Six-mark level of response questions

Mark scheme, with the levels and indicative points

Examiner's comment

Many candidates scored highly on this level of response question, answering all three parts. Generally, candidates found it easier to explain the difference in reactivity between Mg and Ca than the other parts of the question. Some lacked clarity in their responses, describing a decreased “electrostatic attraction” without reference to the nucleus and/or outer electrons. Some muddled the order of reactivity describing the gain of electrons. A few candidates described a general trend down the group without linking to the position of Mg and Ca in the group. A number of candidates had an incorrect equation forming CaO rather than hydroxide and occasional errors in the formula, e.g. CaOH or balancing of the equation were seen. Most gave at least one correct observation. Some candidates suggested that bubbles would be seen but did not have hydrogen pro- duced in their equation. Most candidates were able to recognise that redox meant that oxidation and reduction took place. How- ever, some were not clear here, e.g. “it is redox as calcium is oxidised”. Most understood that to explain why this was a redox reaction they needed to give oxidation number changes and/or number of electrons lost or gained. Some errors were seen in assigning oxidation numbers such as H in water, which was given as +2, rather than +1 for each atom. Lower-attaining candidates often suggested that oxygen was reduced. Candidates should be aware that they do not need to fill the answer space provided. Often longer re- sponses introduce contradictions in responses. An excellent response which achieved Level 3 and gained all 6 marks.

Notes from the mark scheme and report

• Ca + 2H2O → Ca(OH)2 + H2, fizzing and Ca dissolves/white ppt
• Ca 0 to +2 oxidised, H +1 to 0 reduced (error: +2 as initial oxidation state of H)
• compare Mg and Ca directly - larger radius, more shielding, less attraction, lower ionisation energies

Mark scheme, with the levels and indicative points

Examiner's comment

Many understood the necessary steps to purify a liquid and were confident with the calculation. Most knew at least one or two purification steps and could explain their purpose. Reasons marks were lost included: • Not remembering the correct name for equipment i.e. separating funnel • Getting the layers the wrong way round despite being given density information • Unclear which layer an anhydrous salt was added to for drying • Using an inappropriate salt for dry step, e.g. Na2CO3 or MgCO3 • Slips in calculations, e.g. incorrect Mr values, yield inverted or not accounted for • Thinking a solid would be formed • Describing reflux (including drawing apparatus) • Saying to distil at a specific boiling point but not stating the actual value (given in the question) As with all calculations, it is important to set out working clearly as it helps to avoid slips. Remind candi- dates of the importance of checking their Mr values – many made slips here. Candidates should also avoid rounding intermediate values and consider the number of significant figures used for the final re- sponse. Some rounded intermediate values to less than 3 significant figures but gave a final response to 4 or more. The response scored Level 3 and gained all 6 marks. All three purification steps are given in the correct order and explained. The calculation for mass is set out clearly and the final mass is given to 3 signifi- cant figures.

Notes from the mark scheme and report

• separating funnel - organic layer is the top (less dense) layer
• dry with anhydrous MgSO4/CaCl2
• distil collecting fraction at 70 °C
• 0.125 mol × 0.65 × 92.5 = 7.52 g
• IGNORE washing with carbonate/water
• ALLOW one small slip e.g. Mr 73

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG5 Synthesis of an organic liquid

Mark scheme, with the levels and indicative points

Examiner's comment

The first level of response question in the paper was answered well. Almost all candidates were able to draw correct dot cross diagrams. Many candidates scored Level 3 with correct shapes, bond angles and gave good explanations of the electron pair repulsion. The mark scheme allowed for the taught bond angle values and the true bond angle values of these molecules which followed the same trend. Level 2 candidates often had an incorrect shape of either PH3 or H2S. The tetrahedral shape for SiH4 was almost always correct but less successful candidates suggested trigonal planar or bipyramidal for PH3 and linear for H2S. Bond angles were often correct with some candidates linking the incorrect angle to their incorrect shape. Explanations were well described but some candidates were not clear about electron pairs repelling but described atoms or bonds repelling. Some candidates explained the bond angle in terms of being reduced by 2.5o without referencing electron pair repulsion Most candidates were able to access Level 1 often by having the correct dot cross diagrams.

Notes from the mark scheme and report

• SiH4 tetrahedral 109.5° with 4 bonding pairs
• PH3 pyramidal with 3 bonding pairs and 1 lone pair
• H2S non-linear with 2 bonding pairs and 2 lone pairs
• bond angle SiH4 > PH3 > H2S
• lone pairs repel more than bonding pairs
• DO NOT ALLOW atoms repel
• IGNORE 'electrons repel' and 'bonds' for bonded pairs

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, with the levels and indicative points

Examiner's comment

This was our second Level of Response question which proved more challenging for candidates. Only a few candidates scored full marks. Candidates are reminded to structure their answers and make sure they answer all points in the question. Candidates often find pH of buffer solution calculations difficult to process. The most difficult part in this calculation was the concentration of the salt. The majority that calculated salt concentration were able to calculate the mass. We should encourage the candidates to always write the units when calculating any quantity as concentration and moles were often confused within the calculation. Some candidates were given Level 2 for converting what they thought was the salt concentration into a value for the mass of salt needed, using a correct method. Less successful candidates calculated the correct [H+] but then used a weak acid calculation approach. A few responses attempted to use the Henderson-Hasselbalch approach, but most were unsuccessful due to the mathematical inaccuracies. The buffer action description proved to be a more challenging part to the question. Some candidates wrote a clear and concise description, linked to a written equilibrium equation, of the reaction of an alkali and the resulting equilibrium shift. Level 2 candidates often described an equilibrium shift of a non- existent equilibrium equation. Level 1 candidates were often only able to calculate [H+] and tried to explain the action of a buffer.

Notes from the mark scheme and report

• n = 0.168 × 0.125 = 0.0210 mol
• mass = 0.0210 × 96.0 = 2.02 g (2.00–2.04)
• added OH– reacts with H+/CH3CH2COOH and equilibrium CH3CH2COOH ⇌ H+ + CH3CH2COO– shifts right restoring [H+]
• use full formulae rather than HA/A–

Skills: M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.2 Change the subject of an equation; 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, with the levels and indicative points

Examiner's comment

Generally, this question was well answered with most candidates achieving Level 3. Many candidates knew the correct reagents to use to carry out the two-stage synthesis. Some missed out on Level 3 as they had the steps in the wrong order. It was essential to carry out the oxidation prior to substitution otherwise both -OH groups would be oxidised. A significant number struggled with the calculation of the mass of compound D required. Many gave the answer as 7.94g so had only considered the yield, not the difference in molar mass of F compared to D. Other common errors seen in the calculation included errors in Mr values, missing or inverted percentage yield, as well as intermediate rounding of values to two significant figures. Level 2 could still be achieved without the calculation if they had the correct reagents in the correct order with equations mostly correct. To be mostly correct we required them to have at least one of the inorganic reactants or products given. Communication marks were lost if there were errors in reactants or products as well as balancing of equations. Students using skeletal structures generally made fewer mistakes with the balanced equation than those using structural formulae, where additional Hs were inserted alongside the ketone group i.e. CH3CHClCH2CHOCH3. Lower attaining students attempted to write the H+/K2Cr2O7 in the balanced equation and didn’t know to represent this with [O]. Other common errors in balancing the oxidation equation included missing H₂O or using 2[O]. A common error was to see H2SO4 as a ‘catalyst’ added with NaOH in the substitution reaction – this additional and unnecessary reagent lost candidates the communication mark. This response achieved Level 3 and all 6 marks. The calculation is laid out clearly showing both Mr values used in the calculation. Showing the values used in each step makes it much easier to spot where any errors may have been made if the correct value is not obtained. The reagents for each stage are given and the reactions are carried out in the correct order. The equations are given and are correctly balanced. This was an excellent response.

Notes from the mark scheme and report

• m(D) = 9.53 g (0.946 g inverted yield, 3.00 g yield omitted are partly correct)
• oxidation with H+/Cr2O7 2– using [O] then substitution of Cl with NaOH(aq)
• IGNORE conditions
• inappropriate acid (HCl) or extra reagents limit communication

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme, with the levels and indicative points

Examiner's comment

Most candidates achieved Level 3 and were able to gain 5 or 6 marks. The communication mark was given when a clear justification was given for either ester from their NMR data or from the mass spectrum fragment ions. The highest attaining candidates were able to organise their interpretation of the NMR data clearly in a table, showing how they used the information on chemical shift, integration and splitting to produce a fragment and finally linking these together to give a correct structure. However, for some candidates, while information was correctly determined, this was not then used to give a final structure. Some candidates analysed the fragment peaks from the mass spec. The best solutions linked the structure with peaks in the mass spectrum, e.g. CH3CH2+ peak at m/z 29 and C3H7+ peak at m/z 43, even though these fragments were often missing a charge when referred to. Some candidates only achieved Level 2 as although their structures had the correct molecular formula they contained -OH (suggesting N was a hydroxyl ketone or carboxylic acid) rather than H-C-O (found in esters) by misinterpreting the peak at 4.4 ppm as an O-H peak. Candidates should be reminded that the H bonded to an O will not couple with neighbouring Hs and can be present anywhere in the spectrum. This is usually highlighted by running the spectrum in D2O, removing this peak. Some candidates correctly identified parts of the molecule from their analysis of the NMR data but then gave a molecule which didn’t fit this, e.g. identified the (CH3)2CH fragment but then did not show it in the final structure or included a -COOH despite there being no peak at 10-12 ppm. Some, despite understanding that the peak at 2.4 ppm was next to -C=O and showed quartet splitting so had 3 neighbouring H, would assign it as a carbonyl group with the methyl group on one side and a CH2 on the other side. Lower attaining candidates struggled to interpret the NMR data, especially the presence of 6 equivalent Hs – many translated this as being a straight chain rather than 2 x CH3 groups. It is also important that candidates make use of the chemical shift information on the data sheet to give a chemical environment and try to use that to piece together their structure. Almost all candidates attempted to calculate the empirical formula and determine the molecular formula of the unknown compound, N, and therefore secured Level 1. However, many of these candidates gave structures which were not consistent with the molecular formula. This limited them to achieving Level 1, 2 marks only. Also, those without a final structure could only score a maximum of 2 marks. This response achieved Level 2 and 4 marks. The final structure is a hydroxy ketone, not one of the correct esters, but does have the correct molecular formula and C=O as well as an ethyl group. The candidate has analysed some of the NMR data. The communication mark has been given as the analysis shown in their response is easy to follow with clear information given for each NMR peak.

Notes from the mark scheme and report

• empirical formula C3H6O
• M+ at 116 gives C6H12O2
• triplet 3H 1.1, doublet 6H 1.4, quartet 2H 2.4, heptet 1H 4.4 (ALLOW multiplet)
• CH3CH2COOCH(CH3)2 (or (CH3)2CHCOOCH2CH3 justified from m/z 71)

Skills: M0.2 Use ratios, fractions and percentages; M3.1 Translate information between graphical, numerical and algebraic forms

Mark scheme, with the levels and indicative points

Examiner's comment

Most candidates were able to gain some marks for this Level of Response (LoR) question, with very few obtaining no marks. To achieve Level 1, candidates needed to identify at least 2 of the 3 orders with no further correct analysis. For Level 2 candidates were also expected to show the correct rate equation to match their orders or to provide a reason for why the mechanism must proceed via more than one step. Most candidates reaching Level 2 wrote a correct rate equation and usually calculated the rate constant. At this level, comparatively few produced a valid analysis of the mechanism. For a Level 3 response, all three aspects of the analysis were required: correct orders, correct rate equation and rate constant, and a correct explanation of the mechanism. Many candidates were able to provide all three aspects. Candidates usually determined the 1st order with respect to propanone from the initial rates data provided. The 1st order with respect to H+ from the rate–concentration graph was often seen. The zero order with respect to bromine from the concentration–time graph caused more problems, with many candidates incorrectly choosing 1st or 2nd order. The communication strand for this LoR question was given if the candidate had linked their orders with the evidence provided. Most candidates gave the initial rates but the evidence from the graphs was often omitted and vague. Responses such as ‘because of the graph’ are inadequate. The candidate’s response also had to be structured in a comprehensive way. This strand was usually given. beyond the minimum requirements for Level 3 including the communication strand.

