Chemistry 9701/42 February/March 2023: question paper and mark scheme
Cambridge AS & A Level Chemistry (9701), Paper 4 A Level Structured Questions, February/March 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) The most common zinc mineral contains zinc(II) sulfide, ZnS.
(i) Complete the electrons in boxes diagram in Fig. 1.1 to show the electronic configuration of a zinc(II) ion.
3d 4s 4p
[Ar]
Fig. 1.1
[1]
(ii) Complete Fig. 1.2 to show the Born–Haber diagram for the ionic solid ZnS.
Include state symbols of relevant species.
Zn2+(g) + S2–(g)
Zn2+(g) + S(g) + 2e–
EA1 EA2
IE2
Zn2+(g) + S–(g) + e–
IE1
ΔHlatt
ΔHat(S)
ΔHat(Zn)
ΔHf
ZnS(s)
Fig. 1.2
[3]
(iii) Describe the trend in the first electron affinity of the Group 16 elements S to Te.
Explain your answer.
…… [2]
(iv) Explain why the lattice energy, ∆Hlatt, of ZnO is more exothermic than that of ZnS.
…… [2]
(b) Zinc metal can be obtained in a two-step process as shown.
step 1 2ZnS(s) + 3O2(g) 2ZnO(s) + 2SO2(g)
step 2 ZnO(s) + C(s) Zn(l) + CO(g)
The reactions are carried out at 800 °C.
(i) Predict the sign of the entropy change, ∆S o , of the reaction in step 1.
Explain your answer.
…… [1]
(ii) Use the data in Table 1.1 to calculate ∆S o of the reaction shown in step 2.
Table 1.1
chemical ZnO(s) C(s) Zn(l) CO(g)
S o / J K–1 mol–1 43.7 5.7 50.8 197.7
∆S o = …… J K–1 mol–1 [1]
(iii) An equation for the direct reduction of ZnS by carbon is shown.
2ZnS(s) + C(s) 2Zn(l) + CS2(g) ∆H o = +733 kJ mol–1
∆S o = +218 J K–1 mol–1
This reaction is not feasible at 800 °C.
Calculate ∆G o for this reaction at 800 °C.
∆G o = …… kJ mol–1 [2]
(c) Zn(NO3)2 undergoes thermal decomposition when heated. The reaction is similar to the thermal decomposition of Group 2 nitrates.
(i) Construct an equation for the thermal decomposition of Zn(NO3)2.
…… [1]
(ii) The radii of some Group 2 cations and Zn2+ are shown in Table 1.2.
Table 1.2
cation Mg2+ Ca2+ Sr2+ Ba2+ Zn2+ radius / pm 65 99 113 135 74
State and explain the trend in thermal stability of the Group 2 nitrates down the group.
…… [2]
(iii) Use Table 1.2 to suggest which Group 2 nitrates are less thermally stable than zinc nitrate.
…… [1]
[Total: 16]
Mark scheme
1(a)(i) 1
[Ar] ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂
1(a)(ii) 3
one ● for each of the eight species (including state symbol) on the correct line
AND one ● e–
any three [1] any six [2] all nine [3]
1(a)(iii) • EA becomes less negative/ less exothermic (down group / S to Te) 2
• atomic radii increases OR outer shell gets farther from nucleus OR electron added at higher energy level
OR more shielding (of outer shells)
• less nuclear attraction
OR less attraction for incoming/added electron
any two [1] all three [2]
1(a)(iv) M1: O2– (has same charge but) smaller (radius than S2–) ORA 1
OR oxygen has a smaller ion (than S2–)
M2: stronger ionic bond OR greater attraction between Zn2+ and O2– ORA 1
1(b)(i) ΔS negative AND more moles / molecules of gaseous reactants ORA 1
OR ΔS negative AND moles / molecules of gas are reduced (in the reaction)
1(b)(ii) ΔS = 50.8 + 197.7 – 43.7 – 5.7 = (+)199.1 (J K–1 mol–1) 1
1(b)(iii) ΔG = ΔH – TΔS ALLOW G = ΔH – TΔS 1
= +733 – (800 + 273) × 0.218 = (+)499.086 (kJ mol–1) min 3sf 1
1(c)(i) Zn(NO3)2 → ZnO + 2NO2 + ½O2 1
OR 2Zn(NO3)2 → 2ZnO + 4NO2 + O2
1(c)(ii) increases (in thermal stability down the group) 1
AND (cat)ion(ic) radius / ion size increases (down the group)
less polarisation / less distortion 1
of anion/ of nitrate ion/NO3– / NO32–
OR less weakening of N—O bond
1(c)(iii) Mg(NO3)2 only ALLOW Mg2+ / magnesium 1
Hypophosphorous acid is an inorganic acid.
