Paper 42, worked through. Cambridge IGCSE Chemistry, May/June 2026: worked solutions

Eighty marks in an hour and a quarter: just under a minute a mark. The Periodic Table is on the back page and several answers come straight from it. Read the command word: "state" and "name" want a word or two, "explain" wants a reason, and when a question asks for a name, give the name and not the formula.

1. 1(a)(i)Finding an element from its number of protons.[1]
   Show how to do itHide the working

   1. The number of protons is the proton (atomic) number, which is the smaller number in each box of the Periodic Table.
   2. Proton number 11 is sodium, and sodium is in Group I, so it fits the question.

   AnswerNa

   Write symbols exactly as the table prints them: a capital letter, then a small one. NA or na is not sodium.

   Revise this: Inside the atom
2. 1(a)(ii)Finding an element from its number of electrons.[1]
   Show how to do itHide the working

   1. An atom is neutral, so it has as many electrons as protons. 35 electrons means proton number 35.
   2. Proton number 35 is bromine, in Group VII.

   AnswerBr

   Revise this: Inside the atom
3. 1(a)(iii)Finding which element has molecules with a relative molecular mass of 254.[1]
   Show how to do itHide the working

   1. Group I elements are metals: they don't form molecules. Group VII elements do, and their molecules are pairs of atoms: F₂, Cl₂, Br₂, I₂.
   2. Two atoms have a mass of 254, so one has 254 ÷ 2 = 127.
   3. The element with a relative atomic mass of 127 is iodine.

   AnswerI

   The word "molecules" is the clue that this is a halogen, and that you must halve the 254 before you look for it in the table.

   Revise this: Relative formula mass
4. 1(b)(i)The element used to test for unsaturated hydrocarbons.[1]
   Show how to do itHide the working

   1. Unsaturated hydrocarbons, such as alkenes, have a carbon–carbon double bond.
   2. The test is aqueous bromine (bromine water). An unsaturated compound turns it from orange to colourless. A saturated one leaves it orange.

   AnswerBr

   Revise this: Alkenes and cracking
5. 1(b)(ii)The Group I element with a lilac flame.[1]
   Show how to do itHide the working

   1. Flame colours for Group I: lithium is red, sodium is yellow, potassium is lilac.

   AnswerK

   Learn the three as a set. Copper, a common extra, is blue-green.

   Revise this: Tests for ions and gases
6. 1(b)(iii)The halogen that can push bromine out of a bromide.[1]
   Show how to do itHide the working

   1. A more reactive halogen displaces a less reactive one from a solution of its salt.
   2. Reactivity falls as you go down Group VII, so a halogen above bromine is needed. Chlorine is the one you meet in the syllabus: chlorine + bromide ions → chloride ions + bromine.

   AnswerCl

   Iodine is below bromine, so it cannot do it. The mark scheme also accepts fluorine, which is above both.

   Revise this: Group 7: the halogens
7. 2(a)(i)Filling in a table about the strong bonding in four substances: what kind it is, and which particles are being pulled together.[7]
   Show how to do itHide the working

   1. Bromine is a non-metal, so its atoms are joined by covalent bonds. In a Br₂ molecule the two particles held together are two bromine atoms, Br and Br, sharing a pair of electrons. The type of particle is atoms.
   2. Magnesium is a metal, so the bonding is metallic. The row already gives you "and e⁻" and "ions and electrons", so the missing particle is the positive ion. Magnesium is in Group II, so the ion is Mg²⁺.
   3. Silicon(IV) oxide is covalent, as the table says. The particles joined are silicon atoms and oxygen atoms: Si and O. The type of particle is atoms.

   AnswerBromine: covalent; Br (and Br); atoms. Magnesium: metallic; Mg²⁺. Silicon(IV) oxide: Si and O; atoms.

   Use the filled-in sodium chloride column as your model: it shows the level of detail wanted. Ionic bonding attracts oppositely charged ions, covalent bonding holds atoms together with shared electrons, metallic bonding attracts positive ions to delocalised electrons. Give the ion its charge: Mg alone does not score.