Notes from the mark scheme and report

• propanone first order ([ ] × 3, rate × 3)
• Br2 zero order (constant gradient on concentration–time graph)
• H+ first order (straight line through origin)
• rate = k[CH3COCH3][H+]
• k = 4.53 × 10^-4
• Br2 in overall equation but not in rate equation (and H+ in rate equation but not overall) so more than one step

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M3.1 Translate information between graphical, numerical and algebraic forms; PAG10 Rates of reaction – initial rates

Mark scheme, with the levels and indicative points

Examiner's comment

As with the other Level of Response (LoR) question (Question 3(a)), most candidates were able to gain some marks for this question, with very few obtaining no marks. Overall fewer candidates were given Level 3 for Q6 than for Q3(a). To achieve Level 1, candidates needed to calculate a value for the energy change q from mc∆T using at least two of m, c and ∆T. Candidates needed to first process the raw experimental results provided to extract the required values for the calculation. In the mc∆T calculation, the most common incorrect values used were 1.97 (the mass of Na2CO3) for m, 8.314 (the gas constant) for c and 277 (Kelvin) for ∆T. Only the lowest attaining candidates were unable to carry out this task. For Level 2, candidates were expected to determine the energy change per mole using the correct mc∆T value. Most candidates reached Level 2. A Level 3 response proved to be more elusive, requiring candidates to use an energy cycle to determine an enthalpy change. This task was far more testing than the requirements for Level 2. Some leeway was allowed for values used within the response. For m, a value in the range of 30 – 33.5 was allowed to take into account either HNO3(aq) or the mass of solution. Responses calculated per mole of Na2CO3 or HNO3 were allowed. This allowed for the fact that, even though the question described an excess of HNO3 being used in the reaction, the volume and concentration specified (30 cm3 of 1.00 dm3 HNO3) is short of being an excess. The approach taken is fully outlined in the mark scheme for this question and ensured no students were disadvantaged. The communication strand for this LoR question was given for responses that were arranged in a cohesive way. Unfortunately, a significant number of candidates produced responses that were difficult to follow with unlabelled quantities and numbers spread randomly across the page.

Notes from the mark scheme and report

• m = 33.50 g (mass of final solution, not 30 cm3)
• q = 33.50 × 4.18 × 4.0 = 560 J
• n(Na2CO3) = 1.97/106 (or via n(HNO3) = 0.0300 if acid limiting)
• ∆H2 = –30.1 (temperature rise so exothermic)
• ∆H1 = 2∆H3 – ∆H2 = +69.7 kJ mol–1
• energy cycle and signs

Skills: M0.0 Recognise and make use of appropriate units in calculations; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations; PAG3 Enthalpy determination

Mark scheme, with the levels and indicative points

Examiner's comment

The calculation of enthalpy change was generally well-answered and the majority of candidates were able to recall the equation q = mcT. Many candidates forgot the minus sign or gave a positive sign for final enthalpy change. Errors in calculation were most commonly for using an incorrect mass, usually by finding the mass of copper(II) nitrate (from moles and Mr). Some also used the wrong value for the heat capacity, selecting the value for R from the data sheet instead. Many gave the final answer to an inappropriate six significant figures. Candidates often found it challenging to give appropriate assumptions and improvements, limiting the level achieved to Level 2. However, many candidates did correctly give the assumptions that the specific heat capacity and density of the solution was the same as water. This is usually stated for the candidates with these types of questions. The most common improvements suggested were use of a polystyrene cup, adding a lid or using a thermometer with a higher resolution. Quite a few candidates suggested using a larger volume of solution which would indeed reduce the % uncertainty in the volume measurement. However, it would lead to a smaller temperature change, increasing the % uncertainty in the temperature measurement. Some confused the question with a calculation of enthalpy change of combustion and gave improvements accordingly, e.g. ‘use a copper or bomb calorimeter’, ‘draft shields’, ‘heat for longer’, ‘position of flame and supplies of oxygen’. This response achieved Level 3 - 6 marks. There is a correct calculation for H, the final value has a correct negative sign and is given to 4 significant figures. Lots of valid assumptions and improvements are given.

Notes from the mark scheme and report

• q = 100 × 4.18 × 18.6 = 7.77 kJ
• n(Cu(NO3)2) = 0.05
• ∆H = –155 kJ mol–1 with negative sign
• assumptions (density/c same as water, no heat loss, ignore zinc)
• improvements (polystyrene cup, lid, cooling curve, more precise thermometer)

Skills: M0.0 Recognise and make use of appropriate units in calculations; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG3 Enthalpy determination

Mark scheme, with the levels and indicative points

Examiner's comment

Over a third of candidates achieved Level 3, gaining 5 or 6 marks. A correct structure (either aldehyde or ketone) alone was not enough to award Level 3 and candidates were expected to give a comprehensive description of how the evidence helped them determine the structure. The biggest challenge for many candidates was finding the correct empirical formula. The ratio worked out to 1:1.33:1 so many incorrectly rounded this to either 1:1:1 or 1:2:1, which meant they struggled to find a molecular formula that worked and added up to 88. Incorrect molecular formulas seen included C3H3O3, which adds to 87 (often the extra H was just added to make it fit), or C4H8O2, which does add to 88. Most candidates could analyse the IR spectrum, identifying peaks corresponding to C=O or O-H. Candidates should identify bonds present before making conclusions about the functional groups. Many were able to use mass spectra to determine the Mr value from the M+ peak. Some did go on to make use of other peaks, identifying fragments and confirming whether the structure was an aldehyde or ketone depending on analysis. For example, CHO+ at m/z = 29 suggests an aldehyde, or conversely CH3+ at m/z = 15 suggests a ketone. Candidates should always be encouraged to comment on all the data provided. This can be through good annotation of the spectra and notes added to the first page of the question. Many candidates didn’t mention the evidence from the bromine test. If candidates pieced together information to give a structure that is chemically feasible containing either a C=O or COOH group then they could achieve Level 2. Without a structure they were limited to Level 1. The most common incorrect structures seen included butanoic acid, 2-hydroxypropenoic acid or structures with 2 × C=O and an alcohol OH. This response achieved Level 2 - 4 marks. Despite a correct calculation for empirical formula, they rounded incorrectly to CHO. They then state that molecular formula is C4H8O2 which matches the Mr value determined from the mass spectrum but not their empirical formula. They demonstrated good analysis of the IR spectrum and have even looked at potential fragments from the mass spectrum. The response meets the Level 2 descriptor as they have a feasible structure with a COOH group and have some analysis of all three scientific points. The communication is good with a clearly laid out calculation and bonds identified from IR and MS fragments have a positive charge.

Notes from the mark scheme and report

• empirical formula C3H4O3
• M+ at m/z 88 so molecular formula C3H4O3
• IR broad O–H 2500–3300 and C=O 1630–1820 so carboxylic acid
• no C=C (bromine water)
• fragments m/z 43 CH3CO+, 29 CHO+, 15 CH3+ with positive charge
• CH3COCOOH or OHCCH2COOH

Skills: M0.2 Use ratios, fractions and percentages; M3.1 Translate information between graphical, numerical and algebraic forms

Mark scheme, with the levels and indicative points

Examiner's comment

The first Level of Response question in the paper was answered well with the higher-attaining candidates on the paper scoring full marks. These students started with a nice clear analysis of the half-life, referring to labelled sections of the graph, then went on to calculate the rate from a well-drawn tangent with correct indices and were careful to write down the correct units. Then used the rate equation to calculate K and get the correct units. Almost all candidates were able to conclude that it was first order for CV. Most used the half-life approach with others comparing two gradients. Candidates should be advised, especially in LoR questions, that the conclusion needs a clear link to the data. The gradient at 3 minutes was done well and the candidates used the appropriate scale from the graph in their calculations. Clarity of communication does require indication of what is being calculated and how the numbers were obtained. This would allow an initial rate versus a rate at three minutes to be distinguished. Units were particularly important in this question as some candidates used minutes from the graph while others converted time into seconds. A common error was to state the wrong units or leave them out altogether. Most candidates used the rate equation to calculate K and get the correct units. A few approached the value by using k = ln2 t½ . Error carried forward was given for those with incorrect half-lives or rate value. Candidate errors arose from graph readings that caused rate to be wrong, errors in concentrations used in rate or K calculation, and badly drawn tangents causing the rate to be out of tolerance. Some candidates tried to adjust the concentrations as if conducting mole calculations or take the rate from two points on the graph. The candidate scored Level 3. The graph was clearly used to obtain half-lives and gradients. This was communicated on the answer lines, showing calculations and units.

Notes from the mark scheme and report

• half-lives or gradients on the graph
• tangent at 3 min
• k from t½ = ln2/k or from rate/[CV]
• units of k
• M3 section of the Maths Skills handbook contains useful information on use of graphs in chemistry, including M3.5 on drawing and using the slope of the tangent to a curve as a measure of a rate of change: Teach Cambridge (ocr.org.uk)

Skills: M0.0 Recognise and make use of appropriate units in calculations; M1.1 Use an appropriate number of significant figures; M3.1 Translate information between graphical, numerical and algebraic forms; M3.5 Draw and use the slope of a tangent to a curve as a measure of rate of change

Mark scheme, with the levels and indicative points

Examiner's comment

Structure and bonding continue to be a difficult concept for many candidates. High-attaining candidates were able to identify why the element had a certain magnitude of melting point. They clearly linked the structure type with the type of bonding. They then described, in detail, the nature of the bond. The strength of force required to break/overcome the bond/London Force was linked to the melting point. It was very common for ‘giant’ to be omitted in the name of the lattice, especially in Al. Candidates find it particularly challenging to associate the correct terminology with the correct structure, often describing intermolecular forces in giant covalent explanations or use of molecules in giant metallic explanations. London forces were mentioned widely but sometimes not described as being forces between molecules and not linked to the increased number of electrons. A holistic, rather than a point based, approach is used in marking these responses. This allowed Level 2 to be given when the candidate did not use all of the correct terminology throughout the three structure types. Several candidates described the varying melting point going across the period as being due to atoms having more electrons in the outer shell and a greater nuclear charge.

Notes from the mark scheme and report

• giant metallic (Al), giant covalent (Si), simple molecular P4/S8 with London forces
• S8 > P4 because more electrons
• Our bonding delivery guide provides details of common misconceptions students hold relating to this topic, and also includes resources and guidance that can help overcome them: Teach Cambridge (ocr.org.uk)

Mark scheme, with the levels and indicative points

Examiner's comment

A very good proportion of candidates scored all 6 marks, giving well-drawn mechanisms with some additional details such as mechanism names, functional group interconversions or other additional reaction information. Some attempted to ‘describe’ the mechanism using only words rather than drawing it out with a conventional curly arrow mechanism. Candidates may need more clarity on what ‘describe’ means in an organic chemistry context. Equally, a few gave just the mechanisms with no additional details, limiting themselves to Level 2. Candidates were usually more confident with the addition to alkenes using an electrophilic addition mechanism. Some gave additional details about major and minor products, although not always relevant as for a symmetrical alkene. Some represented the addition of hydrogen or water to alkenes via an electrophilic addition mechanism. While not correct it showed an understanding of mechanisms and a correct addition reaction for alkenes, so credit was given. Some candidates included incorrect reagents for reactions, such as acid catalysts with addition of a hydrogen halide, or incorrect conditions, such as the requirement for ultraviolet light on addition of a halogen. The addition to carbonyl compounds was not always as well-described. Some candidates struggled to identify carbonyl compounds, selecting carboxylic acids or their derivatives, with attempts at addition- elimination mechanism i.e. condensation reactions. Some gave incorrect reagents for carbonyls, including H2 and HBr. However, some used off-specification reactions such as the addition of H2O to form a geminal diol which was given but as the mechanism differs from the nucleophilic addition mechanism taught in this specification, full credit was rarely achieved. Some also considered oxidation of aldehyde or ketone to be an addition reaction. Most who presented a correct mechanism for addition to a carbonyl used the reaction with cyanide rather than reduction with NaBH4. Common errors included arrows coming from the N of CN–, a lack of putting dipoles on carbonyl bonds, missing charges on O in intermediates or showing the wrong direction of arrows. Two correct addition mechanisms have been shown, one for an alkene and one for a carbonyl compound. Additional details include the names of the mechanisms, names of the functional groups in the products, and the fact that a C-C bond is formed in the second mechanism. All curly arrows, charges and dipoles are correctly positioned so this response was also given the communication mark.

Notes from the mark scheme and report

• alkene with Br2/HX by electrophilic addition and carbonyl with HCN (NaCN/H+) or NaBH4 by nucleophilic addition
• curly arrows from bonds/lone pairs/negative charges, dipoles and charges on intermediates
• IGNORE addition–elimination reactions of carboxylic acid derivatives
• The OCR Guide to Level of Response questions can be found on Teach Cambridge and can be used to help your students better understand this type of question.

