October/November 2025 Paper 43 Worked Answers (IGCSE Chemistry 0620 Extended)

48 questions · 80 marks · 75 minutes

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1. Question 1(a)

   2 marksRelative charge and relative mass of subatomic particles
   Step 1: A neutron carries no charge, so its relative charge is 0, and its relative mass is 1. Step 2: A proton carries a single positive charge, so its relative charge is $+1$, and its relative mass is also 1. Step 3: Only the electron has the very small relative mass of $\frac{1}{1840}$.
   Method:

   Recall the standard relative charges and masses of the three subatomic particles.
   Examiner tips

   • Protons and neutrons both have relative mass 1; only the electron is much lighter.

2. Question 1(b)

   5 marksNumbers of protons, neutrons and electrons in atoms and ions
   Step 1: A $\text{Ca}^{2+}$ ion has 20 protons but has lost 2 electrons, so it has $20 - 2 = 18$ electrons. Step 2: For $^{19}_{9}\text{F}$, neutrons = mass number $-$ proton number $= 19 - 9 = 10$. Step 3: The third species has 8 protons (so it is oxygen, O) and 9 neutrons, giving a mass number of $8 + 9 = 17$; with 10 electrons it has a charge of $8 - 10 = 2-$, so it is $^{17}_{8}\text{O}^{2-}$.
   Method:

   Use proton number, mass number and charge to count each particle and name the element.
   Examiner tips

   • Identify the element from the proton number, then use mass number and charge for neutrons and electrons.

3. Question 1(c)(i)

   1 marksReference isotope for relative atomic mass
   Relative atomic mass is measured on a scale where the average mass of an element’s isotopes is compared to one atom of carbon-12 ($^{12}\text{C}$).
   Method:

   Recall the standard isotope used to define the relative atomic mass scale.
   Examiner tips

   • Relative atomic masses are defined against carbon-12.

4. Question 1(c)(ii)

   1 marksPercentage abundance from relative atomic mass
   Step 1: Let the abundance of magnesium-24 be $x$ (as a fraction), so magnesium-26 is $(1-x)$. Step 2: The weighted average gives $24x + 26(1-x) = 24.3$, which simplifies to $26 - 2x = 24.3$. Step 3: So $2x = 1.7$, $x = 0.85$, meaning the abundance of magnesium-24 is 85%.
   Method:

   Use the weighted-average equation for relative atomic mass and solve for the unknown abundance.
   Examiner tips

   • The Ar is closer to 24, so the lighter isotope must be the more abundant.

5. Question 2(a)(i)

   1 marksIdentifying a gas from melting and boiling points
   Step 1: A substance is a gas when the temperature is above its boiling point. Step 2: Substance B boils at -183\,^{\circ}\text{C}, which is well below 25\,^{\circ}\text{C}, so at 25\,^{\circ}\text{C} it is already a gas. Step 3: The other substances all have boiling points above 25\,^{\circ}\text{C}, so they are liquids or solids; only B is a gas.
   Method:

   Compare room temperature with each boiling point to decide the physical state.
   Examiner tips

   • Compare 25 C with the boiling point: above the boiling point means gas.

6. Question 2(a)(ii)

   1 marksIdentifying a simple molecular solid
   Step 1: A simple molecular substance has weak forces between molecules, giving a low melting point, and never conducts electricity. Step 2: Substance F is a solid at 25\,^{\circ}\text{C} (melting point 44\,^{\circ}\text{C}) with a low melting point, and is a poor conductor both when solid and when molten. Step 3: So F is the simple molecular solid.
   Method:

   Match low melting point plus no electrical conductivity to simple molecular bonding.
   Examiner tips

   • Simple molecular substances have low melting points and never conduct electricity.

7. Question 2(a)(iii)

   1 marksIdentifying a liquid metal
   Step 1: A metal conducts electricity both when solid and when molten, which rules out B and C. Step 2: To be a liquid at 25\,^{\circ}\text{C} the melting point must be below 25\,^{\circ}\text{C}. Step 3: Substance D melts at -39\,^{\circ}\text{C} and conducts well when solid and molten, so it is a metal that is liquid at room temperature (like mercury); G is a metal but melts at 1085\,^{\circ}\text{C}, so it is solid.
   Method:

   Use conductivity to find the metals, then the melting point to find the liquid one.
   Examiner tips

   • Mercury is the metal that is liquid at room temperature.