   Revise this: Covalent bonding and simple molecules
8. 2(a)(ii)What has to be overcome for bromine to boil.[1]
   Show how to do itHide the working

   1. Bromine is made of small Br₂ molecules. Boiling separates the molecules from each other. It doesn't split the molecules up.
   2. The attractions between one molecule and the next are weak. They are called intermolecular forces.

   AnswerIntermolecular forces.

   The covalent bonds inside each molecule are strong and do not break on boiling. Mixing those two up is the classic mistake, and writing "covalent bonds" here scores nothing.

   Revise this: Covalent bonding and simple molecules
9. 2(b)(i)Describing how the ions are arranged in sodium chloride.[2]
   Show how to do itHide the working

   1. One mark is for the pattern: the ions are in a regular arrangement in which the two kinds alternate.
   2. The other mark is for saying what alternates: positive ions and negative ions (Na⁺ and Cl⁻).

   AnswerA regular arrangement of alternating positive and negative ions.

   "Lattice" is given in the question, so it earns nothing on its own. Two marks, two ideas: regular and alternating, and oppositely charged ions.

   Revise this: Ions and ionic bonding
10. 2(b)(ii)Naming another giant covalent substance that behaves like silicon(IV) oxide.[1]
    Show how to do itHide the working

    1. Silicon(IV) oxide is hard, has a very high melting point and does not conduct electricity, because every atom is held to its neighbours by strong covalent bonds.
    2. Diamond is the same: each carbon atom is bonded to four others throughout the structure.

    AnswerDiamond.

    Graphite is giant covalent too, but it is soft and conducts, so its properties are not similar. The mark scheme rejects it.

    Revise this: Diamond, graphite and giant structures
11. 3(a)Completing the equation for a carbonate reacting with an acid.[1]
    Show how to do itHide the working

    1. Acid + carbonate → salt + water + carbon dioxide. The salt, ZnSO₄, is already there.
    2. So the two missing products are H₂O and CO₂. Count the atoms and the equation balances as it stands.

    AnswerH₂O and CO₂.

    Revise this: Making salts
12. 3(b)Naming another zinc compound that makes zinc sulfate with dilute sulfuric acid.[1]
    Show how to do itHide the working

    1. An acid also forms a salt with a metal oxide or a metal hydroxide. Both are bases.
    2. So zinc oxide or zinc hydroxide would work, each added in excess in the same way.

    AnswerZinc oxide (or zinc hydroxide).

    The question asks for a compound, so zinc metal is not accepted, although it would react.

    Revise this: Making salts
13. 3(c)Another sign that all the acid has reacted.[1]
    Show how to do itHide the working

    1. While acid is left, the carbonate keeps reacting and carbon dioxide bubbles off.
    2. When the acid is used up, the bubbling stops.

    AnswerThe fizzing stops (no more bubbles).

    Say that it stops. "Effervescence" on its own is the sign the reaction is still going, and the mark scheme rejects it.

    Revise this: Making salts
14. 3(d)(i)The word for a solution that can hold no more solute.[1]
    Show how to do itHide the working

    1. A solution with the most solute that will dissolve at that temperature is called saturated.

    AnswerSaturated.

    Revise this: Measuring and apparatus
15. 3(d)(ii)Working out how far a solution can evaporate before crystals appear.[1]
    Show how to do itHide the working

    1. Crystals appear when the solution becomes saturated. Saturated means 60 g in every 100 cm³.
    2. Solution A holds 30 g, which is half of 60 g. So it is saturated when the volume is half of 100 cm³.

    Answer50 cm³

    Evaporation removes water, not zinc sulfate. The 30 g stays put, and the volume shrinks until that 30 g is all the water can hold.

