June 2024: Paper 3, Unified chemistry

Mark scheme

Examiner's comment

Questions 1 (a), 1 (b) and 1 (c) required candidates to calculate the pH of a base, a diluted strong acid and a buffer solution. These different types of pH calculation form the basis of acid-base chemistry, all requiring the relationship pH = –log[H+] at some stage. Most candidates showed competency in their pH calculations. A number of candidates often did not appreciate the type of pH calculation required. Question 1(a) was a standard pH calculation of a strong base and this was an easy start to the paper. Most candidates used Kw and the concentration of NaOH to determine the H+ concentration of 2.5 × 10– 14 mol dm–3, from which the pH of 13.60 can be calculated using pH = –log[H+]. Some candidates first determined pOH as 0.40, and the pH using 14.00 – 0.40. This approach is based on sound chemistry and is acceptable. Several candidates often calculated the pH as –log 0.4 = 0.4 and gave this as their answer. NaOH is a common alkali and cannot have a pH <7 and this should have triggered that the candidate had made an error.

Notes from the mark scheme and report

• ALLOW pOH method
• ECF only for pH > 7

Skills: M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.2 Change the subject of an equation; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude

Mark scheme

Examiner's comment

This question was answered well although nearly a half of candidates made errors. The initial solution had been diluted by 10 times and its concentration had been reduced from 0.750 to 0.0750 mol dm–3. From here pH = –log[H+] gives the correct answer of 1.12 to 2 decimal places, required by the question. Candidates found Question 1 (b) more difficult than 1 (a) or 1 (c). There seemed to be little pattern in candidate errors, the dilution being the difficult part of the question. Some did not dilute the initial concentration of 0.750 mol dm–3, giving 2.12. Others divided 0.750 by 2 instead of 10 or introduced 90 into the calculation as 90 cm3 of water would have been added. The most disappointing error was for a correct calculation to be displayed using the wrong number of decimal places. Two decimal places should be the norm for pH, reflecting the accuracy of most pH meters.

Notes from the mark scheme and report

• dilution ×10 gives 0.0750 mol dm–3
• 2 decimal places required (1.12)

Skills: M0.2 Use ratios, fractions and percentages; M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude

Mark scheme

Examiner's comment

This question was a standard calculation for the pH of a buffer solution and most candidates correctly identified the mixture as being a buffer. Most candidates used the expression for Ka and the concentrations of the weak acid and its salt to determine the H+ concentration of 5.25 × 10–5 mol dm–3. From here, the pH of 4.28 can easily be calculated. A common error was for the concentrations of the weak acid and its salt to be substituted into the Ka expression the wrong way round, producing an answer of 5.23. As there was only one error, this answer was given 1 mark with error carried forward (ECF). A few candidates first subtracted 0.1 from 0.3, using concentration values of 0.2 and 0.1 instead of 0.3 and 0.1. This gave an answer of 4.46 which was also given 1 mark. The substantial number of less successful responses fell into the trap of using the pH method for calculating the pH of a weak acid rather than a buffer, squaring the H+ concentration. This was the wrong method to apply for a pH buffer calculation and was not given any marks.

Notes from the mark scheme and report

• 2 decimal places required
• common errors: 5.23 from inverted [HA] and [A–] (1 mark)
• zero marks for the square-root (weak acid) approach
• ALLOW Henderson–Hasselbalch

Skills: M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude

Mark scheme

Examiner's comment

This question differentiated between candidates extremely well. See Exemplar 1 below. some of the issues encountered in the responses. Unfortunately, nearly a quarter of candidates could not be given any marks at all for their responses. To improve, it is worth studying Exemplar 1. The candidate has communicated the key steps required to prepare the standard solution: • Calculation of the mass of hydrated copper(II) nitrate required. • Dissolving the hydrated copper(II) nitrate in water in a suitable container (a beaker). • Transferring the solid to a 100 cm3 volumetric flask, washing the beaker with water and transferring the washings also to the volumetric flask. • Making the solution up to the 100 cm3 mark in the volumetric flask and inverting the flask to mix the contents thoroughly. Issues with responses which arose by not reading the question closely enough: • Omitting to calculate the mass of hydrated copper(II) nitrate required. • Calculating the mass of anhydrous copper(II) nitrate instead of the hydrated salt. • Dissolving in 100 cm3 of water and then adding more water for rinsing. • Not rinsing out the original container at all. • Making the solution up in the volumetric flask. • Using of a 250 cm3 volumetric flask for preparing 100 cm3 of solution. • Omitting the inversion stage. • Answering a different question, e.g. how to carry out a titration, how to determine an enthalpy change, how to work out the number of waters of crystallisation by heating in a crucible.

