June 2024: AS Paper 2, Depth in chemistry

Mark scheme

Examiner's comment

Many managed to score at least one mark here but most struggled to give a comprehensive definition. A mark was often lost mark for omission of ‘atom’, i.e. not saying what was attracting electrons, or for saying ‘element’ instead. The ability to attract an electron was not accepted as this is describing electron affinity rather than electronegativity. Many described the formation of dipoles or partial charges across the bond due to the difference in electronegativity but didn’t demonstrate a clear understanding of what electronegativity is. Lower-scoring candidates often referred to the formation of ions and/or loss of electrons.

Notes from the mark scheme and report

• ability of an atom to attract electrons
• in a covalent bond
• DO NOT ALLOW 'element' for 'atom'
• DO NOT ALLOW attract 'an electron' (single electron)
• Learning key words and terminology is an important part of understanding Chemistry.
• Encourage students to make their own flash cards for each topic – this can be done on card or electronically using websites such as Quizlet.

Mark scheme

Examiner's comment

Around half of candidates scored both marks with a well-drawn diagram (such as that shown in the exemplar below). Practising drawing hydrogen bonds between different molecules is a really good way of exploring students’ understanding of what they are. The most common errors were missing the lone pair or showing a hydrogen bond that didn’t originate from those drawn. Lower-scoring candidates often didn’t draw a second water molecule, labelling the covalent bond between O and H as a hydrogen bond. Quite a few diagrams were seen with δ+ and δ- only on one molecule or with δ+ on one structure and δ- on the other. It was essential here to show that the O-H bond in both molecules is polar. Diagrams were often unclear so sometimes difficult to interpret. This is an example of a good diagram for hydrogen bonding.

Notes from the mark scheme and report

• δ+ on H and δ– on O on each molecule
• hydrogen bond must go from H to a lone pair on O of the other molecule
• DO NOT ALLOW more than 2 lone pairs on O

Mark scheme

Examiner's comment

Some appropriate answers were seen, mostly discussing the asymmetrical shape of water. More than half didn’t score the mark. However, many were able to describe O-H being a polar bond due to the difference in electronegativity but didn’t realise that this doesn’t always mean that the molecule itself is polar. The relationship between the words electronegative, dipole and polar need more attention and how this all relates to the shapes of molecules. Some candidates said symmetrical when they meant asymmetrical.

Notes from the mark scheme and report

• dipoles do not cancel/overall dipole/unsymmetrical
• IGNORE 'polar bonds do not cancel'
• IGNORE charges uneven

Mark scheme

Examiner's comment

Some good answers were seen referring to hydrogen bonds holding/keeping water molecules further apart or creating an open lattice. However, a number did not score for not mentioning hydrogen bonds or for referring to alternative intermolecular forces. For example, ‘In solid state, the London forces are weaker’ and ‘Ice has less induced dipole-dipole forces’. Some also indicated that ‘air’ was in the structure and this wasn’t accepted as air molecules are too large to fit in the gaps between water molecules in the ice crystal structure. Responses often demonstrated some significant misconceptions in candidate understanding. Many felt that the length of the hydrogen bond changed, for example ‘when ice freezes hydrogen bonds expand’ ‘Atoms are further apart so longer hydrogen bonds’. Some suggested that the strength of the hydrogen bond changed (for example, ‘ice has weaker H-bonds than water’). Some recognised that ice has a crystalline structure but didn’t explain how this is different from other solid structures, which would be denser than their liquid counterparts. Lower-scoring candidates struggled to understand density (for example, ‘ice is more compact’ or ‘molecules closer together in ice’). Terminology was also a problem with candidates often referring to ‘ice’ or ‘particles’ rather than ‘molecules’.

