June 2022: Paper 2, Synthesis and analytical techniques

Answer

B

Mark scheme

Examiner's comment

The majority of candidates identified B as the correct answer.

Answer

B

Mark scheme

Examiner's comment

Candidates found this multiple choice question challenging. While some identified B as the correct answer, many candidates selected C.

Answer

D

Mark scheme

Examiner's comment

This question discriminated well, with higher ability candidates correctly identifying D. Often students overlooked the sigma bonds in the aromatic ring and selected B.

Answer

A

Mark scheme

Examiner's comment

Most candidates correctly selected A. The most common incorrect response was option D as candidates had misinterpreted the amide group as a ketone and an amine.

Answer

D

Mark scheme

Examiner's comment

Many candidates correctly answered D, with those selecting the correct response showing annotations on the given structure to identify the chiral carbons. C proved a good distractor.

Answer

B

Mark scheme

Examiner's comment

The majority of candidates identified B as the correct answer, with candidates frequently showing their calculations alongside each option to aid their choice.

Answer

D

Mark scheme

Examiner's comment

Most candidates correctly identified D as the correct response.

Answer

C

Mark scheme

Examiner's comment

The majority of candidates correctly calculated C as the mass of paracetamol in the tablet.

Answer

C

Mark scheme

Examiner's comment

This question discriminated well, with higher ability candidates correctly identifying C as the volume of hydrogen produced. Many candidates overlooked the hydrogenation of the CN group and incorrectly selected option A.

Answer

B

Mark scheme

Examiner's comment

Most candidates correctly identified B as one of the products. These candidates often drew out the structures of butyl propanoate and the product options to aid in their selection.

Answer

B

Mark scheme

Examiner's comment

Most candidates correctly identified C as the compound after labelling the peaks in the spectrum.

Answer

A

Mark scheme

Examiner's comment

This question proved challenging. Candidates who drew out the different compounds were able to identify A as the correct response.

Answer

D

Mark scheme

Examiner's comment

Most candidates identified D (only 1) for an equation that could be part of a propagation step.

Answer

A

Mark scheme

Examiner's comment

Many candidates did not identify CH3NH2 as a nucleophile and selected option B.

Answer

B

Mark scheme

Examiner's comment

Candidates found this question difficult. Those who drew out the different compounds were able to identify B as the correct response. Section B overview Section B includes a mixture of short answer and extended response questions, including two questions marked using a level of response mark scheme (19e and 21). This section of the paper is worth 85 marks.

Mark scheme

Examiner's comment

The majority of candidates were able to correct name hydrocarbon A as 3-methylhex-2-ene. A number of responses used incorrect numbering or suggested 3-methylhexan-2-ene as the name.

Notes from the mark scheme and report

• DO NOT ALLOW 3-methyhex-2-ene, 3-methhex-2-ene or 3-methylhexan-2-ene
• IGNORE E/Z or cis/trans

Mark scheme

Examiner's comment

Candidates are familiar with the mechanism for the bromination of hydrocarbons. So, the majority of candidates scored 3 marks. Common errors included the use of HBr rather than Br2 or putting a dipole on the carbon-carbon double bond.

Notes from the mark scheme and report

• curly arrow from C=C to Br AND dipole on Br–Br AND arrow breaking Br–Br
• carbocation with full + AND arrow from lone pair/– charge of Br–
• DO NOT ALLOW partial charge on C=C
• DO NOT ALLOW half-headed arrows
• DO NOT ALLOW use of HBr
• ALLOW bromonium ion

Mark scheme

Examiner's comment

The majority of candidates were able to correctly define a structural isomer.

Notes from the mark scheme and report

• 'same formula' insufficient (must be molecular)
• 'different arrangement of atoms' insufficient (must be structural)
• DO NOT ALLOW any reference to spatial/space

Mark scheme

Examiner's comment

This definition was well known by candidates with the majority of responses given the mark. Some candidates omitted the reference to structural formula.

Notes from the mark scheme and report

• same structural formula AND different arrangement in space
• DO NOT ALLOW same empirical/general formula
• IGNORE same molecular formula
• naming E/Z or optical isomerism is not sufficient

Mark scheme

Examiner's comment

This question required candidates to link their knowledge of cis and trans isomers with branched hydrocarbons. Higher ability candidates were able to do this. The majority of candidates scored 1 mark for correctly drawing cis and trans isomers of an unbranched hydrocarbon.