Notes from the mark scheme and report

• mass 4.83 g using M of the hydrated salt (241.5), 2 or more decimal places
• dissolve in less than 100 cm3 of distilled water in a beaker (DO NOT ALLOW 100 cm3 or more)
• transfer to volumetric flask AND rinse washings in
• make up to the mark AND invert

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; PAG2 Acid–base titration

Mark scheme

Examiner's comment

Most candidates were aware that a bond enthalpy is a measure of the energy to break bonds. Most went on to link the positive sign to an endothermic reaction which requires energy. Some candidates linked bond enthalpies to bond formation instead. A significant number of responses described an endothermic reaction, instead of bond enthalpy, in terms of the energy required to break bonds being greater than the energy to make bonds. This was the answer to a different question and could not be given marks.

Notes from the mark scheme and report

• bond breaking is endothermic/requires energy
• IGNORE 'more energy needed to break bonds than released making bonds'

Mark scheme

Examiner's comment

This question differentiated well, with about half the candidates obtaining the correct bond enthalpy of +413 kJ mol–1. With ECF, few candidates scored 0 marks. Most candidates calculated the energy involved in making 3 H–H bonds and 1 C≡O bond as 2385 kJ mol–1. This value had to be incorporated with the energy associated with breaking bonds and the enthalpy change of –195 kJ mol–1. It was this step where problems arose. There were many errors with signs and the lowest attaining candidates sometimes omitted the enthalpy change completely in their calculation. The commonest error of 315.5 kJ mol–1 was the result of using the wrong sign for the enthalpy change. With ECF, this was still given 2 of the 3 available marks.

Notes from the mark scheme and report

• common errors: –413 (sign, 2 marks), 304 (2 mol H2), 316 (wrong sign for 195), 364 (∆H omitted, 1 mark), 529 (1 O–H instead of 2)
• DO NOT ALLOW negative final value

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

Unlike the standard bond enthalpy calculation in Question 2 (a) (ii), this question required candidates to apply their understanding of enthalpy changes to a new context relevant to energy. Candidates also had to interconvert units and provide an answer to 3 significant figures in standard form. Apart from the chemistry, there were various mathematical skills to use. Candidates found this question much more demanding than Question 2 (a) (ii), with only just over a half gaining marks. The first step involved working out that the number of moles of hydrogen gas in 60 m3 at RTP is 2500 mol. Candidates then had to apply their understanding of ∆fH to calculate the energy as 7.15 × 105 kJ. Common mistakes included: • multiplying 60 m3 by the enthalpy change of 285.8 kJ mol–1 (omitting to calculate the moles of H2 – a lack of understanding) • using more than 3 significant figures, e.g. 7.145 instead of 7.15 • making an error with powers of 10 in the mole calculation (usuallysually by not converting 60 m3 into 60,000 dm3, leading to 2.5 moles). This type of question looks simple but makes for an excellent way of developing application and mathematical skills.

Notes from the mark scheme and report

• convert m3 to dm3 (n = 2500 mol)
• 3 sig figs AND standard form required
• 1 mark for 7.15 × 10^x, 7.145 × 10^5 or 715000
• IGNORE sign

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.1 Recognise and use expressions in decimal and ordinary form; M1.1 Use an appropriate number of significant figures

Mark scheme

Examiner's comment

Candidates were required to write a straightforward ionic equation that they would have encountered many times during the A Level Chemistry course. It was surprising that only just over half the candidates produced an equation that could be given. Common errors included the following. • Omission of state symbols or incorrect state symbols, especially (aq) in AgCl(s). • Inclusion of nitrate ions or use of AgNO3 instead of Ag+. • An equation using Cl2 and forming AgCl2. Some candidates used the ideal gas equation to determine the moles of hydrogen, choosing suitable values for temperature and pressure. This approach was allowed, although the exercise would have wasted candidate time compared to the much simpler division by 24 for using RTP, which is stated in the question.