The conjugate base of hypophosphorous acid is H2PO2–.
(a) Give the formula of hypophosphorous acid.
…… [1]
(b) H2PO2– is a strong reducing agent. It can be used to reduce metal cations without the need for electrolysis.
equation 1 HPO32– + 2H2O + 2e– H2PO2– + 3OH– E o = –1.57 V
(i) In an experiment, an alkaline HPO32–/H2PO2– half-cell is constructed with
[H2PO2–] = 0.050 mol dm–3.
All other ions are at their standard concentration.
Predict how the value of E of this half-cell differs from its E o value.
Explain your answer.
…… [2]
(ii) The Cr3+/Cr half-cell has a standard electrode potential of –0.74 V.
An electrochemical cell consists of an alkaline HPO32–/H2PO2– half-cell and a Cr3+/Cr half-cell. o
Calculate the standard cell potential, Ecell .
o
Ecell = …… V [1]
(iii) Complete the diagram in Fig. 2.1 to show how the standard electrode potential of the
Cr3+/Cr half-cell can be measured relative to that of the standard hydrogen electrode.
Identify the chemicals, conditions and relevant pieces of apparatus.
Fig. 2.1
[3]
(iv) Label Fig. 2.1 to show:
• which is the positive electrode
• the direction of electron flow in the external circuit.
[1]
(v) H2PO2– reduces Ni2+ to Ni in alkaline conditions.
Use equation 1 to construct the ionic equation for this reaction.
equation 1 HPO32– + 2H2O + 2e– H2PO2– + 3OH–
…… [1]
(c) H2PO2–(aq) reacts with OH–(aq).
H2PO2–(aq) + OH–(aq) HPO32–(g) + H2(g)
Table 2.1 shows the results of a series of experiments used to investigate the rate of this reaction.
Table 2.1
[H2PO2–(aq)] [OH–(aq)] volume of H2 experiment produced in 60 s
/ mol dm–3 / mol dm–3 / cm3
1 0.40 2.00 6.4
2 0.80 2.00 12.8
3 1.20 1.00 4.8
(i) The volume of H2 was measured under room conditions.
Use the molar volume of gas, Vm, and the data from experiment 1 to calculate the rate of reaction in mol dm–3 s–1.
rate of reaction = …… mol dm–3 s–1 [1]
(ii) The rate equation was found to be:
rate = k [H2PO2–(aq)] [OH–(aq)]2
Show that the data in Table 2.1 is consistent with the rate equation.
…… [2]
(iii) State the units of the rate constant, k, for the reaction.
…… [1]
(iv) The experiment is repeated using a large excess of OH–(aq).
Under these conditions, the rate equation is:
rate = k1 [H2PO2–(aq)]
k1 = 8.25 × 10–5 s–1
Calculate the value of the half-life, t , of the reaction.
1
2
t = …… s [1]
1
2
(v) Describe how an increase in temperature affects the value of the rate constant, k1.
…… [1]
(d) A student suggests that the reaction between H2PO2–(aq) and OH–(aq) might happen more quickly in the presence of a heterogeneous catalyst.
Describe the mode of action of a heterogeneous catalyst.