Mark scheme, with the levels and indicative points

Examiner's comment

This question was well-attempted by most candidates, with the majority of candidates gaining full marks or gaining 4 marks for a top Level 2 response. Many candidates showed excellent recall of how to determine the correct empirical formula from the percentage composition data. Most then went on to use the m/z peak on the mass spectrum to confirm that the Mr was 134, and therefore the molecular formula was identical to the empirical formula. A few also made use of the mass spectrum to identify possible fragment ions including a correct positive charge. Most candidates used the IR spectrum to identify a C=O bond and many also mentioned the absence of O-H or spotted C=C for arenes. Lower attaining candidates sometimes incorrectly mentioned the presence of a carboxylic acid O-H despite the molecular formula only having 1 oxygen atom. Many candidates annotated the NMR spectrum and/or presented their analysis clearly in a table format and were able to identify aldehyde and arene hydrogen environments. The best candidates had fragments built up alongside their NMR analysis clearly building them using chemical shift, integration ratios and splitting patterns. Those that struggled to interpret the splitting patterns correctly suggested incorrect structures but often with correct features so were still able to score Level 2, 4 marks. Some initially identified the multiplet peak at 3.8ppm as being HC-O environment but many realised this did not fit the IR data. However, some changed other evidence to fit this, e.g. the peak at 9.0 ppm being an O-H rather than CHO and the IR having C=C only without C=O as well. A large proportion of candidates were able to correctly determine the structure of compound J, recognising that the peak at 3.8 ppm was shifted up-field as adjacent to both the benzene ring and the aldehyde group. The data sheet refers to this: ‘CH bonded to ‘shifting groups’ on either side, e.g. O– CH2–C=O, may be shifted more than indicated above’. Several candidates who did not get the correct structure gave structures which were chemically unfeasible, e.g. with pentavalent carbons. Many candidates had several structures as part of working but did not always ensure their final structure was clearly highlighted. A very small number of candidates received no credit for this question, as the majority were able to show analysis of 2 aspects, e.g. the calculation of empirical formula and labelling of IR or NMR spectra. There is clear and detailed analysis throughout this response to determine the correct final structure for J. The empirical formula calculation shows how the empirical formula was determined. On the mass spectrum the annotation links to the Mr of 134 and at the end of the response they have identified the fragment responsible for the parent ion. The C=O IR peak is labelled and described in the response. The NMR analysis is clear, with each peak being numbered and linked to a table which shows how the candidate has identified the hydrogens responsible for each peak as well as linking to neighbouring hydrogens from splitting patterns. The final compound is labelled as such to distinguish it clearly from other structures given, which were part of their problem solving to find a structure that fits all of the analysis they had completed.

Notes from the mark scheme and report

• empirical = molecular formula C9H10O (M+ at m/z 134)
• fragments m/z 105 C6H5CHCH3+, 77 C6H5+, 29 CHO+ (positive charge)
• IR C=O about 1700 cm–1
• NMR δ 1.4 doublet 3H, 3.8 quintet 1H, 7.3 5H C6H5, 9.0 doublet 1H CHO
• structure C6H5CH(CH3)CHO

Skills: M0.2 Use ratios, fractions and percentages; M3.1 Translate information between graphical, numerical and algebraic forms

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, 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

Mark scheme, with the levels and indicative points

Examiner's comment

Very few candidates managed to access the higher level on this question. They found the calculation very challenging. In addition to a typical mole calculation, they needed to use the tonnes conversion (given on the data sheet), look at the ratio of I2 to KI and use a mass concentration. Many used the mass concentration as a molar concentration. Even if a balanced equation was given, it didn’t guarantee the correct ratio in calculations and a common error was to find moles of I2 using the relative atomic mass of iodine (126.9). As with previous calculations on this paper, it was often challenging to follow the logic in the steps of the calculation. It is important to encourage candidates to label each step of their calculation, showing them as independent steps and only using = when terms are equal. Explanations of relative reactivities of chlorine and iodine often lacked detail or had extra irrelevant information, such as reference to electronegativity, ionisation energy, bond enthalpy or intermolecular forces. Some candidates gave responses for the general trend in reactivity in the group rather than comparing Cl and I. Confusion around nuclear charge and nuclear attraction was also often seen. A key skill for longer answer questions is to plan out answers to avoid contradictions. Candidates should also be encouraged to re-read their answers to check that what they have written makes sense. This exemplar demonstrates a L3 6 mark response. A comprehensive explanation of relative reactivities of chlorine and iodine is given. The steps in the calculation are logical with each step labelled, making it easy to follow. A balanced equation is also given.

Notes from the mark scheme and report

• comparison required throughout
• IGNORE 'nuclear charge' for nuclear attraction
• 2:1 ratio I–:I2
• n(I2) = 1 × 10^6/253.8
• acceptable errors: 126.9 used for I2 (1.74 × 10^7) or missing × 2 (4.36 × 10^6), not both for L3

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.1 Recognise and use expressions in decimal and ordinary form; M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme, with the levels and indicative points

Examiner's comment

Many candidates limited themselves to only L1 here by not calculating the enthalpy change from the student’s results. Remind candidates to read the question carefully and to use all the information provided (such as tables of results). Some missed out on L3 as their evaluation lacked detail, as they needed to include at least two differences and two suitable improvements. Not all improvements given were suitable, with some suggesting candidates confusing different practicals - for example, ‘use a polystyrene cup’ or ‘heat to a constant mass’. It was evident that some lacked understanding about complete combustion, with suggestions to ‘measure mass until it doesn’t change any further’ or ‘heat water for longer’ to make sure complete combustion had occurred. Some also suggested ‘using a larger mass’ but further information was needed for credit to be given. This exemplar shows a candidate who achieved L3 with all 6 marks given. Working for the calculation of H is clear with all steps shown, in a logical order, and the final answer has correct negative sign. More than 2 differences and 2 improvements are given. The communication is clear, linking potential differences with suggestion for improvements.

Notes from the mark scheme and report

• q = 200 × 4.18 × 31 (mass of water, not fuel)
• n = 1.31/74
• ∆cH = –1460 to –1464
• heat losses, incomplete combustion, evaporation, non-standard conditions
• improvements: draught shield, lid, copper can, cap on wick, bomb calorimeter
• Further information about practical skills assessed on written exams can be found in Section 3 of the practical skills handbook. If using the suggested practical activities developed by OCR, then encourage candidates to answer the extension opportunity questions to help develop a deeper understanding in preparation for written assessments.

Skills: M0.0 Recognise and make use of appropriate units in calculations; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG3 Enthalpy determination

Mark scheme, with the levels and indicative points

Examiner's comment

The first Level of Response question in the paper was answered well. Almost all candidates were able to conclude that the experimental results showed that they were consistent with first order with respect to Br-. Some candidates were able to use a sequential approach to determine the orders with respect to BrO3- and H+, reaching a Level 3, but others found this more problematic. Some did not notice that more than one concentration had been changed between experiments. This led to many determining the rate to be 0 order with respect to [BrO3-] and [H+]. Candidates should focus on the quality of their descriptions when linking data to their conclusions with some candidates creating their own data set to fit their explanations. Having determined orders, nearly all candidates were able to give a corresponding rate equation and could calculate a value for the rate constant, albeit with frequent omission of units. Some candidates confused the rate equation with a Kc expression.

Notes from the mark scheme and report

• first order Br– (expts 1 and 2)
• first order BrO3– (expts 1 and 3, both concentrations change)
• second order H+ (expts 1 and 4, ×5 gives ×25 after allowing for BrO3–)
• rate = k[Br–][BrO3–][H+]2
• k = 16.4 from any row of data

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG10 Rates of reaction – initial rates

Mark scheme, with the levels and indicative points

Examiner's comment

Only the higher-attaining candidates scored full marks. Very few candidates were able to define d-block element correctly without the minor slip of saying outermost electron instead of highest energy or valence electron. Many candidates often did not mention ions for the transition metal definition. Many did not include any of the electron arrangements for d-block elements and transition elements. The majority of candidates who were able to recognise zinc and scandium as d-block elements, but not transition elements, gave their electronic configurations correctly. Common errors included the presence of the 4s electrons in the electron configurations of the ions and incorrect electron configurations of copper and chromium atoms. A few candidates thought chromium and copper were not transition elements due to the 4s1 electron configuration. This candidate has mentioned outer electrons rather than highest energy but was this is a minor slip and they were still given a Level 3 response as everything else is correct. A holistic approach for LoRs is used and not a point-based marking system.

Notes from the mark scheme and report

• d-block: highest energy electron in d sub-shell (DO NOT ALLOW d shell/d block)
• transition element forms an ion with incomplete d sub-shell
• Sc3+ (empty 3d) and Zn2+ (full 3d10) with configurations

Mark scheme, with the levels and indicative points

Examiner's comment

This challenging level of response question was generally well attempted. Many candidates recognised the reagents required in this synthesis but fewer candidates were able to deduce three correct reagents in the right order with equations to achieve Level 3. Most candidates achieved Level 2 4 marks. Many correctly identified suitable reagents but carried out the stages in the wrong order. A common error was to carry out electrophilic addition with HBr first then react with cyanide ion, not realising both Br groups would react and the reaction would not be selective. Alternatively, having introduced the nitrile group then carried out the reduction first, not realising that the C=C would also be reduced. The lowest scoring responses were often incomplete and despite identifying some reagents did not give equations. Candidates are encouraged to read the questions carefully so they can make sure their response covers all the detail required. Many candidates used molecular formula in their equations. It is usually preferrable in organic chemistry to give structures. Some gave surplus information such as the mechanisms and reaction conditions for each reaction. The key to answering this question well was knowing reagents for different functional group interconversions as well as planning each step to make sure of a logical synthesis. Some candidates were seemingly confused by the term ‘intermediate’ and gave an intermediate as in a mechanism, e.g. carbocation.

Notes from the mark scheme and report

• order matters: CN– (ethanol) first (it would substitute both Br atoms after HBr), then HBr across C=C (Markownikoff), then H2/Ni (would reduce C=C if before HBr)
• ALLOW LiAlH4 as stage 2 or 3
• IGNORE conditions
• A useful resource for teaching how to identify functional groups and practice at devising synthetic routes is the Topic exploration pack on Organic synthesis.
• This should be used in conjunction with the reaction pathways summaries.

Mark scheme, with the levels and indicative points

Examiner's comment

Most candidates were able to deduce the empirical and/or molecular formula of the organic compound. Analysis of the IR spectrum was also well attempted, but some candidates assumed the unknown was a carboxylic acid, attributing the sharp peak just below 3000 cm–1 to an OH group. Others misidentified the C=O peak as a C=C group suggesting alkene or arene structure. They were often led to this conclusion as they believed no precipitate with 2,4-DNP suggested no C=O rather than no aldehyde or ketone. Good analysis of the NMR data was crucial for deducing the correct ester. Some candidates opted to annotate the proton NMR spectrum, some produced tables and others gave written details for each peak. It was vital that they were able to interpret all information for each peak i.e. number of proton environments, the type of environment from chemical shift, the number of protons in each environment from relative peak areas and use of splitting patterns to find information about adjacent protons. Many tried to make the data fit their proposed structure rather than the other way round. Some suggested structures that were only partially consistent with the data such as CH3CH2CH2COOC(CH3)3 and were awarded Level 2. Others did not take full note of all the information provided, for example omitting the 2,4-DNP observations, giving the ketone (CH3)3COCH2COCH2CH3 or not checking it matched the molecular formula CH3CH2COOC(CH3)3 so only achieved Level 1. Candidates need to be encouraged to draw a structure as without they can only achieve a maximum of 2 marks despite some excellent analysis of the data. Conversely, it is not sufficient to just give a structure, candidates must give analysis of the data provided. This is a good Level 3 6 mark response. As well as this clearly laid out analysis they also had details written on the question, e.g. no aldehyde or ketone due to no reaction with 2,4-DNP. This response has been selected due to the detailed analysis of NMR data that has been summarised in a table.

Notes from the mark scheme and report

• C4H8O then C8H16O2 from m/z 144
• IR C=O about 1740 but no reaction with 2,4-DNP so not aldehyde/ketone: ester
• singlet 9H C(CH3)3, triplet 3H + quartet 2H CH3CH2CO, singlet 2H OCH2 at 4.1
• CH3CH2COOCH2C(CH3)3
• IGNORE C–O peaks

Skills: M0.2 Use ratios, fractions and percentages; M3.1 Translate information between graphical, numerical and algebraic forms; PAG7 Qualitative analysis of organic functional groups

Mark scheme, with the levels and indicative points

Examiner's comment

Overall, candidates performed well when answering this question. They were required to identify that compound A shows optical isomerism and to choose a reaction for each of the three functional groups. Candidates were also expected to use structures for the organic products. To achieve the highest level of response, a description of optical isomerism should be accompanied by 3D diagrams of the optical isomers. Optical isomerism was usually identified, with associated diagrams with almost all candidates identifying the chiral centre. Most attempted 3D diagrams but candidates do need to take care that the groups attached to the chiral C atom are those in compound A and that no parts of chains are omitted. Optical isomers do also require use bold and dashed wedges to be used. Most candidates showed good knowledge and understanding of reactions for the three functional groups. • For the primary alcohol, most chose H+/Cr2O72–, with distil (→ aldehyde) or reflux (→ carboxylic acid); a significant number chose a concentrated acid (→ alkene) or Br–-/H+ (→ haloalkane) • For the ketone, most chose NaBH4 (→ secondary alcohol) • For the nitrile, most chose either H2/Ni (→ amine) or H+(aq) (→ carboxylic acid). Clear diagrams of the products were usually seen although many omitted a CH2 from the amine branch for hydrolysis of the nitrile or an extra CH2 in the aldehyde or carboxylic acid branch from oxidation of the primary alcohol. Some candidates chose 2,4-DNP for a reaction of the ketone and treated the question as one requiring tests, and then proving that the compound was a ketone from no reaction with Tollens’ reagent. The question asked for the organic product and the 2,4- DNP product is beyond the demands of this specification (although this was seen very rarely). Candidates adopting this reaction were limiting the extent of their response and candidate should have considered this requirement before selecting 2.4- DNP. This exemplar shows a good response that lacks 3D diagrams for the optical isomers. The candidate has clearly given reagents and conditions and has shown the organic products. In the response, you can see that the candidate initially showed an extra CH2 in the –COOH branch, and a mistake in the amine branch. The absence of 3D structures limits the response to Level 2 and 4 marks have been awarded for choosing correct and relevant reagents and conditions, and for the clear communication of the structures.

