Chemistry 9701/41 October/November 2025: question paper and mark scheme
Cambridge AS & A Level Chemistry (9701), Paper 4 A Level Structured Questions, October/November 2025. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) Solutions of Group 2 hydrogencarbonates, M(HCO3)2, decompose on heating to give the corresponding metal carbonate, carbon dioxide and water.
(i) Write an equation for the decomposition of strontium hydrogencarbonate, Sr(HCO3)2.
…… [1]
(ii) The thermal stability of Group 2 carbonates increases down the group.
Explain this trend.
…… [2]
(b) The hydroxides and fluorides of Group 2 elements show similar trends in solubility.
Describe the trend in the solubility of the fluorides of calcium, strontium and barium.
Explain your answer.
least soluble most soluble
explanation ……
…… [4]
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(c) (i) Define enthalpy change of hydration, ΔHhyd.
…… [1]
(ii) State the main factors that affect the magnitude of enthalpy change of hydration.
Explain your answer.
…… [2]
(d) Table 1.1 shows various energy changes.
Table 1.1
energy change value / kJ mol–1 lattice energy of MgF2 –2957 enthalpy change of hydration, ΔHhyd, of Mg2+ –1926 enthalpy change of hydration, ΔHhyd, of F– –505
Use data from Table 1.1 to calculate the enthalpy change of solution, ΔHsol, for MgF2(s).
It may be helpful to draw a labelled energy cycle. Show your working.
ΔHsol of MgF2(s) = …… kJ mol–1 [2]
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(e) Mercury(I) fluoride, Hg2F2, is sparingly soluble in water.
The cation in Hg2F2 exists as the diatomic ion Hg22+ with a covalent Hg–Hg bond.
(i) Write the expression for the solubility product, Ksp, of Hg2F2. Include the units.
Ksp =
units ……
[2]
(ii) The solubility of Hg2F2 is 9.20 × 10–3 mol dm–3 at 298 K.
Calculate the value of Ksp of Hg2F2 at 298 K.
Ksp = …… [1]
[Total: 15]
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Mark scheme
1(a)(i) Sr(HCO3)2 → SrCO3 + CO2 + H2O 1
1(a)(ii) M1: radius of cation / M2+ increases OR charge density of cation / M2+ decreases 2
M2: less polarisation / less distortion of the anion / carbonate ion
1(b) M1: least soluble CaF 2 < SrF 2 < BaF 2 most soluble 4
M2: ΔHlatt and ΔHhyd both become less exothermic / less negative
M3: ΔHhyd changes less / becomes less exothermic by a smaller extent
OR ΔHlatt changes more / becomes less exothermic by a larger extent
M4: ΔHsol becomes more exothermic / more negative
1(c)(i) enthalpy change when one mole of gaseous ions dissolve in water to form a solution 1
1(c)(ii) M1: ionic radii AND ionic charge 2
M2: (ionic) radii increase / charge density decreases AND ∆Hhyd decreases/less exothermic AND less attraction between
water molecules and (gaseous) ions
OR
as ionic charge increases / charge density increases AND ∆Hhyd increases/more exothermic AND more attraction between
water molecules and (gaseous) ions
1(d) M1: use of –2957 / –1926 / –505 AND 2 × (–505) 2
M2: correct signs and evaluation
∆Hsol of MgF 2(s) = –1926 + (2 × –505) – (–2957) = (+)21 kJ mol–1
1(e)(i) M1: Ksp = [Hg22+] [F –]2 2
M2: units = mol3 dm–9
1(e)(ii) Ksp = 22 × (9.20 × 10–3)3 = 3.11 × 10–6 min 2sf 1
(a) Iron can form stable ions in the +2 and +3 oxidation states.
Explain why transition elements have variable oxidation states.
…… [1]
(b) Aqueous solutions of iron(II) salts contain the complex ion [Fe(H2O)6]2+.
Define complex ion.
…… [1]
(c) [Fe(H2O)6]2+ can be converted into [Fe(H2O)4(OH)2].
(i) Suggest a suitable reagent for this conversion. State the type of reaction.
reagent ……
type of reaction ……
[1]
(ii) [Fe(H2O)4(OH)2] is a green precipitate that turns brown on standing in air.
Table 2.1 shows electrode potentials for some electrode reactions.