…… [2]
[Total: 17]
Mark scheme
2(a) H3PO2 1
2(b)(i) electrode potential E would become more positive / less negative (than E⦵) 1
lower [H2PO2–] AND shifts equilibrium to the right-hand side 1
2(b)(ii) +1.57 – 0.74 = (+)0.83 (V) 1
2(b)(iii) 3
• Pt(s) / C / graphite
• Cr(s)
• H+(aq) / any strong acid
• Cr3+(aq)
• H2(g)
• voltmeter / V
• salt bridge labelled
• conditions of 1 atm AND 1 mol dm–3 / 1 M / 1 mol / dm3
• other liquid level (salt bridge) and wire to electrode
any three [1] any six [2] all nine [3]
2(b)(iv) Pt electrode positive AND flow of electrons anticlockwise (to the SHE) 1
2(b)(v) H2PO2– + 3OH– + Ni2+ → HPO32– + 2H2O + Ni 1
2(c)(i) (6.4 / 24000) ÷ 60 = 4.44 × 10–6 (mol dm-3 s–1) min 2sf 1
2(c)(ii) [H2PO2–] doubles / ×2 from experiments 1 to 2 ORA 1
volume of H2 produced doubles / ×2 (∴ first order wrt [H2PO2–])
[H2PO2–] × 3 and [OH–] × ½ from experiments 1 to 3 ORA 1
volume of H2 produced falls to ¾ original
(if first order wrt [H2PO2–] then must be second order wrt [OH–])
ALLOW input data into rate equation and show k is the same
k =2.8 × 10–6 / k = 6.7 × 10–2 (1 / 15) / k = 4 for all experiments [2]
2(c)(iii) mol–2 dm6 s–1 1
2(c)(iv) t½ = 0.693 / 8.25 × 10–5 = 8400 (s) OR t½ = In2 / 8.25 × 10–5 = 8401.8 (s) 1
2(c)(v) (k1) increases (with temperature) 1
2(d) • reactants adsorb (to surface of catalyst) 2
• bonds (in reactant) weaken
• (reaction occurs and the) products are desorbed
OR reaction occurs and substances are desorbed
any two [1] all three [2]
Vanadium is a transition element in Period 4 of the Periodic Table.
(a) Define transition element.
…… [1]
(b) Vanadium shows typical chemical properties of transition elements, including variable oxidation states.
(i) State two other typical chemical properties of transition elements.
1 ……
2 ……
[1]
(ii) Explain why transition elements have variable oxidation states.
…… [1]
(c) VO2+ can be reduced to V2+ by C2O42– in acidic conditions.
equation 2 2VO2+ + 3C2O42– + 8H+ 2V2+ + 6CO2 + 4H2O
(i) In a titration, 25.00 cm3 of 0.0300 mol dm–3 VO2+(aq) is added to 10 cm3 of dilute sulfuric acid.
A solution of 0.0400 mol dm–3 C2O42–(aq) is then added from a burette until the end-point is reached.
The titration is repeated and concordant results obtained, as shown in Table 3.1.
Table 3.1
1 2 volume of C2O42–(aq) added / cm3 28.15 28.10
Show that these results are consistent with the stoichiometry of equation 2.
[2]
(ii) An excess of C2O42– reacts with VO2+ to form a mixture of two octahedral complex ions.
The complex ions are stereoisomers of each other.
Each complex ion contains a V2+ cation and three C2O42– ions.
Complete the diagram to show the three-dimensional structure of one of the complex ions.
Include the charge of the complex ion.
Use O O to represent a C2O42– ion.
V
[2]
(d) V2+(aq) can be oxidised by H2O2(aq).
Table 3.2 gives some relevant data.
Table 3.2
half-equation E o /V
1 H2O2(aq) + 2H+(aq) + 2e– 2H2O(l) +1.77
2 VO2+(aq) + 2H+(aq) + e– VO2+(aq) + H2O(l) +1.00
3 VO2+(aq) + 2H+(aq) + e– V3+(aq) + H2O(l) +0.34
4 V3+(aq) + e– V2+(aq) –0.26
(i) Identify the vanadium species that forms when an excess of H2O2(aq) reacts with V2+(aq) under standard conditions.
Explain your answer with reference to the data in Table 3.2.
…… [1]
(ii) Concentrated acidified H2O2 can react with V2+ to form red VO23+ ions.
VO23+ contains vanadium combined with the peroxide anion, O22–.
Deduce the oxidation state of vanadium in VO23+.