Notes from the mark scheme and report

• must refer to hydrogen bonds holding molecules apart/open lattice
• DO NOT ALLOW 'atoms' for 'molecules'
• DO NOT ALLOW gaps containing air

Mark scheme

Examiner's comment

Despite being told in the question that ammonia contains hydrogen bonds, many gave responses in terms of ammonia having either London forces and permanent dipole-dipole forces which are weaker than hydrogen bonds. For example, ‘ammonia consists of permanent dipole-dipole interactions which are weaker than hydrogen bonding in ice’ and ‘NH3 has 17 electrons and H2O has 18 electrons. Due to NH3 having fewer electrons, there are fewer London forces’. Lower-scoring candidates often confused hydrogen bonds and covalent bonds, consistent with what was seen in 1(b)(i). For example, ‘O-H bond is stronger than N-H bond’ and ‘more energy needed to break O- H bonds rather than N-H bonds’. Some of these candidates did score a mark for recognising that N is less electronegative than O. For others they understood that ammonia has weaker hydrogen bonds but then struggled to give a reason either in terms of lone pairs or electronegativity.

Notes from the mark scheme and report

• comparative: weaker/fewer hydrogen bonds in ammonia
• N has one lone pair, O has two OR N less electronegative
• DO NOT ALLOW breaking N–H/O–H covalent bonds
• IGNORE other intermolecular forces

Mark scheme

Examiner's comment

Less than half of the candidates scored both marks. Most candidates drew 4 x N-H shared covalent bonds and therefore lost the dative bond mark. Some added an additional electron to either N or H. Some drew an additional shaped electron (e.g. using a triangle) on one of the bonding pairs, obviously not realising that both electrons in dative bond originate from N, so have the same symbol. Many diagrams were unclear making it hard distinguish between dots and crosses especially if adding circles for electron shells. A few lower-attaining candidates attempted to draw an ionic dot-cross diagrams.

Notes from the mark scheme and report

• three N–H bonds as dot-and-cross pairs
• dative bond as two dots or two crosses
• charge and brackets not required
• DO NOT ALLOW additional electrons on N or H

Mark scheme

Examiner's comment

Higher-attaining candidates often gave a very detailed responses with all stages, including warming the NaOH, use of damp litmus paper and some included an ionic equation. Quite a few lost a mark as they missed the addition of hydroxide, just warmed, but they still gained mark for testing with indicator paper. Some thought that the indicator paper would turn red or be bleached and a few gave incorrect ion test e.g. add silver nitrate, add acid. Over a third of candidates did not score on this question, with a significant proportion not even attempting it.

Notes from the mark scheme and report

• (warm with) NaOH/hydroxide
• damp litmus/indicator paper turns blue
• DO NOT ALLOW ammonium hydroxide or ammonia as reagent

Skills: PAG4 Qualitative analysis of ions

Mark scheme

Examiner's comment

Just under half of the candidates scored all 3 marks. A range of responses were seen for this question but the most common error was to use the incorrect Mr for the hydrated which included either using the Mr for the anhydrous salt or the Mr for ZnSO4.H2O (179.5). It was evident that candidates struggle to understand what is meant by water of crystallisation. Some scored 2 marks for the correct Mr and finding moles but then were unsure how to proceed with the calculation. For example, some divided moles by 7 before multiplying by 161.5. Many candidates took the more complicated alternative route to find the mass of water before subtracting from mass of hydrated compound. Low-scoring candidates often struggled to correctly work out Mr values. Their working was set out so that individual steps were unclear, often with contradictory calculations.

Notes from the mark scheme and report

• M(ZnSO4•7H2O) = 287.5
• n = 0.04
• × 161.5 = 6.46 g
• no ECF if 161.5 used as molar mass of hydrate
• alternative via mass of water lost

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

Mark scheme

Examiner's comment

A very well answered question with most candidates very confident in giving the correct electron configurations and blocks. Errors were rare but included: 2p5 or 3p6 ending for Cl; using mass number for number of electrons; and assigning group 17 as d block and giving orbital box diagrams rather than block.

