Alkanes Module 4

Mark scheme

Examiner's comment

This was well answered by the majority of candidates. Where the mark was missed, it was either through incorrect products or incorrect balancing of oxygen.

Notes from the mark scheme and report

• ALLOW multiples
• IGNORE state symbols

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

Many candidates were able to score all 4 marks. More successful responses were clear and concise, covering all the detail required in much less space that provided here (see exemplar). Less successful responses often scored a mark for the correct trend but stated that this was due to more (covalent) bonds needing to be broken. Some were able to give well-rehearsed explanations for the dif- ference in boiling point but lost the trend mark as state change by the position in the table only, e.g. “as you go down the table, the boiling point increases”. A few muddled different types of intermolecular forces, suggesting that H-bonds were present. There was evidence that some held the misconception that a C-H-bond constitutes a H-bond. Explanations were sometimes missing the ‘contact’ between molecules or alternatively an increase in the number of electrons. Some did not gain marks as their responses were not comparative, e.g. “has sur- face contact so can form London forces” or” requires a lot of energy to break London forces”. It was fairly common to see “more energy to break the bonds” which was not given as it needed to be clear that it was the intermolecular/London forces which are broken. Some referred to enthalpy of combustion data which was not asked about in the question. Some re- sponses had lots of repetition i.e. explained why longer chains had a high boiling point, then repeated the same but for the reverse argument for smaller chains. A clear and concise response which scores all 4 marks. The diagrams at the bottom confirm that this candidate has understood the concepts completely.

Notes from the mark scheme and report

• comparisons needed throughout
• boiling point increases as chain length increases
• more surface CONTACT (surface area alone insufficient)
• more/stronger London forces
• more ENERGY to break them
• DO NOT ALLOW branching, permanent dipoles, hydrogen bonding or covalent bonds breaking
• IGNORE van der Waals

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

Mark scheme

Examiner's comment

The majority gave a correctly balanced equation here. The most common error was an incorrect number of oxygen molecules. A few candidates gave the equation for propane rather than pentane.

Notes from the mark scheme and report

• C5H12 + 8O2 → 5CO2 + 6H2O
• DO NOT ALLOW multiples (one mole of pentane)

Mark scheme

Examiner's comment

Generally well known, with most recognising that HBr is formed as a product. Incorrect responses sometimes had 2 Br instead of Br2 or included other radicals in their response. Some candidates confused propane and pentane, despite being given them in the table. Some started with a longer C chain, breaking the chain and substituting with Br to give bromopropane as state in the question, e.g. C4H10 + Br2 → C3H7Br + CH3Br. Some used structural formulae rather than the molecular formula which was asked for.

Notes from the mark scheme and report

• C3H8 + Br2 → C3H7Br + HBr

Mark scheme

Examiner's comment

Most scored both marks here. The most common error involved substituting the Br in the first step – this lost both marks. Very few candidates scored 1 mark; this was usually for a correct first step but then re- acting the C3H7 radical with either 2 Br or HBr in second step. Candidates must check that equations balance in terms of atoms. Candidates should also understand that hydrogen radicals are highly reactive and are therefore unlikely to exist under standard conditions.

Notes from the mark scheme and report

• C3H8 + Br• → •C3H7 + HBr then •C3H7 + Br2 → C3H7Br + Br•
• 1 mark if both correct but dots omitted

Mark scheme

Examiner's comment

This question was a slightly different way of asking why a number of different products are formed in rad- ical reactions. However here candidates needed to recognise that it would be due to substitution at dif- ferent positions along the chain. Common incorrect response included ”multiple termination products” or “further substitution”. It was also evident in some responses that candidates had muddled this with nucle- ophilic substitution and described the “stability of the carbocation intermediate”. Some just stated the two different structural isomers that would be formed but didn't gain the mark as needed to say how they are formed.

Notes from the mark scheme and report

• Br can substitute at different positions on the chain (end or middle C) / two different radicals form
• DO NOT ALLOW stability of carbocations
• IGNORE major/minor product

Mark scheme

Examiner's comment

Most scored both marks here, as they were able to use the ‘dot-and-cross’ diagram drawn in (i). Some struggled with recalling the bond angle. A few “trigonal planar” and “120” responses were seen perhaps confusing themselves with the next part of the question.