Notes from the mark scheme and report

• optical isomerism with chiral centre/non-superimposable mirror images and 3D diagrams with wedges
• ketone: NaBH4 or HCN
• –OH: H+/Cr2O72– (reflux gives acid, distil gives aldehyde), NaBr/H+, acid catalyst elimination
• –CN: H2/Ni to amine or H+/H2O to COOH
• reagents and products must be linked

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

Mark scheme, with the levels and indicative points

Examiner's comment

This level of response question required candidates to interpret three pieces of information to identify 3 unknown chemicals, linked by three reactions of sulfuric acid. Levels were assigned based on identifying the three unknowns and writing equations for the reactions. The 3 reactions were tiered in difficulty with the cyclic structure for compound F being the most difficult. The question discriminated extremely well, with comparatively few candidates not scoring any marks. This exemplar shows an excellent response which was awarded Level 3 and 6 marks. The crossing out shows how the candidate progressively solved the problem. Despite the crossing out, working is shown throughout, and the presentation of the response is very clear. For reaction 1, the candidate initially made an error in the stock formula determination calculation. Notice that the candidate crossed out their initial values and wrote them afresh. This approach is recommended. as changing incorrect numbers can lead to ambiguous numbers. In reaction 2, the candidate comfortably identified compound E from its molar mass and wrote a balanced equation for this reaction. For reaction 3, you can see how the candidate’s thought developed during the solve. They initially went for a straight chain structure and then worked out that there must be a cyclic structure. They finally show the cyclic diester as a conventional structural (based on glucose?), and with a balanced equation. This was an excellent response of a high-ability candidate and shows what a well-prepared candidate is capable of achieving.

Notes from the mark scheme and report

• D = NiSO4•6H2O from mole ratio 1:1:10:12
• E = SO2 (2.67 × 24 = 64)
• H2SO4 + 2HBr → Br2 + SO2 + 2H2O
• F = cyclic diester from 2 CH3CH(OH)COOH + 2H2O (ALLOW unsaturated ester/acid or anhydride isomers)
• IGNORE connectivity for F

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme, with the levels and indicative points

Examiner's comment

Only a small number of candidates were given all 6 marks for their response. Many responses made little or no reference to equilibrium, considering only the rate of reaction. A clear understanding of the difference between rate and equilibrium was essential to scoring highly on this question. Excellent descriptions were given regarding the role of the catalyst on the rate of reaction, possibly influenced by the previous question. Some gave detailed responses in terms of the impact of both pressure and temperature on the equilibrium position but then did not give the changes made by industry such as reducing temperature due to energy demand or increasing pressure to increase rate. Some candidates indicated that 15atm was a low pressure including reference to it being “lower than normal atmospheric pressure” which suggested that they did not understand either the measurement in atmospheres and/or scale of pressure. A common incorrect response was to suggest that a higher pressure gives increased yield as more moles on right/products. Some stated that the reaction was exothermic, clearly confusing the enthalpy change signs. Some also described the “endothermic side” demonstrating a lack of understanding that enthalpy changes require a difference between products and reactants. Many did not correctly consider the use of excess of steam to increase concentration of reactant and shift equilibrium to products/right but talked in terms of limiting reagents, e.g. “This is used so all of the methane is reacted” or increased collisions. Many were prone generalisations, e.g. “These conditions are used industrially to achieve the most hydrogen efficiently with the least amount of undesired (waste) products”. Candidates need to be encouraged to plan their answers, even using subheadings to organise information and follow the key approach: point, evidence, explain. Candidates should also be encouraged to re-read their answers to check communication is clear and accurate without any contradictory statements. Many included irrelevant or inaccurate information such as describing the toxicity of carbon monoxide, so reducing yield of that being beneficial, or “using a catalyst improves atom economy”. This exemplar shows a candidate who achieved L3 with all 6 marks given. They have considered both the effect of temperature and pressure on equilibrium and potential compromises used industrially. They have also considered the impact of using a catalyst. Their response is logically structured and has a clear line of reasoning.

Notes from the mark scheme and report

• endothermic so high temperature shifts right
• 2 mol gas to 4 mol so low pressure favours products but reduces rate
• compromise pressure
• high temperature costs energy
• catalyst increases rate without changing equilibrium position
• excess steam shifts right
• This question was good at highlighting candidates understanding of the impact of different conditions on equilibrium reactions. We have produced a delivery guide to help with teaching about equilibrium: https://www.ocr.org.uk/Images/261544-equilibrium.pdf

Mark scheme, with the levels and indicative points

Examiner's comment

About a third of candidates were given Level 3 for this question. The key to answering this style of question well is to make sure all the information provided is used and to avoid contradictory statements, e.g. “structure contains carboxylic acid from IR” but then not present in final structure drawn, or a structure that doesn’t match the molecular formula given. A significant number of candidates did not include C=C despite being told in the question that it was “unsaturated and is a trans stereoisomer”, plus the C=C bond is shown in the IR and molecular formula needed unsaturation. The M+1 peak did confuse some candidates who then tried to add an extra H to final structure. It is very important that any structure given is feasible in terms of bonding; many candidates gave structures with C with 5 bonds (with both C=C and C=O attached), limiting them to achieving only L1. Some candidates gave a cis structure rather than trans. Other candidates ignored the O-H peak (from the carboxylic acid) in the IR spectrum, attributing this to a C-H bond as it was not as smooth or as prominent as they may have seen in other spectra. Some listed the bonds observed in the IR without linking to their position – this can easily be done by annotating the spectra. Many candidates had messy answers with lots of rough working which was then not crossed out – this made their answer very confusing. Note that no marks were given for just the empirical formula calculation and some attempted to produce a structure from the empirical formula without determining the Mr from the M+ peak in mass spectrum. This is an example of a good response which achieved L3 6 marks. Empirical and molecular formula are determined. Spectra have been annotated to aid their interpretation, but they have also included the key IR peaks as well as a MS fragment in their response. The correct structure for A is clearly drawn out. The response is clear and concise, with all information presented being relevant. If any of your students’ results are not as expected, you may wish to consider one of our post-results services. For full information about the options available visit the OCR website. We send a weekly roundup to tell you about important updates. You can also sign up for your subject specific updates. If you haven’t already, sign up here. Attend one of our popular CPD courses to hear directly from a senior assessor or drop in to a Q&A session. Most of our courses are delivered live via an online platform, so you can attend from any location. Please find details for all our courses on the relevant subject page on our website or visit OCR professional development. ExamBuilder is the question builder platform for a range of our GCSE, A Level, Cambridge Nationals and Cambridge Technicals qualifications. Find out more. ExamBuilder is free for all OCR centres with an Interchange account and gives you unlimited users per centre. We need an Interchange username to validate the identity of your centre's first user account for ExamBuilder. If you do not have an Interchange account please contact your centre administrator (usually the Exams Officer) to request a username, or nominate an existing Interchange user in your department. Review students' exam performance with our free online results analysis tool. It is available for all GCSEs, AS and A Levels and Cambridge Nationals. It allows you to: • review and run analysis reports on exam performance • analyse results at question and/or topic level • compare your centre with OCR national averages • identify trends across the centre • facilitate effective planning and delivery of courses • identify areas of the curriculum where students excel or struggle • help pinpoint strengths and weaknesses of students and teaching departments. Find out more. Post-results services Keep up-to-date OCR Professional Development Signed up for ExamBuilder?

Notes from the mark scheme and report

• empirical formula C2H3O, M+ 86 so C4H6O2
• broad O–H 2500–3300 and C=O 1630–1820 so carboxylic acid
• peak at 41 from loss of COOH
• CH3CH=CHCOOH drawn as trans isomer with justification (cis or trans accepted for structure)
• We have put together a range of online resources to support teaching of organic chemistry: https://www.ocr.org.uk/qualifications/as-a-level-gce-chemistry-a-h032-h432-from-2015/delivery- guide/module-cam04-module-4-core-organic-chemistry/delivery-guide-cadg012-organic- chemistry-as

Skills: M0.2 Use ratios, fractions and percentages; M3.1 Translate information between graphical, numerical and algebraic forms; M4.2 Visualise and represent 2-D and 3-D forms including 2-D representations of 3-D objects

Mark scheme, with the levels and indicative points

Examiner's comment

This Level of Response question was generally well answered with many candidates achieving maximum marks by simply considering what was required in the question. Responses were often split between a calculation on the main paper and the conditions explanation on extra pages. The calculation errors included no shift or incorrect shift in the equilibrium values. Not calculating the concentration or incorrectly multiplying by 8 rather than dividing by 8. Some candidates attempted a ‘hybrid’ calculation of Kp by trying to calculate a mole fraction and partial pressures. There was a number of candidates who confidently worked out the value of Kc. There were also some very good analyses of the operational conditions. Many of those who had done well on the calculation treated the explanation as an afterthought, not giving it enough attention to give them an answer that would access Level 3. An example of a complete answer, showing a good level of communication in the description and the calculation layout, achieving Level 3 (6 marks) is shown above. This candidate gave a clear method using an “ICE” table to calculate the number of moles at equilibrium. The calculation of the new concentrations can be seen by the use of [ ] and division by 8. The candidate then shows the Kc expression and substitutes the numerical values before successfully calculating the value and includes the units. This is followed by a concise explanation of the conditions used by industry.

Notes from the mark scheme and report

• equilibrium moles N2 1.12, H2 3.36 (0.08 and 0.24 reacted)
• divide by 8.00 dm3
• Kc = 0.0386 dm6 mol–2
• low temperature/high pressure for yield vs high temperature for rate, cost and safety of high pressure
• catalyst allows lower temperature

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme, with the levels and indicative points

Examiner's comment

Answers were distributed across all 3 levels of achievement. Most of the candidates managed to identify at least some of the substances. Of the equations, the reaction of copper (II) oxide with nitric acid was most regularly seen correct, although many students could also represent the ligand replacement in Experiment 1. Many candidates were able to calculate Mr for gas D but some of those suggesting SO2 as a possible formula preferred to have an equation in experiment 2 producing hydrogen. A few candidates used the Mr to suggest that the gas was 2O2 and as such candidates found the equation between copper and sulphuric acid challenging. A good number of candidates identified F and G, recognising what they had learned from their work on redox titrations, and some were able to reproduce the equation. Incorrect formula of copper (I) iodide (CuI2) was a common error. Many candidates made no attempt at identifying changes in oxidation states. Candidates are advised to address all parts of the question in order to access the higher levels and to allow sufficient time to attempt the LoR questions. The candidate has clearly set out the three experiments, written the equation and followed with a clear identification of each of the formulae B-G. The calculation is well laid out and the response clearly indicates how the final value has been calculated. Oxidation numbers have then been commented on where appropriate. The colours of the transition metal ions and complexes can be reviewed at: A Level Chemistry A - Colours of inorganic ions and complexes (ocr.org.uk) If any of your students’ results are not as expected, you may wish to consider one of our post-results services. For full information about the options available visit the OCR website. We send a weekly roundup to tell you about important updates. You can also sign up for your subject specific updates. If you haven’t already, sign up here. Attend one of our popular CPD courses to hear directly from a senior assessor or drop in to a Q&A session. Most of our courses are delivered live via an online platform, so you can attend from any location. Please find details for all our courses on the relevant subject page on our website or visit OCR professional development. ExamBuilder is the question builder platform for a range of our GCSE, A Level, Cambridge Nationals and Cambridge Technicals qualifications. Find out more. ExamBuilder is free for all OCR centres with an Interchange account and gives you unlimited users per centre. We need an Interchange username to validate the identity of your centre's first user account for ExamBuilder. If you do not have an Interchange account please contact your centre administrator (usually the Exams Officer) to request a username, or nominate an existing Interchange user in your department. Review students' exam performance with our free online results analysis tool. It is available for all GCSEs, AS and A Levels and Cambridge Nationals. It allows you to: • review and run analysis reports on exam performance • analyse results at question and/or topic level • compare your centre with OCR national averages • identify trends across the centre • facilitate effective planning and delivery of courses • identify areas of the curriculum where students excel or struggle • help pinpoint strengths and weaknesses of students and teaching departments. Find out more. Post-results services Keep up-to-date OCR Professional Development Signed up for ExamBuilder?