Notes from the mark scheme and report

• 1 mark per correct row
• blocks s and p
• IGNORE superscripts on block letter

Mark scheme

Examiner's comment

Many found this question challenging despite doing well in Question 2(a)(i). The question states ‘use your answers from (a)(i)’ but not many candidates wrote about the electron configurations they had given. Many gave very simplistic responses in terms of the number of electrons increasing but made no reference to how those electrons are arranged (e.g. ‘number of electrons increases across a period as the electron configuration gets higher’ or ‘atomic number increases’). Some candidates struggled with terminology, often referring to ‘block’ or ‘orbital’ instead of subshell (e.g. ‘outer electrons are in same block’, ‘going across a period the number of orbitals increases’, ‘elements in same group have their highest energy electron in same block’ or ‘orbital’). Candidates need clarity on the terminology used for electron configurations and periodic table i.e. blocks, shells, sub-shells and orbitals. It was rare for candidates to score both marks as this was a question that they were unfamiliar with. However, some did gain a mark for linking the number of outer shell electrons to the group number or stating that elements in the same period have the same number of shells. It was not enough to refer to the highest energy electron being in the s-subshell or p-subshell as this is the link to the block, but all groups in same block will be the same. Some described the trend in other physical or chemical properties. Some examples included: ‘Elements have same chemical and physical properties due to similar electronic configuration’; ‘as you go across period, number of electrons increase and their boiling and melting points increase’; and ‘electrons are more easily lost in a paired orbital, due to greater repulsion and so have lower ionisation energies’.

Notes from the mark scheme and report

• s sub-shell fills then p across Period 2
• same pattern repeated across other periods
• ALLOW same outer-shell configuration within a group
• We have produced a transition guide on the topic of atomic structure. It covers content from KS4 and how this is developed at KS5 with a wide range of suggested resources to support teaching. At KS4, candidates are expected to be able to explain how the position of an element in the Periodic Table is related to the arrangement of electrons in its atoms, with development at KS5 to arrangement in to s, p and d orbitals.
• https://ocr.org.uk/Images/170375-atomic-structure-ks4-ks5.pdf

Mark scheme

Examiner's comment

Generally, this question was well answered with a clear understanding of how and why ions are formed. However, approximately a quarter of students only gained 1 mark as they either didn’t explain electrons being lost by Mg and gained by Cl or gave no justification. A common slip was stating Cl has one electron in its outer shell. Some described bonding between Mg and Cl, which wasn’t what the question asked, but this didn’t prevent them from scoring both marks.

Notes from the mark scheme and report

• Mg loses 2 electrons AND Cl gains 1
• to obtain a full outer shell/noble gas configuration

Mark scheme

Examiner's comment

Many candidates correctly gave the balanced equation here. However, some didn’t balance but had the correct formula. A few gave Mg2 as a reactant or MgO2 as a product. Some had O2 on both sides of the equation.

Notes from the mark scheme and report

• 2Mg + O2 → 2MgO
• ALLOW multiples
• IGNORE state symbols

Mark scheme

Examiner's comment

Explaining redox reactions is a common question in exam papers, however here candidates needed to do it ‘in terms of electron transfer’. Subsequently, many lost a mark as they identified oxidation and reduction in terms of oxidation numbers only. However, many gave responses both in terms of oxidation numbers and electrons. It was necessary to be specific here and say Ca had lost 2 electrons, so a few lost the mark by only referring to ‘Ca losing electrons’. Some lost marks for only describing oxidation of Ca and not reduction of H. There was some evidence that candidates were not sure of Cl’s role in the reaction (i.e. as a spectator ion) with some stating it was reduced and/or accepted electrons from Ca but then gave them to H.