8. Question 2(b)(i)

   3 marksIdentifying a giant covalent substance with reasons
   Step 1: A giant covalent structure has many strong covalent bonds, giving a very high melting and boiling point. Step 2: Apart from graphite, giant covalent substances have no free charge carriers, so they are poor conductors of electricity both when solid and when molten. Step 3: Substance A has a melting point of 1600\,^{\circ}\text{C}, a boiling point of 2230\,^{\circ}\text{C}, and conducts poorly in both states, so it is the giant covalent substance.
   Method:

   Match very high melting points plus no conductivity to giant covalent bonding.
   Examiner tips

   • Giant covalent structures (except graphite) do not conduct electricity at all.

9. Question 2(b)(ii)

   2 marksIdentifying an ionic compound with a reason
   Step 1: In an ionic solid the ions are held in a fixed lattice and cannot move, so it does not conduct when solid. Step 2: When molten the ions are free to move and carry charge, so it conducts. Step 3: Substance C is a poor conductor when solid but a good conductor when molten, which is the signature of an ionic compound.
   Method:

   Use the change from non-conducting solid to conducting liquid to identify ionic bonding.
   Examiner tips

   • Solid ionic: ions fixed, no conduction. Molten ionic: ions free, conducts.

10. Question 3(a)(i)

    1 marksName of the aluminium compound in bauxite
    Bauxite is the ore of aluminium, and the aluminium compound it contains is aluminium oxide ($\text{Al}_2\text{O}_3$).
    Method:

    Recall the aluminium compound present in the ore bauxite.
    Examiner tips

    • Bauxite = impure aluminium oxide.

11. Question 3(a)(ii)

    1 marksSolvent for purified bauxite before electrolysis
    The purified aluminium oxide is dissolved in molten cryolite, which lowers the melting point and so reduces the energy needed for electrolysis.
    Method:

    Recall the molten substance used to dissolve aluminium oxide for electrolysis.
    Examiner tips

    • Cryolite lowers the melting point of aluminium oxide.

12. Question 3(a)(iii)

    1 marksMaterial used for the positive electrodes
    The positive electrodes used in the electrolysis of aluminium oxide are made of carbon (graphite), which is why they slowly burn away and must be replaced.
    Method:

    Recall the electrode material in the aluminium extraction cell.
    Examiner tips

    • Carbon electrodes are used in aluminium extraction.

13. Question 3(a)(iv)

    2 marksHalf-equation for aluminium at the negative electrode
    Step 1: Aluminium ions, $\text{Al}^{3+}$, are attracted to the negative electrode (cathode). Step 2: Each ion gains 3 electrons (reduction) to become a neutral aluminium atom. Step 3: So the half-equation is $\text{Al}^{3+} + 3\text{e}^- \rightarrow \text{Al}$.
    Method:

    Balance the charge by adding the right number of electrons to reduce the metal ion.
    Examiner tips

    • At the negative electrode, positive ions gain electrons (reduction).

14. Question 3(b)(i)

    1 marksWhy aluminium reacts slowly despite its reactivity
    Step 1: Aluminium is more reactive than zinc, so a displacement reaction is expected. Step 2: However, aluminium has a thin, unreactive layer of aluminium oxide on its surface. Step 3: This oxide coating stops the aluminium metal touching the solution at first, so no reaction is seen until the layer is broken down.
    Method:

    Explain the delay using the protective aluminium oxide layer.
    Examiner tips

    • The aluminium oxide layer protects the metal underneath.

15. Question 3(b)(ii)

    1 marksBalancing a displacement ionic equation
    Step 1: Aluminium loses 3 electrons to form $\text{Al}^{3+}$, while each $\text{Zn}^{2+}$ gains 2 electrons to form Zn. Step 2: The electrons lost must equal the electrons gained, so use 2 Al (losing $2\times3 = 6$ electrons) and 3 $\text{Zn}^{2+}$ (gaining $3\times2 = 6$ electrons). Step 3: This gives $2\text{Al} + 3\text{Zn}^{2+} \rightarrow 2\text{Al}^{3+} + 3\text{Zn}$, with balanced atoms and charges.
    Method:

    Balance the electron transfer to find the coefficients in the ionic equation.
    Examiner tips

    • Match total electrons lost to total electrons gained.

16. Question 3(b)(iii)

    2 marksOxidation number of zinc in Zn2+ and Zn
    Step 1: The oxidation number of a simple ion equals its charge, so $\text{Zn}^{2+}$ has an oxidation number of $+2$. Step 2: An uncombined element always has an oxidation number of 0. Step 3: So zinc is $+2$ in $\text{Zn}^{2+}$ and 0 in the metal Zn.
    Method:

    Use the rule that an ion charge is its oxidation number and an element is 0.
    Examiner tips

    • Uncombined elements have oxidation number 0.