    Revise this: Measuring and apparatus
16. 3(d)(iii)What happens to the zinc sulfate when the solution is evaporated down to 20 cm³.[1]
    Show how to do itHide the working

    1. At 20 cm³ the solution is saturated, and 20 cm³ can hold 60 × 20 ÷ 100 = 12 g.
    2. The solution started with 30 g, so the other 30 − 12 = 18 g has come out as crystals.

    Answer12 g (the mark scheme accepts 18 g as well).

    The mark scheme says the wording can be read two ways: the mass still dissolved in solution B (12 g) or the mass of crystals formed in getting there (18 g). Either scored. Working out both, as here, is the safe habit.

    Revise this: Measuring and apparatus
17. 3(e)(i)Choosing two soluble salts that will make insoluble lead(II) sulfate.[2]
    Show how to do itHide the working

    1. To make an insoluble salt you mix two solutions: one supplies the positive ion, the other the negative ion.
    2. All nitrates are soluble, so lead(II) nitrate supplies the lead ions.
    3. Sodium, potassium and ammonium salts are all soluble, so sodium sulfate supplies the sulfate ions.

    AnswerLead(II) nitrate and sodium sulfate.

    Both must be soluble, and the second must be a salt: sulfuric acid is not accepted. Don't pick barium sulfate or calcium sulfate either, since they are insoluble themselves.

    Revise this: Making salts
18. 3(e)(ii)The name of the method.[1]
    Show how to do itHide the working

    1. Two solutions are mixed and a solid forms. A solid made this way is a precipitate.

    AnswerPrecipitation.

    Revise this: Making salts
19. 3(e)(iii)Getting a pure, dry sample of an insoluble salt out of the mixture.[3]
    Show how to do itHide the working

    1. Filter. The lead(II) sulfate is the solid, so it stays in the filter paper as the residue.
    2. Wash the residue with distilled water, to rinse away the solution of the other salt.
    3. Dry it: press it between filter papers or leave it in a warm place.

    AnswerFilter, wash the residue with water, then dry it.

    Don't crystallise. That is how you get a soluble salt out of its solution. Here the salt is already a solid, and the mark scheme gives nothing for evaporating before you filter.

    Revise this: Making salts
20. 3(e)(iv)The ionic equation for the precipitation, with state symbols.[2]
    Show how to do itHide the working

    1. An ionic equation shows only the ions that join up. The sodium and nitrate ions stay in solution unchanged, so leave them out.
    2. Lead ions and sulfate ions come together: Pb²⁺ + SO₄²⁻ → PbSO₄.
    3. The ions are dissolved, so they are (aq). The product is the precipitate, so it is (s).

    AnswerPb²⁺(aq) + SO₄²⁻(aq) → PbSO₄(s)

    One mark is for the equation and one for the state symbols. Don't put charges on the PbSO₄: once the ions have joined, the formula is written without them.

    Revise this: Making salts
21. 3(e)(v)Another insoluble sulfate.[1]
    Show how to do itHide the working

    1. Most sulfates are soluble. The exceptions to learn are barium sulfate, calcium sulfate and lead(II) sulfate.

    AnswerBarium sulfate.

    This is the same reaction as the test for sulfate ions: barium ions give a white precipitate of barium sulfate.

    Revise this: Making salts
22. 4(a)Two physical differences between transition elements and Group I metals.[2]
    Show how to do itHide the working

    1. Group I metals are soft, melt at low temperatures and have low densities: lithium, sodium and potassium float on water.
    2. Transition elements are the opposite: high melting points, high densities, and they are hard and strong.

    AnswerThey have higher melting points and higher densities (harder or stronger also scores).

    The question says physical. Coloured compounds, acting as catalysts and having more than one oxidation number are true, but they are chemical properties and score nothing here.

    Revise this: The transition elements
23. 4(b)Writing formulas from names that include an oxidation number.[2]
    Show how to do itHide the working

    1. The Roman numeral is the charge on the metal ion. Copper(I) is Cu⁺ and nickel(II) is Ni²⁺.
    2. Copper(I) oxide: oxide is O²⁻, so it takes two Cu⁺ to balance one O²⁻. That gives Cu₂O.
    3. Nickel(II) nitrate: nitrate is NO₃⁻, so it takes two of them to balance one Ni²⁺. That gives Ni(NO₃)₂.