Table 2.1
electrode reaction E o /V
Fe(H2O)3(OH)3 + H2O + e– Fe(H2O)4(OH)2 + OH– –0.56
O2 + 2H2O + 4e– 4OH– +0.40
Use the information in Table 2.1 to explain why [Fe(H2O)4(OH)2] turns brown on standing in air.
Include an equation for this reaction.
…… [3]
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(d) The complex [Co(NH3)6]2+ reacts with hydrogen peroxide as shown. o reaction 1 2[Co(NH3)6]2+ + H2O2 2[Co(NH3)6]3+ + 2OH– E cell = +1.67 V
Calculate ΔG o , in kJ mol–1, for reaction 1.
ΔG o = …… kJ mol–1 [2]
[Total: 8]
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Mark scheme
2(a) the (3)d and (4)s sub-shells / orbitals / electrons are close / similar in energy 1
2(b) species or ion formed by a central metal atom/ion AND surrounded by/bonded to (one or more) ligands 1
2(c)(i) NaOH / OH–(aq) AND precipitation / ligand exchange / deprotonation / acid–base 1
2(c)(ii) M1: Fe(H2O)4(OH)2 is oxidised (from + 2 to + 3 on standing) 3
M2: Eo of Fe(H2O)3(OH)3 is more negative than Eo of O2
OR Eocell = 0.40 – (–0.56) = (+)0.96 V
M3: 4Fe(H2O)4(OH)2 + O2 → 4Fe(H2O)3(OH)3 + 2H2O
2(d) ∆Go = –2 × 1.67 × 96 500 = –322 310 J mol–1 2
M1: ∆Go = –nEocellF AND n = 2
OR –2 × 1.67 × 96 500
M2: ∆Go = –322.3 kJ mol–1
(a) Solid manganese(IV) oxide, MnO2, catalyses the decomposition of hydrogen peroxide.
2H2O2(aq) 2H2O(l) + O2(g)
State the type of catalysis for this reaction. Explain your answer.
…… [1]
(b) Hydrogen peroxide reacts with iodide ions in acidic conditions as shown.
H2O2 + 2I– + 2H+ 2H2O + I2
The initial rate of this reaction is investigated with different concentrations of H2O2, I– and H+.
The results obtained are shown in Table 3.1.
Table 3.1
experiment [H2O2] / mol dm–3 [I–] / mol dm–3 [H+] / mol dm–3 initial rate / mol dm–3 s–1
1 0.0450 0.0300 0.0125 2.42 × 10–3
2.42 × 10–3
2 0.0225 0.0600 0.0125
3 0.0225 0.120 0.0125 4.84 × 10–3
4 0.0450 0.120 0.0500 9.68 × 10–3
(i) Use the information in Table 3.1 to deduce the rate equation for this reaction.
Explain your reasoning.
…… [4]
(ii) Use your rate equation from (b)(i) and the data from Experiment 1 to calculate the rate constant, k, for this reaction. Include the units of k.
k = …… units …… [2]
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(c) The rate of the thermal decomposition of azomethane, CH3N=NCH3, is investigated.
CH3N=NCH3 N2 + C2H6
Fig. 3.1 shows the results obtained. The reaction is first order with respect to CH3N=NCH3.
0.06
0.05
0.04
[CH3N=NCH3] / mol dm–3 0.03
0.02
0.01
0
0 50 100 150 200 250 300 350 400 time / s
Fig. 3.1
(i) Use Fig. 3.1 to calculate two half-lives, t 1 , to show that the reaction is first order.
2
…… [2]
(ii) Use your answer to (c)(i) to calculate the rate constant, k, for the decomposition of
azomethane.
k = …… s–1 [1]
(d) Describe the effect of increasing temperature on the rate constant and on the rate of a reaction.
