Deduce what information can be obtained from the ¹H NMR spectrum.

Identify the functional group that shows stretching at 1710 cm^–1 in the infrared spectrum of this compound using section 26 of the data booklet and the ¹H NMR.

Suggest the structural formula of this compound.

Bromine was added to hexane, hex-1-ene and benzene. Identify the compound(s) which will react with bromine in a well-lit laboratory.

Deduce the structural formula of the main organic product when hex-1-ene reacts with hydrogen bromide.

State the reagents and the name of the mechanism for the nitration of benzene.

Outline, in terms of the bonding present, why the reaction conditions of halogenation are different for alkanes and benzene.

Below are two isomers, A and B, with the molecular formula C₄H₉Br.

Explain the mechanism of the nucleophilic substitution reaction with NaOH(aq) for the isomer that reacts almost exclusively by an S_N2 mechanism using curly arrows to represent the movement of electron pairs.

Number of hydrogen environments: 3

Ratio of hydrogen environments: 2:3:9

Splitting patterns: «all» singlets

Accept any equivalent ratios such as 9:3:2.

Accept “no splitting”.

[3 marks]

carbonyl
OR
C=O

Accept “ketone” but not “aldehyde”.

[1 mark]

Accept (CH₃)₃CCH₂COCH₃.

Award [1] for any aldehyde or ketone with C₇H₁₄O structural formula.

[2 marks]

hexane AND hex-1-ene

Accept “benzene AND hexane AND hex-1-ene”.

[1 mark]

CH₃CH₂CH₂CH₂CHBrCH₃

Accept displayed formula but not molecular formula.

[1 mark]

Reagents: «concentrated» sulfuric acid AND «concentrated» nitric acid

Name of mechanism: electrophilic substitution

[2 marks]

benzene has «delocalized» $\pi$ bonds «that are susceptible to electrophile attack» AND alkanes do not

Do not accept “benzene has single and double bonds”.

[1 mark]

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

curly arrow showing Br leaving

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

Accept OH^– with or without the lone pair.

Do not allow curly arrows originating on H in OH^–.

Accept curly arrows in the transition state.

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

Do not award M3 if OH–C bond is represented.

Award [2 max] if wrong isomer is used.

[3 marks]

State the full electron configuration of the chlorine atom.

State, giving a reason, whether the chlorine atom or the chloride ion has a larger radius.

Outline why the chlorine atom has a smaller atomic radius than the sulfur atom.

The mass spectrum of chlorine is shown.

NIST Mass Spectrometry Data Center Collection © 2014 copyright by the U.S. Secretary of Commerce on behalf of the United States of America. All rights reserved.

Outline the reason for the two peaks at $m/z=35$ and $37$.

Explain the presence and relative abundance of the peak at $m/z=74$.

Calculate the amount, in $\mathrm{mol}$, of manganese(IV) oxide added.

Determine the limiting reactant, showing your calculations.

Determine the excess amount, in $\mathrm{mol}$, of the other reactant.

Calculate the volume of chlorine, in ${\mathrm{dm}}^3$, produced if the reaction is conducted at standard temperature and pressure (STP). Use section 2 of the data booklet.

State the oxidation state of manganese in ${\mathrm{MnO}}_2$ and ${\mathrm{MnCl}}_2$.

Deduce, referring to oxidation states, whether ${\mathrm{MnO}}_2$ is an oxidizing or reducing agent.

Hypochlorous acid is considered a weak acid. Outline what is meant by the term weak acid.

State the formula of the conjugate base of hypochlorous acid.

Calculate the concentration of ${\mathrm{H}}^{+} \left(\mathrm{aq}\right)$ in a $\mathrm{HClO} \left(\mathrm{aq}\right)$ solution with a $\mathrm{pH}=3.61$.

State the type of reaction occurring when ethane reacts with chlorine to produce chloroethane.

Predict, giving a reason, whether ethane or chloroethane is more reactive.

Explain the mechanism of the reaction between chloroethane and aqueous sodium hydroxide, $\mathrm{NaOH} \left(\mathrm{aq}\right)$, using curly arrows to represent the movement of electron pairs.

Ethoxyethane (diethyl ether) can be used as a solvent for this conversion.
Draw the structural formula of ethoxyethane

Deduce the number of signals and chemical shifts with splitting patterns in the ¹H NMR spectrum of ethoxyethane. Use section 27 of the data booklet.

Calculate the percentage by mass of chlorine in ${\mathrm{CCl}}_2{\mathrm{F}}_2$.

Comment on how international cooperation has contributed to the lowering of $\mathrm{CFC}$ emissions responsible for ozone depletion.

$\mathrm{CFC}$s produce chlorine radicals. Write two successive propagation steps to show how chlorine radicals catalyse the depletion of ozone.

$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^5$ ✔

Do not accept condensed electron configuration.

${\mathrm{Cl}}^{-}$ AND more «electron–electron» repulsion ✔

Accept $C{l}^{\mathit{-}}$ AND has an extra electron.

$\mathrm{Cl}$ has a greater nuclear charge/number of protons/${\mathrm{Z}}_{\mathrm{eff}}$ «causing a stronger pull on the outer electrons» ✔

same number of shells
OR
same «outer» energy level
OR
similar shielding ✔

«two major» isotopes «of atomic mass $35$ and $37$» ✔

«diatomic» molecule composed of «two» chlorine-37 atoms ✔

chlorine-37 is the least abundant «isotope»
OR
low probability of two ${}_{{}_{37}}\mathrm{Cl}$ «isotopes» occurring in a molecule ✔

$«\frac{2.67 \mathrm{g}}{86.94 \mathrm{g} {\mathrm{mol}}^{-1}}=»0.0307 «\mathrm{mol}»$ ✔

$«{\mathrm{n}}_{\mathrm{HCl}}=2.00 \mathrm{mol} {\mathrm{dm}}^{-3}\times 0.2000 {\mathrm{dm}}^3»=0.400 \mathrm{mol}$✔

$«\frac{0.400}{4}=» 0.100 \mathrm{mol}$ AND ${\mathrm{MnO}}_2$ is the limiting reactant ✔

Accept other valid methods of determining the limiting reactant in M2.