    AnswerCu₂O and Ni(NO₃)₂

    The 2 after the bracket multiplies the whole nitrate. NiNO₃₂ or Ni(NO₃)2 without the bracket means something else. Leave the Roman numerals out of the formula.

    Revise this: Formulae and equations
24. 4(c)(i)Why turning chromate into dichromate is not a redox reaction.[1]
    Show how to do itHide the working

    1. In a redox reaction something is oxidised and something is reduced, which shows up as a change in oxidation number.
    2. Work out chromium in each ion. In CrO₄²⁻: Cr + 4 × (−2) = −2, so Cr is +6. In Cr₂O₇²⁻: 2Cr + 7 × (−2) = −2, so Cr is +6 again.
    3. Hydrogen stays at +1 and oxygen at −2 as well.

    AnswerNo oxidation number changes.

    Two different chromium ions looks like a change, which is why it is worth doing the sum.

    Revise this: Oxidation and reduction
25. 4(c)(ii)What is true about rates and concentrations at equilibrium.[2]
    Show how to do itHide the working

    1. Rates: the forward reaction and the reverse reaction are going at the same rate.
    2. Concentrations: because each substance is being made as fast as it is used up, the concentrations stop changing. They stay constant.

    AnswerThe two rates are equal, and the concentrations stay constant.

    Constant does not mean equal. The concentrations of reactants and products are usually different from each other, and saying they are equal is rejected.

    Revise this: Reversible reactions and equilibrium
26. 4(c)(iii)Why pressure does not shift this equilibrium.[1]
    Show how to do itHide the working

    1. Changing the pressure only moves an equilibrium when gases are involved, because only gases can be squeezed into a smaller space.
    2. Look at the state symbols: everything is (aq) or (l). There are no gases.

    AnswerThere are no gases in the equilibrium.

    Revise this: Reversible reactions and equilibrium
27. 4(c)(iv)Why adding alkali pushes the equilibrium to the left.[2]
    Show how to do itHide the working

    1. An alkali contains OH⁻ ions. They react with the H⁺ ions: H⁺ + OH⁻ → H₂O.
    2. That lowers the concentration of H⁺, which is on the left of the equation. The equilibrium moves in the direction that replaces what was removed, so it moves left, and dichromate turns back into chromate.

    AnswerOH⁻ ions react with H⁺ ions, so the concentration of H⁺ falls and the equilibrium moves left to replace them.

    Two marks, two steps: what the alkali does, then what the equilibrium does about it. An equilibrium always moves to oppose the change.

    Revise this: Reversible reactions and equilibrium
28. 4(d)(i)The alloy of copper and zinc.[1]
    Show how to do itHide the working

    1. An alloy is a mixture of a metal with other elements. Copper mixed with zinc is brass.

    AnswerBrass.

    Bronze is copper and tin. The two are easy to swap.

    Revise this: Alloys
29. 4(d)(ii)The steel made from iron, chromium, nickel and one non-metal.[2]
    Show how to do itHide the working

    1. Iron mixed with chromium and nickel resists rusting: that is stainless steel.
    2. All steels are iron with a little carbon in them. Carbon is the non-metal.

    AnswerStainless steel; carbon.

    Revise this: Alloys
30. 5(a)(i)Where the hydrogen and the nitrogen for the Haber process come from.[2]
    Show how to do itHide the working

    1. Hydrogen is made by reacting methane (natural gas) with steam.
    2. Nitrogen is taken from the air, which is 78% nitrogen.

    AnswerHydrogen: methane (natural gas). Nitrogen: air.