…… [1]
[Total: 11]
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o
Mark scheme
3(a) heterogeneous AND MnO2 is in a different state / phase to the reactants / H2O2 1
3(b)(i) M1: (expt 2 and 3) [I–] × 2 initial rate × 2 so 1st order wrt [I–] 4
M2: (expt 1 and 2) [I–] × 2, [H2O2] × 0.5 initial rate × 1 so 1st order wrt [H2O2]
OR (expt 1 and 4) [I–] × 4 initial rate × 4 AND [H+] × 4 so zero order wrt [H+]
OR (expt 1 and 3) [I–] × 4, [H2O2] × 0.5 initial rate × 2 so 1st order wrt [H2O2]
M3: (expt 1 and 4) [I–] x 4 , [H+] × 4 initial rate × 4 so zero order wrt [H+]
OR (expt 3 and 4) [H2O2] × 2 initial rate × 2AND [H+] x 4 so zero order wrt [H+]
OR (expt 3 and 4) [H2O2] × 2, [H+] × 4 initial rate × 2 so 1st order wrt [H2O2]
M4: rate = k [H2O2] [I–] OR rate = k [H2O2] [I–] [H+]0
3(b)(ii) M1: k = 2.42 × 10–3 ÷ (0.045 × 0.030) = 1.79 min 2sf 2
M2: units = dm3 mol-1 s-1
3(c)(i) M1: calculation of one t1/2 = 60 s ± 5 s 2
M2: two t1/2 calculated that are constant
3(c)(ii) k = In2 ÷ 60 = 0.0116 OR k = 0.693 ÷ 60 = 0.0116 min 2sf 1
3(d) increases k and increases the rate of reaction 1
(a) Define standard cell potential, E cell . Include a description of standard conditions.
…… [2]
(b) The Daniell cell is an electrochemical cell consisting of a Cu2+(aq) / Cu(s) electrode and a
Zn2+(aq) / Zn(s) electrode.
(i) Draw a labelled diagram of this electrochemical cell.
Include all necessary substances and relevant pieces of apparatus needed to measure o the E cell .
It is not necessary to state the conditions used.
[3]
(ii) State the charge carriers that transfer current through the solutions and through the wire.
the solutions …… the wire …… [1]
(iii) The standard electrode potential, E o , for the Zn2+(aq) / Zn(s) electrode is –0.76 V.
Water is added to a standard Zn2+(aq) / Zn(s) electrode.
The new concentration of Zn2+(aq) is 0.25 mol dm–3.
Use the Nernst equation to calculate the electrode potential, E, for this new Zn2+(aq) / Zn(s) electrode.
E (Zn2+(aq) / Zn(s)) = …… V [2]
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(c) An electrochemical cell consists of a ZnO / Zn electrode and a MnO2 / Mn2O3 electrode in an alkaline electrolyte. o
The standard cell potential, E cell , for this cell is +1.47 V.
The half-equation at each electrode when this cell is discharging is shown.
Zn + 2OH– ZnO + H2O + 2e–
2MnO2 + H2O + 2e– Mn2O3 + 2OH–
(i) Use this information to determine the change in oxidation state of manganese when this cell is discharging.
from …… to …… [1]
(ii) Write the equation for the overall reaction that occurs when this cell is discharging.
…… [1]
(iii) The E o for the ZnO / Zn electrode is –1.28 V.
Calculate the standard electrode potential, E o , for the MnO2 / Mn2O3 electrode.
E o (MnO2 / Mn2O3) = …… V [1]
[Total: 11]
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Mark scheme
4(a) M1: potential difference / voltage / EMF AND between two half-cells / two electrodes (in a cell) 2
M2: (at concentration of) 1 mol dm–3 AND (pressure of) 1 atm / 101 kPa AND (temperature of) 298 K / 25 °C
4(b)(i) 3
M1: salt bridge labelled AND voltmeter / V AND wire to electrode and liquid level shown [complete circuit]
M2: Cu(s) AND Cu2+(aq)
M3: Zn(s) AND Zn2+(aq)
4(b)(ii) ions AND electrons 1
4(b)(iii) M1: Nernst equation E = Eo + (0.059/z) log(Zn2+/Zn) 2
M2: E = –0.76 + (0.059 / 2) log(0.25) = –0.778 V
OR E = –0.76 + (8.31 × 298) / (96 500 × 2) ln(0.25) = –0.778 V min 2sf
4(c)(i) (from) +4 (to) +3 1
4(c)(ii) Zn + 2MnO2 → ZnO + Mn2O3 1
4(c)(iii) Eo (MnO2 / Mn2O3) = 1.47 – 1.28 = (+)0.19 V 1
(a) Copper shows typical properties of transition elements, including its behaviour as a catalyst.
Complete Table 5.1 to show the total number of unpaired electrons in the 3d and 4s orbitals of an isolated gaseous Cu atom and a Cu2+ ion.