$«0.0307 \mathrm{mol}\times 4=0.123 \mathrm{mol}»$

$«0.400 \mathrm{mol}–0.123 \mathrm{mol}=»0.277 «\mathrm{mol}»$ ✔

$«0.0307 \mathrm{mol}\times 22.7 {\mathrm{dm}}^3 {\mathrm{mol}}^{-1}=» 0.697 «{\mathrm{dm}}^3»$ ✔

Accept methods employing $pV=nRT$.

${\mathrm{MnO}}_2: +4$ ✔

${\mathrm{MnCl}}_2: +2$ ✔

oxidizing agent AND oxidation state of $\mathrm{Mn}$ changes from $+4$ to $+2$/decreases ✔

partially dissociates/ionizes «in water» ✔

${\mathrm{ClO}}^{-}$ ✔

$«\left[{\mathrm{H}}^{+}\right]={10}^{-3.61}=» 2.5\times {10}^{-4} «\mathrm{mol} {\mathrm{dm}}^{-3}»$ ✔

«free radical» substitution/${\mathrm{S}}_{\mathrm{R}}$ ✔

Do not accept electrophilic or nucleophilic substitution.

chloroethane AND C–Cl bond is weaker/$324 \mathrm{kJ} {\mathrm{mol}}^{-1}$ than C–H bond/$414 \mathrm{kJ} {\mathrm{mol}}^{-1}$
OR
chloroethane AND contains a polar bond ✔

Accept “chloroethane AND polar”.

curly arrow going from lone pair/negative charge on $\mathrm{O}$ in ⁻OH to $\mathrm{C}$ ✔

curly arrow showing $\mathrm{Cl}$ leaving ✔

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

Accept $O{H}^{\mathit{-}}$ with or without the lone pair.

Do not accept curly arrows originating on $H$ in $O{H}^{-}$.

Accept curly arrows in the transition state.

Do not penalize if $HO$ and $Cl$ are not at 180°.

Do not award M3 if $OH-C$ bond is represented. 

 / ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{OCH}}_2{\mathrm{CH}}_3$ ✔

Accept $\mathit{(}C{H}_{\mathit{3}}C{H}_{\mathit{2}}{\mathit{)}}_{\mathit{2}}O$.

2 «signals» ✔

0.9−1.0 AND triplet ✔

3.3−3.7 AND quartet ✔

Accept any values in the ranges.

Award [1] for two correct chemical shifts or two correct splitting patterns.

$«M\left({\mathrm{CCl}}_2{\mathrm{F}}_2\right) =» 120.91 «\mathrm{g} {\mathrm{mol}}^{-1}»$ ✔

$\frac{2\times 35.45 \mathrm{g} {\mathrm{mol}}^{-1}}{120.91 \mathrm{g} {\mathrm{mol}}^{-1}}\times 100\%=» 58.64 «\%»$ ✔

Award [2] for correct final answer.

Any of:

research «collaboration» for alternative technologies «to replace $\mathrm{CFC}$s»
OR
technologies «developed»/data could be shared
OR
political pressure/Montreal Protocol/governments passing legislations ✔

Do not accept just “collaboration”.

Do not accept any reference to $CFC$ as greenhouse gas or product of fossil fuel combustion.

Accept reference to specific measures, such as agreement on banning use/manufacture of $CFC$s.

${\mathrm{O}}_3+\mathrm{Cl}·\to \mathrm{O}2+\mathrm{ClO}·$ ✔

$\mathrm{ClO}·+\mathrm{O}·\to {\mathrm{O}}_2+\mathrm{Cl}·$
OR
$\mathrm{ClO}·+{\mathrm{O}}_3\to \mathrm{Cl}·+2{\mathrm{O}}_2$ ✔

Penalize missing/incorrect radical dot (∙) once only.

Well answered question with 90% of candidates correctly identifying the complete electron configuration for chlorine.

Most candidates could correctly explain the relative sizes of chlorine atom and chloride ion.

Fairly well answered though some candidates missed M2 for not recognizing the same number of shells affected.

More than 80% could identify that the two peaks in the MS of chlorine are due to different isotopes.

Not well answered. Some candidates were able to identify m/z 74 being due to the m/z of two Cl-37 atoms, however fewer candidates were able to explain the relative abundance of the isotope.

Stoichiometric calculations were generally well done and over 90% could calculate mol from a given mass.

90% of candidates earned full marks on this 2-mark question involving finding a limiting reactant.

Surprisingly, quite a number of candidates struggled with the quantity of excess reactant despite correctly identifying limiting reactant previously.

Most candidates could find the volume of gas produced in a reaction under standard conditions.

More than 90% could identify the oxidation number of manganese in both MnO₂ and MnCl₂.

Most candidates stated that MnO₂ is an oxidizing agent in the reaction but many did not get the mark because there was no reference to oxidation states.

Another well answered 1-mark question where candidates correctly identified a weak acid as an acid which partially dissociates in water. 

Roughly ⅓ of the candidates failed to identify the conjugate base, perhaps distracted by the fact it was not contained in the equation given.

Vast majority of candidates could calculate the concentration of H⁺ (aq) in a HClO (aq) solution with a pH =3.61.

Many identified the reaction of chlorine with ethane as free-radical substitution, or just substitution, with some erroneously stating nucleophilic or electrophilic substitution.

The underlying reasons for the relative reactivity of ethane and chloroethane were not very well known with a few giving erroneous reasons and some stating ethane more reactive.

Few earned full marks for the curly arrow mechanism of the reaction between sodium hydroxide and chloroethane. Mistakes being careless curly arrow drawing, inappropriate –OH notation, curly arrows from the hydrogen or from the carbon to the C–Cl bond, or a method that missed the transition state.

Approximately 60% could draw ethoxyethane however many demonstrated little knowledge of structure of an ether molecule.

A poorly answered question with some getting full marks on this 1HNMR spectrum of ethoxyethane question. Very few could identify all 3 of number of signals, chemical shift, and splitting pattern.

Another good example of candidates being well rehearsed in calculations with 90% earning 2/2 on this question of calculation percentage by mass composition. 

Somewhat disappointing answers on this question about how international cooperation has contributed to the lowering of CFC emissions. Many gave vague answers and some referred to carbon emissions and global warming.