Notes from the mark scheme and report

• ALL 3 state symbols required

Mark scheme

Examiner's comment

Most candidates determined the moles of AgNO3 and hence Ag+ as 1.50 × 10–3 mol. This was given 1 mark, but candidates then needed to use this amount to predict the identity of compound A. Most candidates could not see the way forward and many received only 1 mark. Many candidates had worked out ‘something’ from the supplied data, without knowing where this initial step would take them. Candidates needed to spot that the ratio of the element : Cl in compound A was 5 × 10–4 : 1.50 × 10–3 or 1 : 3. The correct formula of AlCl3 then follows. MgCl2 was a common error obtained by subtracting 5 × 10–4 from 1.50 × 10–3 to obtain a 1 : 2 ratio. This question would be a good exercise for improving the application skills of candidates.

Notes from the mark scheme and report

• n(AgNO3) = 1.50 × 10–3 is the essential mark
• ratio A:Cl = 1:3 so AlCl3 (ALLOW Al2Cl6 or PCl3)
• ALLOW ECF from formula of silver chloride in (b)(i)

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

Most candidates made a good attempt at working out the molecular formula of the structure as being C13H19N3O7. N and O were usually correct with mistakes most common with carbon (especially 12) and hydrogen (especially 17–20).

Notes from the mark scheme and report

• C13H19N3O7
• elements in any order

Mark scheme

Examiner's comment

This question was answered well with the correct answer of 4 being seen on most scripts, reflecting good understanding of chiral carbon centres. The commonest incorrect response was 5, presumably by including the C atom on the bottom right of the structure within the –CH(CH3)2 group.

Notes from the mark scheme and report

• 4 chiral carbons

Skills: M4.2 Visualise and represent 2-D and 3-D forms including 2-D representations of 3-D objects

Mark scheme

Examiner's comment

This question was answered extremely well with about three-quarters of candidates securing both marks. Most candidates calculated the molecular masses of compounds C and D as 380 and 441.7 respectively, to obtain a difference of 61.7. Some candidates adopted a simpler different approach which gives the same correct answer, working out the difference between the masses of nitrogen and phosphorus (for C) and oxygen and sulfur (for D). ECF was applied to any incorrect molecular formulae from Question 2 (c) (i) from which both marks could be obtained.

Notes from the mark scheme and report

• 61.7 from Mr(D) 441.7 – Mr(C) 380, or 7 × (32.1 – 16) – 3 × (31 – 14)
• ALLOW ECF from incorrect formula in (c)(i)

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

The term ‘bidentate ligand’ was well-known with most candidates gaining some credit. A mark was sometimes not given if ‘lone pairs’ had been omitted. A significant number of candidates just wrote ‘a pair of electrons’, which could have been a bonded pair.

Notes from the mark scheme and report

• two lone pairs AND forms dative/coordinate bonds to a metal ion
• IGNORE 'donates two pairs of electrons' alone

Mark scheme

Examiner's comment

The diagrams and 3– charge were usually correct. The standard of 3D diagrams has improved over the years with most candidates using the wedges requirements outlined in the mark scheme. Some candidates often did not use wedges and their diagrams were not given marks. It should be noted that ‘impossible’ 3D diagrams showing wedges in unrealistic positions were also not given marks. Examples are shown in the published mark scheme.

Notes from the mark scheme and report

• overall 3– charge shown on at least one isomer
• bonds MUST go to the O– atoms of the ligands
• 3D wedges required
• DO NOT ALLOW simplified loop for ethanedioate
• MAX 2 for incorrect central element

Skills: M4.1 Use angles and shapes in regular 2-D and 3-D structures; M4.2 Visualise and represent 2-D and 3-D forms including 2-D representations of 3-D objects; M4.3 Understand the symmetry of 2-D and 3-D shapes

Mark scheme

Examiner's comment

Most candidates were given a mark for one of their half-equations. The half-equation for the oxidation of iron(II) ions was correct more often than that for ethanedioate ions. Strangely, a significant number of half-equations started with Fe metal instead of Fe2+ ions. The mark scheme allowed for variations in the oxidation of ethanedioate ions, with several alternative equations being given from the correct equation forming CO2 receiving a mark.