Notes from the mark scheme and report

• tetrahedral
• 109.5° (ALLOW 109–110)

Skills: M4.1 Use angles and shapes in regular 2-D and 3-D structures

Mark scheme

Examiner's comment

Very few candidates did not score any marks here. Most gained a mark for the correct molecular formulae in the first row, although some candidates gave structural formulae instead of molecular formulae or gave both. Candidates found counting the different types of bonds quite challenging and we would recommend drawing out displayed formulae to make it easier to label and count each type of bond. Many gave 5 and 6 respectively for the  bonds, forgetting to count the C-H bonds that are not shown in the skeletal structure. More were able to correctly count the  bonds but some counted the number of atoms involved in  bonds instead. A few had  and  bonds the wrong way round.

Notes from the mark scheme and report

• both C6H10
• σ bonds 15 and 16
• π bonds 2 and 1
• one mark per row
• 1 mark for a correct column if no other marks

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

Mark scheme

Examiner's comment

Most candidates were unable to gain credit here. The most common incorrect response given was C6H14 + 6O2 → 6CO + 7H2O which doesn’t balance in terms of oxygen. A significant number recognised that this was not balanced so gave C6H14 + 6.5O2 → 6CO + 7H2O but this misses the crucial 1:6 ratio of fuel/hexane to oxygen required in the question. Some gave H2 as a product of incomplete combustion, e.g. C6H14 + 6.5O2 → 6CO + 6H2O + H2. Candidates need to be reminded that C or CO will form in a limited supply of O2 but not H2. Candidates seemed unaware that multiple C products of combustion can be formed in one reaction. A few gave O2 as a product as well as a reactant.

Notes from the mark scheme and report

• C6H14 + 6O2 → 5CO + C + 7H2O (or other CO/CO2/C combinations in a 1:6 ratio)
• DO NOT ALLOW H2 as a product

Skills: M0.2 Use ratios, fractions and percentages

Answer

C

Mark scheme

Examiner's comment

Some candidates chose B as the correct option. The other options were chosen randomly, suggesting that many had not learnt this specification content and had guessed.

Mark scheme

Examiner's comment

Most candidates were given 3 or 4 marks. The most common omission was the idea of surface contact. Most candidates identified London forces or induced dipole interactions as the relevant intermolecular force. A few candidates gave a general comment in terms of ‘intermolecular’ forces without specifying the type of intermolecular forces. There has been a general improvement in candidate responses to this type of question with fewer candidates than in previous exams suggesting the breaking of hydrogen bonds or covalent bonds. main factors responsible for the trend in boiling points. This response was given the full 4 marks.

Notes from the mark scheme and report

• boiling point decreases with more branching
• less surface contact between molecules
• weaker London forces
• less energy to break intermolecular forces
• DO NOT ALLOW different numbers of electrons
• IGNORE just 'bonds'
• surface area alone not sufficient

Mark scheme

Examiner's comment

This question was answered extremely well with most candidates obtaining the full 5 marks. It was encouraging to see the widespread correct use of dots to indicate radicals, with relatively few omissions. Of the three steps, initiation and termination were answered better than the equations for propagation.

Notes from the mark scheme and report

• correct dots required for all marks
• UV AND Br2 → 2Br•
• propagation C2H6 + Br• → C2H5• + HBr then C2H5• + Br2 → C2H5Br + Br•
• DO NOT ALLOW charged formulae

Mark scheme

Examiner's comment

The bond angles and shapes rewarded the well-prepared candidates, with many being given both available marks for this part of the question. This part discriminated very well. For the explanation, most candidates identified 4 and 3 for C1 and C2, but candidates often linked 4 and 3 to atoms, rather than to electron pairs or bonded pairs for C1 and to bonding regions for C2. A mark was available for stating that ‘electron pairs repel’, but this important fact was often omitted despite being the main principle that determines molecular shapes. The question discriminated well, giving a good spread of marks across the five available.