    Revise this: Making ammonia and sulfuric acid
31. 5(a)(ii)The equation for making ammonia.[2]
    Show how to do itHide the working

    1. Nitrogen and hydrogen are both made of two-atom molecules, and ammonia is NH₃: N₂ + H₂ ⇌ NH₃.
    2. Balance it. Two nitrogen atoms make 2NH₃, and that needs six hydrogen atoms, which is 3H₂.

    AnswerN₂ + 3H₂ ⇌ 2NH₃

    One mark is for the right formulas and one for balancing them. The reaction is reversible, so use the two-way arrow.

    Revise this: Making ammonia and sulfuric acid
32. 5(a)(iii)The pressure used in the Haber process, in kilopascals.[1]
    Show how to do itHide the working

    1. The conditions to learn: 450 °C, 20 000 kPa (200 atmospheres) and an iron catalyst.

    Answer20 000 kPa

    The unit here is kPa, so the number is 20 000, not 200. The mark scheme allows anything from 15 000 to 25 000.

    Revise this: Making ammonia and sulfuric acid
33. 5(b)(i)What aqueous ammonia does to a solution of chromium(III) ions.[2]
    Show how to do itHide the working

    1. Aqueous ammonia is alkaline, so it contains hydroxide ions. Metal ions form their hydroxides, and most metal hydroxides are insoluble.
    2. Chromium(III) ions give a green precipitate. It is chromium(III) hydroxide, Cr(OH)₃.

    AnswerGreen; chromium(III) hydroxide.

    The colours come from the table of tests in the syllabus: chromium(III) green, copper(II) light blue, iron(II) green, iron(III) red-brown, and white for aluminium, calcium and zinc.

    Revise this: Tests for ions and gases
34. 5(b)(ii)Why the order of mixing zinc ions and aqueous ammonia changes what you see.[2]
    Show how to do itHide the working

    1. A few drops of ammonia into zinc ions: zinc ions react with the hydroxide ions in the ammonia and form zinc hydroxide, which is insoluble. That is the white precipitate.
    2. A few drops of zinc ions into ammonia: now the ammonia is in excess. Zinc hydroxide dissolves in excess aqueous ammonia, so any precipitate dissolves at once and none is seen.

    AnswerFirst: insoluble zinc hydroxide forms. Second: the ammonia is in excess, and zinc hydroxide dissolves in excess aqueous ammonia.

    This is the usual test run backwards. In the normal test you add ammonia a little at a time and see "white precipitate, soluble in excess". Here the excess is there from the start.

    Revise this: Tests for ions and gases
35. 5(c)(i)Reading "exothermic" and "reversible" off an equation.[2]
    Show how to do itHide the working

    1. Exothermic: the enthalpy change, ΔH, is negative. Energy is given out to the surroundings.
    2. Reversible: the equation is written with the two-way arrow, ⇌, between the two sides.

    AnswerExothermic: ΔH is negative. Reversible: the ⇌ arrow.

    Say what is negative. "Because it is −870" does not score unless you tie it to ΔH or the enthalpy change.

    Revise this: Energy changes in reactions
36. 5(c)(ii)Finishing a reaction profile: labelling the two levels, and adding arrows for the activation energy and the enthalpy change.[3]
    Show how to do itHide the working

    1. The profile starts at the reactants and ends at the products. So NH₃ + 3F₂ goes on the higher line at the left and NF₃ + 3HF on the lower line at the right. Lower is right for an exothermic reaction.
    2. Activation energy, Eₐ: an arrow pointing up, from the level of the reactants to the top of the hump.
    3. Enthalpy change, ΔH: an arrow pointing down, from the level of the reactants to the level of the products.

    AnswerReactants on the left line, products on the right line; Eₐ arrow up from the reactant level to the peak; ΔH arrow down from the reactant level to the product level.

    Both arrows start at the reactant level, and both are vertical. Give the ΔH arrow one arrowhead, pointing down. The question asks for formulas, so the words "reactants" and "products" are not enough.