…… [1]
[Total: 9]
Mark scheme
3(a) (a d-block element forms one or more) 1
stable ions with incomplete filled d-subshell
3(b)(i) • they behave as catalysts 1
• they form complex ions / complexes
• they form coloured compounds / salts / ions
any two
3(b)(ii) the d and s sub-shells/orbitals 1
are close/similar in energy
3(c)(i) 0.02500 × 0.0300 OR 7.50 × 10–4 mol VO2+ 1
OR ½(28.15 + 28.10)/1000 × 0.0400 OR 1.13 × 10–3 mol C2O42–
Use of their values to show ratio of VO2+ : C2O42– = 1:1.5 1
ALLOW any viable approach
3(c)(ii) 1
3D structure
charge 1
3(d)(i) VO2+ AND E⦵ of H2O2 is largest / most positive value 1
OR VO2+ AND E⦵ is less positive than H2O2
3(d)(ii) +5 1
Ethylamine and phenylamine are primary amines.
ethylamine phenylamine
NH2
CH3CH2NH2
Fig. 4.1
These two compounds are synthesised by different methods.
(a) Several methods can be used to form ethylamine.
(i) Ethylamine forms when ethanamide, CH3CONH2, is reduced by LiAl H4.
Write an equation for this reaction. Use [H] to represent one atom of hydrogen from the reducing agent.
…… [1]
(ii) Ethylamine is a product of the reaction of bromoethane with ammonia.
Name the mechanism of this reaction and state the conditions used.
mechanism ……
conditions ……
[2]
(iii) The reaction in (a)(ii) also forms secondary and tertiary amines.
Suggest the identity of a secondary or tertiary amine formed by the reaction in (a)(ii).
…… [1]
(b) Ethylamine is a weak base.
State the relative basicities of ammonia, ethylamine and phenylamine.
Explain your answer.
…… < …… < ……
least basic most basic
…… [4]
(c) Pure phenylamine, C6H5NH2, can be prepared from benzene in two steps.
Draw the structure of the intermediate compound.
Suggest reagents and conditions for each step.
…… [3]
(d) Fig. 4.2 shows some reactions of phenylamine.
phenylamine reaction 1
NH2 excess Br2(aq)
W room temperature
reaction 2
benzenediazonium chloride
N2Cl reaction 3
X
reaction 4
Y
OH
N
N
Fig. 4.2
(i) Draw the structure of W, the organic product of reaction 1.
[1]
(ii) State the reagents used in reaction 2.
…… [1]
Benzenediazonium chloride, C6H5N2Cl, and X react together in reaction 4 to form Y, an azo compound.
(iii) Name X, the organic product of reaction 3.
…… [1]
(iv) State the necessary conditions for reaction 4 to occur.
…… [1]
(v) Suggest a use for Y.
…… [1]
(e) Methylamine, CH3NH2, is another primary amine. CH3NH2 can act as a monodentate ligand.
(i) Define monodentate ligand.
…… [2]
(ii) Cu2+(aq) reacts with CH3NH2 to form [Cu(CH3NH2)2(H2O)4]2+.
Draw three-dimensional diagrams to show the two geometrical isomers of
[Cu(CH3NH2)2(H2O)4]2+.
Cu Cu
[2]
(iii) State the coordination number of copper in [Cu(CH3NH2)2(H2O)4]2+.
…… [1]
(f) Cd2+(aq) ions form tetrahedral complexes with CH3NH2, OH– and Cl – ions, as shown in equilibria 1, 2 and 3.
equilibrium 1 Cd2+(aq) + 4CH3NH2(aq) [Cd(CH3NH2)4]2+(aq) Kstab = 3.3 × 106
equilibrium 2 Cd2+(aq) + 4OH–(aq) Cd(OH)42–(aq) Kstab = 5.0 × 108
equilibrium 3 Cd2+(aq) + 4Cl –(aq) CdCl 42–(aq) Kstab = 6.3 × 102
(i) Give the units of Kstab for equilibrium 1.
…… [1]
(ii) Write an expression for Kstab for equilibrium 3.
Kstab =
[1]
(iii) A solution of Cl –(aq) is added to Cd2+(aq) and allowed to reach equilibrium. The equilibrium concentrations are given.