Notes from the mark scheme and report

• Ca loses 2 electrons AND is oxidised
• H(+) gains electron AND is reduced
• IGNORE oxidation numbers (question asks for electron transfer)

Mark scheme

Examiner's comment

Most candidates correctly calculated the amount of HCl as 0.012 mol. However, many struggled with demonstrating that Ca is in excess. Responses often highlighted misconceptions here in terms of candidates’ understanding about excess and limiting reagents. For example, ‘Ca has a lower concentration than HCl so becomes the limiting reagent’ and ‘Not all the HCl had reacted’ Many compared moles of HCl calculated (i.e. 0.012) directly to moles of Ca (i.e. 0.01) saying that HCl was in excess, despite being told otherwise in the question. Some had the 2:1 ratio of HCl to Ca the wrong way around. Some attempted to calculate mass of Ca and HCl to use for comparison. Only a small proportion of candidates were able to access the third mark and correctly suggest that Ca was also reacting with water. Some other suggestions that were seen included: • ‘Ca reacted with oxygen or was impure’. In both cases this would mean that we would expect solid to remain. • ‘Higher concentration of HCl added’, or ‘HCl is a strong acid’, or ‘acid acts as a catalyst’. • ‘H2 evolved’ or ‘Ca reacts with hydrogen formed’. • ‘Human error’, ‘didn’t weigh Ca correctly’, ‘measured volume of HCl incorrectly’.

Notes from the mark scheme and report

• n(HCl) = 0.012 mol
• must show 2:1 ratio (0.006 mol Ca needed or 0.02 mol HCl needed)
• excess Ca reacts with the water
• Candidates often struggle to understand the concepts around limiting reagents and those in excess. Using a simple baking analogy can help to relate this to everyday life.
• For example: To make 10 pancakes you need 100 g flour, 2 eggs and 300 ml milk How many pancakes can I make if I have only 50 g flour, 2 eggs and 300 ml milk? Which is the limiting ingredient and which are in excess? The number of pancakes we can make is the theoretical yield.

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

Mark scheme

Examiner's comment

Approximately a quarter of candidates scored both marks here. Many candidates omitted to plot a point for Mg or positioned the point for Mg at 900 or above so higher than Be.

Notes from the mark scheme and report

• B below Be but above Li (about 800)
• Mg above Na but below Be (about 700)
• DO NOT ALLOW Mg on the 900 line

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

Mark scheme

Examiner's comment

More than half of candidates scored both marks here. Errors seen included missing or incorrect state symbols, especially (s), but also (aq) was seen. Some had electrons on the left hand side of the equation, i.e. ‘B+ + e- → Be2+’. Some included negatively charged ions and occasionally the wrong element was used, e.g. Mg or Be.

Notes from the mark scheme and report

• B+(g) → B2+(g) + e–
• state-symbol mark dependent on correct equation except for listed near-misses (e.g. B(g) → B2+(g) + 2e–)

Mark scheme

Examiner's comment

The term was not well known, with only a small proportion of candidates scoring this mark. Many gave the definition for enthalpy change of combustion or formation or bond enthalpy. Many referred to one mole of either reactants or products or bonds, or sometimes a combination of these. Lower-attaining candidates often said it was the temperature change.

Notes from the mark scheme and report

• enthalpy change for the molar quantities in the stated equation
• DO NOT ALLOW 'energy released'
• IGNORE standard conditions

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

Examiner's comment

There was a lot of misunderstanding associated with this question, with many candidates failing to score any marks. Many said that nothing would change as the concentration was still the same or because the same bonds were being broken and formed. Under a quarter of students scored 1 mark, usually for making the link between the drop in volume to a change in the q and n values. A few did state that the temperature didn't change. Only a small proportion scored both marks, usually by showing by calculation that the temperature change was the same, moles was half and energy was half. Some did believe that the temperature changed, either that it decreased as less reacted or increased as there was less volume to heat. A wide variety of alternative responses were given including: ‘Enthalpy change the same regardless of mass used’ ‘Number of moles doesn’t impact energy required as it is the same bonds breaking’ ‘Energy to break and form bonds will still be the same with any volume’ ‘Amount of energy required to make the new bond would be the same’ ‘Only concentration has an effect on bond enthalpy values not volume’ ‘Decrease in volume increases concentration’ ‘Cu(NO3)2 would still run out first so enthalpy change is the same’ ‘Zn is in excess so it doesn’t matter how much volume we use because Zn and Cu(NO3)2 still 1:1 ratio.’ ‘Mole ratio is still the same’ or ‘same molar ratio’ wasn’t enough.