Notes from the mark scheme and report

• C1 109.5° tetrahedral with 4 bonded pairs
• C2 120° trigonal planar with 3 bonded regions
• electron pairs repel
• DO NOT ALLOW 'atoms repel'
• IGNORE 'electrons repel' and 'bonds repel'
• Many students think that molecular shapes are determined solely by lone pairs or by repulsion between bonded atoms. The principle behind molecular shapes is called electron pair repulsion theory because it is based on repulsion between electron pairs, which may be bonded pairs or lone pairs, but not atoms.

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

Answer

B

Mark scheme

Examiner's comment

A large majority of candidates were able to correctly identify shape at x as being trigonal planar and y as being tetrahedral. The most common incorrect responses seen were for getting one of these incorrect i.e. D incorrect for x or C incorrect for y.

Mark scheme

Examiner's comment

This question differentiated well, with a wide mix of responses given. Just over 40% scored all 3 marks. Equation: The most common error was to omit HBr as a product. Some gave H2 or just Br as a product. Some misunderstood the question and attempted to give either a partial mechanism, such as the propagation step or a complete mechanism, rather than the overall equation. Some gave molecular formula rather than structures as asked for. Those using structural or displayed formula were more prone to errors, such as missing hydrogens or incorrect chain length, than those that were confident using skeletal formula. Candidate should avoid giving equations in two formats, e.g. skeletal and structural because slips in one will lose marks. Mechanism: Many candidates were able to identify the radical substitution mechanism, but a significant number did not score here. Most common incorrect responses were electrophilic or nucleophilic substitution but there were also those that thought it was an addition reaction. Bond fission: A significant number identified this as heterolytic, even if the recognised mechanism was radical. Some struggled with the spelling or even suggested homogeneous or heterogenous.

Notes from the mark scheme and report

• overall equation (CH3)3CH + Br2 → (CH3)3CBr + HBr (IGNORE mechanism steps)
• radical substitution
• homolytic fission
• Many struggled to identify the mechanism and then to link to bond fission. Try to introduce key terminology early on in teaching organic chemistry so that it can then revisited with each topic. Relevant mechanism terminology: - substitution / addition / elimination - electrophile / nucleophile / radical - homolytic / heterolytic

Mark scheme

Examiner's comment

Most candidates scored one mark here, with approximately a third of candidates scoring both marks. Most recognised that further or multiple substitution would occur or that there would be a variety of termination products. The second mark was harder to achieve as although many responses suggested a different position, it was not always clearly conveyed that a hydrogen on a different carbon along the chain was being substituted, e.g. ‘substitution can occur anywhere in the molecule’. The best responses gave examples to clarify their point, e.g. ‘the bromine radical can be substituted anywhere along the hydrocarbon chain making other products such as 1-bromo-2-methylpropane’. However, many gave vague answers such as poor yield or mixture of products, without explanation of how or what they might be. Some focused on HBr being formed as a product, suggesting it is toxic, or lowers the atom economy. Others highlighted ultraviolet radiation as being a limitation due to it being ‘expensive’, ‘hard to achieve’, ‘lack of sources’ or ‘hazardous’. Others suggested that the reaction needed ‘high temperatures’.

Notes from the mark scheme and report

• further substitution (or different termination products)
• substitution at different positions along the chain
• IGNORE separation of products, atom economy and yield

Mark scheme

Examiner's comment

Most candidates were able to construct a balanced equation for the combustion of heptane. Most were aware that CO2 and H2O would be the products although some generated CO, C6H12 or unusual compounds such as C7H14O. The hardest part was the formula of heptane itself with use of hexane instead being a common error; candidates who made this error were given 1 mark, provided that their equation was balanced.

Notes from the mark scheme and report

• 11O2
• ALLOW multiples
• 1 mark ECF for a balanced combustion equation of a different alkane

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

Most candidates were given 1 or 2 marks, with some omitting the idea of surface area or surface contact. Most candidates identified London forces or induced dipole interactions as the relevant intermolecular force. A few candidates gave a general comment in terms of ‘intermolecular’ forces without specifying the type of intermolecular forces. There has been a general improvement in candidate responses to this type of question with fewer candidates than in previous exams suggesting the breaking of hydrogen bonds or covalent bonds.

