Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
CHEMISTRY 9701/41
Paper 4 A Level Structured Questions May/June 2017
MARK SCHEME
Maximum Mark: 100
Published
This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the
examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the
details of the discussions that took place at an Examiners’ meeting before marking began, which would have
considered the acceptability of alternative answers.
Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for
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Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE®,
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This document consists of 12 printed pages.
© UCLES 2017 [Turn over
9701/41 Cambridge International AS/A Level – Mark Scheme May/June 2017
PUBLISHED
Question Answer Marks
1(a) solubility increases down the group 1
∆Hlatt and ∆Hhyd both decrease 1
or ∆Hlatt and ∆Hhyd both become less exothermic / more endothermic
∆Hlatt decreases / changes more (than ∆Hhyd as OH– being smaller than M2+) 1
∆Hsol becomes more exothermic / more negative / less endothermic / less positive 1
1(b)(i) ∆Hr1 – (538 + 2x230 + 394) = –(1216 + 286) 1
∆Hr1 – 1392 = –1502
∆Hr1 = –110 1
1(b)(ii) let ∆Hf(HCO3–(aq)) = y 1
2y – 538 = –1216 – 394 – 286 – 26
y = –692 1
1(b)(iii) ∆Hr3 –538 – 2(230 + 394) = –538 – 2(692) 1
∆Hr3 = –136
1(b)(iv) ∆Hr3 will be identical to ∆Hr4, / unchanged 1
as the reaction is the same, or: 1
2OH–(aq) + 2CO2(g) → 2HCO3–(aq) or
metal ions stay in solution/metal ions are unchanged / are spectators
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9701/41 Cambridge International AS/A Level – Mark Scheme May/June 2017
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Question Answer Marks
1(c) more gaseous moles are being consumed (in reaction 3) 1
or more CO2 moles are being consumed (in reaction 3)
∆S is therefore expected to be more negative/less positive for reaction 3. 1
Total: 13
Question Answer Marks
2(a)(i) 1+1
O C N H N C O
H
16 electrons on each diagram 1
2(a)(ii) HNC = 115–125° AND NCO = 180° 1
2(a)(iii) cyanic acid, because it’s a stronger / higher bond enthalpy / triple / C≡N / more electrons involved bond 1
2(b)(i) [H+] = √([HNCO]Ka) = √(0.1 × 1.2 × 10–4) or 3.46 × 10–3 1
pH = log [H+] = 2.5 (2.46) 1
2(b)(ii) Na2CO3 + 2(NH2)2CO → 2NaNCO + CO2 + 2NH3 + H2O 1
2(c)(i) (n(OH–) at start = (2 × 0.1 × 30) / 1000 = 6 × 10–3 mol)
(n(OH–) reacted = (0.1 × 20) / 1000 = 2 × 10–3 mol)
n(OH–) remaining = (6–2) × 10–3 = 4 × 10–3 mol, (in 50 cm3) 1
so [OH–]end = (4 × 10–3 × 1000) / 50 = 0.08 mol dm–3 1
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Question Answer Marks
2(c)(ii) [H+] = Kw / [OH–] = (1 × 10–14) / 0.08 = 1.25 × 10–13 mol dm–3 1
so pH = –log(1.25 × 10–13) = 12.9 1
2(c)(iii) curve starts at 2.46 / 2.5 1
vertical portion (end point) at vol added = 10.0 cm3 1
finishes at pH = 12.9 1
2(d)(i) monodentate: (a species that) forms one dative / coordinate bond 1
ligand: a species that uses a lone pair of electrons to form a dative / coordinate bond to a metal atom / metal ion 1
2(d)(ii) [Ag(NCO)2]– or [Ag(OCN)2]– correct formula 1
correct charge 1
2(e)(i) n(BaCO3) =1.66 / 197.3 = 8.4(1) × 10–3 mol 1
2(e)(ii) n(RNCO) = 8.41 × 10–3 mol, so Mr = 1 / (8.41 × 10–3) = 119 1
2(e)(iii) molecular formula = C7H5NO 1
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Question Answer Marks
2(e)(iv) 1
Total: 23
Question Answer Marks
3(a)(i) +3 or Co3+ 1
3(a)(ii) oxidation 1
ligand displacement / replacement / exchange / substitution 1
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Question Answer Marks
3(a)(iii) 1+1
NH3 NH3 Cl NH3
H3N NH3 H3N Cl H3N NH3 H3N Cl
Co or Co Co or Co
H3N Cl H3N Cl H3N NH3 Cl NH3
Cl NH3 Cl NH3
cis trans
geometrical or cis-trans 1
3(b)(i) The number of bonds / atoms bonded to an atom / ion / species / metal 1
3(b)(ii) C 6 [Cr(CN)6] – 6
D – [Ni(NH2CH2CH2NH2)3] 2+/+2
E 4 [PtCl4] –
F 3 – 3–/–3