    Revise this: Energy changes in reactions
37. 5(c)(iii)Using bond energies and ΔH to find the energy of a bond that is not in the table.[4]
    Show how to do itHide the working

    1. Bonds broken, in the reactants: NH₃ has three N–H bonds and there are three F–F bonds. 3 × 390 + 3 × 150 = 1170 + 450 = 1620 kJ.
    2. ΔH = energy to break bonds − energy released making bonds. So −870 = 1620 − energy released, and the energy released is 1620 + 870 = 2490 kJ.
    3. The products are NF₃, with three N–F bonds, and three HF molecules, with one H–F bond each. The N–F bonds account for 3 × 270 = 810 kJ, which leaves 2490 − 810 = 1680 kJ for the three H–F bonds.
    4. One H–F bond: 1680 ÷ 3 = 560 kJ/mol.

    Answer1620 kJ; 2490 kJ; 560 kJ/mol

    In an exothermic reaction more energy is released making bonds than is taken in breaking them, so the second number must be bigger than the first. If yours is 750, you subtracted the 870 when you should have added it. Count the bonds from the drawn-out structures, not from the formulas.

    Revise this: Energy changes in reactions
38. 5(d)Completing a dot-and-cross diagram for nitrogen trifluoride.[3]
    Show how to do itHide the working

    1. Each N–F bond is one shared pair: put one dot (from nitrogen) and one cross (from fluorine) in each of the three places where the outer shells overlap.
    2. Nitrogen is in Group V, so it has five outer electrons. Three are in the bonds. The other two go on nitrogen's outer shell as a pair that is not shared.
    3. Fluorine is in Group VII, so each has seven outer electrons. One is in the bond. Add the other six to each fluorine's outer shell, as three pairs.

    AnswerThree shared pairs (a dot and a cross each), one unshared pair of dots on N, and six unshared crosses on each F.

    Check at the end that every atom has eight in its outer shell, counting the shared ones for both atoms. The inner shells are already drawn in, with nitrogen's electrons as dots and fluorine's as crosses: keep to that.

    Revise this: Covalent bonding and simple molecules
39. 6(a)Why the members of a homologous series react in similar ways.[1]
    Show how to do itHide the working

    1. How an organic compound reacts is decided by its functional group. Every alcohol has the same one, –OH.

    AnswerThey have the same functional group.

    "Same general formula" is true as well, but it is not the reason for the chemistry, and the mark scheme rejects an answer that adds wrong reasons to the right one.

    Revise this: Families of organic compounds
40. 6(b)(i)Counting the electrons in a molecule of methanol.[1]
    Show how to do itHide the working

    1. The number of electrons in an atom equals its proton number: carbon 6, hydrogen 1, oxygen 8.
    2. Methanol is CH₃OH: one carbon, four hydrogens, one oxygen. 6 + 4 × 1 + 8 = 18.

    Answer18

    Count all four hydrogens: three on the carbon and one on the oxygen.

    Revise this: Families of organic compounds
41. 6(b)(ii)The molecular formula of ethanol.[1]
    Show how to do itHide the working

    1. A molecular formula just counts the atoms of each element. CH₃CH₂OH has 2 carbons, 3 + 2 + 1 = 6 hydrogens and 1 oxygen.

    AnswerC₂H₆O

    C₂H₅OH is a structural formula, because it shows the –OH group. The molecular formula gathers all the hydrogens together.

    Revise this: Families of organic compounds
42. 6(b)(iii)Naming the three-carbon alcohol with its –OH on an end carbon.[1]
    Show how to do itHide the working

    1. Three carbons: prop-. An alcohol: -ol. The –OH is on carbon 1, so the number goes in: propan-1-ol.

    AnswerPropan-1-ol.

    Revise this: Naming organic compounds
43. 6(b)(iv)Naming the three-carbon alcohol with its –OH on the middle carbon.[1]
    Show how to do itHide the working

    1. CH₃CH(OH)CH₃ has the –OH on the second carbon of the three.

    AnswerPropan-2-ol.