Few could construct the propagation equations showing how CFCs affect ozone, and many lost marks by failing to identify ClO· as a radical.

Ethane, ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$, reacts with chlorine in sunlight. State the type of this reaction and the name of the mechanism by which it occurs.

Formulate equations for the two propagation steps and one termination step in the formation of chloroethane from ethane.

Deduce the splitting patterns in the ¹H NMR spectrum of C₂H₅Cl.

Explain why tetramethylsilane (TMS) is often used as a reference standard in ¹H NMR.

One possible product, X, of the reaction of ethane with chlorine has the following composition by mass:

carbon: 24.27%, hydrogen: 4.08%, chlorine: 71.65%

Determine the empirical formula of the product.

The mass and ¹H NMR spectra of product X are shown below. Deduce, giving your reasons, its structural formula and hence the name of the compound.

When the product X is reacted with NaOH in a hot alcoholic solution, C₂H₃Cl is formed. State the role of the reactant NaOH other than as a nucleophile.

Chloroethene, ${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl}$, can undergo polymerization. Draw a section of the polymer with three repeating units.

substitution AND «free-»radical

OR

substitution AND chain

Award [1] for “«free-»radical substitution” or “S_R” written anywhere in the answer.

[1 mark]

Two propagation steps:

${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}+\bullet \text{Cl}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\bullet +\text{HCl}$

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\bullet +\text{C}{\text{l}}_{\text{2}}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{Cl}+\bullet \text{Cl}$

One termination step:

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\bullet +{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\bullet \to {\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$

OR

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\bullet +\bullet \text{Cl}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{Cl}$

OR

$\bullet \text{Cl}+\bullet \text{Cl}\to \text{C}{\text{l}}_{\text{2}}$

Accept radical without $\bullet$ if consistent throughout.

Allow ECF for incorrect radicals produced in propagation step for M3.

[3 marks]

triplet AND quartet

[1 mark]

chemical shift/signal outside range of common chemical shift/signal

strong signal/12/all H atoms in same environment
OR
singlet/no splitting of the signal

volatile/easily separated/easily removed
OR
inert/stabl

contains three common NMR nuclei/¹H and ¹³C and ²⁹Si

Do not accept chemical shift = 0.

[2 marks]

$\text{C}=\frac{24.27}{12.01}=2.021$ AND $\text{H}=\frac{4.08}{1.01}=4.04$ AND $\text{Cl}=\frac{71.65}{35.45}=2.021$

«hence» CH₂Cl

Accept $\frac{24.27}{12.01}$ : $\frac{4.08}{1.01}$ : $\frac{71.65}{35.45}.$

Do not accept C₂H₄Cl₂. 

Award [2] for correct final answer.

[2 marks]

molecular ion peak(s) «about» m/z 100 AND «so» C₂H₄Cl₂ «isotopes of Cl»

two signals «in ¹H NMR spectrum» AND «so» CH₃CHCl₂
OR
«signals in» 3:1 ratio «in ¹H NMR spectrum» AND «so» CH₃CHCl₂
OR
one doublet and one quartet «in ¹H NMR spectrum» AND «so» CH₃CHCl₂

1,1-dichloroethane

Accept “peaks” for “signals”.

Allow ECF for a correct name for M3 if an incorrect chlorohydrocarbon is identified.

[3 marks]

base
OR
proton acceptor

[1 mark]

Continuation bonds must be shown.

Ignore square brackets and “n”.

Accept  .

Accept other versions of the polymer, such as head to head and head to tail.

Accept condensed structure provided all C to C bonds are shown (as single).

[1 mark]

(i) Calculate ΔH^θ, in kJ, for this similar reaction below using $\mathrm{\Delta }{H}_{\mathrm{f}}^{\theta }$ data from section 12 of the data booklet. $\mathrm{\Delta }{H}_{\mathrm{f}}^{\theta }$ of HOCH₂CH₂OH(l) is –454.8kJmol^-1.

2CO (g) + 3H₂ (g) $\rightleftharpoons$ HOCH₂CH₂OH (l)

(ii) Deduce why the answers to (a)(iii) and (b)(i) differ.

(iii) ΔS^θ for the reaction in (b)(i) is –620.1JK^-1. Comment on the decrease in entropy.

(iv) Calculate the value of ΔG^θ, in kJ, for this reaction at 298 K using your answer to (b)(i). (If you did not obtain an answer to (b)(i), use –244.0 kJ, but this is not the correct value.)

(v) Comment on the statement that the reaction becomes less spontaneous as temperature is increased.

Predict the ¹HNMR data for ethanedioic acid and ethane-1,2-diol by completing the table.

i
«ΔH = Σ ΔH_f products – ΣΔH_f reactants = –454.8 kJ mol^-1 – 2(–110.5 kJ mol^-1) =» –233.8 «kJ»

ii
in (a)(iii) gas is formed and in (b)(i) liquid is formed
OR
products are in different states
OR
conversion of gas to liquid is exothermic
OR
conversion of liquid to gas is endothermic
OR
enthalpy of vapourisation needs to be taken into account

Accept product is «now» a liquid.
Accept answers referring to bond enthalpies being means/averages.

iii
«ΔS is negative because five mols of» gases becomes «one mol of» liquid
OR
increase in complexity of product «compared to reactants»
OR
product more ordered «than reactants»

Accept “fewer moles of gas” but not “fewer molecules”.

iv
ΔS = $\left(\frac{-620.1}{1000}\right)$«kJ K^-1»
ΔG = –233.8 kJ – (298 K $\left(\frac{-620.1}{1000}\right)$ kJ K^-1) = –49.0 «kJ»

Award [2] for correct final answer.
Award [1 max] for «+»185 × 10³.

If –244.0 kJ used, answer is:
ΔG = –244.0 kJ – (298 K $\left(\frac{-620.1}{1000}\right)$kJ K^-1) = –59.2 «kJ»
Award [2] for correct final answer.

v
increasing T makes ΔG larger/more positive/less negative
OR
–TΔS will increase

Accept “none/no splitting” for singlet.

State the class of compound to which ethene belongs.

State the molecular formula of the next member of the homologous series to which ethene belongs.

Justify why ethene has only a single signal in its ¹H NMR spectrum.