Table 5.1
number of unpaired electrons species
3d 4s
Cu
Cu2+
[1]
(b) The 3d orbitals in an isolated Cu2+ ion are degenerate.
Complete the diagram to show the relative energies of the 3d orbitals in an isolated Cu2+ ion and in Cu2+ in a tetrahedral complex.
energy
isolated Cu2+ ion Cu2+ in a tetrahedral complex
[2]
(c) Explain why transition elements behave as catalysts.
…… [2]
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(d) CN– is a monodentate ligand.
Table 5.2 shows information about two complex ions that contain only CN– ions as ligands.
Complete Table 5.2.
Table 5.2
metal ion coordination number formula of complex ion charge of complex ion
Ag+ 2
Fe2+ 4–
[2]
(e) The complex ion [Au(CN)2Br2]– displays geometrical (cis / trans) isomerism.
Draw the structure of trans-[Au(CN)2Br2]–. State its shape and the Br-Au-Br bond angle.
shape …… Br-Au-Br bond angle = ……
[2]
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(f) An impure sample of a vanadium(V) compound of mass 0.250 g is dissolved in aqueous acid.
This solution contains VO3– ions.
An excess of zinc is added to this solution. All the VO3– ions are reduced to V2+ ions and Zn atoms are oxidised to Zn2+ ions.
The unreacted zinc is removed and the resulting solution is titrated with acidified MnO4–.
The end-point is reached when 22.5 cm3 of 0.0750 mol dm–3 MnO4– is added.
A redox reaction takes place and all the V2+ reacts forming VO3–.
3MnO4– + 5V2+ + 3H2O 3Mn2+ + 5VO3– + 6H+
(i) Calculate the percentage by mass of vanadium in the 0.250 g of impure sample.
Assume the impurities do not contain any vanadium ions.
Show your working.
percentage of vanadium = …… [3]
(ii) Complete the equation for the reaction between acidified VO3– ions and Zn metal.
…… VO3– + …… Zn + …… …… V2+ + …… Zn2+ + ……
[2]
[Total: 14]
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Mark scheme
5(a) 1
5(b) • five 3d orbitals (lines, boxes) in the isolated Cu 2+ ion of the same energy 2
• splitting: three higher and two lower d orbitals
• energy of all five d orbitals in complex higher than all d orbitals in isolated ion
any two [1], all three [2]
5(c) M1: more than one (stable) oxidation state / exist in variable oxidation states 2
M2: vacant / empty (d) orbitals are energetically accessible
OR vacant / empty (d) orbitals can form dative bonds with ligands
5(d) 2
each row [1]
5(e) M1: 2
M2: (shape) square planar AND (bond angle) 180o
5(f)(i) M1: moles MnO4– = 0.075 × 0.0225 = 1.6875 × 10–3 3
M2: moles V2+ = 5 / 3 × 1.6875 × 10–3 = 2.8125 × 10–3
M3: mass V = 50.9 × 2.8125 × 10–3 = 0.143 g
% of V = 0.143 / 0.250 × 100 = 57.3 min 2sf
5(f)(ii) 2VO3– + 3Zn + 12H+ → 2V2+ + 3Zn2+ + 6H2O 2
M1: 2 : 3 ratio on both sides
M2: rest of the equation correct
(a) Thin-layer and gas / liquid chromatography can be used to separate mixtures into their individual components.
(i) Define the following terms used in chromatography.
Rf value ……
retention time ……
[2]
(ii) Each type of chromatography makes use of a stationary phase and a mobile phase.
Complete Table 6.1 with a description of each of these.
Table 6.1
stationary phase mobile phase
thin-layer chromatography
gas / liquid chromatography
[1]
(b) A mixture of two substances A and B is analysed by thin-layer chromatography.
The Rf value of substance A is larger than that of substance B.
Suggest why substance A has a larger Rf value.
…… [1]
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(c) The two isomeric compounds Y and Z are analysed by proton (1H) NMR spectroscopy.
Y Z
O O
O O
(i) Complete Table 6.2 to predict the number of peaks observed in the proton (1H) NMR spectra for Y and Z.
Table 6.2
compound number of peaks observed
Y
Z
[1]
(ii) Name all the different splitting patterns observed in the proton (1H) NMR spectra for Y
and Z.