Notes from the mark scheme and report

• B = [CuCl4]2–
• D: n = 45/24000, M = 64 so SO2, C = CuSO4 with Cu 0 to +2 and S +6 to +4
• E = Cu(NO3)2
• F = CuI (white) and G = I2 with Cu +2 to +1 and I –1 to 0
• Level 3 needs Mr calculation, most equations and some oxidation number changes
• The AS delivery guide ‘Theme: Identifying Unknowns’ https://www.ocr.org.uk/Images/208563-identifying-unknowns.pdf ‘Transition elements’: Delivery Guide for OCR AS/A Level Chemistry A

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme, with the levels and indicative points

Examiner's comment

This question was marked using a level of response mark scheme. Most candidates gave an answer worth of at least Level 2 (3-4 marks) by providing the synthetic steps with reagents and equations for the synthesis of compound I. Exemplar 2, below, shows a frequent Level 2 response. The best performing candidates correctly determined the mass attempting to calculate the mass and showed the synthesis efficiently, using equations to communicate the preparation of compound I, with these responses being given Level 3 (5-6 marks). A number of responses omitted the mass calculation, such responses received Level 2 (1-2 marks). In this response the candidate has attempted to calculate the starting mass but has made little progress. Two stages of the synthesis have been covered with the reagents and most of the conditions identified. Both equations are complete. This is a Level 2 response and 4 marks have been given as the repones is logical and well communicated

Notes from the mark scheme and report

• mass of CH3CHClCOOH = 13.56 g (0.0800 mol × 100/64 × 108.5)
• partly correct e.g. 5.56 g from inverted yield
• excess NH3 in ethanol substitutes Cl
• ethanol with acid catalyst (H2SO4) to esterify (ALLOW ammonium salt of I)

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

Mark scheme, with the levels and indicative points

Examiner's comment

This question differentiated well. Candidates who were given Level 3 (5-6 marks) understood the term aliphatic and were able to provide two different mechanisms that produced a C-C bond. The most common responses seen involved the nucleophilic substitution of a halogenoalkane and a cyanide ion and the nucleophilic addition of a carbonyl with a cyanide ion. Some candidates offered radical substitution mechanisms, detailing initiation, propagation and termination steps, with the termination step producing a C-C bond. Candidates who scored Level 2 (3-4 marks) frequently detailed a reaction involving aromatic compounds or polymerisation of alkenes for which they were unable to give a mechanism. If any of your students’ results are not as expected, you may wish to consider one of our post-results services. For full information about the options available visit the OCR website. We send a weekly roundup to tell you about important updates. You can also sign up for your subject specific updates. If you haven’t already, sign up here. Attend one of our popular CPD courses to hear directly from a senior assessor or drop in to a Q&A session. Most of our courses are delivered live via an online platform, so you can attend from any location. Please find details for all our courses on the relevant subject page on our website or visit OCR professional development. ExamBuilder is the question builder platform for a range of our GCSE, A Level, Cambridge Nationals and Cambridge Technicals qualifications. Find out more. ExamBuilder is free for all OCR centres with an Interchange account and gives you unlimited users per centre. We need an Interchange username to validate the identity of your centre's first user account for ExamBuilder. If you do not have an Interchange account please contact your centre administrator (usually the Exams Officer) to request a username, or nominate an existing Interchange user in your department. Review students' exam performance with our free online results analysis tool. It is available for all GCSEs, AS and A Levels and Cambridge Nationals. It allows you to: • review and run analysis reports on exam performance • analyse results at question and/or topic level • compare your centre with OCR national averages • identify trends across the centre • facilitate effective planning and delivery of courses • identify areas of the curriculum where students excel or struggle • help pinpoint strengths and weaknesses of students and teaching departments. Find out more. Post-results services Keep up-to-date OCR Professional Development Signed up for ExamBuilder?

Notes from the mark scheme and report

• haloalkane with NaCN in ethanol by nucleophilic substitution
• aldehyde/ketone with NaCN/H+ (HCN) by nucleophilic addition
• curly arrows from lone pair/– charge on C of CN–, dipoles, intermediate with O– then protonation

Mark scheme, with the levels and indicative points

Examiner's comment

Showed limited knowledge and understanding of practical scenarios and drawing labelled diagrams of apparatus, e.g. • Question 3 (a): Diagram of reflux • Question 3 (a): Purification of an organic liquid There was no evidence that any time constraints had led to a candidate underperforming or of scripts where there were no responses to many questions.

Notes from the mark scheme and report

• reflux diagram: condenser above flask, water in at bottom and out at top
• mass = 6.10 × 1.268 = 7.73 g, yield 45.2% (35.6% if density omitted)
• separating funnel - organic layer is the BOTTOM layer (denser), dry with named anhydrous salt, redistil at 102 °C
• bare 'separate, dry, redistil' not detailed

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG5 Synthesis of an organic liquid

Mark scheme, with the levels and indicative points

Examiner's comment

This question was assessed by level of response (LoR). Candidates were required to process raw experimental results to determine one enthalpy change, and then to determine a second enthalpy change by using an energy cycle. Levels were determined by the accuracy of the candidates’ processing of the results, calculations and use of the energy cycle. Marks within a level were determined by communication. This question discriminated extremely well. Level 3 candidates used the mass of solution as 106.6 g with mc∆T to obtain an energy change of 8.24 kJ. They then divided this value by the moles of CaCl2 that reacted (0.0835 mol) to obtain the enthalpy change of –98.7 kJ mol–1. Finally, they constructed an energy cycle which they then used to obtain the second enthalpy change of –1566 kJ mol–1. Level 2 candidates determined the first enthalpy change but may have used the approximate mass of 100 g for the mass of solution from the experimental method. Their energy cycle was often incorrect or absent, with the second enthalpy change calculated incorrectly. Level 1 candidates often calculated the initial energy change using mc∆T but made little further correct progress. Less successful responses used the solid mass of calcium chloride (9.28 g) instead of the mass of the solution in their mc∆T calculation. Overall, mathematical skills were displayed well but some basic errors were made, particularly with subtractions. This may have been the result of mis-keying values into a calculator and believing the answer displayed. calcium chloride and the first enthalpy change using a correct method. This response has not been given marks for the communication strand of Level 3 because there is nothing to indicate what the numbers refer to. The response could be summarised as a mass of numbers scrawled across the page. Unfortunately, this is the pattern of many responses. So, this response was given 3/6 marks.

Notes from the mark scheme and report

• q = 106.6 × 4.18 × 18.5 = 8.24 kJ (mass of final solution)
• n(CaCl2) = 9.28/111.1 = 0.0835
• ∆solH = –98.7
• ∆hydH(Ca2+) = –2223 + (–98.7) – 2(–378) = –1566
• 2 × hydration of Cl–
• one small error allowed e.g. using 97.32 g or 115.88 g for m

Skills: M0.0 Recognise and make use of appropriate units in calculations; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations; PAG3 Enthalpy determination

Mark scheme, with the levels and indicative points

Notes from the mark scheme and report

• mass of solution 103.01 g from weighings (using 100 g cannot reach Level 3)
• q = mc∆T = 2153 J
• n(MgCO3) = 0.0500
• ∆H2 = –43.06 with sign
• ∆H1 = ∆H2 – ∆H3 = +93.04 kJ mol–1

Skills: M0.0 Recognise and make use of appropriate units in calculations; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations; PAG3 Enthalpy determination

Mark scheme, with the levels and indicative points

Notes from the mark scheme and report

• separating funnel – organic layer is the TOP layer (lower density)
• dry with anhydrous MgSO4/CaCl2
• distil collecting at 50 °C
• mass = 7.70/74.0 × 0.76 × 92.5 = 7.3 g
• inverse ratio 100/76 gives 12.7 g
• IGNORE washing with carbonate/water

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG5 Synthesis of an organic liquid

Mark scheme, with the levels and indicative points

Notes from the mark scheme and report

• A = Cr2(SO4)3•12H2O
• B = [Cr(H2O)2(C2O4)2]–
• trans isomer plus two cis optical isomers drawn in 3D with wedges
• charges and brackets

Skills: M0.2 Use ratios, fractions and percentages; 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, with the levels and indicative points

Notes from the mark scheme and report

• HNO2 in excess, partial neutralisation gives HNO2/NO2– mixture
• n(HNO2) left 0.0350 and n(NO2–) 0.0150
• pH = 2.97
• N2O3 + H2O → 2HNO2 so 0.0250 mol = 1.90 g

Skills: M0.2 Use ratios, fractions and percentages; 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, with the levels and indicative points

Examiner's comment

Question 19(b): This question differentiated well.

Notes from the mark scheme and report

• 20.2 g (n = 0.0400 × 100/25 = 0.160)
• partly correct: 1.265 g (yield inverted), 5.06 g (yield omitted)
• stage 1 CN– in ethanol to C6H5CH2CN
• stage 2 H+/H2O acid hydrolysis giving NH4+

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme, with the levels and indicative points

Notes from the mark scheme and report

• D no C=O, oxidised so primary/secondary alcohol, saturated, 3 carbon environments with C–O at 73 (alicyclic alcohol)
• E and F ketones (not oxidised so not aldehydes)
• E quartet 2.4 + triplet 1.1 (CH3CH2CO)
• F heptet 2.6, singlet 2.1 (CH3CO), doublet 1.1

Skills: M3.1 Translate information between graphical, numerical and algebraic forms; PAG7 Qualitative analysis of organic functional groups

Mark scheme, with the levels and indicative points

Examiner's comment

Q1(a) Partial pressure calculation Q1(d) Formula determination from percentage composition by mass Q2(b) Converting mass and volume into moles Q3(b) Unstructured titration calculation Q5(a)(ii) Ideal gas calculation Q5(b)(i) Calculation of the pH of a weak acid • Produced clear and concise responses for a Level of Response question based on a familiar context i.e., Q4(b) rates by continuous monitoring and graphical analysis. • Had difficulty when answering questions requiring application of knowledge and understanding on questions set in novel and unfamiliar contexts, e.g. Q4(b) is a Level of Response question which is commented on after this section. Q5(c) is a Level of Response question which is commented on after this section Level of response questions Q4(b): Candidates were required to analyse the graphical results of a rates investigation to determine the reaction order, rate and rate constant.

Notes from the mark scheme and report

• tangent at t = 0 gives initial rate 4.5–6.5 × 10–6 mol dm–3 s–1
• constant half-life of about 2500 s (two half-lives shown) so first order
• k = rate/[I2] or ln2/t½ with units s–1

Skills: M0.0 Recognise and make use of appropriate units in calculations; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M3.1 Translate information between graphical, numerical and algebraic forms; M3.5 Draw and use the slope of a tangent to a curve as a measure of rate of change; PAG9 Rates of reaction – continuous monitoring

Mark scheme, with the levels and indicative points

Examiner's comment

Q2(c), Q2(d): Predicting the effect of modifications to an experiment Q5(c): Interpreting information in a Level of Response question provided in an unfamiliar contexts • Had difficulty when writing equations for unfamiliar reactions, e.g. Q5(c) is a Level of Response question which is commented on after this section Level of response questions Q4(b): Candidates were required to analyse the graphical results of a rates investigation to determine the reaction order, rate and rate constant. Q5(c): Candidates were supplied with information about two reactions, supported by numerical data and an IR spectrum.

Notes from the mark scheme and report

• E = Cu (M = 3.81/0.06 = 63.5), F = H2O, G = N2 (M = 0.560/0.02 = 28)
• 3CuO + 2NH3 → 3Cu + 3H2O + N2
• I has C=O (~1700) and N–H (3200–3400) so CH3CONH2
• H = CH3COCl, J = NH4Cl
• CH3COCl + 2NH3 → CH3CONH2 + NH4Cl

Skills: M0.2 Use ratios, fractions and percentages; M3.1 Translate information between graphical, numerical and algebraic forms

Mark scheme, with the levels and indicative points

Examiner's comment

1(a) A considerable number of candidates did not score the mark here despite knowing that it was an oxidation reaction. They either omitted to give the number of electrons lost or gave their justification using oxidation numbers. This question highlights the importance of reading the question carefully and answering the question asked, not one that may have been seen previously. 1(b)(i) Most candidates were able to make some progress with this calculation. Common errors included omitting to multiply by two to find the number of moles of HCl or not multiplying their volume in dm3 by 1000 to give an answer in cm3. 1(b)(ii) More candidates struggled with this part of calculation. A significant number of candidates used the number of moles rounded to 3 SF but then gave a final answer to 4 SF, giving an answer of 78.48 cm3. This was given credit as correct calculation had been carried out for one mark. A couple of candidates correctly used the ideal gas equation and were given credit for this, but made it more complicated than necessary for one mark. 1(b)(iii) Very few candidates managed to gain marks here. Most did not recognise that HCl was the limiting reagent in the reaction and suggested that a greater volume of H2 would be produced, as same mass of Mg would give a greater number of moles. 1(c)* Very few candidates managed to score full marks for this question. Even the highest-attaining candidates struggled with writing balanced chemical equations. The most successful candidates used ionic equations with state symbols in their responses. A large proportion of candidates gave unnecessary details such as testing for CO2 using limewater or the colours of other silver halide precipitates. The best responses broken down their response to cover each test in turn, giving clear and concise details for each.

