides react with water too, but slowly and controllably, and produce no HCl.
Aspirin & Required Practical 10
Aspirin is made by acylating salicylic acid (2-hydroxybenzenecarboxylic acid). Its phenol –OH is the nucleophile, and industry uses ethanoic anhydride as the acylating agent:
salicylic acid + (CH3CO)2O → aspirin + CH3COOH
Ethanoic anhydride is less corrosive than ethanoyl chloride, is less vulnerable to hydrolysis, and does not form HCl. The only by-product is (weak) ethanoic acid.
Those three are the creditworthy reasons. AQA mark schemes have explicitly ignored answers about cost, safety in general, volatility or “less violent”. Give a specific chemical advantage, not a vague one.
The solid half is aspirin: made by the acylation above, then purified by recrystallisation:
- Dissolve the impure solid in the minimum volume of hot solvent. The minimum volume means the solution is saturated, so crystals form when it cools. Heat on a water bath.
- Filter hot (through a warmed funnel). This removes the insoluble impurities; hot, so the product stays dissolved instead of crystallising in the funnel.
- Cool the filtrate in an ice bath. The pure product crystallises out of the saturated solution.
- Filter under reduced pressure (Büchner funnel and side-arm flask): fast, and pulls the impure solution away from the crystals.
- Wash the crystals with a little cold solvent, then leave them to dry.
- Check purity by melting point: pure aspirin melts sharply at the data-book value; an impure sample melts lower and over a range.
| Impurity | Removed at | How |
|---|---|---|
| Insoluble | the hot filtration | never dissolved, so trapped on the filter paper while the product passes through in solution |
| Soluble | the cold filtration + cold wash | stay dissolved in the filtrate: there is too little of them ever to saturate the cooled solution, so they never crystallise |
The liquid half of the practical is the ester preparation in the esters section. Full method notes on the required practicals page.
1 · impure solid
The product plus both kinds of impurity: insoluble (filled) and soluble (rings).
2 · dissolve · minimum hot solvent
Hot, so everything soluble dissolves; minimum volume, so the solution is saturated. That is what makes crystals form on cooling. Heat on a water bath.
3 · filter hot
Insoluble impurities removed: they never dissolved, so they are trapped on the paper. Filter hot so no product crystallises in the funnel.
4 · cool in an ice bath
The product crystallises out of the saturated solution. The soluble impurities stay dissolved, too little of them ever to saturate the solution.
5 · filter under reduced pressure
Soluble impurities removed: they leave in the filtrate. Wash the crystals with a little cold solvent: cold, so the product does not redissolve.
6 · dry
Leave to dry, then check the melting point: sharp and at the data-book value = pure.
Exam questions
Silver nitrate solution can be used to distinguish between propanoyl chloride and 1-chloropropane. Give the observations you would expect when a few drops of silver nitrate solution are added to separate samples of propanoyl chloride and 1-chloropropane.
Show answer
M1. Propanoyl chloride: misty / white / steamy fumes or an (immediate) white precipitate forms. 1 mark
M2. 1-chloropropane: no visible change (or a white precipitate forms only slowly). 1 mark
The contrast is the point: the C–Cl bond in the acyl chloride is attacked at once by the water in the reagent, releasing Cl−; the chloroalkane needs prolonged warming.
Which compound reacts with warm dilute aqueous sodium hydroxide?
Aspirin can be produced by reacting salicylic acid with ethanoic anhydride. An incomplete method to determine the yield of aspirin is shown.
- Add about 6 g of salicylic acid to a weighing boat.
- Place the weighing boat on a 2 decimal place balance and record the mass.
- Tip the salicylic acid into a 100 cm3 conical flask.
- ______________________________________________
- Add 10 cm3 of ethanoic anhydride to the conical flask and swirl.
- Add 5 drops of concentrated phosphoric acid.
- Warm the flask for 20 minutes.
- Add ice-cold water to the reaction mixture and place the flask in an ice bath.
- Filter off the crude aspirin from the mixture and leave it to dry.
- Weigh the crude aspirin and calculate the yield.