Notes from the mark scheme and report

• 4-methylpent-2-ene cis and trans (ALLOW 3-methylpent-2-ene in either box)
• C3H7 not sufficient (could be unbranched)
• 1 mark if in wrong boxes or for cis/trans of an unbranched C6H12

Skills: M4.2 Visualise and represent 2-D and 3-D forms including 2-D representations of 3-D objects; M4.3 Understand the symmetry of 2-D and 3-D shapes

Mark scheme

Examiner's comment

This question discriminated well. Candidates were required to identify the groups around a chiral carbon and then draw the two corresponding optical isomers. Incorrect responses frequently had incorrect connectivity around the chiral carbon, bond angles of 180 or 2D structures.

Notes from the mark scheme and report

• C is 3-methylpent-1-ene
• four bonds to chiral C with wedges
• mirror images with correct connectivity
• DO NOT ALLOW a bond angle of 180°

Skills: M4.2 Visualise and represent 2-D and 3-D forms including 2-D representations of 3-D objects; M4.3 Understand the symmetry of 2-D and 3-D shapes

Mark scheme

Examiner's comment

Most candidates were able to correctly draw the structure of D and E. Many candidates did not explain their answers in terms of the number of different hydrogen and carbon environments or the presence/ absence of a carbon-carbon double bond.

Notes from the mark scheme and report

• D is cyclohexane (all H and all C equivalent, no C=C)
• E is 2,3-dimethylbut-2-ene (all H equivalent, 2 C environments, C=C present)
• 1 mark for structures if in the wrong boxes

Skills: M4.3 Understand the symmetry of 2-D and 3-D shapes

Mark scheme

Examiner's comment

Most candidates were able to score 1 mark for correctly drawing the structures of the two aldehyde products of the first reaction. The second reaction proved more challenging, with most candidates incorrectly drawing two products.

Notes from the mark scheme and report

• BOTH structures required for each mark: propanal and ethanal from pent-2-ene
• a single open-chain dicarbonyl from the cyclic alkene

Mark scheme

Examiner's comment

Few candidates were given this mark. A common error was to produce multiple products (rather than a ring structure) or to put positive/negative charges on the oxygen atoms within the ring structure.

Notes from the mark scheme and report

• product must follow the curly arrows given (five-membered ring intermediate)

Mark scheme

Examiner's comment

This question proved difficult and discriminated well. Higher ability candidates correctly used SI units and showed each step of their calculation and then using this to correctly identify a structure of compound X. Candidates frequently used the wrong interconversions and gave structures that lacked a chiral centre. A small number of candidates used molar gas volume rather than PV=nRT for their calculation.

Notes from the mark scheme and report

• convert cm3 to m3 and °C to K
• M = 168 so C3H2F6O
• structure must contain F6O, C–O–C and a chiral carbon (DO NOT ALLOW no chiral carbon)
• use of 24 dm3 loses the calculation mark but ECF (171, C3H5F6O)

Skills: M0.0 Recognise and make use of appropriate units in calculations; M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

The use of 2,4-dinitrophenylhydrazine as a test for the carbonyl group is well known by candidates at this level. The majority of the cohort correctly identified this test and the subsequent analysis of the melting point of the products as a method of identifying each compound. Lower ability candidate responses made reference to analysis of the boiling points or omitted the reference to analysis of the melting points.

Notes from the mark scheme and report

• 2,4-DNP(H) AND orange/yellow/red precipitate
• take melting point of the derivative
• compare with known values
• DO NOT ALLOW boiling points or chromatograms

Skills: PAG6 Synthesis of an organic solid; PAG7 Qualitative analysis of organic functional groups

Mark scheme

Examiner's comment

Almost all candidates were able to correctly describe the use of Tollens’ reagent as a test for an aldehyde fictional group

Notes from the mark scheme and report

• Tollens' reagent AND silver mirror
• ALLOW acidified dichromate turning green

Skills: PAG7 Qualitative analysis of organic functional groups

Mark scheme

Examiner's comment

Most candidates were able to calculate the initial ratios and then use this to calculate the molecular formula, linking to the m/z peak at 58. The most common error seen was for the structures of F and G to the in the incorrect boxes, suggesting that candidates have not interpreted the fragment ions to work out F was the aldehyde.

Notes from the mark scheme and report

• C3H6O with evidence of 58 (molecular ion)
• F is propanal from fragment m/z 29 (CH3CH2+ or CHO+)
• G is propanone
• IGNORE m/z 15 (not unique) and 43

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

Mark scheme

Examiner's comment

This question proved difficult and although most candidates score some marks, only the very best responses secured all 4 marks. Candidates often did not link the doublet to two CH3 groups and many only scored 1 mark for their suggested structure as the carbonyl groups were not side by side within the molecule.