Notes from the mark scheme and report

• acid then fizzing for carbonate
• Ba2+(aq) white precipitate for sulfate
• AgNO3(aq) cream precipitate for bromide
• sequence carbonate then sulfate then halide
• equations ionic or full, IGNORE state symbols

Skills: PAG4 Qualitative analysis of ions

Mark scheme, with the levels and indicative points

Examiner's comment

5(a) Generally this was well answered. However, some candidates only gave two responses where three were required, presumably because it was worth two marks. 5(b) A significant number of candidates did not gain the mark here as they incorrectly balanced the number of oxygens, forgetting that the alcohol contains an O atom i.e. 6.5 O2 given rather than 6. 5(c) Nomenclature in organic chemistry takes lots of practice. Approximately half of the candidates did not score this mark. It was more challenging as they needed to work with skeletal structures but evidence of drawing out displayed formula was often seen. 5(d)(i) Most candidates were able to recall the formula to calculate atom economy, however a number made errors in working out Mr values. However, some left this blank or just gave an answer without any working. 5(d)(ii) Mechanisms were often seen showing curly arrows going in the wrong direction and between the wrong bonds and atoms, charges and dipoles were often incorrect, and partial changes used where full charges were required. Writing mechanisms is an important skill in organic chemistry so it is vital that time is spent practising writing them out and fully understanding the significance of the curly arrow. 5(d)(iii) Many candidates gave the correct mechanism here, with common incorrect responses being other types of mechanism, substitution only, or isomerism. 5(e) Very few candidates gained both marks here. It was not enough here to just state faster/slower without specifying which haloalkane they were referring to. It was also important to specify the C-X bond not just vague reference to the chlorine or bromine bond. Many described the difference in bond polarity so had the wrong order. Some were more general and discussed the reactivity of bromine and chlorine themselves. 5(f)* This was a challenging problem-solving question, relying on candidates to make use of all of the information provided to determine the structure. Very few candidates made no attempt at all at this question. The most common error was incorrectly identifying the peak at 3000 cm-1 as O-H from a carboxylic acid, despite it being indicated on the data sheet that this would be a broad peak. It suggested that candidates were not familiar with looking at spectra and understanding what data it gave. Candidates should also be reminded when answering these types of question that they should give the structures, not just molecular formulae, where possible. If any of your students’ results are not as expected, you may wish to consider one of our review of results services. For full information about the options available visit the OCR website. If university places are at stake you may wish to consider priority service 2 reviews of marking which have an earlier deadline to ensure your reviews are processed in time for university applications. Our priority is supporting you and your students this autumn and to support you as you prepare for summer 2021 exams. We’ll update our website information regularly with resources, guidance and key information. • Teachers • Students • Exams officers • Assessment specialists We are sending a weekly roundup to tell you about important updates. You can also sign up for your subject specific updates. If you haven’t already, sign up here. Attend one of our popular CPD courses to hear directly from a senior assessor or drop in to a Q&A session. All our courses for the academic year 2020-2021 are being delivered live via an online platform, so you can attend from any location. Please find details for all our courses on the relevant subject page on our website or visit OCR professional development. ExamBuilder is the question builder platform for a range of our GCSE, A Level, Cambridge Nationals, Cambridge Technicals and Functional Skills qualifications. See the full list of available qualifications in the sign up form. ExamBuilder is free for all OCR centres with an Interchange account and gives you unlimited users per centre. We need an Interchange username to validate the identity of your centre's first user account for ExamBuilder. If you do not have an Interchange account please contact your centre administrator (usually the Exams Officer) to request a username, or nominate an existing Interchange user in your department. Review of results

Notes from the mark scheme and report

• M+ = 86, m/z 43 = CH3CO+ or C3H7+
• C=O at about 1700 and no broad O–H so ketone not carboxylic acid
• X is secondary alcohol E (3-methylbutan-2-ol), Y 3-methylbutan-2-one
• equation with [O] forming H2O

Skills: M3.1 Translate information between graphical, numerical and algebraic forms

Mark scheme, with the levels and indicative points

Notes from the mark scheme and report

• Mg giant metallic with mobile delocalised electrons
• Br2 simple molecular with London forces between molecules (DO NOT ALLOW between atoms)
• MgBr2 giant ionic – ions fixed in solid, mobile when liquid (DO NOT ALLOW electrons)
• metallic and ionic bonds stronger than London forces
• IGNORE 'carry' charge

Mark scheme, with the levels and indicative points

Notes from the mark scheme and report

• B = FeSO4, D = Fe(OH)2, G = BaSO4
• C = CrCl3, E = Cr(OH)3, F = [Cr(NH3)6]3+, H = AgCl
• ionic equations for D–H
• Level 3 needs at least six identified and equations mostly correct

Skills: PAG4 Qualitative analysis of ions

Mark scheme, with the levels and indicative points

Examiner's comment

Level of response questions Question 20(a) This question differentiated well.

Notes from the mark scheme and report

• mass of cyclopentanol 7.90 g (errors: inverse ratio 3.24 g, ignoring yield 5.06 g)
• neutralise e.g. Na2CO3
• separating funnel with organic layer on top
• dry with anhydrous salt e.g. MgSO4
• redistil at about 44 °C

Skills: M0.2 Use ratios, fractions and percentages; M2.2 Change the subject of an equation; PAG5 Synthesis of an organic liquid

Mark scheme, with the levels and indicative points

Examiner's comment

Question 22(b) Candidates achieving Level 2 had most of the key features, e.g.

Notes from the mark scheme and report

• C6H9NO2 from composition and m/z 127
• IR: C=O at 1750 and C≡N at 2280 cm–1
• NMR: quartet 2H at 4.2 (CH3CH2O), quartet 1H at 2.9, doublet 3H at 1.7, triplet 3H at 1.3

Skills: M0.2 Use ratios, fractions and percentages; M3.1 Translate information between graphical, numerical and algebraic forms

Mark scheme, with the levels and indicative points

Examiner's comment

This question was assessed by level of response (LoR). The question required candidates to demonstrate their knowledge and understand of some important terms used in transition metal chemistry. Level 3 candidates showed complete definitions of ligand, coordination number and ligand substitution, supported by suitable equations with clear diagrams of complex ions. The responses were concise with 3D diagrams of 6- and 4-coordinate complex ions displayed with wedges, correct connectivity to ligand atoms and showing the role of lone pairs in the formation of the coordinate bonds. 4-coordination was shown as either or both of tetrahedral (usually CuCl42–) or square planar (e.g. platin). Ligand substitution was accompanied by a correct balanced equation, most commonly between [Cu(H2O6]2+ and NH3. Level 2 candidates usually gave definitions of ligand, coordination number and ligand substitution. There was usually a balanced equation for ligand substitution and one correct 3D diagram. A second diagram often had an unsuitable shape for the complex ion chosen (often CuCl42– shown as square planar). Level 1 candidates did produce definitions, but these were often incomplete. There was usually an attempt to show a 3D diagram or equation, but these often contained unsuitable examples. This question rewarded the candidates who had learnt their chemistry and the levels enabled the amount of knowledge and understanding to be assessed. The question discriminated well.

Notes from the mark scheme and report

• ligand donates a lone pair forming a coordinate bond to the metal ion
• coordination number = number of coordinate bonds
• octahedral e.g. [Cu(H2O)6]2+, tetrahedral e.g. CuCl4 2– or square planar Pt(NH3)2Cl2 with 3D diagrams
• balanced ligand substitution equation

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

Mark scheme, with the levels and indicative points

Examiner's comment

3(a)(i) Most candidates analysed the raw practical results supplied to correctly calculate the energy change as 10,241 J using q = mc∆T, and the amount of propane burnt as 0.0550 mol. Candidates usually determined the energy change per mole of propane as 1862 kJ. The question asked for the enthalpy change of combustion to 3 significant figures, which gives –1860 kJ mol–1. The mark for this final value was often not given because either the – sign was omitted or the answer was not quoted to the required 3 significant figures. Candidates are advised to read the requirements of each question more closely to eliminate such errors. A double-check on significant figure requirements is recommended before moving on to the next question. Lower-attaining candidates often extracted the wrong data from the results table. Instead of the mass of water (100 cm3), the mass of propane burnt (0.242 g) was often used in the q = mc∆T calculation. Candidates are advised that the thermometer measures the temperature change of the material in which it is placed (the water). 3(a)(ii) One mark was available for identifying two reasons for the difference in enthalpy change. It was expected that candidates would respond with ‘heat loss’ and ‘incomplete combustion’. Most candidates suggested heat loss but the other reason often suggested was equipment error, e.g. the thermometer did not reading correctly’, ‘the balance gave incorrect readings’. Candidates are advised that responses of faulty apparatus are unlikely to be creditworthy. 3(b) This question was assessed by LoR. Level 3 candidates calculated the enthalpy change for Equation 3.1 correctly as –2054 kJ mol–1. They linked 4 × ∆vapH with the enthalpy change for Equation 3.1 in an energy cycle and calculated the enthalpy change of combustion as –2054 – (4 × 40.65) = –2216.6 kJ mol–1. Most Level 3 candidates drew out bonds in the molecules involved in Equation 3.1, which allowed them to visually count the bonds broken and bonds made. This strategy is recommended to all candidates for any calculation based on bond enthalpies and is much more likely to generate the correct enthalpy change than visualising the relevant bonds solely from the equation. Level 2 candidates often derived the correct enthalpy change for Equation 3.1 but then tried to link this value with ∆vapH instead of 4 × ∆vapH. Other Level 2 candidates calculated ∆H correctly for either the bonds broken or bonds made and combine this value with 4 × ∆vapH. A common error was the omission of the 2 C–C bonds in propane, use of 3 C–C, instead of 2 C–C, bonds, or use of 4, rather than 8 O–H bonds. These errors could have been avoided by drawing out the bonds as described above. Level 1 candidates typically attempted to use the bond enthalpies to calculate energies for bonds broken and bonds made but made several errors. They then combined this value with ∆vapH. This question discriminated extremely well across different abilities.

Notes from the mark scheme and report

• bonds broken 6488, bonds made 8542, ∆rH = –2054
• errors: 3 C–C used (–1707), C–C omitted (–2748)
• ∆cH needs H2O(l) so ∆cH = ∆rH – 4 × ∆vapH = –2217
• adding 4 × 40.65 (–1891) is incorrect

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme, with the levels and indicative points

Examiner's comment

Most candidates focused on removing the water of crystallisation, often going to great depths of explanation, with apparatus diagrams, of how to remove it. Some candidates then went on to explain how to make a standard solution and could be given marks. Calculations for the mass required were often correct or could be given some marks. This in a response of a common approach to this question, it was given 4 marks. Despite calculating the mass properly, the candidate then put all their efforts into describing removing the water of crystallisation and left out the valuable fine detail required for a Level 3 response.

Notes from the mark scheme and report

• mass 25.63 g of Mg(NO3)2•6H2O (14.83 g if water of crystallisation omitted)
• dissolve in distilled water, transfer with washings to 250 cm3 volumetric flask, make up to mark with bottom of meniscus on mark, invert to mix
• IGNORE removing water of crystallisation

Skills: M0.0 Recognise and make use of appropriate units in calculations; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG2 Acid–base titration

Mark scheme, with the levels and indicative points

Examiner's comment

This question discriminated well and most were able to attempt to calculate a mass and explain the purification steps with some fine detail. It was evident that most candidates were aware of the apparatus required but the logic in the order was sometimes out of sequence. Drying agents were mentioned, some of the examples used were incorrect and errors were made with the boiling points if candidates mentioned distillation. This in a response of an excellent approach to this question, it was given 6 marks. The correct mass has been calculated and all of the fine detail, except distilling at 63°C

Notes from the mark scheme and report

• 0.0500 mol hex-1-ene needs 0.0800 mol hexan-1-ol = 8.16 g
• partly correct: 5.1 g (% yield omitted), 3.19 g (% yield inverted)
• separating funnel (upper organic layer), dry with anhydrous MgSO4/CaCl2, distil collecting fraction at 63 °C
• incorrect purification method not creditworthy

Skills: M0.2 Use ratios, fractions and percentages; M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG5 Synthesis of an organic liquid

Mark scheme, with the levels and indicative points

Examiner's comment

This Level of Response question was generally well answered with many candidates achieving maximum marks by simply considering what was required in the question. The calculation of the [HCO3 –] : [H2CO3] ratio was well described although sometimes the final expression of the ratio left ambiguity as it was hard to tell whether the ratio given referred to the [HCO3 –] : [H2CO3] ratio or the [H2CO3] : [HCO3 –] ratio The buffer reactions on separate addition of acid (H+ ions) and alkali (OH– ions) were explained and, better still, shown in equation form. The direction of shift on the H2CO3 ⇌ H+ + HCO3 – equilibrium was invariably correct, but many candidates did not achieve credit for responses such as ‘Addition of H+ ions shifts the equilibrium to the left’ because they did not give the equilibrium which was undergoing shift. An example of a complete answer gaining L3 (6 marks) is given. The candidate clearly writes the equilibrium at the top. The first bullet point gives a chemical equation for the reaction occurring when H+ ions are added as well as the shift in equilibrium. The second bullet point gives a chemical equation for the reaction occurring when OH– ions are added as well as the shift in equilibrium. The steps in the calculation are clearly shown and the ratio is clear.

Notes from the mark scheme and report

• H2CO3 ⇌ H+ + HCO3–
• added H+ removed by HCO3– (shift left), added OH– removed by H+/H2CO3 (shift right)
• ratio = Ka/[H+] = 10.5:1

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, with the levels and indicative points

Examiner's comment

The second Level of Response question in the paper was also answered very well. Almost all candidates determined the order with respect to H+ to be first order and gave suitable explanations. A very high proportion of candidates determined the order with respect to CH3COCH3 to be first order and related this to the direct proportionality shown on the graph. The zero order with respect to Br2 proved a little more problematic with many candidates just giving an order with no attempted explanation. Having determined orders, nearly all candidates were able to give a corresponding rate equation and could calculate a value for the rate constant, albeit with frequent omission of units and forgetting that the initial rates given were in terms of 10–5. An example of a complete answer achieving L3 (6 marks) is given. This candidate gave each order clearly followed by a concise explanation of each choice. The rate equation (based on the orders given) is clearly stated, the calculation clearly shows working and correct units are given.