Notes from the mark scheme and report

• comparisons needed
• longer chain so more surface contact
• stronger/more London forces need more energy to break
• 'branching' is a CON
• IGNORE van der Waals' forces

Mark scheme

Examiner's comment

This question discriminated very well at the top end of the ability range, but many ignored the instruction to use skeletal formula and obtained no marks as a result. Of those that did use skeletal formula, many placed the dot on the wrong carbon atom or produced 1- brombutane, rather than 2-bromobutane, stated in the question. A mark was still available by ECF for misplaced or absent dots or formation of 1-brombutane with dots. This question was aimed to be demanding and so it proved to be.

Notes from the mark scheme and report

• skeletal formulae required (no credit for molecular formulae)
• radical dot on C2
• 1 mark ECF if dots omitted/misplaced or for forming 1-bromobutane correctly

Answer

B

Mark scheme

Examiner's comment

This was a demanding question. Candidates needed to calculate the moles of oxygen and then determine the ratio of alkane to oxygen to find the correct response. The majority of successful candidates clearly showed their working to help them to arrive at the correct answer. The most common incorrect answer was C.

Mark scheme

Examiner's comment

Most candidates attempted this question, gaining at least 1 mark, with over half scoring 3 or more marks. Responses often lacked clarity as many candidates struggled to articulate their ideas. It was common to see lengthy responses often with unnecessary repetition and sometimes even contradictions. A good strategy adopted by some was to draw skeletal formulae for the compounds next to the data provided. This enabled them to focus their response more easily on the extent of branching. Many candidates were unable to give a clear trend for the first marking point, as asked for in the question, but were able gain credit by a lengthier comparison of all three as indicated in the extra guidance. However, this mark was often lost through incomplete explanation, not referring to boiling point at all or an attempt to compare just chain length. The most common error for the second mark was omission of ‘contact’ or ‘interaction’ with reference only to surface area or ‘packing’ of molecules. Some lost this mark for a change in number of electrons. The third marking point was the most frequently awarded. Some candidates lost the mark for not explicitly naming the intermolecular forces as London forces/induced dipole-dipole interactions or for incorrectly using van der Waals. Some lost the mark for not explicitly indicating how increased or decreased contact would affect the strength or magnitude of London forces, e.g. ‘less contact to form London forces’. The final mark was harder to obtain as it needed to be clear that energy was required to break intermolecular forces. For example, ‘less energy to break bonds’ or ‘easier to separate the molecules’ or ‘more energy to boil’ were not sufficient. This exemplar shows a clear, concise response. The candidate has drawn skeletal structures next to the table. The trend is stated first followed by a detailed explanation, presented as a bullet point list, with all 4 marking points awarded.

Notes from the mark scheme and report

• comparisons needed throughout
• must rank all three by branching
• surface contact not just surface area
• DO NOT ALLOW different numbers of electrons (all the same)
• more energy to break London forces
• IGNORE 'bonds' alone and chain length
• Responses often highlighted that candidates lacked understanding about what London forces are, e.g. indicating that they form ‘between atoms’ or referring to induced dipole-dipole forces as something else. Intermolecular forces are difficult to fully comprehend as they can’t be visualised making this a challenging topic to teach.
• OCR have produced a ‘Bonding’ teaching guide with lots of useful suggestions and resources. This includes a link to this Salters A Level chemistry revision activity on intermolecular bonding

Mark scheme

Examiner's comment

Many candidates tackled this question confidently, especially when using skeletal formula following the format of the structure given in the question. Over half the candidates scored 5 or 6 marks. Only the highest attaining candidates were able to provide all three correct termination steps. Many lost a mark for the combination of the two alkyl radicals, typically either by simply joining the ends of the chains or by missing the connecting C-C bond. Those that attempted to use structural formula often lost marks due to missing Hs. Other common errors included the incorrect positioning of the radical dot, most typically on the terminal carbon, addition of Br in the first propagation step or use of molecular formula. Lower attaining candidates were often able to score a mark for the initiation step and the termination step involving two Br radicals. However, for some this was not a well-known mechanism, with attempts to break up the chain or form hydrogen radicals or charged species. Errors were also seen with correct balancing of equations such as truncated C chains or extra Br atoms added.