17. Question 3(b)(iv)

    1 marksDefinition of reduction in terms of oxidation number
    In terms of oxidation number, reduction is defined as a decrease in oxidation number.
    Method:

    Recall the oxidation-number definition of reduction.
    Examiner tips

    • Reduction = reduction (decrease) in oxidation number.

18. Question 3(c)(i)

    1 marksMeaning of the symbol delta H
    The symbol $\Delta H$ represents the enthalpy change (energy change) of the reaction, measured in kJ/mol.
    Method:

    Recall what the thermochemical symbol delta H stands for.
    Examiner tips

    • Delta H = enthalpy (energy) change of reaction.

19. Question 3(c)(ii)

    1 marksMeaning of a negative enthalpy change
    Step 1: A negative $\Delta H$ means the products have less energy than the reactants. Step 2: The energy difference is released to the surroundings. Step 3: Releasing energy means the reaction is exothermic.
    Method:

    Link the sign of delta H to energy released or absorbed.
    Examiner tips

    • Negative delta H = exothermic; positive delta H = endothermic.

20. Question 3(d)

    1 marksMeaning of an amphoteric oxide
    Step 1: An amphoteric oxide shows both acidic and basic behaviour. Step 2: This means it reacts with acids (acting as a base) and with bases (acting as an acid). Step 3: In each case it forms a salt and water, so an amphoteric oxide reacts with both acids and bases.
    Method:

    Recall the definition of an amphoteric oxide and what it reacts with.
    Examiner tips

    • Amphoteric = reacts with both acids AND bases.

21. Question 3(e)

    2 marksEquation for an amphoteric oxide reacting with alkali
    Step 1: The product $\text{NaGaO}_2$ contains one Na and one Ga, but $\text{Ga}_2\text{O}_3$ has two Ga, so 2 $\text{NaGaO}_2$ are formed, needing 2 NaOH. Step 2: Counting oxygen: left has $2 + 3 = 5$ O; right has $2\times2 = 4$ O in the salt, leaving 1 O for water. Step 3: Counting hydrogen: 2 H from 2 NaOH gives 1 $\text{H}_2\text{O}$, so the balanced equation is $2\text{NaOH} + \text{Ga}_2\text{O}_3 \rightarrow 2\text{NaGaO}_2 + \text{H}_2\text{O}$.
    Method:

    Balance gallium and sodium first, then hydrogen and oxygen for the water.
    Examiner tips

    • Balance Ga first (gives the salt and NaOH coefficients), then H and O for water.

22. Question 3(f)

    2 marksDeducing formulae from ionic charges
    Step 1: A $\text{Ga}^{3+}$ ion needs three $\text{Br}^-$ ions to balance the charge, giving $\text{GaBr}_3$. Step 2: For sulfate, $\text{Ga}^{3+}$ and $\text{SO}_4^{2-}$ balance when there are 2 Ga ($2\times3 = +6$) and 3 sulfate ($3\times2 = -6$). Step 3: This gives $\text{Ga}_2(\text{SO}_4)_3$.
    Method:

    Balance the 3+ gallium ion against the bromide and sulfate ion charges.
    Examiner tips

    • Cross over the ion charges to find the subscripts.

23. Question 4(a)(i)

    2 marksDescribing a reversible reaction at equilibrium
    Step 1: At equilibrium the forward and reverse reactions are both still happening, but at the same rate. Step 2: Because the rates are equal, the amount of each substance formed equals the amount used up. Step 3: So the concentrations of reactants and products stay constant (they do not change), even though both reactions continue.
    Method:

    State both the rate condition and the concentration condition for equilibrium.
    Examiner tips

    • Equilibrium: rates equal, concentrations constant (not necessarily equal).

24. Question 4(a)(ii)

    3 marksEffects of pressure and catalyst on equilibrium
    Step 1: Lowering the pressure spreads the gas particles out, so they collide less often and the forward rate decreases. Step 2: The forward reaction makes fewer gas molecules (2 to 1), so lowering the pressure shifts the position of equilibrium towards the side with more molecules (the reactants), decreasing the equilibrium concentration of $\text{PCl}_5$. Step 3: A catalyst speeds up the forward and reverse reactions equally, so it does not change the position of equilibrium: the equilibrium concentration of $\text{PCl}_5$ shows no change.
    Method:

    Apply collision theory for the rate and Le Chatelier reasoning for the equilibrium shift.
    Examiner tips

    • A catalyst never changes the position of equilibrium, only the rate.