Notes from the mark scheme and report

• zero order in Br2 (straight line concentration–time, constant gradient)
• first order in CH3COCH3 (straight line through origin)
• first order in H+ (doubling doubles rate)
• rate = k[CH3COCH3][H+]
• k = 3.9 × 10–5 dm3 mol–1 s–1

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M3.1 Translate information between graphical, numerical and algebraic forms; PAG10 Rates of reaction – initial rates; PAG9 Rates of reaction – continuous monitoring

Mark scheme, with the levels and indicative points

Examiner's comment

This question was marked using a level of response mark scheme. Most candidates gave a response worthy of at least Level 2 (3-4 marks) by either providing the synthetic steps with reagents and/or equations for the synthesis of ester C as well as calculating the mass of 2-methylpropanal required. determined the mass correctly and showed the synthesis efficiently, using equations to communicate the preparation of ester C and received Level 3 (5-6 marks). Some responses focused solely on the 2- methylpropanal and suggested it would be both oxidised and reduced with each of the products being used to form an ester. Others omitted the mass calculation or gave only a partial synthesis. Such responses received Level 1 (1-2 marks). A variety of responses was seen in this demanding question. In general candidates adopted one of two approaches. The most common was to multiply the Mr of 2-hydroxybenzoic acid by 200 and then subtract the mass of the 199 water molecules removed during polymerisation. The other approach used the Mr of the repeat unit. This was multiplied by 200 and the mass of H and OH at each end of the polymer was added to give the final answer. Many candidates were successful with the first step of their approach, but the best responses included the second step taking into account the Mr of water. A significant number of candidates used an incorrect value for the Mr in their first step. Candidates are advised to draw out the structure of a compound before determining the Mr. This is particularly important when only a name is given in the question. This response includes an excellent description of the synthesis of ester C. Each step is labelled clearly with the most of the reagents and conditions stated and correct equations. The candidate makes little progress with the calculation of the mass of 2-methylpropanal required, completing only the first step. This response satisfies the Level 2 criteria. The information is provided is relevant and presented with a clear logical structure and was therefore given the upper mark within the level and achieved four marks. The presentation of the synthesis by equations rather than words is an excellent strategy. Candidates are advised to adopt such an approach in questions like this as it allows them develop and clear line of reasoning without the need to write lots of unnecessary text.

Notes from the mark scheme and report

• 22.5 g (n = 0.125 × 100/40)
• partly correct: 3.60 g (yield inverted), 9.00 g (yield omitted)
• step 1 Cr2O72–/H+ reflux with [O] equation
• step 2 methanol with acid catalyst, heat/reflux
• IGNORE attempts to make methanol

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme, with the levels and indicative points

Examiner's comment

Candidates are confident in tackling questions requiring spectral analysis. Most candidates were able to determine the empirical and molecular formula of the unknown compound and analyse the IR and NMR data. A number of excellent responses were seen which included interpretation of the peak splitting in the NMR spectrum. However, many candidates were unable to suggest a structure that matched their spectral analysis. The majority of structures seen were aromatic but not all contained the carboxylic acid group. Some candidates focused on the NMR data and suggested a structure that was consistent with some of these peaks and included a two substituents of the ring or a degree of chain branching. Lower ability responses often showed a simpler structure of a monosubstituted ring with a straight chain. Stronger responses were able to use all the data to suggest a correct structure as demonstrated in significant proportion of responses were given six marks. This candidate starts their analysis by determining the empirical and molecular formula of the unknown compound. Next the IR and NMR data is analysed in a clear, logical fashion. Although not shown in this report, the candidate had annotated the spectra, a good strategy. At the end of the response the candidate has determined a possible structure. Notice how the candidate has crossed out the first attempt at a structure. This is much better than rubbing out a pencil diagram and re-drawing over the space as this often leads to unclear images that are difficult for examiners to read. This candidate received full marks in this question.

Notes from the mark scheme and report

• C5H6O then C10H12O2 from m/z 164
• IR broad O–H 2300–3700 and C=O about 1720 = COOH
• doublet 3H CH3–CH, singlet 3H Ar–CH3, quartet 1H, multiplet 4H C6H4
• CH3C6H4CH(CH3)COOH, 2-, 3- or 4-substituted

Skills: M0.2 Use ratios, fractions and percentages; M3.1 Translate information between graphical, numerical and algebraic forms

Mark scheme, with the levels and indicative points

Examiner's comment

This question was assessed by level of response (LoR). Candidates were required to process raw experimental results to determine one enthalpy change, and then to determine a second enthalpy change by constructing and using an energy cycle. Levels were determined by the accuracy of the candidates’ processing, calculations and use of the energy cycle. This question discriminated extremely well. Level 3 candidates used the mass of solution as 50.7 g with mc∆T to obtain an energy change of 2861 J. They then divided this value by the moles of CuSO4 that reacted (0.05 mol) to obtain the enthalpy change of –57.22 kJ mol–1. Finally, they constructed an energy cycle which they then used to obtain the second enthalpy change of –65.65 kJ mol–1. Level 2 candidates determined the first enthalpy change but may have used the approximate mass of 50 g for the mass of solution from the experimental method. Their energy cycle was often incorrect or absent, with the second enthalpy change calculated incorrectly. Level 1 candidates often calculated the initial energy change using mc∆T but made little further correct progress. A significant number of lower ability candidates used the solid mass of copper sulfate in their calculation. Overall, mathematical skills were displayed well but some basic errors were made, particularly with subtractions. This may have been the result of mis-keying values into a calculator and believing the answer displayed. An example was the mass of the solution (74.13 – 23.43) being seen as 49.7 rather than 50.7. A significant number of candidates added or subtracted the mass of copper sulfate from 50.7 or 50 for their value of m in mc∆T, using for example 57.98 and 42.72. This limited the level that they could reach. initial energy change using mc∆T, the first enthalpy change using the energy change with the moles of copper sulfate, and the second enthalpy change including a clear energy cycle. Numbers are clearly shown and unrounded until the final values. This is an excellent response.

Notes from the mark scheme and report

• mass of solution 50.70 g (not 50.0) and ∆T 13.5 °C
• n(CuSO4) = 7.98/159.6 = 0.0500 mol
• ∆solH = –57.22 kJ mol–1 with – sign
• ∆rH = ∆solH(CuSO4) – ∆solH(CuSO4•5H2O) = –65.65
• energy cycle expected

Skills: M0.0 Recognise and make use of appropriate units in calculations; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M3.1 Translate information between graphical, numerical and algebraic forms; PAG3 Enthalpy determination

Mark scheme, with the levels and indicative points

Examiner's comment

This question presented a practical scenario in which candidates were asked to identify 5 unknown chemicals. There are many routes that lead to correct identifications of the unknowns. Most candidates identified G as oxygen and made some headway towards identifying C and D by determining that 9 waters of crystallisation were present in C. The formulae of C and D sometimes followed but many candidates found it difficult to link 9H2O to their formulae. The best responses showed the nitrate ion separately in the formula, e.g. Fe(NO3)3, but many showed the empirical formula instead, e.g. FeN3O9. Gas G (NO2) proved to be the most difficult unknown to identify as it required two pieces of data for its determination. There were some very competent attempts at writing equations, with the decomposition of compound D in Stage 2 to form E, F and G being the most difficult. 1:9 molar ratio of C: H2O. The candidate then writes the correct equation for Stage 1, using NO3 for the nitrate ion. The candidate then identifies E, F and G using a methodical approach with clear working throughout. Finally, the candidate writes the correct equation for the reaction in Stage 2. This is an excellent Level 3 response, given 6/6 marks.

Notes from the mark scheme and report

• n(H2O) = 0.027 so 9H2O
• C = Fe(NO3)3•9H2O, D = Fe(NO3)3, E = Fe2O3 (M 159.6), F = NO2 (0.414/0.00900 = 46), G = O2 (relights glowing splint)
• 2Fe(NO3)3 → Fe2O3 + 6NO2 + 1½O2

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme, with the levels and indicative points

Examiner's comment

Candidates were not prepared to answer this type of question and the diagrams were hard to give credit to. Many had significant safety implications such as open beakers of butan-1-ol being heated by a Bunsen burner. Most mis-read the question and just outlined the method for purification and struggled to recall the practical details. Very few candidates mentioned the use of anhydrous salts, referring instead to ‘boiling off’ the water. This candidate was credited 4 marks for this level 2 answer. Although they have drawn distillation apparatus instead of reflux, they have considered the boiling point of the product, detailed using a separating funnel, a drying agent and that the lower organic layer would be drawn off first.

Notes from the mark scheme and report

• reflux with water in at bottom of condenser and open system
• separating funnel – organic layer is the lower (denser) layer
• dry with anhydrous salt
• redistil collecting at 102 °C

Skills: PAG5 Synthesis of an organic liquid

Mark scheme, with the levels and indicative points

Examiner's comment

This question was a good discriminator. Most candidates were able to deduce the empirical formula and the C=O peak on the IR spectrum. However, many mistook the C-H peaks around 3000 cm-1 for an alcohol O-H peak or assumed from the empirical formula that it was an alcohol so made the spectra ‘fit’ their hypothesis. At AS, the exposure of candidates to IR and MS spectra is not as comprehensive as in the second year of A Level and this was evident. There was very little annotation of the spectra and candidates should be encouraged to do this as it is helpful to them in their deductions (and to the examiners for crediting marks). Analysis of the IR spectrum was much more detailed than the mass spectrum. Most candidates just referred to the molecular ion peak and made no attempt, or an incorrect attempt, at discerning the peak at 41. Those that did quickly realised what the structure was and gained 5 or 6 marks. Some candidates, despite ascertaining that a trans stereoisomer should be drawn, drew the cis version instead. This candidate has very logically worked through all the information provided and has come up with the correct structure from the deductions. Like most candidates, the crucial information about F being a trans isomer was not picked up so they drew the double bond in the wrong place. Copyright acknowledgements Q7, graph 1 © SDBSWeb : https://sdbs.db.aist.go.jp/sdbs/cgi-bin/cre_index.cgi (National Institute of Advanced Industrial Science and Technology, accessed 31 January 2017). Q7, graph 2 © SDBSWeb : https://sdbs.db.aist.go.jp/sdbs/cgi-bin/cre_index.cgi (National Institute of Advanced Industrial Science and Technology, accessed 31 January 2017).

Notes from the mark scheme and report

• empirical formula C4H6O
• M+ at m/z 70 so molecular formula C4H6O
• IR C=O 1630–1820 cm–1
• m/z 41 = C3H5+ (loss of CHO)
• trans-but-2-enal (cis accepted for Level 3)

Skills: M0.2 Use ratios, fractions and percentages; M3.1 Translate information between graphical, numerical and algebraic forms

Mark scheme, with the levels and indicative points

Examiner's comment

Most candidates were able to use the information in the table to determine the order of reaction with respect to Fe3+ ions and I– ions, which were 1 and 2 respectively. Most candidates were then able to use this knowledge and further information from the table to determine the rate constant, including units (22.5 dm6 mol–1 s–1). Many candidates appreciated that the mechanism involved a stoichiometric ratio of 1 : 2 with respect to Fe3+ ions and I– ions, but only the more able were able to suggest a suitable possible two step mechanism. Common errors in proposing a mechanism included equations such as Fe3+(aq) + 2I–(aq)  Fe2+(aq) + ½I2(aq) + I–(aq) which gives a net ratio of 1 : 1 for the reactants. or Fe3+(aq) + 2I–(aq)  Fe2+(aq) + I2(aq) + e– which although introducing a correct ratio of reactants also introduces a ‘floating’ electron which in all reality would have attached itself to one of the product species. In questions such as 17a where the quality of extended responses is assessed candidates need to be aware of the need for explanation of their answers. For instance, just giving a correct order of a species in the rate equation is not as strong an answer as one which explains how altering, say doubling, the concentration of a reactant effects the overall rate of reaction, leading to determination of the order. This exemplifies how a concise response can gain full marks on a Level of Response question. The candidate has clearly identified which data they have referred to in the table, and explained the conclusions that can be drawn. The rate equation is then clearly shown, along with subsequent working to determine the value of the rate constant. Finally, the suggested mechanism is given, along with an indication of which is the slow step. The line breaks clearly identify each stage in the response, and it fully satisfies the level 3 criteria in the mark scheme.