Notes from the mark scheme and report

• Br2 → 2Br• AND UV
• dot required throughout and on C-3 of the radical
• 1 mark for propagation if both correct but dot omitted or misplaced
• three terminations incl. two alkyl radicals
• no ECF from incorrect radical in termination

Mark scheme

Examiner's comment

Most responses gained at least 1 mark for this question giving the correct molecular formula of C6Br14. However many hadn’t assimilated that when a hydrogen atom is substituted in an alkane it requires one mole of a halogen and produces one mole of the hydrogen halide. So many gave this incorrect equation instead: C6H14 + 7Br2 → C6Br14 + 7H2. Some lost marks for C5H14 or for use of structural formulae.

Notes from the mark scheme and report

• C6Br14
• C6H14 + 14Br2 → C6Br14 + 14HBr (HBr, not H2)
• 1 mark if balanced but not in molecular formulae

Skills: M0.2 Use ratios, fractions and percentages

Answer

D

Mark scheme

Examiner's comment

Candidates found this question difficult with very many choosing option B rather than the correct option D. Candidates are advised to draw out all bonds displayed when tackling such as question as the answer of B (3) results from considering just the three bonds shown in the skeletal formula and omitting the other 6 C–H bonds.

Answer

D

Mark scheme

Examiner's comment

This question discriminated extremely well with most candidates choosing the correct option. Most candidates showed some working. The key to success was to identify the balanced equation that produced CO2 and H2O in a 3 : 4 molar ratio.

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.

Mark scheme

Examiner's comment

Most candidates gave the correct response to this question. Incorrect responses included use of high temperatures and/or catalyst.

Notes from the mark scheme and report

• UV
• ALLOW sunlight
• IGNORE temperature

Mark scheme

Examiner's comment

Candidates always find radical mechanisms tricky and this one had the added complexity of forming 2- bromo isomer. However, a majority of students still gained marks. Many candidates formed the incorrect radical removing H from C-1 i.e. CH3CH2CH2CH2• therefore scoring only 1 mark. Some responses were a little messy making it very easy to miss off a dot or H or Br. Many candidates reacted with Br• in the first step but added Br to the radical intermediate (as well as forming HBr). Candidates should always check equations so that they balance in terms of atoms.

Notes from the mark scheme and report

• radical must be CH3CH2•CHCH3 (1 mark if CH3CH2CH2CH2• used or both equations correct with molecular formulae)
• dots required

Mark scheme

Examiner's comment

Candidates found this question very challenging and few scored both marks. Many responses considered only the formation of HBr (other product) and/or general statements about other products with no indication of how they were formed. Some described losses due to the purification method or incomplete reaction (due to conditions such as T and P) or low atom economy. Some referred to the stability of the radical intermediate, showing possible confusion with electrophilic addition. Candidates who understood the mechanism were more confident in answering this question, at least recognising that further substitution was possible.

Notes from the mark scheme and report

• further substitution/di- and tribromobutanes or other termination products
• substitution at a different position (1-bromobutane)

Answer

D

Mark scheme

Examiner's comment

Most candidates correctly identified D 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.

Mark scheme

Examiner's comment

Most candidates attempted to write an equation for the combustion of the 1:1 molar mixture of octane and ethanol. The formulae of C8H18 and C2H5OH were usually seen although some candidates combined these as a ‘mixture formula’ of C10H24O (which was accepted). The balancing of the equation using 15½O2 was the hardest part of the equation and many different balancing numbers for O2 were seen (10CO2 and 12H2O where usually correct). Less successful responses often attempted a combustion equation using octane OR ethanol, but not both. This is not an easy equation to construct, and the context was novel. Overall candidates made a good attempt at this question.