Notes from the mark scheme and report

• half the energy/volume AND half the moles
• temperature change the same
• ALLOW shown by calculation

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

This is a familiar question with around half of candidates scoring all 4 marks. Common errors included drawing 2 lines, as you would have with different temperatures, but labelling one with catalyst and other without. Some had incorrect or missing labels on the axes. The most frequently gained mark was for knowing that the activation energy was lowered by a catalyst; this could be given by correct lines and labelling shown on the distribution, although care needed to be taken as some contradicted their answers with their diagram. Some labelled Ea and Ec lines but didn’t have energy on x-axis scale. Some struggled with the final marking point, not recognising that more molecules have the required activation energy. For example, ‘more frequent successful collisions’ with no reference to the activation energy, i.e. why they are successful.

Notes from the mark scheme and report

• one curve only, starting at origin and not touching x-axis (DO NOT ALLOW two curves – confusion with temperature)
• axes: number of molecules vs energy (DO NOT ALLOW atoms/enthalpy)
• Ec below Ea
• more molecules with energy above activation energy
• IGNORE 'more collisions'

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

Mark scheme

Examiner's comment

Most scored both marks here. Some didn’t circle the anomalous result and some lost a mark for a poorly drawn curve. Candidates must ensure they have a sharp pencil and draw a single line through all the points (except the anomalous point). Some didn’t start at the origin or didn’t level off at around 90 cm3.

Notes from the mark scheme and report

• smooth curve through all points except 100 s
• anomaly at 100 s circled
• curve levels off at about 90 cm3

Skills: M3.2 Plot two variables from experimental or other data; PAG9 Rates of reaction – continuous monitoring

Mark scheme

Examiner's comment

More candidates were able to correctly draw a tangent than seen in previous years with similar questions. A generous range was given for both tangent and gradient so many scored both marks. The most common reasons for losing marks was for having a gap between the curve and the tangent or calculating the gradient incorrectly, e.g. misreading scales, dx/dy, or by using interpolation rather than a tangent.

Notes from the mark scheme and report

• tangent at 50 s (±10 s)
• rate = gradient about 0.67 cm3 s–1
• DO NOT ALLOW interpolation
• ALLOW ECF from tangent or straight line of best fit

Skills: 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

Examiner's comment

Over half of candidates scored all 3 marks here. However, around a quarter did not gain any credit at all. Some confused the volume of H2O2 for a volume for a gas and attempted to find moles using molar gas volume (i.e. used 50cm3 rather than 90cm3). This often lost all 3 marks as they then divided by 2 and to find concentration divided by 90/1000 instead. A few attempted to use the ideal gas equation but this rarely yielded a correct value. Candidates must be encouraged to set out working clearly, showing logical steps and preferably labelling each step with what is being calculated and giving units. Many wrote the calculation as series of steps which all equalled the previous. e.g. 90  24000 = 3.75 x 10-3 x 2 = 7.5 x 10-3  0.05 = 0.15

Notes from the mark scheme and report

• n(O2) = 90/24000
• × 2 for H2O2
• ÷ 0.0500 dm3 = 0.15
• common errors 0.075 (no × 2) and 150 (no cm3 to dm3)
• ALLOW ideal gas equation with sensible p and T
• Encourage students to assign information in the question to the correct chemical. One way to do this is to write out the equation and then underneath each species put correct volumes as given. It also helps to highlight ratios shown in the equation. It is important here to pay close attention to state symbols as it helps identify correct calculations to use.
• For example; 2H2O2 (aq) → 2H2O (l) + O2(g) 50.0cm3 90 cm3 Solution Gas at RTP Conc?