3(c)(i) Kstab(1) = [FeSCN2+]/([Fe3+][SCN–]) mol–1 dm3 3
Kstab(2) = [FeCl4–]/([Fe3+][Cl –]4) mol–4 dm12
3(c)(ii) Keq(3) = Kstab(1) / Kstab(2) 1
3(c)(iii) Keq(3) = 1750 1
mol3 dm–9 1
Total: 19
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Question Answer Marks
4(a)(i) optical, because it contains a / one chiral C-atom or chiral C-atoms or chiral atom / centre or C* indicated or C with 4 different 1
groups
4(a)(ii) C10H14O + 3H2 → C10H20O correct formulae 1
balancing 1
4(b)(i) electrophilic substitution 1
4(b)(ii) step 3 reduction 1
step 5 substitution / hydrolysis 1
4(b)(iii) step 1 (CH3)2CHCl + Al Cl3 / Al Br3 / FeCl3 / FeBr3 1+1
step 2 HNO3 + H2SO4 conc (T < 55 °C) 1
step 3 Sn + HCl 1
step 4 HNO2 (or NaNO2 + HCl ) (at T < 10 °C) 1
the two temperatures for steps 2 and 4 1
4(c)(i) H2 + Pt or H2 + Ni + heat or pressure 1
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Question Answer Marks
4(c)(ii) H CH3 1
H
OH
H
CH(CH3)2
(CH3)2CH, CH3 and OH on the correct ring atoms i.e. structure is correct
all Hs on the same side of the ring 1
Total: 15
Question Answer Marks
5(a) J K L M
amine aromatic amine amine
amide
methyl ketone aldehyde methyl ketone
J and L correct 1+1
K correct 1+1
M correct 1
5(b)(i) hydrolysis 1
5(b)(ii) P is C6H5NH2 1
Q is CH3CH2CO2Na 1
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Question Answer Marks
5(c) O 1
NH NHCH3 NH2
J is or or
O O
1
CHO
K is
NH2
NH2 1
L is O
H 1
N
M is O
K&L only: two chiral atoms shown 1
5(d) W is C6H5CO2Na 1
Total: 14
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Question Answer Marks
6(a) 4
Any of the three methods possible. Any 4 of the 5 points for each method available for maximum 4 marks.
Method 1
1 Ensure both solutions (A and B) at 40 °C before mixing
2 mix known volumes of A and B and start the clock
3 at known time take out a sample / X and add it to ice-cold solvent
4 titrate against HCl
5 repeat at time at known time intervals
Method 2
1 Ensure both solutions (A and B) at 40 °C before mixing
2 mix known volumes of A and B and start the clock
3 at known time pour into ice-cold solvent or pour ice-cold solvent in
4 titrate against HCl
5 repeat with different concentrations of either A or B, or repeat using different times
Method 3
1 Ensure both solutions (A and B) at 40 °C before mixing
2 mix known volumes of A and B and start the clock and add pH meter
3 at a known time . . . .
4 . . . . record the pH
5 repeat pH readings at known time intervals
6(b)(i) from 1 and 3: when [RCl ] is trebled, so is rate, so order w.r.t. [RCl ] = 1 1
from 1 and 2: when both concentrations are doubled, rate doubles so [OH–] has no effect on rate, so order w.r.t.[OH–] = 0 1
6(b)(ii) rate = k[RCl ] AND units: sec–1 1 / s 1
6(b)(iii) relative rate = 2.0 1
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Question Answer Marks
6(c)(i)
C-Cl dipole and first curly arrow 1
intermediate cation 1
OH– with lone pair and curly arrow 1
6(c)(ii) Beginning with candidate’s mechanism in (c)(i): 1
If SN1: racemate / mixture of / two optical isomers will be formed, because:
the intermediate is planar / has a plane of symmetry / OH– can approach from top or bottom or from any direction
If SN2: one optical isomer because attack always from fixed direction / from same side / the “configuration” always inverts /
there is an asymmetric transition state
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Question Answer Marks
6(d)(i) δ value number of H atoms group splitting result with D2O
1.4 3 CH3 / methyl doublet peak remains
2.7 1 OH / hydroxyl / alcohol singlet peak disappears
4.0 1 CH quartet peak remains
the three groups are in their correct places wrt the δ values 1
no. of H atoms for each peak agrees with group column 1
splitting patterns doublet, singlet and quartet are assigned to correct groups 1
peak identified as OH disappears with D2O, no other peak disappears 1
Total: 16
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