Deduce the chemical shift of this signal. Use section 27 of the data booklet.

Suggest two possible products of the incomplete combustion of ethene that would not be formed by complete combustion.

A white solid was formed when ethene was subjected to high pressure.

Deduce the type of reaction that occurred.

Sketch the mechanism for the reaction of propene with hydrogen bromide using curly arrows.

Explain why the major organic product is 2-bromopropane and not 1-bromopropane.

Explain why the major organic product is 2-bromopropane and not 1-bromopropane.

2-bromopropane can be converted directly to propan-2-ol. Identify the reagent required.

Propan-2-ol can also be formed in one step from a compound containing a carbonyl group.

State the name of this compound and the type of reaction that occurs.

alkene ✔

C₃H₆ ✔

Accept structural formula.

hydrogen atoms/protons in same chemical environment ✔

Accept “all H atoms/protons are equivalent”.
Accept “symmetrical”

4.5 to 6.0 «ppm» ✔

Accept a single value within this range.

carbon monoxide/CO AND carbon/C/soot ✔

«addition» polymerization ✔

curly arrow going from C=C to H of HBr AND curly arrow showing Br leaving ✔

representation of carbocation ✔

curly arrow going from lone pair/negative charge on Br⁻ to C⁺ ✔

Award [2 max] for mechanism producing 1-brompropane.

«2-bromopropane involves» formation of more stable «secondary» carbocation/carbonium ion/intermediate
OR
1-bromopropane involves formation of less stable «primary» carbocation/carbonium ion/intermediate ✔

«increased» positive inductive/electron-releasing effect of extra–R group/–CH₃/methyl «increases stability of secondary carbocation» ✔

Award [1] for “more stable due to positive inductive effect”.

Do not award marks for quoting Markovnikov’s rule without any explanation. 

«2-bromopropane involves» formation of more stable «secondary» carbocation/carbonium ion/intermediate
OR
1-bromopropane involves formation of less stable «primary» carbocation/carbonium ion/intermediate ✔

«increased» positive inductive/electron-releasing effect of extra–R group/–CH₃/methyl «increases stability of secondary carbocation» ✔

Award [1] for “more stable due to positive inductive effect”.
Do not award marks for quoting Markovnikov’s rule without any explanation.

sodium hydroxide/NaOH/potassium hydroxide/KOH ✔

Accept «aqueous» hydroxide ions/OH⁻

Name of carbonyl compound:
propanone ✔

Type of reaction:
reduction ✔

Accept other valid alternatives, such as “2-propyl ethanoate” for M1 and “hydrolysis” for M2.

Calculate the percentage by mass of nitrogen in urea to two decimal places using section 6 of the data booklet.

Suggest how the percentage of nitrogen affects the cost of transport of fertilizers giving a reason.

The structural formula of urea is shown.

Predict the electron domain and molecular geometries at the nitrogen and carbon atoms, applying the VSEPR theory.

Urea can be made by reacting potassium cyanate, KNCO, with ammonium chloride, NH₄Cl.

KNCO(aq) + NH₄Cl(aq) → (H₂N)₂CO(aq) + KCl(aq)

Determine the maximum mass of urea that could be formed from 50.0 cm³ of 0.100 mol dm⁻³ potassium cyanate solution.

State the equilibrium constant expression, K_c.

Predict, with a reason, the effect on the equilibrium constant, K_c, when the temperature is increased.

Determine an approximate order of magnitude for K_c, using sections 1 and 2 of the data booklet. Assume ΔG^Θ for the forward reaction is approximately +50 kJ at 298 K.

Suggest one reason why urea is a solid and ammonia a gas at room temperature.

Sketch two different hydrogen bonding interactions between ammonia and water.

The combustion of urea produces water, carbon dioxide and nitrogen.

Formulate a balanced equation for the reaction.

Calculate the maximum volume of CO₂, in cm³, produced at STP by the combustion of 0.600 g of urea, using sections 2 and 6 of the data booklet.

Describe the bond formation when urea acts as a ligand in a transition metal complex ion.

The C–N bonds in urea are shorter than might be expected for a single C–N bond. Suggest, in terms of electrons, how this could occur.

The mass spectrum of urea is shown below.

Identify the species responsible for the peaks at m/z = 60 and 44.

The IR spectrum of urea is shown below.

Identify the bonds causing the absorptions at 3450 cm⁻¹ and 1700 cm⁻¹ using section 26 of the data booklet.

Predict the number of signals in the ¹H NMR spectrum of urea.

Predict the splitting pattern of the ¹H NMR spectrum of urea.

Outline why TMS (tetramethylsilane) may be added to the sample to carry out ¹H NMR spectroscopy and why it is particularly suited to this role.

molar mass of urea «4 $\times$ 1.01 + 2 $\times$ 14.01 + 12.01 + 16.00» = 60.07 «g mol^_-1»

«% nitrogen = $\frac{\text{2}\times \text{14.01}}{\text{60.07}}$ $\times$ 100 =» 46.65 «%»

Award [2] for correct final answer.

Award [1 max] for final answer not to two decimal places.

[2 marks]

«cost» increases AND lower N% «means higher cost of transportation per unit of nitrogen»

OR

«cost» increases AND inefficient/too much/about half mass not nitrogen

Accept other reasonable explanations.

Do not accept answers referring to safety/explosions.

[1 mark]

Note: Urea’s structure is more complex than that predicted from VSEPR theory.

[3 marks]

n(KNCO) «= 0.0500 dm³ $\times$ 0.100 mol dm^–3» = 5.00 $\times$ 10^–3 «mol»

«mass of urea = 5.00 $\times$ 10^–3 mol $\times$ 60.07 g mol^–1» = 0.300 «g»

Award [2] for correct final answer.