Notes from the mark scheme and report

• Fe3+ first order and I– second order supported by quantitative comparison of experiments
• k = 22.5 dm6 mol–2 s–1
• two steps must add up to overall equation, slow step matches rate equation, each step balanced for species and charge with no electrons

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG10 Rates of reaction – initial rates

Mark scheme, with the levels and indicative points

Examiner's comment

Many candidates had a stab at identifying C–F but neglected to include equations for the three reactions described or to show relevant working. Most candidates recognised C as the ammoniacal copper(II) ion but the formula was frequently incorrect and correct attempts at a ligand substitution equation from [Cu(H2O)6]2+ was rarely seen. Diagrams showing the trans isomer were attempted but often of poor quality due to incorrect linking. Candidates recognised D as being CuSO4 but often did not identify E as Cu due to a lack of familiarity with this common disproportionation reaction. Cu(OH)2(s) was a common incorrect identification of E. Only the best responses described the oxidation number changes which made this a disproportionation reaction. F was identified by a percentage by mass calculation to determine an empirical formula and then by deduction to produce Cu(NO3)2.3H2O. Having done this, many candidates did not give the relatively simple equation for reaction 3 between copper(II) oxide and dilute nitric acid. This exemplifies how considered structuring of candidate responses can enhance their clarity. The clear labelling of C-F in the response, along with numbering which corresponds to the reactions in the question stem, make the candidate’s line of reasoning easy to follow. The underlining and annotations in the question stem show good practice in picking out and interpreting key information.

Notes from the mark scheme and report

• C = [Cu(NH3)4(H2O)2]2+ drawn as trans isomer with correct wedges and connectivity
• D = CuSO4, E = Cu with Cu(+1) → Cu(+2) + Cu(0)
• F = Cu(NO3)2•3H2O from mole ratio CuH6N2O9
• formulae with missing brackets or numbers limit the level

Skills: M0.2 Use ratios, fractions and percentages; M4.2 Visualise and represent 2-D and 3-D forms including 2-D representations of 3-D objects

Mark scheme, with the levels and indicative points

Examiner's comment

This question was marked using a level of response mark scheme. Most candidates gave an answer worthy of at least level two (3-4 marks) by providing a suitable mechanism and identifying the major product. The strongest candidates identified both products and were able to describe which was most likely. Such responses received level 3 (5-6 marks) as shown in Exemplar 9. Lower ability candidate responses seemed to ignore the reference to electrophilic addition in the question and tried reacting ICl with either the benzene ring or the aldehyde group. This response starts with a clear outline of the electrophilic addition mechanism showing correct partial charges and accurately drawn curly arrows. The structures of the two possible organic products are shown and the most likely product to be formed is identified correctly. Notice that the candidate has labelled these structures as ‘Product A’ and ‘Product B’ and refers to the labels later in the response. This is a good strategy, which enables the candidate to provide a clear and easy to follow answer. The response concludes with a detailed explanation, referring to carbocation stability to justify why product A is most likely to be formed. This response therefore satisfies the Level 3 criteria. The response is logically structured with a well-developed line of reasoning and was therefore credited the upper mark within the level and achieved six marks.

Notes from the mark scheme and report

• dipole Iδ+–Clδ– from electronegativity so I is the electrophile
• curly arrow from C=C to I and from I–Cl bond to Cl
• carbocation with full + charge
• arrow from lone pair/– charge on Cl–
• major product via the more stable (tertiary) carbocation

Mark scheme, with the levels and indicative points

Examiner's comment

Most candidates were able to determine the empirical and molecular formula of the unknown compound. A number of excellent and clear response were seen, where the NMR data was explained, including interpretation of the additional peaks observed without D2O. However many candidates were unable to suggest a structure that matched their NMR interpretation. Some candidates used the quartet, doublet and singlet to suggest a structure that would give rise to this splitting pattern, but which was not consistent with the chemical shifts, see Exemplar 12. Such responses received a level 2 mark (3-4). Stronger responses were able to use all the data to suggest a correct structure. The most common was CH3CH(OH)C(CH3)2COOH although other viable structures, including CH3CH(OH)COC(CH3)2OH, were also seen. Examiners were impressed with the problem solving ability shown by candidates and a significant proportion of responses were credited six marks. This logically presented Level 2 response uses the elemental analysis and mass spectrum data to determine the correct empirical and molecular formula of the unknown compound. The peaks in the NMR spectrum are analysed in detail, with a clear explanation of the splitting patterns. A comment about the two additional peaks observed when the spectrum is run without D2O is also provided. The response concludes with a structure of C6H12O3 that would show a singlet, doublet and quartet in its 1H NMR spectrum. However, this structure is not consistent with the chemical shift values shown in the spectrum provided. In particular this structure would produce a quartet between 2.0–2.9 ppm, rather than at 4.0 as in the spectrum shown. Consequentially this response does not achieve Level 3. When tackling questions of this type candidates are advised to check that a proposed structure would produce peaks in the correct region of the NMR spectrum to ensure it is totally consistent with the data analysed. Copyright acknowledgements Section A, Q9 © SDBS, National Institute of Advanced Industrial Science and Technology. Reproduced by kind permission of SDBS. Section A, Q12 © SDBS, National Institute of Advanced Industrial Science and Technology. Reproduced by kind permission of SDBS.

Notes from the mark scheme and report

• empirical formula C2H4O then C6H12O3 from m/z 132
• quartet (1H) at 4.0 CH–O coupled to doublet (3H) at 1.2
• singlet (6H) at 1.3 (CH3)2C
• peaks at 11.0 (COOH) and 3.6 (OH) disappear in D2O

Skills: M0.2 Use ratios, fractions and percentages; M3.1 Translate information between graphical, numerical and algebraic forms

Mark scheme, with the levels and indicative points

Examiner's comment

This part assessed practical aspects of a two-stage organic synthesis. Overall, candidates responded well, and this part was discriminating. Many candidates produced well-structured responses although lower ability candidates do have problems with constructing a cohesive response. Most candidates identified the correct reagents (Sn and concentrated HCl) and the intermediate (4- aminophenol), which was usually shown as its structure. Able candidates usually showed that 11.5 g of 4-nitrophenol is needed for the synthesis. A common error used the ‘inverse percentage’ ratio of 40/100, resulting in an incorrect mass of 1.84 g. Candidates are recommended to check whether a calculated answer looks sensible. Looking at the structures and with a percentage yield of 40%, 1.84 g does not look to be enough of the starting chemical. Some lower ability candidate responses assumed that 5.00 g was 40% of the required mass and responded with 5.00 × 100/40 = 12.5 g. There were some good descriptions of purification, although finer details such as using a minimum volume of hot solvent, washing with cold solvent, and drying) were often omitted. Candidates needed to respond with more than just ‘recrystallisation’. In the purification, common errors were showing the correct steps but in the wrong order and use of a drying agent such as CaCl2 (confusion with part of the purification of an organic liquid). These candidates seemed unaware that adding a solid drying agent to an organic solid would result in impure paracetamol rather than purifying. In comparison, Exemplar 12 is much less detailed: concentrated HCl has not been shown as a reagent for step 1, the candidate has not shown that they know how to carry out a percentage yield calculation, and the purification is confused, and lacks detail.

Notes from the mark scheme and report

• 5.00/151 = 0.0331 mol, ×100/40 = 0.0828 mol, × 139 = 11.5 g (inverted ratio gives 1.84 g)
• Sn + conc HCl then NaOH
• intermediate 4-aminophenol
• recrystallise: dissolve in MINIMUM volume of HOT solvent, cool, filter, wash with cold solvent, dry
• 'recrystallisation' alone is not a detailed description

Skills: M0.2 Use ratios, fractions and percentages; M2.2 Change the subject of an equation; PAG6 Synthesis of an organic solid

Mark scheme, with the levels and indicative points

Examiner's comment

Overall, this part was answered well. Good answers were well organised, showing clear mechanisms with precisely positioned curly arrows. Most candidates included a clear definition of an electrophile and were able to select appropriate reactions. Most candidates were familiar with the mechanisms for electrophilic addition and electrophilic substitution. Mechanisms of an alkene with HBr or Br2 and benzene with NO2 + or Br+ were the most commonly seen. Most candidates showed the role of a catalyst in electrophilic substitution. Common errors/omissions were the direction of the curly arrow from the aromatic ring to the electrophile, the position of the curly arrow when H+ is lost from an aromatic intermediate, and not showing the lone pair when Br- attacks a carbocation. Some answers lacked detail and gave only two mechanisms with minimal supporting words. Lower ability candidates described nucleophilic addition or substitution as one of their mechanisms or had curly arrows going in the wrong direction. A few candidates answered in prose without including equations or diagrams. Candidates are advised that mechanisms must always be communicated in the usual diagrammatic way. and electrophilic substitution, including the role of the FeCl3 catalyst. Curly arrows are precisely positioned, with correct use of lone pairs and charges. The candidate has demonstrated excellent knowledge and understanding. The response in Exemplar 17 is clearly at a different level. The candidate has chosen an alkane rather than an alkene and has used curly arrows and charges incorrectly. This candidate appears to have been poorly prepared.

Notes from the mark scheme and report

• electrophile as electron pair acceptor
• aliphatic e.g. ethene + Br2 electrophilic addition with dipole, curly arrows and carbocation
• aromatic e.g. nitration/halogenation of benzene electrophilic substitution with generation of electrophile and regeneration of catalyst
• accurately positioned curly arrows and charges

Mark scheme, with the levels and indicative points

Notes from the mark scheme and report

• 72/24000 = 0.00300 mol H2 so minimum 0.20 g Zn and 0.00600 mol HCl
• acid volume must suit the apparatus (e.g. 60 cm3 of 0.1 mol dm–3)
• plot volume against time, tangent at t = 0, gradient = initial rate

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M3.2 Plot two variables from experimental or other data; M3.5 Draw and use the slope of a tangent to a curve as a measure of rate of change; PAG1 Moles determination; PAG9 Rates of reaction – continuous monitoring

Mark scheme, with the levels and indicative points

Notes from the mark scheme and report

• acidified dichromate
• distil for butanal (removed to prevent further oxidation), reflux for butanoic acid
• equations with [O] (2[O] for the acid)
• distillation diagram with condenser and water flow

Skills: PAG5 Synthesis of an organic liquid

Mark scheme, with the levels and indicative points

Notes from the mark scheme and report

• tangent at t = 0 (gradient 1.5–2.0 × 10–3 mol dm–3 s–1)
• two constant half-lives (800–1000 s) or gradient halves at c/2 gives first order
• k = rate/[H2O2] or ln2/t½ with units s–1

Skills: M0.0 Recognise and make use of appropriate units in calculations; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M3.1 Translate information between graphical, numerical and algebraic forms; M3.5 Draw and use the slope of a tangent to a curve as a measure of rate of change

Mark scheme, with the levels and indicative points

Notes from the mark scheme and report

• B = NiC6N6H24Cl2
• A = B•2H2O from mass loss
• C = [Ni(H2NCH2CH2NH2)3]2+
• D = H2NCH2CH2NH2
• both optical isomers in 3D with wedges and bonding from N

Skills: M0.2 Use ratios, fractions and percentages; 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; PAG1 Moles determination

Mark scheme, with the levels and indicative points

Notes from the mark scheme and report

• recrystallise: dissolve in minimum volume of hot solvent, cool, filter, wash with cold solvent, dry
• yield 71.3% must go via moles of benzoic acid
• purity: melting point compared with data book value (sharp), ALLOW TLC or spectroscopy

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG6 Synthesis of an organic solid

Mark scheme, with the levels and indicative points

Notes from the mark scheme and report

• triplet 3H + quartet 2H = CH3CH2
• singlet 9H = (CH3)3C
• singlet 2H at 4.0 = CH2–O
• M broad O–H and C=O = carboxylic acid, N O–H = alcohol, so L is an ester: CH3CH2COOCH2C(CH3)3 or (CH3)3CCH2COOCH2CH3
• IGNORE C–O peaks

Skills: M3.1 Translate information between graphical, numerical and algebraic forms

Mark scheme, with the levels and indicative points

Notes from the mark scheme and report

• mass of solution 25.75 g and ∆T 35.0 °C from raw data
• q = mc∆T = 3767 J
• n(Na2O) = 0.0200 mol so ∆H(3.2) = –188 kJ mol–1
• ∆H(3.1) = ∆H(3.2) + 2∆H(3.3) = –303.6
• signs and SF

Skills: M0.0 Recognise and make use of appropriate units in calculations; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG3 Enthalpy determination

Mark scheme, with the levels and indicative points

Notes from the mark scheme and report

• bromine decolourised by B, C, D, G
• acidified dichromate turns green with C, D, F
• 2,4-DNP orange precipitate with C, E, G
• Tollens' silver mirror with C only
• no credit for tests on products, melting points or spectra

Skills: PAG7 Qualitative analysis of organic functional groups

Mark scheme, with the levels and indicative points

Notes from the mark scheme and report

• two products 2-bromo-2-methylbutane and 2-bromo-3-methylbutane
• curly arrow from C=C to H of H–Br, dipole on H–Br, arrow from H–Br bond to Br, carbocation, arrow from Br– lone pair to C+
• major product from more stable tertiary carbocation

Mark scheme, with the levels and indicative points

Notes from the mark scheme and report

• empirical formula C2H4O and M+ at 88 gives C4H8O2
• broad O–H 2500–3300 and C=O about 1700 so carboxylic acid
• m/z 43 is secondary carbocation (CH3)2CH+ so C is (CH3)2CHCOOH not butanoic acid
• IGNORE C–O peaks

Skills: M0.2 Use ratios, fractions and percentages; M3.1 Translate information between graphical, numerical and algebraic forms