Notes from the mark scheme and report

• doublet at 1.1 (6H) = 2 × CH3
• singlet at 2.2 (3H) = CH3–C=O
• multiplet at 2.9 (1H) = CH(CH3)2 next to C=O
• structure must be C6H10O2 with two carbonyl groups
• ALLOW δ ± 0.2 ppm

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

Mark scheme

Examiner's comment

This question proved challenging for candidates. The first equation was often answered correctly, although some candidates used sodium hydroxide rather than potassium hydroxide in their response. The second equation was frequently incorrect. Candidates frequently missed a hydrogen from the structure for methanolic acid or did not recognise that hydrogen was a product. Many candidates did not account for magnesium having a 2+ charge when working out the product. For the third equation, the majority of candidates identified that carbon dioxide and water would be produced but were unable to give the correct formula of the salt as they did not interpret the information given regarding the R group.

Notes from the mark scheme and report

• 2HCOOH + Mg → (HCOO)2Mg + H2
• both COOH groups of aspartic acid react with excess Na2CO3 giving H2O AND CO2 (ALLOW H2CO3)
• DO NOT ALLOW –COO–K covalent bond or missing charge (penalise once)

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

The majority of candidates were able to identify at least one of the structures. A significant number of candidates did not check the number of bonds of each atom in their structures and frequently had too many or too few hydrogen atoms attached. Most candidates identified that acidic conditions were required but some missed the aqueous condition that was also required for the mark.

Notes from the mark scheme and report

• with C6H5COOH the OH group is esterified
• with (CH3)2CHOH the COOH group is esterified
• nitrile to acid needs H+/H2O (ALLOW HCl/H2O, H2SO4/H2O)
• NaBH4 box is the ketone CH3CH2COCOOH

Mark scheme

Examiner's comment

Most candidates were given at least 1 mark with many correctly drawing the structure of pent-3-enoic acid. Common errors included drawing pent-2-enoic acid or propenoic acid, suggesting a lack of knowledge of prefixes. The second mark required candidates to draw two repeat units, frequently candidates tried to draw repeat units linking the carboxylic acid groups rather than identifying that it is the carbon-carbon double bond that breaks.

Notes from the mark scheme and report

• CH3CH=CHCH2COOH (E or Z)
• two repeat units with side bonds
• ECF only from pent-2-enoic or pent-4-enoic acid
• IGNORE brackets and n

Mark scheme

Examiner's comment

This question differentiated well. Candidates who scored 1 mark had often shown an ester link but their structure was missing hydrogen atoms from the carbon chain (as shown in exemplar 1) or the methyl group was missing from the ring. This type of response was seen frequently by examiners. The candidate has correctly drawn the ester link but has omitted the hydrogen atoms from the carbon chain.

Notes from the mark scheme and report

• ester link shown
• ONE repeat unit only
• end bonds MUST be shown
• CH3 may be on position 2 or 3 of the ring

Mark scheme

Examiner's comment

This question proved difficult for candidates with the majority of candidates not scoring the mark. A significant proportion of candidate had identified the monomer but drew structures that combined skeletal and displayed formulae. This resulted in ambiguous structures being given that had missing hydrogen atoms on the carbons. Drawing of organic structures Candidates need practice at drawing structures that are not ambiguous. They should check the number of bonds on each atom and make sure the appropriate number of hydrogen atoms are drawn.

Notes from the mark scheme and report

• single hydroxy-acid monomer containing both OH and COOH

Mark scheme

Examiner's comment

Few candidates were given the mark for this question. Frequently candidates drew structures with two repeat units or the did not remove the oxygen atom from the OH group.

Notes from the mark scheme and report

• ONE repeat unit only with amide link
• end bonds MUST be shown

Mark scheme

Examiner's comment

A variety of responses were seen in this demanding question. In general candidates adopted one of two approaches. The most common was to multiply the Mr of the repeat unit by 200 and then add the mass of H and OH at each end of the polymer. The other approach used the Mr of the monomer by 200 and then subtract the mass of the 199 water molecules removed in the polymerisation. 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.

Notes from the mark scheme and report

• 28418: 400 × 89 minus 399 × 18 (water lost)
• 28400 scores 1 mark (end groups ignored)

Skills: M0.2 Use ratios, fractions and percentages

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

Examiner's comment

Few candidates knew the test for a phenol group. Frequent incorrect responses involved the production of a gas with a carbonate or stating that bromine water is decolourised but failing to state that a white precipitate is also formed. A significant number of candidates also stated that the reaction with sodium hydroxide confirms the presence of the phenol group.