Describe the instruction that is missing from step 4 of the method. Justify why this step is necessary.
Show answer
M1: (re)weigh the empty boat. 1 mark
M2: in order to calculate the (exact) mass of salicylic acid added to the reaction mixture. 1 mark
Suggest a suitable piece of apparatus to measure out the ethanoic anhydride in step 5.
Show answer
A 10 cm3 measuring cylinder, or a 10 cm3 pipette, burette / graduated pipette, or a 10 cm3 syringe. 1 mark
Identify a hazard of using concentrated phosphoric acid in step 6.
Show answer
Corrosive. 1 mark
“Skin burn / permanent eye damage” is allowed; “irritant” and “toxic” are ignored.
Complete the equation for the reaction of salicylic acid with ethanoic anhydride to produce aspirin.
Show answer
Left-hand side: + (CH3CO)2O Right-hand side: + CH3COOH 1 mark
A 6.01 g sample of salicylic acid (Mr = 138.0) is reacted with 10.5 cm3 of ethanoic anhydride (Mr = 102.0). In the reaction the yield of aspirin is 84.1%. The density of ethanoic anhydride is 1.08 g cm−3. Show by calculation which reagent is in excess. Calculate the mass, in g, of aspirin (Mr = 180.0) produced.
Show answer
M1: n(salicylic acid) = 6.01138 = 4.36 × 10−2 mol 1 mark
M2: mass of (CH3CO)2O = 10.5 × 1.08 = 11.34 g 1 mark
M3: n((CH3CO)2O) = 11.34102 = 1.11 × 10−1 mol 1 mark
M4: (CH3CO)2O is in excess. 1 mark
M5: mass of aspirin = 4.36 × 10−2 × 0.841 × 180 = 6.59 g 1 mark
You must show and state that the ethanoic anhydride is in excess; 2 s.f. or more is allowed for the final mass.
Suggest two ways in which the melting point of the crude aspirin collected in step 9 would differ from the melting point of pure aspirin.
Show answer
M1: the value is lower. 1 mark
M2: it melts over a range of values / the melting point is not sharp. 1 mark
The crude aspirin can be purified by recrystallisation using hot ethanol (boiling point = 78 °C) as the solvent. Describe two important precautions when heating the mixture of ethanol and crude aspirin.
Show answer
M1: (ethanol is flammable so) use a water bath to heat / do not use a Bunsen burner. 1 mark
M2: heat to a temperature below the boiling point (so the ethanol does not boil away) / use the minimum volume of solvent. 1 mark
M1 must give the practical step. Just stating the hazard scores nothing.
The pure aspirin is filtered under reduced pressure. A small amount of cold ethanol is then poured through the Buchner funnel. Explain the purpose of adding a small amount of cold ethanol.
Show answer
To remove any soluble impurities (washing them away). The ethanol is cold and little is used so that the aspirin itself does not dissolve. 1 mark
A sample of the crude aspirin is kept to compare with the purified aspirin. Describe one difference in appearance you would expect to see between these two solid samples.
Show answer
The pure product will have (larger) crystals / needle-like crystals / be lighter in colour. 1 mark
Whiter, less grey, more crystalline, less powdery or shinier are all allowed, but the difference must be tied to the pure (or crude) product.
Aspirin is produced from 2-hydroxybenzenecarboxylic acid by reaction with ethanoic anhydride in the presence of concentrated phosphoric acid. The flask is heated to 85 °C for 10 minutes.
Aspirin can also be produced by reacting 2-hydroxybenzenecarboxylic acid with ethanoyl chloride. State why ethanoic anhydride is preferred to ethanoyl chloride for this preparation.
Show answer
Ethanoic anhydride is less / not corrosive; or it does not form a strong acid / HCl (it only forms weak ethanoic acid); or it is less / not vulnerable to hydrolysis. 1 mark
The reverse argument about ethanoyl chloride is allowed. Cost, volatility, “safer / less toxic” and “less exothermic / violent / vigorous” are all ignored.