25. Question 4(a)(iii)

    1 marksDeducing the energetics from a temperature change
    Step 1: Raising the temperature shifts an equilibrium in the endothermic direction. Step 2: Here, raising the temperature reduces the product $\text{PCl}_5$, so the equilibrium has shifted backwards (towards the reactants). Step 3: If the reverse reaction is favoured by heating, it must be endothermic, which means the forward reaction is exothermic.
    Method:

    Use Le Chatelier reasoning to link the temperature change to the energetics.
    Examiner tips

    • Heat favours the endothermic side; the product fell, so forward is exothermic.

26. Question 4(b)

    3 marksEnthalpy change from bond energies
    Step 1: Bonds broken in the reactants $= (4\times410) + 610 + 240 = 1640 + 610 + 240 = 2490$ kJ. Step 2: Bonds formed in the product $= (4\times410) + 350 + (2\times340) = 1640 + 350 + 680 = 2670$ kJ. Step 3: Enthalpy change $=$ bonds broken $-$ bonds formed $= 2490 - 2670 = -180$ kJ/mol (exothermic).
    Method:

    Sum the bond energies broken and formed, then subtract to find delta H.
    Examiner tips

    • Enthalpy change = (energy to break bonds) - (energy released forming bonds).

27. Question 4(c)

    3 marksDot-and-cross diagram of nitrogen trichloride
    Step 1: Nitrogen has 5 outer electrons and shares one with each of three chlorine atoms, forming three N–Cl bonding pairs and leaving one lone pair on nitrogen (to complete its octet). Step 2: Each chlorine has 7 outer electrons, sharing one in the bond and keeping three lone pairs (to complete its octet). Step 3: So there are three bonding pairs, one lone pair on N, and three lone pairs on each Cl.
    Method:

    Use the outer electron counts of N (5) and Cl (7) to place bonding and lone pairs to full octets.
    Examiner tips

    • Complete every atom to a full octet of 8 outer electrons.

28. Question 4(d)(i)

    2 marksObservations when a carbonate reacts with acid
    Step 1: The acid reacts with the magnesium carbonate, so the solid is used up and appears to disappear (dissolve). Step 2: Carbon dioxide gas is produced, seen as bubbling. Step 3: So the two observations are that the solid disappears and there is effervescence.
    Method:

    Predict the observable changes from the products: a soluble salt and carbon dioxide gas.
    Examiner tips

    • Carbonate + acid: effervescence (CO2) and the solid dissolves.

29. Question 4(d)(ii)

    3 marksVolume of gas from a mole calculation
    Step 1: Moles of HCl $=$ concentration $\times$ volume in $\text{dm}^3 = 0.100 \times \frac{50.0}{1000} = 0.005$ mol. Step 2: The equation shows 2 HCl give 1 $\text{CO}_2$, so moles of $\text{CO}_2 = \frac{0.005}{2} = 0.0025$ mol. Step 3: Volume of $\text{CO}_2 = 0.0025 \times 24\,000 = 60\ \text{cm}^3$.
    Method:

    Find moles of acid, apply the reacting ratio, then convert moles of gas to a volume.
    Examiner tips

    • Use the 2:1 mole ratio of HCl to CO2 from the balanced equation.

30. Question 5(a)(i)

    3 marksIonic equation for precipitation with state symbols
    Step 1: The reacting ions are $\text{Pb}^{2+}(\text{aq})$ and $\text{SO}_4^{2-}(\text{aq})$, both in solution. Step 2: Their charges ($2+$ and $2-$) cancel exactly, so they combine in a 1:1 ratio to give $\text{PbSO}_4$. Step 3: Lead(II) sulfate is insoluble, so it forms as a solid: $\text{Pb}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{PbSO}_4(\text{s})$.
    Method:

    Combine the aqueous ions in the ratio that balances their charges and give the precipitate state symbols.
    Examiner tips

    • Only the insoluble product gets the (s) state symbol; the reacting ions are (aq).

31. Question 5(a)(ii)

    1 marksA soluble salt providing lead(II) ions
    Step 1: To make an aqueous solution of lead(II) ions, the lead salt must be soluble. Step 2: Almost all nitrates are soluble, so lead(II) nitrate dissolves to give $\text{Pb}^{2+}$ ions in solution. Step 3: Lead(II) sulfate, carbonate and chloride are insoluble, so lead(II) nitrate is the correct choice.
    Method:

    Recall the solubility rules and pick the soluble lead salt.
    Examiner tips

    • Use a nitrate to get a soluble source of a metal ion.