[Cd2+(aq)] = 0.043 mol dm–3
[Cl –(aq)] = 0.072 mol dm–3
Use your expression in (f)(ii) to calculate the concentration of CdCl 42–(aq) in the equilibrium mixture.
[CdCl 42–(aq)] = …… mol dm–3 [1]
(iv) When CH3NH2(aq) is added to Cd2+(aq), a mixture of [Cd(CH3NH2)4]2+(aq) and
[Cd(OH)4]2–(aq) forms.
Suggest how the [Cd(OH)4]2–(aq) is formed.
…… [1]
(v) Cd2+(aq) exists as a complex ion, [Cd(H2O)6]2+(aq).
Identify the most stable and the least stable of the complexes in Table 4.1 by placing one tick (3) in each column.
Explain your answer.
Table 4.1
complex most stable least stable
[Cd(H2O)6]2+(aq)
[Cd(OH)4]2–(aq)
[Cd(CH3NH2)4]2+(aq)
[CdCl4]2–(aq)
explanation ……
…… [2]
[Total: 27]
Mark scheme
4(a)(i) CH3CONH2 + 4[H] → CH3CH2NH2 + H2O 1
4(a)(ii) nucleophilic substitution 1
(ammonia with) ethanol AND heat under pressure 1
OR ethanol AND heat in a sealed tube
4(a)(iii) (CH3CH2)2NH OR (CH3CH2)3N 1
4(b) (least) phenylamine < ammonia < ethylamine (most) 1
explanation 3
• (order of basicity) ability of base AND to accept a proton
OR donate its lone pair (to a proton)
phenylamine
• lone pair / p-orbital from N delocalised / overlaps with (π-)ring / benzene
ethylamine
• alkyl / ethyl group is electron donating group / +I group
• increases electron density on N (ethylamine) ORA
any two [1] any three [2] all four [3]
4(c) 1
concentrated HNO3 and H2SO4 (and 25–60 °C) 1
(reduction with) Sn and concentrated HCl (heat) 1
4(d)(i) 1
4(d)(ii) nitrous acid / HNO2 1
OR NaNO2 AND dilute HCl
4(d)(iii) phenol 1
4(d)(iv) NaOH / alkali 1
4(d)(v) dyestuffs / dyes 1
4(e)(i) species that uses one lone pair of electrons 1
that forms a single dative covalent bond to a central metal atom / ion 1
4(e)(ii)
correct 3D cis isomer 1
correct 3D trans isomer 1
4(e)(iii) 6 1
4(f)(i) units = mol–4 dm12 1
4(f)(ii) CdCl 4 2− 1
(Kstab =)
4
Cd2+ Cl −
4(f)(iii) [CdCl42–] = Kstab × 0.043 × 0.0724 = 7.28 / 7.3 × 10–4 (mol dm–3) min 2sf 1
4(f)(iv) CH3NH2 (is basic so) reacts with water to produce OH– that reacts with Cd2+ 1
OR
CH3NH2 acts as a base and accepts a proton from Cd[H2O)6]2+ / water
OR
[Cd(H2O)6]2+ + 4CH3NH2 → [Cd(OH)4(H2O)2]2– + 4CH3NH3+ OWTTE
4(f)(v) Cd(OH)42– = most stable (2nd box) AND [Cd(H2O)6]2+ = least stable (first box) 1
Cd(OH)42– has highest Kstab (and all Kstab values given > 1) 1
Tulobuterol is used in some medicines.
tulobuterol
Cl OH
H
N
Fig. 5.1
(a) Tulobuterol contains a benzene ring in its structure.
Describe and explain the shape of benzene.
In your answer, include:
• the bond angle between carbon atoms
• the hybridisation of the carbon atoms
• how orbital overlap forms v and r bonds between the carbon atoms.
…… [3]
(b) In a synthesis of tulobuterol, the first step involves the formation of chlorobenzene. Benzene reacts with Cl2 in the presence of an Al Cl3 catalyst.
Cl
step 1
Cl2 and AlCl3
Fig. 5.2
(i) Write an equation to show how Cl2 reacts with Al Cl3 to generate an electrophile.