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

Examiner's comment

More than half of candidates achieved both marks and most scored at least 1. The most common reason for losing marks was for not being specific, by just saying ‘concentration’ or ‘volume’, or suggesting volume and concentration for metal oxide. Some suggested use of the same gas syringe without realising that the apparatus used should not affect volume of gas produced, just how easy and accurate it is to measure.

Notes from the mark scheme and report

• any two: amount/mass/surface area of catalyst, temperature, volume or concentration of H2O2
• DO NOT ALLOW concentration/volume of catalyst

Skills: PAG9 Rates of reaction – continuous monitoring

Mark scheme

Examiner's comment

Most candidates scored all 4 marks here. Candidates seemed very knowledgeable about the addition reactions of alkenes. Some of the most common errors seen included: • 2,3-dibromobutane, rather than 1,2-dibromobutane • Repeating the same alcohol in bottom boxes • Missing H atoms from structures • Keeping double bonds • Missing CH2CH3 group (with H instead) • H2 as a second product on reaction with water and some gave oxidation products rather than alcohols e.g. aldehyde, ketone, carboxylic acid here • Ni added to the organic molecule for hydrogenation Incorrect connectivity from vertical bonds to CH2CH3 and OH groups was not penalised in this question but would have had a significant impact on marks if it had. Some lost the mark for incorrect connectivity for OH groups when drawn horizontally i.e. OH-C. Drawing organic structures Ensure students get lots of practice drawing organic structures using different types of formula including displayed, structural and skeletal. It is important to check structures carefully to ensure C has 4 bonds, O 2 bonds and H only 1 bond. Under pressure it is very easy to make slips. For exams, remind candidates not to draw in pencil then rub out – the structure beneath often shows up when the paper is scanned. If corrections are needed, cross out the incorrect structure and redraw clearly.

Notes from the mark scheme and report

• 1,2-dibromobutane
• butane
• butan-1-ol and butan-2-ol (mixture)
• DO NOT ALLOW structure with H missing
• DO NOT ALLOW –HO connectivity

Mark scheme

Examiner's comment

Approximately half of candidates scored all 4 marks. Very few scored no marks as they were able to give the correct structure of the named product (2-bromobutane). The most common reasons for losing marks included: • Adding dipoles to C=C • Missing dipoles on H-Br or reversed dipole on H-Br (i.e. + Br) • Missing charge on bromide ion or adding - • Arrows the wrong way round or not coming from a bond or lone pair (or negative charge for bromide ion) Some attempted a radical mechanism.

Notes from the mark scheme and report

• arrow from C=C to H of H–Br
• dipole on H–Br and arrow from bond to Br
• arrow from Br– (lone pair or charge) to C+ of secondary carbocation
• DO NOT ALLOW δ+ on carbocation or partial charge on C=C
• product mark independent
• ALLOW ECF for product from incorrect carbocation
• For ideas on teaching this topic please look at our Topic Exploration Pack: Electrophilic Addition and Markownikoff’s rule: https://teachcambridge.org/item/b4220e86-bc04-492c-b354-8103687ce594

Mark scheme

Examiner's comment

Over half of candidates didn’t gain any credit for their response here. Answers often focused on Markownikoff’s rule, i.e. adding H to C with most H atoms already attached, or described the stability of the product. Students need more support in understanding that the more stable carbocation will be formed in preference and result in forming more product. Many referred to the ‘secondary haloalkane’ being more stable, rather than the ‘secondary carbocation’. Some stated that 2-bromobutane is a ‘secondary carbocation’, showing some misunderstanding about the terminology used here.

Notes from the mark scheme and report

• forms via the more stable (secondary) carbocation/intermediate
• IGNORE statements of Markownikoff's rule about H joining C with most H
• IGNORE stability of the product
• We have a useful PowerPoint Presentation for teaching about Markownikoff’s rule and carbocation stability: https://ocr.org.uk/Images/250388-markownikoff-s-rule-presentation.ppt

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