Y ……
Z ……
[2]
[Total: 7]
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Mark scheme
6(a)(i) M1: (Rf value) is distance moved by a component / spot / solute AND divided by distance moved by solvent 2
OR (Rf value) is the ratio of distance moved by a component / spot / solute AND by distance moved by solvent
M2: (retention time) is (the time) between injection and detection (of a component)
6(a)(ii) thin-layer chromatography: polar solvent OR non-polar solvent OR named solvent 1
gas-liquid chromatography: non-volatile liquid OR high boiling point liquid
6(b) A is more soluble in the mobile phase OR A has less adsorption to the stationary phase 1
6(c)(i) 1
6(c)(ii) M1: Y (two) singlet(s) 2
M2: Z triplet AND quartet
(a) State the relative acidities of bromoethanoic acid, BrCH2COOH, chloroethanoic acid,
Cl CH2COOH, ethanoic acid, CH3COOH and ethanol, CH3CH2OH.
Explain your answer.
most acidic least acidic
[4]
(b) Fig. 7.1 shows the reaction of methylbenzene and ethanedioic acid with KMnO4.
Predict the major carbon-containing product for each of these reactions.
methylbenzene
hot alkaline KMnO4
ethanedioic acid
O
hot acidified KMnO4
OH
HO
O
Fig. 7.1
[2]
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(c) Polyamide X can be synthesised from ethanedioic acid and benzene-1,4-diamine.
benzene-1,4-diamine
H 2N NH2
(i) Draw the repeat unit of polyamide X in the box.
The new functional group formed should be shown displayed.
polyamide X
[2]
(ii) Benzene-1,4-diamine can be formed by reduction of 1,4-dinitrobenzene.
Complete the equation for this reduction.
[H] represents one atom of hydrogen from a reducing agent.
+ …… [H] H 2N NH2 + ……
[1]
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(d) Fig. 7.2 shows the two-step synthesis of the azo compound W.
V C6H4N4Cl2
step 1
H 2N NH2
OH step 2
and NaOH(aq)
W C18H14N4O2
Fig. 7.2
(i) Suggest structures for compounds V and W and draw them in the boxes in Fig. 7.2. [2]
(ii) Give the reagents and conditions for step 1.
…… [1]
[Total: 12]
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Mark scheme
7(a) M1: Cl CH2COOH BrCH2COOH CH3COOH CH3CH2OH 4
most acidic least acidic
M2: electron withdrawing groups / electronegative groups / negative inductive effect
AND weakens O–H bond / more stable anion
OR
electron donating groups / positive inductive effect AND strengthens O-H bond / less stable anion
M3 M4:
• chlorine is more electron-withdrawing / electronegative than bromine (related to XCH 2COOH)
• electron withdrawing of C=O / negative inductive effect of C=O (related to COOH)
• positive inductive effect / electron donating of alkyl / ethyl / R group (related to ROH)
any two [1], all three [2]
7(b) M1: 2
M2: CO2
7(c)(i) 2
M1: correct displayed amide bond with C=O, (C6H5)−N and (C–)C=O
M2: rest of the structure correct with continuation bonds
7(c)(ii) 1
7(d)(i) 2
M1:
M2:
7(d)(ii) HNO2 (and HCl) AND ⩽10 °C 1
OR
NaNO2 AND HCl AND ⩽10 °C
(a) In the electrophilic substitution of arenes, different substituents can direct to different ring positions.
(i) Describe the directing effect of the –NO2 group. Explain your answer.
…… [1]
(ii) The nitration of arenes uses a mixture of concentrated HNO3 and concentrated H2SO4 to generate the NO2+ electrophile.
Write an equation for the formation of the NO2+ electrophile.
…… [1]
(b) Carbon-carbon bond formation is an important reaction in organic synthesis.
Fig. 8.1 shows the synthesis of compound Q from benzene in two reaction steps.
compound Q compound P O
reaction 1 reaction 2
Fig. 8.1
(i) Draw the structure of compound P in the box in Fig. 8.1. [1]
(ii) Suggest reagents and conditions for reactions 1 and 2 in Fig. 8.1.
reaction 1 ……
reaction 2 ……
[2]
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(c) Separate samples of C6H5Br and C6H5CH2Br are added to warm AgNO3(aq).
State the expected observations, if any. Explain your answer.