Give the name of the mechanism for the reaction of 2-hydroxybenzenecarboxylic acid with ethanoic anhydride.
Show answer
(Nucleophilic) addition–elimination. 1 mark
“Esterification” and “acylation” are ignored: the question asks for the mechanism.
Suggest the role of the concentrated phosphoric acid.
Show answer
Catalyst. 1 mark
“Speeds up the reaction / lowers activation energy” is allowed; “proton donor” is ignored.
Suggest why reflux is not essential when the flask is heated to 85 °C for 10 minutes.
Show answer
The boiling points (of the reactants and products) are above 85 °C: nothing would boil / no volatile reagents. 1 mark
Source: AQA A-Level Chemistry past papers.
The derivatives map
Pull the whole topic together around the acyl group. Reactivity runs acyl chloride > acid anhydride > ester > carboxylic acid/amide. The better the leaving group, the more reactive the acylating agent. Reading the map is the exam skill: name the reagent that makes each product, and vice versa.
- Name the mechanism nucleophilic addition–elimination, not substitution, not “esterification” (mark schemes ignore both when the mechanism is asked for).
- Examiner reports note that “hydrolysis” itself is poorly known. Be able to define it (splitting with water) and to give the different products of the acid route (acid + alcohol, reversible) and the alkaline route (carboxylate salt + alcohol, one-way).
- In purification steps, say what each stage removes and how: washing (removes soluble impurities) is not drying (removes water), and a drying agent never “dehydrates”.
- Purity of aspirin is checked by melting point, never boiling point: pure = sharp and matches the data-book value; impure = lower and over a range.
- Remember the second equivalent of NH3 / amine that removes the HCl. The equation must balance.
- For an acyl chloride and water: violent reaction, misty HCl fumes, and an immediate white precipitate with AgNO3.
| Add this nucleophile | Get this product | Product type |
|---|---|---|
| Water | CH3COOH | carboxylic acid |
| Alcohol | CH3COOR | ester |
| Ammonia | CH3CONH2 | amide |
| Primary amine | CH3CONHR | N-substituted amide |
Capstone quiz: six past-paper questions
Six real AQA multiple-choice questions on this topic, in the style that opens Paper 3. Pick one answer each. The reasoning appears once you commit.
Which statement about (CH3)2CHCH2COOH is correct?
Which compound is formed when phenyl benzenecarboxylate is hydrolysed under acidic conditions?
Which compound is an amide?
Compound Y has the structural formula CH3COOCH2CH(CH3)2. Which compound is a position isomer of Y?
Which reaction involves addition–elimination?
Which compound forms a white precipitate when added to aqueous silver nitrate?
Source: AQA A-Level Chemistry past papers.
The acylation reactions return in organic synthesis (3.3.14) as the highest-yield route to esters and amides, and the amines that attack the acyl group get their own page at 3.3.11.
- Carboxylic acids (–COOH) are weak acids: they liberate CO2 from carbonates (the distinguishing test) and form carboxylate salts with metals and bases; they hydrogen-bond, so the short-chain acids are soluble.
- Esterification: carboxylic acid + alcohol ⇌ ester + water (conc H2SO4 catalyst, reversible). Esters are named alkyl alkanoate; used as solvents, plasticisers, perfumes and food flavourings.
- Ester hydrolysis: acid hydrolysis is reversible (→ acid + alcohol); alkaline hydrolysis goes to completion (→ carboxylate salt + alcohol).
- Fats and oils are triesters of glycerol (propane-1,2,3-triol). Alkaline hydrolysis gives soap (salts of long-chain carboxylic acids) + glycerol; biodiesel is the methyl esters, made with methanol + catalyst.
- Acyl chlorides (RCOCl) and acid anhydrides ((RCO)2O) react with the four nucleophiles by nucleophilic addition–elimination: + H2O → acid; + alcohol → ester; + 2NH3 → amide; + 2 amine → N-substituted amide.
- Aspirin (Required practical 10) is made from salicylic acid + ethanoic anhydride: less corrosive, less easily hydrolysed and no HCl formed; purify by recrystallisation and check purity by melting point.