Notes from the mark scheme and report

• C8H18 + C2H5OH + 15½O2 → 10CO2 + 12H2O
• ALLOW multiples

Skills: M0.2 Use ratios, fractions and percentages

Answer

D

Mark scheme

Answer

B

Mark scheme

Answer

C

Mark scheme

Notes from the mark scheme and report

• ALLOW 4

Mark scheme

Notes from the mark scheme and report

• 2-bromo-2-methylpropane and 1-bromo-2-methylpropane
• ALLOW any unambiguous formula type

Mark scheme

Notes from the mark scheme and report

• 2Br2 and 2HBr needed to balance
• correct structure of 1,3-dibromo-2-methylpropane

Answer

C

Mark scheme

Mark scheme

Examiner's comment

Question 16(a)(i): These definitions were known by few candidates.

Notes from the mark scheme and report

• σ: overlap of orbitals directly between the bonding atoms
• π: sideways overlap of p-orbitals
• DO NOT ALLOW pi-orbital

Mark scheme

Notes from the mark scheme and report

• 9 σ-bonds and 2 π-bonds

Answer

C

Mark scheme

Examiner's comment

Less than half the candidates selected the correct response of C for this direct recall question. There was no key distractor, suggesting that many candidates may have guessed.

Mark scheme

Examiner's comment

Other questions Question 16(c) Often candidates did not use skeletal formula, as required by the question, and did not place the radical dot in the correct position.

Notes from the mark scheme and report

• dot REQUIRED at the correct position on the chain
• 1 mark if both propagation steps correct but dot missing/on wrong C or not skeletal
• IGNORE dots in initiation

Mark scheme

Notes from the mark scheme and report

• further substitution
• substitution at different positions along the chain
• IGNORE separation of products, atom economy and yield

Mark scheme

Examiner's comment

1a(i) Most candidates labelled some of the apparatus correctly and identified the reflux technique. A significant number showed water flowing in the wrong direction and ‘distillation’ was given as the name of the technique. The ‘condenser’ was sometimes labelled incorrectly, e.g. ‘condensation tube, ‘distillation tube’ and ‘water jacket’. Only just over half the candidates were given both marks. 1a(ii) Many diagrams were incomplete and it was comparatively rare for both of the two available marks to be given. Important labels were often missing. Some candidates drew diagrams of other techniques, such as distillation. Many responses were not credited with marks and this question was often omitted. Candidates need practice in recognising practical techniques and in drawing acceptable diagrams. 1b(i) Candidates were expected to explain the different boiling points of three amines. Most candidates linked increased branching with less surface contact between molecules, leading to weaker induced dipole interactions (London forces). Fewer candidates linked the decreasing boiling points to less energy being required to break the intermolecular bonds. The highest performing candidates recognised the large difference between the boiling points of the primary and tertiary amines. They then identified the cause: primary amines form hydrogen bonds but tertiary amines do not. This question was answered reasonably well with fewer candidates than in the past describing relative strengths of covalent bonds, a common misconception. 1(b)(ii) Most candidates rearranged the ideal gas equation correctly to make n the subject. They then substituted correct values for p, V, R and T into the equation, with correct unit conversions. Most candidates then calculated n correctly as 2.32 × 10–3 and combined this value with the mass of 0.202 g to derive the molecular formula as 87. Lower-attaining candidates often made an error in their conversion of cm3 into m3, by multiplying by 10–3 rather than 10–6. This error resulted in a value of n as 2.32 and the unrealistic molecular mass of 8.7 × 10–3, from which a molecular formula is impossible. Candidates in this position are advised to check back at the unit conversions in the working. Very few candidates did not convert 100ºC into 373 K. Candidates with the correct molecular formula of 87 usually suggested C5H13N as the molecular formula. A significant number quoted C5H11NH2. This response was not credited as it is not a ‘molecular’ formula. There are several possible structures of C5H13N with three 13C NMR peaks and any were credited. Error carried forward was applied from a different molecular mass for both the molecular formula and a structure. This emphasises the importance of working within calculation; credit will always be given for a correct method. Instead of using the ideal gas equation, some candidates used the molar gas constant at room temperature and pressure, 24.0 dm3, which gives a molecular mass of 67. This simplified approach could only gain the final 2 marks for this question: a molecular formula (with an error carried forward value of 67, e.g. C4H5N) and a structure of an amine with this molecular formula and 3 peaks in the 13C NMR spectrum. 1(c) Candidates were supplied with information about an unfamiliar reaction of an amino acid and asked to predict a possible equation. Many candidates suggested H2O as one product, being the difference in the formula of the amino acid and the C3H3NO cyclic organic product. Any cyclic structure of C3H3NO that met the bonding rules for C, H, N and O was credited. Examples included a 4-membered ring lactam and substituted cyclopropenes. A significant number of candidates showed an equation for the reaction of two molecules of the amino acid to form 2 H2O and a cyclic dipeptide. Although chemically feasible, the dipeptide could not be credited because the molecular formula was C3H3NO in the question. This error could have been avoided if the information in the question had been used.