[2 marks]

$K_{\text{c}}=\frac{[{({\text{H}}_2\text{N})}_2\text{CO}]\times [{\text{H}}_2\text{O}]}{{[\text{N}{\text{H}}_3]}^2\times [\text{C}{\text{O}}_2]}$

[1 mark]

«K_c» decreases AND reaction is exothermic

OR

«K_c» decreases AND ΔH is negative

OR

«K_c» decreases AND reverse/endothermic reaction is favoured

[1 mark]

ln K « = $\frac{-\mathrm{\Delta }{G}^{\mathrm{\Theta }}}{RT}=\frac{-50\times {10}^3\text{ J}}{8.31\text{ J }{\text{K}}^{-1}\text{ mo}{\text{l}}^{-1}\times 298\text{ K}}$ » = –20

«K_c =» 2 $\times$ 10^–9

OR

1.69 $\times$ 10^–9

OR

10^–9

Accept range of 20-20.2 for M1.

Award [2] for correct final answer.

[2 marks]

Any one of:

urea has greater molar mass

urea has greater electron density/greater London/dispersion

urea has more hydrogen bonding

urea is more polar/has greater dipole moment

Accept “urea has larger size/greater van der Waals forces”.

Do not accept “urea has greater intermolecular forces/IMF”.

[1 mark]

Award [1] for each correct interaction.

If lone pairs are shown on N or O, then the lone pair on N or one of the lone pairs on O MUST be involved in the H-bond.

Penalize solid line to represent H-bonding only once.

[2 marks]

2(H₂N)₂CO(s) + 3O₂(g) → 4H₂O(l) + 2CO₂(g) + 2N₂(g)

correct coefficients on LHS

correct coefficients on RHS

Accept (H₂N)₂CO(s) + $\frac{3}{2}$O₂(g) → 2H₂O(l) + CO₂(g) + N₂(g).

Accept any correct ratio.

[2 marks]

«V = $\frac{\text{0.600 g}}{\text{60.07 g mo}{\text{l}}^{-1}}$ $\times$ 22700 cm³ mol^–1 =» 227 «cm³»

[1 mark]

lone/non-bonding electron pairs «on nitrogen/oxygen/ligand» given to/shared with metal ion

co-ordinate/dative/covalent bonds

[2 marks]

lone pairs on nitrogen atoms can be donated to/shared with C–N bond

OR

C–N bond partial double bond character

OR

delocalization «of electrons occurs across molecule»

OR

slight positive charge on C due to C=O polarity reduces C–N bond length

[1 mark]

60: CON₂H₄⁺

44: CONH₂⁺

Accept “molecular ion”.

[2 marks]

3450 cm^–1: N–H

1700 cm^–1: C=O

Do not accept “O–H” for 3450 cm^–1.

[2 marks]

1

[2 marks]

singlet

Accept “no splitting”.

[1 mark]

acts as internal standard

OR

acts as reference point

one strong signal

OR

12 H atoms in same environment

OR

signal is well away from other absorptions

Accept “inert” or “readily removed” or “non-toxic” for M1.

[2 marks]

Identify the wavenumber of one peak in the IR spectrum of benzoic acid, using section 26 of the data booklet.

Identify the spectroscopic technique that is used to measure the bond lengths in solid benzoic acid.

Outline one piece of physical evidence for the structure of the benzene ring.

Draw the structure of the conjugate base of benzoic acid showing all the atoms and all the bonds.

Outline why both C to O bonds in the conjugate base are the same length and suggest a value for them. Use section 10 of the data booklet.

The pH of an aqueous solution of benzoic acid at 298 K is 2.95. Determine the concentration of hydroxide ions in the solution, using section 2 of the data booklet.

Formulate the equation for the complete combustion of benzoic acid in oxygen using only integer coefficients.

The combustion reaction in (f)(ii) can also be classed as redox. Identify the atom that is oxidized and the atom that is reduced.

Suggest how benzoic acid, M_r = 122.13, forms an apparent dimer, M_r = 244.26, when dissolved in a non-polar solvent such as hexane.

State the reagent used to convert benzoic acid to phenylmethanol (benzyl alcohol), C₆H₅CH₂OH.

Any wavenumber in the following ranges:
2500−3000 «cm⁻¹» [✔]
1700−1750 «cm⁻¹» [✔]
2850−3090 «cm⁻¹» [✔]

X-ray «crystallography/spectroscopy» [✔]

Any one of:

«regular» hexagon

OR

all «H–C–C/C-C-C» angles equal/120º [✔]
all C–C bond lengths equal/intermediate between double and single

OR

bond order 1.5 [✔]

     [✔]

Note: Accept Kekulé structures.
Negative sign must be shown in correct position.

electrons delocalized «across the O–C–O system»

OR

resonance occurs [✔]

122 «pm» < C–O < 143 «pm» [✔]

Note: Accept “delocalized π-bond”.
Accept “bond intermediate between single and double bond” or “bond order 1.5” for M1.
Accept any answer in range 123 to 142 pm.

ALTERNATIVE 1:
[H⁺] «= 10^−2.95» = 1.122 × 10⁻³ «mol dm⁻³» [✔]

«[OH⁻] = $\frac{1.00\times {10}^{-14}\text{ mo}{\text{l}}^2\text{ d}{\text{m}}^{-6}}{1.22\times {10}^{-3}\text{ mol d}{\text{m}}^{-3}}$ =» 8.91 × 10⁻¹² «mol dm⁻³» [✔]

ALTERNATIVE 2:
pOH = «14 − 2.95 =» 11.05 [✔]
«[OH⁻] = 10^−11.05 =» 8.91 × 10⁻¹² «mol dm⁻³» [✔]

Note: Award [2] for correct final answer.
Accept other methods.

2C₆H₅COOH (s) + 15O₂ (g) → 14CO₂ (g) + 6H₂O (l)
correct products    [✔]
correct balancing    [✔]

Oxidized:

C/carbon «in C₆H₅COOH»

AND

Reduced:

O/oxygen «in O₂»      [✔]

«intermolecular» hydrogen bonding    [✔]

Note: Accept diagram showing hydrogen bonding.

lithium aluminium hydride/LiAlH₄    [✔]

Most candidates could identify a wavenumber or range of wavenumbers in the IR spectrum of benzoic acid.

Less than half the candidates identified x-ray crystallography as a technique used to measure bond lengths. There were many stating IR spectroscopy and quite a few random guesses.

Again less than half the candidates could accurately give a physical piece of evidence for the structure of benzene. Many missed the mark by not being specific, stating ‘all bonds in benzene with same length’ rather than ‘all C-C bonds in benzene have the same length’.