Notes from the mark scheme and report

• Fe2+ → Fe3+ + e–
• C2O4 2– → 2CO2 + 2e–
• ALLOW multiples
• IGNORE state symbols

Mark scheme

Examiner's comment

Most candidates were given all 3 marks for their titration tables. One mark was allocated to the readings, the other 2 for the correct titres. Candidates are expected to record burette readings to 2 decimal places with the last figure being 0 or 5, reflecting the accuracy of the burette, which is stated in the question. Two issues were seen: • Some candidates ignored the terminal 0 in their readings, e.g. 1.9 for 1.90 cm3. • A small number of candidates recorded the initial and final readings the wrong way round despite these terms being used above each burette diagram. Titres were usually correct.

Notes from the mark scheme and report

• all readings to 2 decimal places with last figure 0 or 5 AND final/initial readings in the correct rows
• titres 21.65, 21.85, 21.55 (2 correct for 1 mark)

Skills: M1.1 Use an appropriate number of significant figures

Mark scheme

Examiner's comment

Candidates are much better at calculating percentage uncertainties than in previous series. Almost all candidates realised that the uncertainty in the titre resulted from two readings and that the overall uncertainty in the titre would be 2 × 0.05 = 0.1 cm3, resulting in a percentage uncertainty of 0.46%. Comparatively few candidates ignored the ‘2’ and gave 0.23% as their answer. Some candidates worked out the percentage uncertainty from the mean titre but this was still given marks.

Notes from the mark scheme and report

• 2 × 0.05 because two readings
• 0.46%
• 2 decimal places minimum
• DO NOT ALLOW 0.50%

Skills: M1.3 Identify uncertainties in measurements and use simple techniques to determine uncertainty when data are combined

Mark scheme, with the levels and indicative points

Examiner's comment

This unstructured titration problem was assessed by Level of Response (LOR). Candidates answered this stock titration calculation well. The key stages are listed below: 1. Determination of the mean titre from the two closest titres in the candidate’s response to Question 3 (b) (ii). If the titres were correct, this would be 21.60 cm3, the mean of 21.55 cm3 and 21.65 cm3. 2. Calculation of the number of moles of MnO4– as 4.32 × 10–4 mol. 3. Calculation of the number of moles of FeC2O4 in 25 cm3 as 5/3 × 4.32 × 10–4 = 7.20 × 10–4 mol. Scaling up this number of moles by 10 for the moles in 250 cm3 as 7.20 × 10–3 mol. 4. Determination of the number of waters of crystallisation. Most candidates completed Stages 1 and 2 correctly, securing a minimum of a Level 1 response. A significant number of candidates then completed Stage 3, with most including scaling by 10 to secure a minimum of Level 2. Many of these candidates compared the moles of FeC2O4 to the mass of FeC2O4•xH2O used to determine the value of x as 2. Such candidates would have reached Level 3. The communication strand of the LOR mark was determined by the clarity of the response, particularly whether the numbers in the calculation had been labelled. Unfortunately, many ‘correct’ responses had not done this. Over half the candidates were given 5 or 6 marks for this stock calculation. This exemplar demonstrates the hazards of not communicating well. The candidate has followed the four stages described above and the chemistry behind the titration analysis is correct. However, the response is mostly a page of numbers, with no explanation about what the numbers mean. The response is clearly at Level 3 but it is impossible to credit the communication strand, receiving a total of 5 marks.