32. Question 5(a)(iii)

    1 marksA soluble salt providing sulfate ions
    Step 1: To provide aqueous sulfate ions the sulfate salt must be soluble. Step 2: Sodium (and potassium and ammonium) sulfates are soluble, so sodium sulfate dissolves to give $\text{SO}_4^{2-}$ ions. Step 3: Barium, lead(II) and calcium sulfates are insoluble or only slightly soluble, so sodium sulfate is the correct choice.
    Method:

    Apply solubility rules to find a soluble source of sulfate ions.
    Examiner tips

    • Sodium/potassium/ammonium salts are always soluble.

33. Question 5(a)(iv)

    1 marksGeneral term for the solid kept on filter paper
    The general term for the insoluble solid that remains on the filter paper during filtration is the residue (the liquid that passes through is the filtrate).
    Method:

    Recall the filtration term for the trapped solid.
    Examiner tips

    • Residue = solid on the paper; filtrate = liquid through the paper.

34. Question 5(a)(v)

    2 marksPurifying an insoluble salt after filtration
    Step 1: Washing the solid on the filter paper with distilled water rinses away soluble impurities. Step 2: The clean solid is then dried, for example in a warm oven or between sheets of filter paper. Step 3: This gives a pure, dry sample of the insoluble salt.
    Method:

    Describe washing away soluble impurities then drying the insoluble solid.
    Examiner tips

    • Wash with distilled water then dry: the standard way to purify an insoluble salt.

35. Question 5(b)(i)

    2 marksWays copper is similar to Group I metals
    Step 1: Group I metals are typical metals: they conduct electricity well, form basic oxides and form soluble (often colourless) salts. Step 2: Copper shares these general metal properties: good electrical conductivity, basic oxides and soluble salts. Step 3: So good electrical conductivity and forming basic oxides are two ways copper is similar to Group I metals (variable oxidation states and coloured compounds are special transition-metal features).
    Method:

    Separate the general metal properties (shared) from the special transition-metal properties.
    Examiner tips

    • Similarities = general metal properties shared with Group I.

36. Question 5(b)(ii)

    2 marksWays copper differs from Group I metals
    Step 1: Group I metals have only one oxidation state (+1), form white/colourless compounds and are not typical catalysts. Step 2: As a transition element, copper has variable oxidation states, forms coloured compounds and acts as a catalyst. Step 3: So variable oxidation states and forming coloured compounds are two ways copper differs from Group I metals.
    Method:

    Pick the special transition-metal properties that Group I metals lack.
    Examiner tips

    • Differences = variable oxidation states, coloured compounds, catalysis.

37. Question 6(a)

    3 marksEmpirical formula from percentage composition
    Step 1: Divide each percentage by the $A_r$: C $= \frac{50.00}{12} = 4.17$, H $= \frac{5.56}{1} = 5.56$, O $= \frac{44.44}{16} = 2.78$. Step 2: Divide each by the smallest (2.78): C $= 1.5$, H $= 2$, O $= 1$. Step 3: Multiply by 2 to get whole numbers: C 3, H 4, O 2, so the empirical formula is $\text{C}_3\text{H}_4\text{O}_2$.
    Method:

    Convert masses to moles, find the simplest ratio, then scale to whole numbers.
    Examiner tips

    • Always scale the smallest-ratio numbers up to whole numbers.

38. Question 6(b)

    1 marksMolecular formula from empirical formula and Mr
    Step 1: The mass of the empirical unit $\text{CHO} = 12 + 1 + 16 = 29$. Step 2: Divide the relative molecular mass by this: $\frac{116}{29} = 4$. Step 3: Multiply the empirical formula by 4 to get the molecular formula $\text{C}_4\text{H}_4\text{O}_4$.
    Method:

    Find the empirical formula mass, divide into the Mr, and scale the formula.
    Examiner tips

    • Molecular formula = empirical formula x (Mr / empirical mass).

39. Question 6(c)(i)

    1 marksOther product of esterification
    When a carboxylic acid reacts with an alcohol to form an ester, the only other product is water.
    Method:

    Recall the products of the reaction between a carboxylic acid and an alcohol.
    Examiner tips

    • Acid + alcohol -> ester + water.