Notes from the mark scheme and report

• less branching gives more points of contact and stronger London forces
• primary amines have hydrogen bonds but (CH3)3N has none
• H bonds stronger than London forces
• more energy to break
• DO NOT ALLOW 'more electrons' (same number)
• DO NOT ALLOW breaking covalent bonds
• IGNORE primary/secondary/tertiary comparison

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

Mark scheme

Examiner's comment

Many candidates scored all 3 marks for this part, showing that most had thoroughly learnt the mechanism for radical substitution. The equation and conditions for the initiation step were well-known but the equations for the propagation steps often included errors. It was common for dots to be omitted for some radicals and C4H9Cl•, rather than C4H9•, was often shown for one of the products in the first propagation stage. H• was then shown as the other product.

Notes from the mark scheme and report

• UV needed with initiation
• dots required in each propagation equation (1 mark for both propagation equations with dots missing/extra)
• DO NOT ALLOW charges

Mark scheme

Examiner's comment

Only the highest attaining candidates were able to write the correct equation. Although most candidates did identify the organic product as C4Cl10, the other product was usually seen as 5H2 rather than 10HCl.

Notes from the mark scheme and report

• 10 Cl2 and 10 HCl
• ALLOW structural formulae

Skills: M0.2 Use ratios, fractions and percentages

Answer

A

Mark scheme

Answer

D

Mark scheme

Answer

D

Mark scheme

Examiner's comment

This is a demanding question. The best responses drew the first step of the mechanism near the table. This showed both the π bond in ethene and the σ bond in bromine breaking in addition to a σ bond forming between carbon and bromine atoms. Although many candidates correctly selected D, option B was a good distractor and was the most commonly seen incorrect response.

Answer

D

Mark scheme

Examiner's comment

Most candidates selected A or D, with D being the correct option. Presumably, A was chosen by halving the ‘8’ in C4H8 without considering that each H2O molecule contains two H atoms. The successful answer of D usually resulted from the candidate constructing equations.

Mark scheme

Examiner's comment

This question discriminated well and resulted in a full range of marks. Most candidates were aware that structural isomers have different structural formulae but the same molecular formulae. It was common though for candidates to refer to different arrangements of atoms in space, clearly confusing with stereoisomerism. The best candidates used the structures (as in the question) to show that the common molecular formula was C5H12. Candidates were expected to link the amount of surface contact between molecules with induced dipole–dipole forces or London forces. ‘Contact’ or the name of the intermolecular forces was often omitted. Finally, candidates were expected to link the amount of branching to the strength of the intermolecular forces and the energy needed to change state. Lower ability candidates often let themselves down by being unable to construct a well-reasoned response. There was often a gulf between the clear responses of able candidates and those of lower ability candidates.

Notes from the mark scheme and report

• same molecular formula AND different structural formulae ('same formula' not sufficient)
• C5H12 for all three
• more branching gives less surface contact and weaker London forces
• IGNORE van der Waals
• IGNORE just 'bonds'
• DO NOT ALLOW reference to spatial arrangement

Mark scheme

Examiner's comment

Most candidates identified this reaction as radical substitution.