Notes from the mark scheme and report

• mean of concordant titres 21.60 cm3
• n(MnO4–) = 4.32 × 10–4
• 5:3 ratio then ×10 for 250 cm3
• M = 179.9 so x = 2
• using 25.0 cm3 instead of titres limited to Level 1

Skills: M0.2 Use ratios, fractions and percentages; M1.2 Find arithmetic means; M2.2 Change the subject of an equation

Mark scheme

Examiner's comment

Candidates were expected to link proton-transfer behaviour in acids and bases to the provided equilibria. The question differentiated between candidates well. Some candidates just stated that an acid is a proton donor and a base a proton acceptor without referring to the provided equilibria. This was the answer to a much simpler question and could not be given marks. The best responses demonstrated excellent understanding within the context of the equilibria. Such candidates clearly explained how H2PO4– behaves as an acid in the forward reaction of Equilibrium 2 and as a base in the reverse reaction of Equilibrium 1.

Notes from the mark scheme and report

• equilibrium 1 (reverse): H2PO4– accepts a proton so acts as a base
• equilibrium 2: donates a proton so acts as an acid
• IGNORE references to HPO4 2– and equilibrium 3

Mark scheme

Examiner's comment

Most candidates used the information in the question to draw a correct displayed formula of H3PO3. Another acceptable approach was to show a ‘dot-and-cross’ diagram. Candidates usually chose 104.5º for the P–O–H bond angles although a significant number suggested 180º. The O–P–O bond angle proved to be more difficult. Many suggested 120º by ignoring the lone pair of electrons on the P atom. The shape was analogous with NH3 giving a bond angle of 107º. Overall, candidates answered this question well. Candidates are advised to assess the number of bonded pairs and lone pairs around each atom when suggesting bond angles. This would have reduced the number of incorrect bond angles such as 180º for P–O–H and 120º for O–P–O.

Notes from the mark scheme and report

• P with 3 bonds only, each O bonded to P and H
• O–P–O = 107° (ALLOW 106–108)
• P–O–H = 104.5° (ALLOW 104–105)
• IGNORE geometry drawn

Skills: M4.1 Use angles and shapes in regular 2-D and 3-D structures; M4.2 Visualise and represent 2-D and 3-D forms including 2-D representations of 3-D objects

Mark scheme

Examiner's comment

Most candidates wrote the correct systematic name of phosphorus(III) acid and the clue given in the question for the name of H3PO4 should have helped. Common errors included phosphorus(IV) acid, the same as for H3PO4, and the (III) oxidation number being placed after ‘acid’ in the name. The commonest error though, was hydrogen phosphate. Candidates are advised to use any information provided in the question, which often contains clues. This certainly would have prevented hydrogen phosphate as a response.

Notes from the mark scheme and report

• phosphoric(III) acid
• DO NOT ALLOW phosphoric acid (III) or phosphorous acid

Mark scheme

Examiner's comment

Candidates found this question quite challenging, with only about one-third writing a correct equation. The question gave the reactants and products with only the formula of phosphorus(V) oxide having to be worked out. The actual reaction does produce P4O10 but P2O5 was shown in almost all equations, and this was acceptable. Various incorrect formulae were seen for phosphorus(V) oxide including PO, PO2, P5O, HPO, etc. Unfortunately a significant number of candidates could not balance the equation, despite using correct formulae.

Notes from the mark scheme and report

• 4PH3 + 8O2 → P4O10 + 6H2O
• ALLOW P2O5 version and multiples
• IGNORE state symbols

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

This question generated a wide range of responses, testing many important chemical skills. Candidates often used oxidation numbers correctly to show that silver is reduced and phosphorus oxidised, with silver being the easier. Hydrogen was sometimes incorrectly chosen for oxidation. The oxidation number change of +1 to 0 for silver was usually correct although +9 and +11 were common errors for silver in AgNO3, presumably by choosing the oxidation number of nitrogen as –3 or – 5. Candidates usually recognised that phosphorus started with an oxidation number of –3 but the oxidation number of +5 was a common error in H3PO3. Balancing the equation was the most difficult part of this question with numbers being added almost at random. It is easier to balance equations for redox reactions by balancing the oxidation number changes first.

Notes from the mark scheme and report

• 6AgNO3 + PH3 + 3H2O → 6Ag + H3PO3 + 6HNO3
• Ag reduced +1 to 0, P oxidised –3 to +3
• + signs required
• 1 mark if both changes correct but oxidised/reduced omitted or reversed
• IGNORE electrons
• Ag+ and NO3– are among the common ions that students should know (see also Question 4 (c) (i)).
• For a NO3– ion to have a charge of 1–, the oxidation number of nitrogen must be +5. By choosing –5, the charge on NO3 would be –11 and silver would have an oxidation number of +11. This is completely unrealistic and should be rejected as it points to a serious error.

