Outline two differences between the bonding of carbon atoms in C₆₀ and diamond.

Explain why C₆₀ and diamond sublime at different temperatures and pressures.

State two features showing that propane and butane are members of the same homologous series.

Describe a test and the expected result to indicate the presence of carbon–carbon double bonds.

Draw the full structural formula of (Z)-but-2-ene.

Write the equation for the reaction between but-2-ene and hydrogen bromide.

State the type of reaction.

Suggest two differences in the ¹H NMR of but-2-ene and the organic product from (d)(ii).

Predict, giving a reason, the major product of reaction between but-1-ene and steam.

Explain the mechanism of the reaction between 1-bromopropane, CH₃CH₂CH₂Br, and aqueous sodium hydroxide, NaOH (aq), using curly arrows to represent the movement of electron pairs.

Deduce the splitting pattern in the ¹H NMR spectrum for 1-bromopropane.

Calculate the enthalpy change of the reaction, ΔH, using section 11 of the data booklet.

Draw and label an enthalpy level diagram for this reaction.

Any two of:

C₆₀ fullerene: bonded to 3 C AND diamond: bonded to 4 C ✔

C₆₀ fullerene: delocalized/resonance AND diamond: not delocalized / no resonance ✔

C₆₀ fullerene: sp² AND diamond: sp³ ✔

C₆₀ fullerene: bond angles between 109–120° AND diamond: 109° ✔

Accept "bonds in fullerene are shorter/stronger/have higher bond order OR bonds in diamond longer/weaker/have lower bond order".

diamond giant/network covalent AND sublimes at higher temperature ✔

C₆₀ molecular/London/dispersion/intermolecular «forces» ✔

Accept “diamond has strong covalent bonds AND require more energy to break «than intermolecular forces»” for M1.

same general formula / C_nH_2n+2 ✔

differ by CH₂/common structural unit ✔

Accept "similar chemical properties".

Accept “gradation/gradual change in physical properties”.

ALTERNATIVE 1:

Test:

add bromine «water»/Br₂ (aq) ✔

Result:

«orange/brown/yellow» to colourless/decolourised ✔

Do not accept “clear” for M2.

ALTERNATIVE 2:

Test:

add «acidified» KMnO₄ ✔

Result:

«purple» to colourless/decolourised/brown ✔

Accept “colour change” for M2.

ALTERNATIVE 3:

Test:

add iodine /${\text{I}}_2$ ✔

Result:

«brown» to colourless/decolourised ✔

Accept

CH₃CH=CHCH₃ + HBr (g) → CH₃CH₂CHBrCH₃

Correct reactants ✔

Correct products  ✔

Accept molecular formulas for both reactants and product

«electrophilic» addition/EA ✔

Do not accept nucleophilic or free radical addition.

ALTERNATIVE 1: Any two of:

but-2-ene: 2 signals AND product: 4 signals ✔

but-2-ene: «area ratio» 3:1/6:2 AND product: «area ratio» 3:3:2:1 ✔

product: «has signal at» 3.5-4.4 ppm «and but-2-ene: does not» ✔

but-2-ene: «has signal at» 4.5-6.0 ppm «and product: does not» ✔

ALTERNATIVE 2:

but-2-ene: doublet AND quartet/multiplet/4 ✔

product: doublet AND triplet AND quintet/5/multiplet AND sextet/6/multiplet ✔

Accept “product «has signal at» 1.3–1.4 ppm «and but-2-ene: does not»”.

CH₃CH₂CH(OH)CH₃ ✔

«secondary» carbocation/CH₃CH₂CH⁺CH₃ more stable ✔

Do not accept “Markovnikov’s rule” without reference to carbocation stability.

curly arrow going from lone pair/negative charge on O in HO^– to C ✔

curly arrow showing Br breaking ✔

representation of transition state showing negative charge, square brackets and partial bonds ✔

formation of organic product CH₃CH₂CH₂OH AND Br– ✔

Do not allow curly arrow originating on H in HO^–.

Accept curly arrow either going from bond between C and Br to Br in 1-bromopropane or in the transition
state.

Do not penalize if HO and Br are not at 180° to each other.