Very poorly answered with only 1 in 5 getting this question correct. Many did not show all the bonds and all the atoms or either forgot or misplaced the negative sign on the conjugate base.

This question was a challenge. Candidates were not able to explain the intermediate bond length and the majority suggested the value of either the bond length of C to O single bond or double bond.

Generally well done with a few calculating the pOH rather than the concentration of hydroxide ion asked for.

Most earned at least one mark by correctly stating the products of the reaction.

Another question where not reading correctly was a concern. Instead of identifying the atom that is oxidized and the atom that is reduced, answers included formulas of molecules or the atoms were reversed for the redox processes.

The other question where only 10 % of the candidates earned a mark. Few identified hydrogen bonding as the reason for carboxylic acids forming dimers. There were many G2 forms stating that the use of the word “dimer” is not in the syllabus, however the candidates were given that a dimer has double the molar mass and the majority seemed to understand that the two molecules joined together somehow but could not identify hydrogen bonding as the cause.

Very few candidates answered this part correctly and scored the mark. Common answers were H₂SO₄, HCl & Sn, H₂O₂. In general, strongest candidates gained the mark.

Suggest why the three-membered ring in methyloxirane is unstable.

Draw two structural isomers of methyloxirane.

State, giving a reason, whether methyloxirane can form cis-trans isomers.

Predict the chemical shift and splitting pattern of the signal produced by the hydrogen atoms labelled X in the ¹H NMR spectrum of the polymer. Use section 27 of the data booklet.

angle between bonds is 60°/strained/smaller than 109.5° ✔

Any two of:
CH₃COCH₃ ✔

CH₃CH₂CHO ✔

CH₂=CHCH₂OH ✔

CH₃OCH=CH₂ ✔

Accept displayed or condensed structural formulas or skeletal formulas.

Accept CH(OH)=CHCH₃ and CH₂=C(OH)CH₃.

no AND only one «axial/methyl/CH₃» substituent «at the ring»

OR

no AND two «axial» substituents required «for cis/trans-isomers» ✔

Accept “no AND «O in the ring and» one carbon has two H atoms”.

Chemical shift:
3.7–4.8 «ppm» ✔

Splitting pattern:
doublet ✔

Draw arrows in the boxes to represent the electron configuration of a nitrogen atom.

Deduce a Lewis (electron dot) structure of the nitric acid molecule, HNO₃, that obeys the octet rule, showing any non-zero formal charges on the atoms.

Explain the relative lengths of the three bonds between N and O in nitric acid.

State a technique used to determine the length of the bonds between N and O in solid HNO₃.

Write an equation for the reaction between the acids to produce the electrophile, NO₂⁺.

Draw the structural formula of the carbocation intermediate produced when this electrophile attacks benzene.

Deduce the number of signals that you would expect in the ¹H NMR spectrum of nitrobenzene and the relative areas of these.

Accept all 2p electrons pointing downwards.

Accept half arrows instead of full arrows.

bonds and non-bonding pairs correct ✔

formal charges correct ✔

Accept dots, crosses or lines to represent electron pairs.

Do not accept resonance structures with delocalised bonds/electrons.

Accept + and – sign respectively.

Do not accept a bond between nitrogen and hydrogen.

For an incorrect Lewis structure, allow ECF for non-zero formal charges.

Any three of:

two N-O same length/order ✔
delocalization/resonance ✔

N-OH longer «than N-O»
OR
N-OH bond order 1 AND N-O bond order 1½ ✔

Award [2 max] if bond strength, rather than bond length discussed.

Accept N-O between single and double bond AND N-OH single bond.

X-ray crystallography ✔

HNO₃ + 2H₂SO₄ $\rightleftharpoons$ NO₂⁺ + H₃O⁺ + 2HSO₄^- ✔

Accept “HNO₃ + H₂SO₄ $\rightleftharpoons$ NO₂⁺ + H₂O + HSO₄^-”.

Accept “HNO₃ + H₂SO₄ $\rightleftharpoons$ H₂NO₃⁺ + HSO₄^-” AND “H₂NO₃⁺ $\rightleftharpoons$ NO₂⁺ + H₂O”.

Accept single arrows instead of equilibrium signs.

Accept any of the five structures.

Do not accept structures missing the positive charge.

Number of signals: three/3 ✔

Relative areas: 2 : 2 : 1 ✔

Drawing arrows in the boxes to represent the electron configuration of a nitrogen atom was done extremely well.

Drawing the Lewis structure of HNO₃ was performed extremely poorly with structures that included H bonded to N, no double bond or a combination of single, double and even a triple bond or incorrect structures with dotted lines to reflect resonance. Many did not calculate non-zero formal charges.

Poorly done; some explained relative bond strengths between N and O in HNO₃, not relative lengths; others included generic answers such as triple bond is shortest, double bond is longer, single longest.

A majority could not state the technique to determine length of bonds; answers included NMR, IR, and such instead of X-ray crystallography.

Many had difficulties writing the balanced equation(s) for the formation of the nitronium ion.

Again, many had difficulty drawing the structural formula of the carbocation intermediate produced in the reaction.

Deducing the number of signals in the 1H NMR spectrum of nitrobenzene, which depend on the number of different hydrogen environments, was done poorly. Also, instead of relative areas, the common answer included chemical shift (ppm) values.

Deduce the structural formulas of the two possible isomers.

Mass spectra A and B of the two isomers are given.

Explain which spectrum is produced by each compound using section 28 of the data booklet.

State the type of bond fission that takes place in a S_N1 reaction.

State the type of solvent most suitable for the reaction.

Draw the structure of the intermediate formed stating its shape.

Suggest, giving a reason, the percentage of each isomer from the S_N1 reaction.

Nitrobenzene, C₆H₅NO₂, can be converted to phenylamine via a two-stage reaction.

In the first stage, nitrobenzene is reduced with tin in an acidic solution to form an intermediate ion and tin(II) ions. In the second stage, the intermediate ion is converted to phenylamine in the presence of hydroxide ions.

Formulate the equation for each stage of the reaction.

Accept condensed formulas.