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

This question again required candidates to construct an equation. Candidates were provided with the formula of all species reactants and products except for that of ammonium chloride. Candidates are expected to know that the ammonium ion is NH4+ but many incorrect equations showed NH3Cl. About half the candidates were able to construct a correctly balanced equation with the ‘12’ balancing number for HCl being the hardest part. This links back to the ‘assessment for learning’ callout added to Question 4 (b) (ii) in this report. As with other questions requiring equations to be written, this question differentiated very well. Writing formulae and balancing equations are fundamentals for mastering chemistry and candidates are advised to practise these skills throughout the course.

Notes from the mark scheme and report

• 3PCl5 + 3NH4Cl → P3N3Cl6 + 12HCl
• ALLOW multiples
• The specification states the following.
• Formulae and equations 2.1.2(a) the writing of formulae of ionic compounds from ionic charges, including: (i) prediction of ionic charge from the position of an element in the periodic table (ii) recall of the names and formulae for the following ions: NO3 –, CO3 2–, SO4 2–, OH–, NH4 +, Zn2+ and Ag+ This section will be studied at the start of the two-year course and form the backbone for chemical literacy. For success in chemistry, the common ions should be learnt.

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

In contrast to equation writing, candidates found this simple calculation far easier with the majority obtaining both marks for 26.72. Common incorrect percentages were 26.7 (wrong number of decimal places) and 8.91 (using 31 rather than 3 × 31 for the numerator).

Notes from the mark scheme and report

• 2 decimal places required (26.72
• 26.7 scores 1)
• 8.91 scores 1 (omitting the 3 P atoms)

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures

Mark scheme

Examiner's comment

About half the candidates suggested a range of creditworthy responses with ‘different bond lengths’ and ‘decolorises bromine’ being the most common.

Notes from the mark scheme and report

• (P–N) bond lengths different, or more exothermic hydrogenation, or reacts with bromine by addition
• IGNORE C–C bond lengths
• IGNORE 'alternating single and double bonds'
• IGNORE 'more reactive' without example

Mark scheme

Examiner's comment

This was another question where valuable information: ’all N and Cl atoms are bonded to P atoms’ had been provided. Many of the structures seen ignored this information with chlorine often been shown bonded to a nitrogen atom. Nitrogen atoms were often shown with 1 bond only and chlorine atoms in the ring structure with 2 or more bonds. Most structures contained 6 or 3-membered rings. This was a difficult question, requiring candidates to use the supplied information to come up with realistic structures that met chemical bonding rules. Only about a quarter of candidates could be given any mark. The Kekulé theme in Questions 4 (c) (i) - (iv) should have prompted candidates that a Kekulé structure was likely here. Several other structures were allowed providing that they met normal chemistry bonding rules.

Notes from the mark scheme and report

• six-membered alternating P–N ring
• each P bonded to 2 Cl
• each N with 3 bonds and each P with 5 (or 3) bonds for second mark
• zero marks if N bonded to Cl or N with only 1 bond

Skills: M4.2 Visualise and represent 2-D and 3-D forms including 2-D representations of 3-D objects

Mark scheme

Examiner's comment

Candidates are well versed with calculating an Rf value, with nearly all candidates obtaining a value in the acceptable range of 0.12–0.18.

Notes from the mark scheme and report

• working required
• measure from origin line to solvent front (about 9.1 cm), not from the bottom of the plate
• ALLOW 0.12–0.18

Skills: M0.2 Use ratios, fractions and percentages; M3.1 Translate information between graphical, numerical and algebraic forms; PAG6 Synthesis of an organic solid

Mark scheme

Examiner's comment

This question was answered well by candidates, with most showing correct structures for the organic reactant and its product, propan-1-ol, and skeletal formulae mostly used. The question asked for an equation for alkaline hydrolysis and candidates were expected to use an alkali. Acceptable answers would include NaOH/KOH and NaBr/KBr, or OH– and Br–. Equations including H2O and HBr were not given a mark, a common error for alkaline hydrolysis.