40. Question 6(c)(ii)

    1 marksCatalyst for esterification
    The catalyst used in the reaction between a carboxylic acid and an alcohol is an acid catalyst (for example concentrated sulfuric acid).
    Method:

    Recall the catalyst type for the carboxylic-acid-plus-alcohol reaction.
    Examiner tips

    • Concentrated sulfuric acid is the usual esterification catalyst.

41. Question 6(c)(iii)

    2 marksIdentifying the acid and alcohol from an ester structure
    Step 1: The ester splits at the C–O of the COO group: the acyl part $\text{CH}_3\text{CH}_2\text{COO}–$ comes from the acid and the $–\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3$ part comes from the alcohol. Step 2: $\text{CH}_3\text{CH}_2\text{COOH}$ has 3 carbons, so it is propanoic acid. Step 3: The alcohol $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}$ has 4 carbons with the OH on the end carbon, so it is butan-1-ol.
    Method:

    Split the ester at the COO group and name each fragment as the acid and the alcohol.
    Examiner tips

    • Count carbons on each side of the ester linkage to name the acid and alcohol.

42. Question 6(d)(i)

    1 marksType of polymerisation
    A polymer formed from unsaturated alkene monomers joining together, with no small molecule lost, is made by addition polymerisation.
    Method:

    Recall the polymerisation type for alkene monomers with C=C bonds.
    Examiner tips

    • Addition polymerisation: alkene monomers, no other product.

43. Question 6(d)(ii)

    1 marksWhy a polymer has no single molecular formula
    Step 1: A polymer is made by joining a very large but variable number of monomer units. Step 2: Different polymer molecules in the same sample contain different numbers of repeat units. Step 3: Because the chain length (and so the number of atoms) is not fixed, no single molecular formula can be written; only the repeat unit is shown.
    Method:

    Explain that variable chain length prevents a fixed molecular formula.
    Examiner tips

    • Polymers vary in chain length, so only the repeat unit is given.

44. Question 6(d)(iii)

    1 marksCounting monomer units in a polymer section
    Step 1: In addition polymerisation of an alkene, each monomer contributes two carbon atoms to the polymer backbone. Step 2: The section shown has a backbone of six carbon atoms. Step 3: So the number of monomer units $= \frac{6}{2} = 3$.
    Method:

    Divide the number of backbone carbons by two carbons per monomer.
    Examiner tips

    • Backbone carbons / 2 = number of alkene monomer units.

45. Question 6(d)(iv)

    2 marksDisplayed formula of the monomer of an addition polymer
    Step 1: In the polymer, each backbone carbon carries a $\text{CH}_3$ group and an H atom, so the monomer must have two such carbons. Step 2: Restoring the $\text{C}{=}\text{C}$ double bond between these two carbons gives $\text{CH}_3\text{CH}{=}\text{CHCH}_3$. Step 3: This is but-2-ene, where each double-bond carbon has one $\text{CH}_3$ and one H.
    Method:

    Identify the repeat unit, then restore the double bond to find the alkene monomer.
    Examiner tips

    • Reverse the polymerisation: put the C=C back between the two repeat-unit carbons.

46. Question 6(e)(i)

    2 marksFunctional groups in an amino acid
    Step 1: Every amino acid has a basic amine group, $-\text{NH}_2$, on the central carbon. Step 2: It also has an acidic carboxylic acid group, $-\text{COOH}$. Step 3: These two functional groups ($-\text{NH}_2$ and $-\text{COOH}$) let amino acids join to form polyamides (proteins).
    Method:

    Recall the two functional groups that define an amino acid.
    Examiner tips

    • The name amino acid tells you: amine group plus acid group.

47. Question 6(e)(ii)

    1 marksIdentifying the amide linkage in a polyamide
    Step 1: A polyamide is formed when amino acids join, losing water. Step 2: The new bond joins the carbon of one $-\text{COOH}$ group to the nitrogen of the next $-\text{NH}_2$ group. Step 3: This $-\text{CO}{-}\text{NH}-$ group is the amide (peptide) linkage that should be circled.
    Method:

    Identify the carbon-to-nitrogen linkage formed when amino acids join.
    Examiner tips

    • Amide/peptide linkage = -CO-NH-.

48. Question 6(e)(iii)

    1 marksName for natural polyamides from amino acids
    Natural polyamides made by joining amino acids together are called proteins.
    Method:

    Recall the name for the natural polymer made from amino acids.
    Examiner tips

    • Natural polyamide = protein.