Notes from the mark scheme and report

• indicator shows acidity AND no reaction with carbonate
• ALLOW bromine giving white precipitate
• ALLOW FeCl3 giving violet/purple colour

Skills: PAG7 Qualitative analysis of organic functional groups

Mark scheme

Examiner's comment

This question proved challenging to candidates with few scoring all 3 marks. Where no marks were given, this was frequently because candidates did not state the number of carbon environments in compounds J, K and L. Candidates who were given 1 or two marks frequently stated the incorrect number of peaks that would be observed.

Notes from the mark scheme and report

• J has 6 peaks, K has 5 peaks, L has 8 peaks
• IGNORE chemical shifts

Skills: M4.3 Understand the symmetry of 2-D and 3-D shapes

Mark scheme

Examiner's comment

This question required candidates to apply their knowledge of the mechanism of electrophilic substitution. Examiners were encouraged by the number of excellent responses to this question, with the majority of candidates securing 3 out of 4 marks. Common errors included the omission of HCl as product from the regeneration of the catalyst or candidates attempting to substitute at the 2 position.

Notes from the mark scheme and report

• Cl2 + AlCl3 → Cl+ + AlCl4– AND H+ + AlCl4– → AlCl3 + HCl (both for 1 mark)
• arrow from ring to Cl+
• horseshoe over more than half the ring with gap towards C–Cl
• DO NOT ALLOW intermediates substituted at positions 3 or 5
• arrow from C–H bond into ring

Mark scheme

Examiner's comment

The most able candidates scored well on what proved a difficult question for many. Although the vast majority of candidates knew about the reasons behind compounds K and L’s increased reactivity many were unable to express themselves clearly to gain credit. Often answers lacked the specific detail about the lone pair on the oxygen atom in the –OH group being delocalised into the ring. Less successful answers discussed electrons being supplied to the ring, the -OH group providing the electrons to the ring or just that –OH is an activator. More successful answers expressed the increase in electron density and the subsequent increase in compound K and L’s susceptibility to electrophilic attack. This response correctly states that the electron density is higher which result in K and L being more susceptible to electrophilic attack and receives 2 marks. To score the third mark, the candidate needed to identify that that the lone pair is on the oxygen of the OH group.

Notes from the mark scheme and report

• a lone pair on O delocalised into the ring (DO NOT ALLOW two lone pairs)
• electron density higher
• ring more susceptible to electrophilic attack/polarises Cl2 more
• IGNORE activating, charge density
• IGNORE 'react more readily' (given in question)

Mark scheme

Examiner's comment

Most candidates were able to score at least 1 mark for this question. Common errors included candidates producing two water molecules or failing to balance [O]. A significant proportion of candidates did not score any marks, frequently due to the organic product having too many carbon atoms in it.

Notes from the mark scheme and report

• product C6H5CH2COOH (reflux)
• 2[O] and H2O

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

Candidates were familiar with the reagents required in these two reactions.

Notes from the mark scheme and report

• HNO3 with H2SO4 for nitration
• Sn and HCl for reduction
• IGNORE concentrations
• IGNORE NaOH after Sn/HCl
• IGNORE H2

Mark scheme

Examiner's comment

The most able candidates were able to identify the use of 6[H] as the reducing agent and the production of 2 water molecules. Incorrect responses commonly included the use of HCl and NaBH4 as a reactant.

Notes from the mark scheme and report

• 6[H] on the left and 2H2O on the right

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

This question proved challenging with only the most able being given full marks. The reagents and conditions were not well known and candidates did not include water in their equations to make sure they were balanced.

Notes from the mark scheme and report

• stage 1: acid catalyst H2SO4/H3PO4 (heat) eliminates H2O to give phenylethene (DO NOT ALLOW HCl)
• stage 2: steam with acid catalyst gives 1-phenylethanol (Markownikoff)
• equations with structures needed

Mark scheme

Examiner's comment

Most candidates did not secure a mark in this question. Many candidates used butan-1-ol in their equations or used benzoic acid rather than benzoic anhydride as the reactant. The most able candidates suggested that the benzoic acid product would then further react with butan-2-ol to produce a second ester molecule and water. This was an acceptable alternative response.

Notes from the mark scheme and report

• (C6H5CO)2O + butan-2-ol → C6H5COOCH(CH3)CH2CH3 + C6H5COOH
• ester structure 1 mark, balanced equation 1 mark

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