Notes from the mark scheme and report

• structures required (DO NOT ALLOW C3H7Br/C3H7OH)
• OH– and Br– or KOH and KBr
• DO NOT ALLOW H2O and HBr
• DO NOT ALLOW K–OH covalent

Mark scheme

Examiner's comment

This novel question assessed whether candidates realised why chemists used TLC when carrying out organic reactions. A good response would identify the following key features after 20 minutes: • The C–Cl bond energy is greater than C–Br and so the reaction would be slower. • The haloalkane spot would be larger and the propan-1-ol spot smaller. • The propan-1-ol is spotted on the chromatogram to monitor the progress of the reaction. The question differentiated very well between candidates, but many did not seem to know where to start with many candidates not scoring any marks. This suggested that candidates recognised chromatography as a technique but did not appreciate its relevance in organic chemistry. Some candidates referred to pigments, recalling their early chromatography experiments in finding the colours in ink.

Notes from the mark scheme and report

• propan-1-ol spot smaller/1-chloropropane spot bigger
• C–Cl bond stronger than C–Br so reacts slower (C–Cl bond required
• DO NOT ALLOW chloride/bromide
• IGNORE halogen reactivity)
• propan-1-ol run to show product formation/monitor progress
• IGNORE 'as a control/comparison' alone

Skills: PAG6 Synthesis of an organic solid

Mark scheme

Examiner's comment

Although high attaining candidates responded with the formulae of chromium-containing species, it was common to see organic compounds being suggested. Consequently, a large proportion of candidates did not score either of the 2 marks. Many candidates seem to expect to only give organic species in their responses on this paper and would benefit from understanding that inorganic species may also need to be provided.

Notes from the mark scheme and report

• green Cr3+ (or [Cr(H2O)6]3+)
• orange Cr2O7 2–
• formulae AND charges must be correct
• DO NOT ALLOW Cr6+
• 1 mark if the wrong way round

Skills: PAG7 Qualitative analysis of organic functional groups

Mark scheme, with the levels and indicative points

Examiner's comment

This Level of Response question was answered well with many candidates identifying compounds F–I correctly to reach Level 3. Structures were usually shown skeletally and this practice is to be recommended. Not only is it far quicker and clearer, it eliminates writing every atom in a displayed or structural formula. Some candidates were not given marks for missing hydrogen atoms or for ‘sticks’ being shown. In these structures, the chemical meaning of a stick is a terminal CH3 group. Candidates were also asked to show how the results of the chemical tests helped the identification of the unknown compounds and this formed the basis of the communication strand of the LOR mark. Candidates answered this part of the analysis extremely well and most were given marks for their good communication skills. This question differentiated very well between well-prepared and less confident candidates. The latter often did not know how the results of these organic tests can be used to identify the functional groups present. It was common for such candidates to identify only one of the four compounds, scoring within Level 1 only. This exemplar is concise and very clear. The candidate has clearly linked the result of each test to the functional groups that must be present. The candidate has drawn skeletal formulae and clearly has experimented with many possible structures before deciding on which must be correct. Notice that the candidate has crossed out the structures that they have rejected. This is an important exam technique – if two structures are drawn, with one correct and the other incorrect, the correct structure cannot be given marks. The response is clearly at Level 3 for the four correct structures and the good communication ensures that the communication strand can be given. This response received all 6 marks.

Notes from the mark scheme and report

• H and I carbonyl (2,4-DNP)
• I aldehyde (Tollens')
• F, G, I oxidised by dichromate
• F has C=C (bromine): F unbranched unsaturated primary/secondary alcohol, G alicyclic alcohol (DO NOT ALLOW tertiary OH), H pentan-2-one or pentan-3-one, I pentanal
• To better prepare candidates, we recommend using either the digital multiple choice quizzes on Teach Cambridge or creating targeted practise materials using ExamBuilder. If you are unsure of how to access these or ways to make the most of them, get in touch via science@ocr.org.uk.

Skills: PAG7 Qualitative analysis of organic functional groups