…… [1]
(ii) Complete the mechanism in Fig. 5.3 for the reaction of benzene with the electrophile generated in (b)(i).
Include all relevant curly arrows and charges.
Draw the structure of the intermediate.
intermediate
Cl
Fig. 5.3
[3]
(c) The second step of the synthesis involves the reaction of chlorobenzene with Cl COCH2Cl, also in the presence of an Al Cl 3 catalyst, forming compound Q.
Q
Cl Cl O step 2
Cl
O
Cl
Cl and AlCl3
Fig. 5.4
(i) Name the mechanism of the reaction in step 2.
…… [1]
(ii) Draw the structure of an isomer of Q that forms as an organic by-product of the reaction in step 2.
[1]
(iii) The reactants used in step 2 contain acyl chloride, alkyl chloride and aryl chloride functional groups.
State and explain the relative ease of hydrolysis of acyl chlorides, alkyl chlorides and aryl chlorides.
…… , …… , ……
easiest to hydrolyse hardest to hydrolyse
…… [3]
(d) Tulobuterol is produced from Q as shown in Fig. 5.5.
Q tulobuterol
Cl O Cl OH step 3 step 4
H
Cl Z N
Fig. 5.5
Suggest reagents and conditions for steps 3 and 4.
Draw the structure of Z in the box.
step 3 ……
step 4 ……
Z
[3]
(e) The synthesis produces two enantiomers of tulobuterol.
(i) Define enantiomers.
…… [1]
(ii) Suggest one disadvantage of producing two enantiomers in this synthesis.
…… [1]
(iii) Suggest a method of adapting the synthesis to produce a single enantiomer.
…… [1]
tulobuterol
Cl OH
H
N
(f) (i) Predict the number of peaks that would be seen in the carbon-13 NMR spectrum of tulobuterol.
…… [1]
(ii) The proton (1H) NMR spectrum of tulobuterol dissolved in D2O shows peaks in four different types of proton environment.
The peak for the —CH2N— environment is a doublet in the chemical shift range d = 2.0–3.0 ppm.
Give details for each of the other three peaks in the proton NMR spectrum of tulobuterol, to include:
• chemical shift
• environment of the proton
• splitting pattern
• number of 1H atoms responsible.
Table 5.1 gives information about typical chemical shift values.
…… [3]
Table 5.1
environment of proton example chemical shift range d / ppm alkane –CH3, –CH2–, >CH– 0.9–1.7 alkyl next to C=O CH3–C=O, –CH2–C=O, 2.2–3.0
>CH–C=O alkyl next to aromatic ring CH3–Ar, –CH2–Ar, 2.3–3.0
>CH–Ar alkyl next to electronegative atom CH3–O, –CH2–O, 3.2–4.0
–CH2–Cl attached to alkene =CHR 4.5–6.0 attached to aromatic ring H–Ar 6.0–9.0 aldehyde HCOR 9.3–10.5 alcohol ROH 0.5–6.0 phenol Ar–OH 4.5–7.0 carboxylic acid RCOOH 9.0–13.0 alkyl amine R–NH– 1.0–5.0 aryl amine Ar–NH2 3.0–6.0 amide RCONHR 5.0–12.0
[Total: 22]
Mark scheme
5(a) any three points from: 3
• bond angle = 120°
AND shape is hexagonal ring planar / trigonal planar
• (carbons are) sp2 hybridised
• contains delocalised electrons in the π bonds / system
• (sp2 orbitals) overlap end-on-end/ head-on to form σ bonds
• a p orbital (from each carbon atom) overlaps sideways
(with each other above and below the ring) forming π bonds
5(b)(i) Cl2 + AlCl3 → Cl+ + AlCl4– 1
5(b)(ii) 3
+ H+
M1 curly arrow from inside hexagon of aromatic ring to Cl +
CON going to a lone pair on Cl
M2 intermediate
M3 curly arrow from C—H bond to inside the ring AND H+
5(c)(i) electrophilic substitution 1
5(c)(ii) 1
OR
4-substituted acyl / alkyl derivative
ALLOW
3-substituted acyl / alkyl derivative