[2 marks]

A:

CH₃CH₂COCH₂CH₃ AND «peak at» 29 due to

(CH₃CH₂)⁺/(C₂H₅)⁺/(M – CH₃CH₂CO)⁺

OR

CH₃CH₂COCH₂CH₃ AND «peak at» 57 due to

(CH₃CH₂CO)⁺/(M – CH₃CH₂)⁺/(M – C₂H₅)⁺

B:

CH₃COCH₂CH₂CH₃ AND «peak at» 43 due to

(CH₃CH₂CH₂)⁺/(CH₃CO)⁺/(C₂H₃O)⁺/(M – CH₃CO)⁺

Penalize missing “+” sign once only.

Accept “CH₃COCH₂CH₂CH₃ by elimination since fragment CH₃CO is not listed” for M2.

[2 marks]

heterolytic/heterolysis

[1 mark]

polar protic

[1 mark]

Shape: triangular/trigonal planar

[2 marks]

«around» 50% «each»

OR

similar/equal percentages

nucleophile can attack from either side «of the planar carbocation»

Accept “racemic mixture/racemate” for M1.

[2 marks]

Stage one:

C₆H₅NO₂(l) + 3Sn(s) + 7H⁺(aq) → C₆H₅NH₃⁺(aq) + 3Sn²⁺(aq) + 2H₂O(l)

Stage two:

C₆H₅NH₃⁺(aq) + OH^–(aq) → C₆H₅NH₂(l) + H₂O(l)

[2 marks]

Determine the empirical formula of the compound using section 6 of the data booklet.

Determine the molecular formula of this compound if its molar mass is 88.12 g mol⁻¹. If you did not obtain an answer in (a) use CS, but this is not the correct answer.

Identify each compound from the spectra given, use absorptions from the range of 1700 cm⁻¹ to 3500 cm⁻¹. Explain the reason for your choice, referring to section 26 of the data booklet.

Predict the number of ¹H NMR signals, and splitting pattern of the –CH₃ seen for propanone (CH₃COCH₃) and propanal (CH₃CH₂CHO).

Predict the fragment that is responsible for a m/z of 31 in the mass spectrum of propan‑1‑ol. Use section 28 of the data booklet.

«$\frac{8.802 \mathrm{g}}{44.01 \mathrm{g} {\mathrm{mol}}^{-1}}=$» 0.2000 «mol of C/CO₂»

AND «$\frac{3.604 \mathrm{g}}{18.02 \mathrm{g} {\mathrm{mol}}^{-1}}=$» 0.2000 «mol of H₂O»/0.4000 «mol of H»

OR

 «$\frac{8.802 \mathrm{g}}{44.01 \mathrm{g} {\mathrm{mol}}^{-1}}\times 12.01 \mathrm{g} {\mathrm{mol}}^{-1}$» 2.402 «g of C»

OR

«$\frac{3.604 \mathrm{g}}{18.02 \mathrm{g} {\mathrm{mol}}^{-1}}\times 2\times 1.01 \mathrm{g} {\mathrm{mol}}^{-1}=$» 0.404 «g of H» ✔

«4.406 g − 2.806 g» = 1.600 «g of O» ✔

«$\frac{2.402 \mathrm{g}}{12.01 \mathrm{g} {\mathrm{mol}}^{-1}}=0.2000 \mathrm{mol} \mathrm{C}; \frac{0.404 \mathrm{g}}{1.01 \mathrm{g} {\mathrm{mol}}^{-1}}=0.400 \mathrm{mol} \mathrm{H}; \frac{1.600 \mathrm{g}}{16.00 \mathrm{g} {\mathrm{mol}}^{-1}}=0.1000 \mathrm{mol} \mathrm{O}$»

C₂H₄O ✔

Award [3] for correct final answer.

«$\frac{88.12 \mathrm{g} {\mathrm{mol}}^{-1}}{44.06 \mathrm{g} {\mathrm{mol}}^{-1}}=2$» C₄H₈O₂ ✔

C₂S₂ if CS used.

Award [1 max] for correctly identifying all 3 compounds without valid reasons given.

Accept specific values of wavenumbers within each range.

CH₃O⁺ ✔

Accept any structure i.e. “CH₂OH⁺”.

Write an equation for the complete combustion of ethyne.

Deduce the Lewis (electron dot) structure of ethyne.

Compare, giving a reason, the length of the bond between the carbon atoms in ethyne with that in ethane, C₂H₆.

Identify the type of interaction that must be overcome when liquid ethyne vaporizes.

State the name of product B, applying IUPAC rules.

Determine the enthalpy change for the reaction, in kJ, to produce A using section 11 of the data booklet.

The enthalpy change for the reaction to produce B is −213 kJ.

Predict, giving a reason, which product is the most stable.

The IR spectrum and low resolution ¹H NMR spectrum of the actual product formed are shown.

Deduce whether the product is A or B, using evidence from these spectra together with sections 26 and 27 of the  data booklet.

Identity of product:

One piece of evidence from IR:

One piece of evidence from ¹H NMR:

Deduce the splitting pattern you would expect for the signals in a high resolution ¹H NMR spectrum.

2.3 ppm:

9.8 ppm:

Suggest the reagents and conditions required to ensure a good yield of product B.

Reagents: 

Conditions:

Deduce the average oxidation state of carbon in product B.

Explain why product B is water soluble.

C₂H₂ (g) + 2.5O₂ (g) → 2CO₂ (g) + H₂O (l)

OR

2C₂H₂ (g) + 5O₂ (g) → 4CO₂ (g) + 2H₂O (l)    [✔]

    [✔]

Note: Accept any valid combination of lines, dots and crosses.

«ethyne» shorter AND a greater number of shared/bonding electrons

OR

«ethyne» shorter AND stronger bond     [✔]

London/dispersion/instantaneous dipole-induced dipole forces    [✔]

ethanal    [✔]

«sum of bond enthalpies of reactants =» 2(C—H)+C ≡ C + 2(O—H)
OR
2 × 414 «kJ mol^-1» + 839 «kJ mol^-1» + 2 × 463 «kJ mol^-1»
OR
2593 «kJ»    [✔]

«sum of bond enthalpies of A =» 3(C—H) + C=C + C—O + O—H
OR
3 × 414 «kJ mol^-1» + 614 «kJ mol^-1» + 358 «kJ mol^-1» + 463 «kJ mol^-1»
OR
2677 «kJ»     [✔]
«enthalpy of reaction = 2593 kJ – 2677 kJ» = –84 «kJ»     [✔]

Note: Award [3] for correct final answer.