C6H5Br with AgNO3(aq) ……
C6H5CH2Br with AgNO3(aq) ……
explanation ……
[3]
(d) Acyl bromides, RCOBr, react readily with H2O.
The mechanism of this reaction is similar to that of the reaction of H2O with acyl chlorides,
RCOCl.
(i) Name the mechanism of this reaction.
…… [1]
(ii) Complete the mechanism in Fig. 8.2 for the reaction of RCOBr with H2O.
Include all relevant lone pairs of electrons, curly arrows, charges and dipoles.
Draw the structure of the intermediate.
intermediate
O
C products
R Br
O
H H
Fig. 8.2
[4]
[Total: 13]
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Mark scheme
8(a)(i) NO2 group directs to 3 (and 5) / meta position AND due to being an electron-withdrawing / electronegative group 1
8(a)(ii) HNO3 + H2SO4 → NO2+ + HSO4– + H2O 1
OR HNO3 + 2H2SO4 → NO2+ + 2HSO4– + H3O+
8(b)(i) 1
8(b)(ii) M1: (reaction 1) ethanoyl chloride / CH3COCl AND Al Cl 3 2
M2: (reaction 2) 2-bromopropane / (CH3)2CHBr AND FeBr3
8(c) M1: 3
with C6H5Br no change / no precipitate AND
with C6H5CH2Br cream precipitate
in C6H5Br
M2: lone pair / p-orbital on Br delocalised / overlaps with ring / π system
M3: C–Br bond stronger / has partially double bond character
8(d)(i) (nucleophilic) addition–elimination 1
8(d)(ii) 4
M1 M2:
• lone pair on O
• curly arrow from (lone pair) O (in H2O) to C (of C=O)
• correct dipole on C=O
• curly arrow from the C=O bond to O atom
any two [1], all four [2]
M3: correct intermediate
M4: curly arrow from (lone pair on) O (–) to C–O bond AND curly arrow from C–Br to Br
(a) Explain why amides are much weaker bases than amines.
…… [2]
(b) Fig. 9.1 shows the preparation of 2-phenylethylamine, C6H5CH2CH2NH2, by three different
routes.
NH2
reaction 1
O
2-phenylethylamine
M
H2 / Ni NH2
NH3
N
Fig. 9.1
(i) Suggest structures for compounds M and N and draw them in the boxes in Fig. 9.1. [2]
(ii) Give the reagents and conditions for reaction 1.
…… [1]
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(c) Fig. 9.2 shows compound H which is a useful starting material in organic synthesis.
H
O
N
H
Fig. 9.2
H contains an alkene and an amine functional group.
Name the other functional group and give the classification of the amine group in H.
other functional group in H ……
classification of amine ……
[1]
(d) Ozonolysis involves the oxidative cleavage of a C=C bond in alkenes using ozone, O3, as shown in Fig. 9.3.
R1 R3 R1 R3
O3
O + O
R2 R4
R2 R4
Fig. 9.3
Fig. 9.4 shows the first step in this reaction which involves the formation of an ozonide intermediate.
ozonide
R2 R3
R1 R3
R1
R4
R2 R4 O O
O
- O
O
O
+
Fig. 9.4
(i) On Fig. 9.4, draw three curly arrows to complete the mechanism of this step. [2]
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(ii) L is formed from alkene K, C8H14, by a similar reaction to that shown in Fig. 9.3.
L
O O
Suggest the structure of K.
K C8H14
[1]
[Total: 9]
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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 = 6.02 × 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
DFD
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
138.9 140.1 140.9 144.2 – 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 – – – – – – – – – – –
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Mark scheme
9(a) M1: (basicity linked to) ability of lone pair / p-orbital to 2
AND accept / coordinate with a proton / H+
M2: (lone pair of) electrons on N is delocalised into C=O group (and make amides neutral)
9(b)(i) M1: M 2
M2: N
9(b)(ii) LiAl H4 1
9(c) ketone / carbonyl AND secondary / 2o 1
9(d)(i) 2
curly arrow 1: from O lone pair to C atom (in C=C)
curly arrow 2: from C=C bond to the right-hand O atom in O=O
curly arrow 3: from O=O bond to O +
9(d)(ii) 1
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Chemistry 9701/41 for October/November 2025 is out of 100 marks, with 2 hours to answer it, across 9 questions.
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