Award [3 max] for S_N1 mechanism.

triplet/3 AND multiplet/6 AND triplet/3 ✔

bond breaking: C–H + Cl–Cl / 414 «kJ mol^–1» + 242 «kJ mol^–1»/656 «kJ»
OR
bond breaking: 4C–H + Cl–Cl / 4 × 414 «kJ mol^–1» + 242 «kJ mol^–1» / 1898 «kJ» ✔

bond forming: «C–Cl + H–Cl / 324 kJ mol^–1 + 431 kJ mol^–1» / 755 «kJ»
OR
bond forming: «3C–H + C–Cl + H–Cl / 3 × 414 «kJ mol^–1» + 324 «kJ mol^–1» + 431 kJ mol^–1» / 1997 «kJ» ✔

«ΔH = bond breaking – bond forming = 656 kJ – 755 kJ» = –99 «kJ» ✔

Award [3] for correct final answer.

Award [2 max] for 99 «kJ».

reactants at higher enthalpy than products ✔

ΔH/-99 «kJ» labelled on arrow from reactants to products
OR
activation energy/E_a labelled on arrow from reactant to top of energy profile ✔

Accept a double headed arrow between reactants and products labelled as ΔH for M2.

A challenging question, requiring accurate knowledge of the bonding in these allotropes (some referred to graphite, clearly the most familiar allotrope). The most frequent (correct) answer was the difference in number of bonded C atoms and hybridisation in second place. However, only 30% got a mark.

Again, this was a struggle between intermolecular forces and covalent bonds and this proved to be even harder than (a)(i) with only 25% of candidates getting full marks. The distinction between giant covalent/covalent network in diamond and molecular in C60 and hence resultant sublimation points, was rarely explained. There were many general and vague answers given, as well as commonly (incorrectly) stating that intermolecular forces are present in diamond. As another example of insufficient attention to the question itself, many candidates failed to say which would sublime at a higher temperature and so missed even one mark.

This easy question was quite well answered; same/similar physical properties and empirical formula were common errors.

Candidates misinterpreted the question and mentioned CH3⁺, i.e., the lost fragment; the other very common error was -COOH which shows a complete lack of understanding of MS considering the question is about butane so O should never appear.

Well answered by most, but some basic chemistry was missing when reporting results, perhaps as a result of little practical work due to COVID. A significant number suggested IR spectrometry, very likely because the question followed one on H NMR spectroscopy, thus revealing a failure to read the question properly (which asks for a test). Some teachers felt that adding "chemical" would have avoided some confusion.

Most were able to draw this isomer correctly, though a noticeable number of students included the Z as an atom in the structural formula, showing they were completely unfamiliar with E/Z notation.

Well done in general and most candidates wrote correct reagents, eventually losing a mark when considering H₂ to be a product alongside 2-bromobutane.

Very well answered, some mentioned halogenation which is a different reaction.

A considerable number of students (40%) got at least 1 mark here, but marks were low (average mark 0.9/2). Common errors were predicting 3 peaks, rather than 4 for 2 -bromobutane and vague / unspecific answers, such as ‘different shifts’ or ‘different intensities’. It is surprising that more did not use H NMR data from the booklet; they were not directed to the section as is generally done in this type of question to allow for more general answers regarding all information that can be obtained from an H NMR spectrum.

Product was correctly predicted by many, but most used Markovnikov's Rule to justify this, failing to mention the stability of the secondary carbocation, i.e., the chemistry behind the rule.

As usual, good to excellent candidates (47.5%) were able to get 3/4 marks for this mechanism, while most lost marks for carelessness in drawing arrows and bond connectivity, issues with the lone pair or negative charge on the nucleophile, no negative charge on transition state, or incorrect haloalkane. The average mark was thus 1.9/4.

Another of the very poorly answered questions where most candidates (90%) failed to predict 3 peaks and when they did, considered there would be a quartet instead of multiplet/sextet; other candidates seemed to have no idea at all. This is strange because the compound is relatively simple and while some teachers considered that predicting a sextet may be beyond the current curriculum or just too difficult, they could refer to a multiplet; a quartet is clearly incorrect.

Only the very weak candidates were unable to calculate the enthalpy change correctly, eventually missing 1 mark for inverted calculations.

Most candidates drew correct energy profiles, consistent with the sign of the energy change calculated in the previous question. And again, only very weak candidate failed to get at least 1 mark for correct profiles.