B AND it has a more negative/lower enthalpy/«potential» energy

OR

B AND more exothermic «enthalpy of reaction from same starting point»     [✔]

Identity of product: «B»

IR spectrum:
1700–1750 «cm^–1 band» AND carbonyl/CO group present
OR
no «band at» 1620–1680 «cm^–1» AND absence of double bond/C=C
OR
no «broad band at» 3200–3600 «cm^–1 » AND absence of hydroxyl/OH group    [✔]

¹H NMR spectrum:
«only» two signals AND A would have three
OR
«signal at» 9.4–10.0 «ppm» AND «H atom/proton of» aldehyde/–CHO present
OR
«signal at» 2.2–2.7 «ppm» AND «H atom/proton of alkyl/CH next to» aldehyde/CHO present
OR
«signal at» 2.2–2.7 «ppm» AND «H atom/proton of» RCOCH_2- present
OR
no «signal at» 4.5–6.0 «ppm» AND absence of «H atom/proton next to» double bond/C=C ✔

Note: Accept a specific value or range of wavenumbers and chemical shifts.

Accept “two signals with areas 1:3”.

2.3 ppm: doublet    [✔]

9.8 ppm: quartet    [✔]

Reagents:
acidified/H⁺ AND «potassium» dichromate«(VI)»/K₂Cr₂O₇/Cr₂O₇^2-    [✔]

Conditions:
distil «the product before further oxidation»       [✔]

Note: Accept “«acidified potassium» manganate(VII)/KMnO₄/MnO₄^-/permanganate”.

Accept “H₂SO₄” or “H₃PO₄” for “H⁺”.

Accept “more dilute dichromate(VI)/manganate(VII)” or “excess ethanol”.

Award M1 if correct reagents given under “Conditions”.

–1     [✔]

Any three of:

has an oxygen/O atom with a lone pair      [✔]

that can form hydrogen bonds/H-bonds «with water molecules»     [✔]

hydrocarbon chain is short «so does not disrupt many H-bonds with water molecules»     [✔]

«large permanent» dipole-dipole interactions with water      [✔]

All candidates were able to write the correct reactants/products for combustion of ethyne, but a few failed to balance correctly.

Most drew correct Lewis structures for ethyne, though some drew ethene.

Surprisingly very few explained the difference in bond length/strength looking at electrons shared and just gave the shorter/triple or longer/single bond answer.

Good to see that most candidates identified the specific IMF correctly.

Most candidates gave the correct IUPAC name.

Candidates were able to calculate the ΔH of the given reaction correctly; a few inverted the calculations or made mathematical errors.

Generally well done, most common error was stating that the enthalpy change was “larger” without the indication that it was an exothermic change or the sign.

Interpretation of spectra was very good and the few candidates that lost marks with ¹H NMR data rather than IR, for example simply mentioning two signals for B. However, most candidates that attempted this question got full marks.

The stronger candidates were able to predict the splitting pattern correctly, others inverted the answer, but many others repeated the information for protons with the given chemical shift, which is unexpected since wording was straightforward.

Candidates seemed to be confused by the prompts, reagent and conditions, so often included the acid among conditions. Careless errors were common such as the wrong charge on the dichromate ion. Few candidates suggest permanganate as an option.

Most candidates were able to calculate oxidation state of carbon in B.

Candidates did not understand that they must mention the IMF responsible for the solubility. Most candidates explained the polarity of the aldehyde and water but did not mention that this results in permanent dipole-dipole interactions; many did mention H-bonding. The mention of the lone pair on O atom and short hydrocarbon chain were very rare.

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.

Outline why metals, like iron, can conduct electricity.

Justify why sulfur is classified as a non-metal by giving two of its chemical properties.

Sketch the first eight successive ionisation energies of sulfur.

Describe the bonding in this type of solid.

State a technique that could be used to determine the crystal structure of the solid compound.

State the full electron configuration of the sulfide ion.

Outline, in terms of their electronic structures, why the ionic radius of the sulfide ion is greater than that of the oxide ion.

Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.

Write the equation for this reaction.

Deduce the change in the oxidation state of sulfur.

Suggest why this process might raise environmental concerns.

Explain why the addition of small amounts of carbon to iron makes the metal harder.

mobile/delocalized «sea of» electrons

Any two of:

forms acidic oxides «rather than basic oxides» ✔

forms covalent/bonds compounds «with other non-metals» ✔

forms anions «rather than cations» ✔

behaves as an oxidizing agent «rather than a reducing agent» ✔

Award [1 max] for 2 correct non-chemical properties such as non-conductor, high ionisation energy, high electronegativity, low electron affinity if no marks for chemical properties are awarded.

two regions of small increases AND a large increase between them✔

large increase from 6th to 7th ✔

Accept line/curve showing these trends.

electrostatic attraction ✔

between oppositely charged ions/between Fe²⁺ and S²⁻ ✔

X-ray crystallography ✔

1s² 2s² 2p⁶ 3s² 3p⁶ ✔

Do not accept “[Ne] 3s² 3p⁶”.

«valence» electrons further from nucleus/extra electron shell/ electrons in third/3s/3p level «not second/2s/2p»✔

Accept 2,8 (for O^2–) and 2,8,8 (for S^2–)

allows them to explain the properties of different compounds/substances
OR
enables them to generalise about substances
OR
enables them to make predictions ✔

Accept other valid answers.

4FeS(s) + 7O₂(g) → 2Fe₂O₃(s) + 4SO₂(g) ✔

Accept any correct ratio.

+6
OR
−2 to +4 ✔

Accept “6/VI”.
Accept “−II, 4//IV”.
Do not accept 2- to 4+.

sulfur dioxide/SO₂ causes acid rain ✔

Accept sulfur dioxide/SO₂/dust causes respiratory problems
Do not accept just “causes respiratory problems” or “causes acid rain”.

disrupts the regular arrangement «of iron atoms/ions»
OR
carbon different size «to iron atoms/ions» ✔

prevents layers/atoms sliding over each other ✔