    "Propanol" is rejected for both of these. With three carbons there are two places for the –OH, so the number is part of the name.

    Revise this: Naming organic compounds
44. 6(c)Why C₃H₇OH is not enough to say which alcohol you mean.[1]
    Show how to do itHide the working

    1. C₃H₇OH fits both C and D. They have the same atoms and differ only in where the –OH is: they are structural isomers.
    2. The formula does not show which carbon the –OH is on.

    AnswerC₃H₇OH could be propan-1-ol or propan-2-ol: it does not show where the –OH group is.

    With two carbons there is only one possible alcohol, which is why C₂H₅OH is safe for ethanol.

    Revise this: Families of organic compounds
45. 6(d)The equation for burning ethanol completely.[2]
    Show how to do itHide the working

    1. Complete combustion of a compound of carbon, hydrogen and oxygen gives carbon dioxide and water only: C₂H₅OH + O₂ → CO₂ + H₂O.
    2. Balance carbon and hydrogen first: 2 carbons give 2CO₂, and 6 hydrogens give 3H₂O.
    3. Now oxygen. The right-hand side has 4 + 3 = 7 oxygen atoms. Ethanol brings 1 of its own, so 6 more are needed: 3O₂.

    AnswerC₂H₅OH + 3O₂ → 2CO₂ + 3H₂O

    Don't forget the oxygen atom already in the ethanol. Leaving it out gives 3½O₂, the usual wrong answer.

    Revise this: Alcohols
46. 6(e)(i)The oxidising agent that turns ethanol into an acid.[1]
    Show how to do itHide the working

    1. Ethanol is oxidised by heating it with acidified aqueous potassium manganate(VII).

    AnswerPotassium manganate(VII).

    The Roman numeral is part of the name and must be in the right place: after manganate.

    Revise this: Carboxylic acids
47. 6(e)(ii)What ethanol is oxidised to.[1]
    Show how to do itHide the working

    1. Oxidising an alcohol gives the carboxylic acid with the same number of carbon atoms. Two carbons: ethanoic acid.

    AnswerEthanoic acid.

    Revise this: Carboxylic acids
48. 6(e)(iii)The food made when bacteria oxidise ethanol.[1]
    Show how to do itHide the working

    1. Bacteria in the air oxidise the ethanol in wine or cider to ethanoic acid. A dilute solution of ethanoic acid is vinegar.

    AnswerVinegar.

    Revise this: Carboxylic acids
49. 6(f)(i)Drawing the ester made from propan-2-ol and ethanoic acid.[2]
    Show how to do itHide the working

    1. An ester has the link –COO–: a carbon with a double bond to one oxygen and a single bond to another, and that second oxygen carries on to the next carbon.
    2. The acid gives the CH₃–C(=O)– end. The alcohol gives the rest, and it joins through the carbon that had the –OH on it.
    3. In D that is the middle carbon. So the oxygen is bonded to a carbon that has one hydrogen and two CH₃ groups on it: CH₃–C(=O)–O–CH(CH₃)–CH₃.
    4. Draw it out with every atom and every bond shown, including each C–H.

    AnswerCH₃COOCH(CH₃)₂, fully displayed: the ester oxygen is joined to the middle carbon of the three-carbon chain.

    One mark is for a correct ester link, with the C=O drawn as a double bond. The other is for the whole molecule. Joining the oxygen to an end carbon would be the ester of propan-1-ol, which is alcohol C.

    Revise this: Esters
50. 6(f)(ii)The second product of making an ester.[1]
    Show how to do itHide the working

    1. Carboxylic acid + alcohol → ester + water.

    AnswerWater.

    Revise this: Esters

These explanations are Papermunch’s own, written to teach the method. The answers have been checked against the exam board’s mark scheme, which has the final say: open it above. The question paper and mark scheme belong toCambridge University Press & Assessment: each question shown here is drawn from the paper itself as you read, and is not kept on this site.