Notes from the mark scheme and report

• ALLOW free radical substitution

Mark scheme

Examiner's comment

Most candidates achieved at least one mark, particularly for isomer A. Successful candidates often drew structures of the isomers alongside the table to help with their response.

Notes from the mark scheme and report

• A gives 3, B gives 4

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

Mark scheme

Examiner's comment

Many candidates correctly drew the structure of compound D but comparatively few were able to construct a correct equation. For this equation, candidates needed to apply their knowledge and understanding of monosubstitution of alkanes to substitution of three H atoms by three Cl atoms. This task proved to be one of the most difficult questions on this paper. The exemplar shows an excellent response. The candidate has drawn a trisubstituted structure that fits the molar mass of 175.5 g mol–1 and a correct equation for its formation. Many attempts at this equation showed H2 as the second product rather than HCl.

Notes from the mark scheme and report

• any trichloro isomer of A
• molecular formulae required in equation (C5H12 + 3Cl2 → C5H9Cl3 + 3HCl)
• no ECF from incorrect structure of D

Answer

B

Mark scheme

Examiner's comment

This question discriminated well, with higher ability candidates correctly identifying B as a non-planar molecule. Common incorrect responses included C and D. Some candidates drew structures alongside each option to aid their choice.

Mark scheme

Examiner's comment

The majority of candidates were able to provide a correct equation for the incomplete combustion of heptane.

Notes from the mark scheme and report

• product must be CO and/or C with H2O
• ALLOW multiples and mixtures with CO2

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

This question was answered well with most candidates scoring three or four marks. Examiners were impressed by the number of responses that accurately referred to induced dipole-dipole interactions or London forces rather than van der Waals’ forces, which is ambiguous. Some responses lacked detail, as demonstrated in Exemplar 10. This response attributes the higher boiling point of pentan-1-ol to the amount energy required to break hydrogen bonds. However, it does not refer the relative strength of this type of interaction. Consequently, the first paragraph only scores marking point four and not marking point three. The higher boiling point of heptane compared to hexane is explained by a correct comparison of the induced dipole-dipole interactions present in these compounds, so marking point two was achieved. However, the justification for the difference in intermolecular forces lacks precision. Candidates should be encouraged to focus on surface contact or surface interaction between molecules rather than referring to surface area alone.

Notes from the mark scheme and report

• heptane: more surface contact/more electrons so stronger London forces
• pentan-1-ol: hydrogen bonds
• more energy to break the forces
• IGNORE van der Waals/VDW/IDID
• IGNORE 'harder to break' without energy
• IGNORE 'more energy to break bonds'

Answer

A

Mark scheme

Mark scheme

Notes from the mark scheme and report

• tetrahedral AND 109.5
• four bonded pairs/bonds repel
• IGNORE 'four electron pairs repel'
• DO NOT ALLOW atoms repel

Skills: M4.1 Use angles and shapes in regular 2-D and 3-D structures

Mark scheme

Notes from the mark scheme and report

• UV
• ALLOW sunlight
• IGNORE temperature

Mark scheme

Notes from the mark scheme and report

• one Cl substituted and HCl formed
• ALLOW molecular formulae

Mark scheme

Notes from the mark scheme and report

• Cl• abstracts H to give HCl
• organic radical + Cl2 regenerates Cl•
• 1 mark if both correct but dots omitted

Mark scheme

Notes from the mark scheme and report

• two or more Cl on the alkyl chain with at least one at C-2
• DO NOT ALLOW molecular formula C3H4Cl2O2

Answer

C

Mark scheme

Mark scheme

Notes from the mark scheme and report

• both points must be comparisons
• fewer points of contact so weaker London forces needing less energy to break
• IGNORE van der Waals
• IGNORE 'less energy to break the bonds'

Mark scheme

Notes from the mark scheme and report

• for fuel or energy
• ALLOW chemical feedstock
• IGNORE cooking

Answer

D

Mark scheme

Answer

D

Mark scheme

Mark scheme

Notes from the mark scheme and report

• σ direct/head-on overlap between nuclei, π sideways overlap above and below
• IGNORE bond length
• IGNORE type of orbital for σ bond