5(c)(iii) (most) acyl chloride > alkyl chloride > aryl chloride (least) 1
any two from: 2
• acyl chlorides: carbon (in C—Cl) is more electron deficient
AND it is also attached to an oxygen atom / two electronegative atoms
OR C—Cl bond is weakest / weakened in acyl chlorides
AND it is also attached to an oxygen atom / two electronegative atoms
• aryl chlorides (no hydrolysis) C—Cl bond is part of delocalised system / partially double bond character so C—Cl
bond is stronger
OR lone pair on Cl delocalises with π ring so C—Cl bond is stronger
• alkyl chlorides carbon atom has a smaller δ+
AND due to the carbon being only attached to one electronegative atom
OR C—Cl bond strengthened
AND by electron donating effect / positive inductive effect of alkyl / R group
5(d) (CH3)3CNH2 (in ethanol) [substitution] } in either order 1
LiAlH4 OR NaBH4 [reduction ] } in either order 1
1
[substitution first] OR [reduction first]
5(e)(i) rotate the plane of polarised light in the opposite direction 1
OR stereoisomers / molecules that are non-superimposable mirror images
5(e)(ii) need to separate the optical isomers to form the pure active isomer 1
OR reduced / different biological activity of ‘other’ enantiomer
OR lower yield of biologically active molecule / desired molecule
5(e)(iii) chiral catalyst OR use of an enzyme 1
5(f)(i) ten / 10 1
5(f)(ii) (δ = 0.9–1.7) 3
• 9H
• singlet
• –CH3 / alkane
(δ = 3.2–4.0) / (δ = 2.3–3.0)
• 1H
• triplet
• –CHO / alkyl next to electronegative atom
OR Ar-CH / alkyl next to aromatic ring
(δ = 6.0–9.0)
• 4H
• multiplet
• H–Ar / attached to aromatic ring
any three [1] any six [2] all nine [3]
A student uses thin-layer chromatography (TLC) to analyse a mixture containing different metal cations. The student repeats the experiment using different solvents.
Fig. 6.1 shows the chromatogram obtained by the student using water as a solvent.
solvent front
5
4
3
M
2
1 baseline
0 cm
Fig. 6.1
(a) (i) Suggest a compound that could be used as the stationary phase in this experiment.
…… [1]
(ii) Table 6.1 shows the Rf values for different metal cations when separated by TLC using water as a solvent.
Table 6.1
cation Rf value (water)
Cd2+(aq) 0.40
Co2+(aq) 0.77
Cu2+(aq) 0.32
Fe3+(aq) 0.12
Hg2+(aq) 0.23
Ni2+(aq) 0.75
Suggest the identity of the cation that causes the spot at M in Fig. 6.1.
Explain your answer.
…… [1]
(b) The student repeats the experiment using butan-1-ol as a solvent.
The metal cations do not travel as far up the TLC plate in this experiment.
Suggest why the metal cations do not move as far up the TLC plate with butan-1-ol as a solvent.
…… [1]
(c) The student sprays the TLC plate in Fig. 6.1 with KSCN(aq).
The colour of some of the spots changes, as some of the metal cations undergo a ligand exchange reaction.
Identify the ligands involved in the ligand exchange reaction.
…… exchanges with ……
[1]
(d) In a third experiment, the pH of the mixture of metal ions is kept constant using a buffer solution.
The student prepares the buffer solution by mixing 20.0 cm3 of 0.150 mol dm–3 KOH(aq) and
50.0 cm3 of 0.100 mol dm–3 C8H5O4K(aq).
C8H5O4K is a weak carboxylic acid that has pKa = 5.40.
C8H5O4K
O
OH
O–K+
O
Fig. 6.2
(i) Complete the equation for the reaction of C8H5O4K(aq) with KOH(aq).
(ii) Calculate the pH of the buffer solution. Show all your working.
pH = …… [4]
[Total: 9]
Important values, constants and standards
molar gas constant R = 8.31 J K–1 mol–1
Faraday constant F = 9.65 × 104 C mol–1
Avogadro constant L = 6.022 × 1023 mol–1 electronic charge e = –1.60 × 10–19 C molar volume of gas Vm = 22.4 dm3 mol–1 at s.t.p. (101 kPa and 273 K)
Vm = 24.0 dm3 mol–1 at room conditions ionic product of water Kw = 1.00 × 10–14 mol2 dm–6 (at 298 K (25 °C)) specific heat capacity of water c = 4.18 kJ kg–1 K–1 (4.18 J g–1 K–1)
The Periodic Table of Elements
Group
1 2 13 14 15 16 17 18
1 2
H He hydrogen helium
Key 1.0 4.0
3 4 atomic number 5 6 7 8 9 10
Li Be atomic symbol B C N O F Ne lithium beryllium name boron carbon nitrogen oxygen fluorine neon
6.9 9.0 relative atomic mass 10.8 12.0 14.0 16.0 19.0 20.2
11 12 13 14 15 16 17 18
Na Mg Al Si P S Cl Ar sodium magnesium aluminium silicon phosphorus sulfur chlorine argon
23.0 24.3 3 4 5 6 7 8 9 10 11 12 27.0 28.1 31.0 32.1 35.5 39.9
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
at www.cambridgeinternational.org after the live examination series.
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr potassium calcium scandium titanium vanadium chromium manganese iron cobalt nickel copper zinc gallium germanium arsenic selenium bromine krypton
39.1 40.1 45.0 47.9 50.9 52.0 54.9 55.8 58.9 58.7 63.5 65.4 69.7 72.6 74.9 79.0 79.9 83.8
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
28
Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe rubidium strontium yttrium zirconium niobium molybdenum technetium ruthenium rhodium palladium silver cadmium indium tin antimony tellurium iodine xenon
85.5 87.6 88.9 91.2 92.9 95.9 – 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3
55 56 57–71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 lanthanoids
Cs Ba Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn caesium barium hafnium tantalum tungsten rhenium osmium iridium platinum gold mercury thallium lead bismuth polonium astatine radon
132.9 137.3 178.5 180.9 183.8 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 – – –
87 88 89–103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 actinoids
Fr Ra Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og francium radium rutherfordium dubnium seaborgium bohrium hassium meitnerium darmstadtium roentgenium copernicium nihonium flerovium moscovium livermorium tennessine oganesson
– – – – – – – – – – – – – – – – –
57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 lanthanoids La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu lanthanum cerium praseodymium neodymium promethium samarium europium gadolinium terbium dysprosium holmium erbium thulium ytterbium lutetium
138.9 140.1 140.9 144.4 – 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.1 175.0
89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 actinoids Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr actinium thorium protactinium uranium neptunium plutonium americium curium berkelium californium einsteinium fermium mendelevium nobelium lawrencium
– 232.0 231.0 238.0 – – – – – – – – – – –
Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download
To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge
Mark scheme
6(a)(i) SiO2 OR Al2O3 OR silica OR alumina OWTTE 1
6(a)(ii) Cd2+ AND Rf of M (= 2/5) = 0.40 / same Rf as in the Table 6.1 1
6(b) metal cations are less soluble in butan-1-ol (than in water) 1
OR metal cations weaker ion-dipole forces with butan-1-ol
6(c) H2O AND SCN– 1
6(d)(i) C8H5O4K + KOH → C8H4O4K2 + H2O OR C8H5O4K + OH– → C8H4O4K– + H2O 1
6(d)(ii) M1 initially: moles of KOH = 0.150 × 20.0 ÷ 1000 = 3.0 × 10–3 1
AND moles of acid HA = 0.100 × 50.0÷1000 = 5.0 × 10–3
M2 at equilibrium: moles of salt KA = 3.0 × 10–3 ecf 1
AND moles of acid HA = 5.0 × 10 – moles of NaOH
–3
= 2.0 × 10–3 ecf
M3 Ka = 10–5.40 = 3.98 × 10–6 1
3.98 × 10–6 = [H+] [3.0 × 10–3] ÷ [2.0 × 10–3]
[H+] = 2.65 × 10–6 ecf
M4 pH = –log(2.65 × 10–6) = 5.58 / 5.6 ecf on a calculated and identified [H+] 1
ALLOW alternative approach using Henderson-Hasselbalch equation for M3 and M4
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