A Level Chemistry 9701 Paper 42, May/June 2026

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9701/42 · May/June 2026

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1
Figure 1 of 2 for question 1
Figure 2 of 2 for question 1
a(i)

Transition elements have variable oxidation states and form complex ions. State two other typical chemical properties of a transition element. 1 ........................................................................................................................................ 2 ........................................................................................................................................

[1]
a(ii)

Explain why transition elements have variable oxidation states.

[1]
b(i)

An aqueous solution of chromium(III) ions, Cr3+, contains the [Cr(H2O)6]3+ complex ion. Complete Figure 1.1 to show the electronic configuration for Cr3+.

[1]
b(ii)

State the type of bonding that exists between H2O and Cr3+ in [Cr(H2O)6]3+.

[1]
c

Explain why transition elements form complex ions.

[1]
d(i)

[Cr(H2O)6]3+ can be converted into the [Cr(H2O)3(OH)3] complex. Suggest a suitable reagent for this conversion. State the type of reaction. reagent .............................................................................................................................. type of reaction ..................................................................................................................

[1]
d(ii)

[Cr(H2O)3(OH)3] reacts with an excess of NH3(aq) to form [Cr(NH3)6]3+. Complete the equation for this reaction. State the type of reaction. [Cr(H2O)3(OH)3]+ ........................................................................................................... type of reaction ..................................................................................................................

[2]
e

[Cr(H2O)3(OH)3] shows stereoisomerism. Draw three-dimensional diagrams in Figure 1.2 to suggest the two stereoisomers of [Cr(H2O)3(OH)3].

[2]
f(i)

An aqueous solution of iron(III) ions, Fe3+, contains the [Fe(H2O)6]3+ complex ion. This solution is acidic in a similar way to an aqueous solution of Al3+. Suggest why a solution of Fe3+ is acidic. Include an equation in your answer.

[2]
f(ii)

An aqueous solution of iron(II) ions, Fe2+, contains the [Fe(H2O)6]2+ complex ion. This solution is also acidic. Predict how the pH of 1.00 mol dm−3 Fe2+(aq) differs from the pH of 1.00 mol dm−3 Fe3+(aq). Explain your answer.

[1]
g

Suggest why solutions of silver(I) salts are colourless.

[2]
h

A mixture contains sodium ethanedioate, Na2C2O4, ethanedioic acid dihydrate, H2C2O4⋅2H2O, and an unreactive solid. 5.00 g of this mixture is dissolved in distilled water and made up to 500 cm3 in a volumetric flask to form solution A. Two different titration experiments are carried out using A. In the first experiment, 25.0 cm3 of A requires 21.80 cm3 of acidified 0.0250 mol dm−3 MnO4−(aq) to reach the end-point. 2MnO4−+5C2O42−+16H+→2Mn2++10CO2+8H2O In the second experiment, 25.0 cm3 of A requires 13.60 cm3 of 0.150 mol dm−3 OH−(aq) to reach the end-point. H2C2O4+2OH−→C2O42−+2H2O Calculate the percentage by mass of Na2C2O4 in the mixture. [Mr: Na2C2O4, 134.0] percentage by mass of Na2C2O4= ......................................................%

[5]
2

Group 2 carbonates decompose on heating to form the corresponding metal oxide and carbon dioxide.

Figure 1 of 1 for question 2
a

The mechanism for this decomposition involves the carbonate ion breaking down to form carbon dioxide and oxide ions. Add two curly arrows to the carbonate ion in Figure 2.1 to suggest a mechanism for this decomposition.

[1]
b

The radii of some cations are shown in Table 2.1. Table 2.1

cationCa2+Cu2+Pb2+
radius/nm0.0990.0730.120

Use the data in Table 2.1 to predict the trend in thermal stability of the metal carbonates containing these cations. Explain your answer. .............................. > .............................. > .............................. most stable least stable explanation ............................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...................................................................................................................................................

[2]
c

Suggest why the solubility of the Group 2 sulfates decreases down the group.

[3]
d(i)

Magnesium hydroxide, Mg(OH)2, is sparingly soluble in water at 298 K. Write the expression for the solubility product, Ksp, of Mg(OH)2. State the units for Ksp. Ksp= units ...............................................................

[2]
d(ii)

A saturated solution of Mg(OH)2 has a pH of 10.35 at 298 K. Calculate the concentration of hydroxide ions, OH−, in mol dm−3, in this solution. concentration of OH−= .......................................... mol dm−3

[1]
d(iii)

Use your answers to 2(d)(i) and 2(d)(ii) to calculate the value of Ksp for Mg(OH)2. Ksp for Mg(OH)2= .........................................................

[1]
3
a(i)

Ethanal, CH3CHO, reacts with methanol, CH3OH, under acidic conditions to form compound B, as shown in reaction 1. reaction 1 CH3CHO+2CH3OH→H+CH3CH(OCH3)2+H2O The initial rate of reaction is investigated using different starting concentrations of CH3CHO, CH3OH and H+. The results obtained are shown in Table 3.1. Table 3.1

experiment[CH3CHO]/mol dm−3[CH3OH]/mol dm−3[H+]/mol dm−3initial rate /mol dm−3 s−1
10.5000.0400.1002.500×10−3
20.5000.0600.1003.750×10−3
30.2500.1200.3001.125×10−2
40.2500.1200.4001.500×10−2

Use the information in Table 3.1 to deduce the rate equation for this reaction. Explain your reasoning. State clearly which experiments you use in your explanation. rate =

[4]
a(ii)

State the units of the rate constant, k, for this reaction.

[1]
a(iii)

Another experiment to investigate reaction 1 is carried out as shown in Table 3.2. Table 3.2

experiment[CH3CHO]/mol dm−3[CH3OH]/mol dm−3[H+]/mol dm−3
50.2500.2500.250

Calculate the initial rate of experiment 5. initial rate = ..................................... mol dm−3 s−1

[1]
b

The reaction of propanone, CH3COCH3, with bromine, Br2, in the presence of an acid catalyst is shown in reaction 2. reaction 2 CH3COCH3+Br2→H+CH3COCH2Br+HBr The rate of reaction is first order with respect to [CH3COCH3] and first order with respect to [H+] under certain conditions. A three-step mechanism for reaction 2 is suggested. Complete Table 3.3 to suggest steps 1 and 3 of the mechanism.

[2]
c

Catalysts can be described as homogeneous or heterogeneous. Describe the mode of action of a heterogeneous catalyst.

[2]
4
a

State what is meant by partition coefficient, Kpc.

[1]
b

Ethanedioic acid, H2C2O4, is more soluble in water than in the organic solvent ethoxyethane. 2.70 g of ethanedioic acid is shaken with a mixture of 50 cm3 of ethoxyethane and 50 cm3 of water at 298 K and left until there is no further change in concentrations. The ethoxyethane layer requires 16.5 cm3 of 0.200 mol dm−3 NaOH(aq) for complete neutralisation. H2C2O4+2NaOH→Na2C2O4+2H2O Calculate the partition coefficient, Kpc, for ethanedioic acid between ethoxyethane and water. Kpc= .........................................................

[3]
c

The formula of ethanedioic acid is HOOCCOOH. The formula of butanedioic acid is HOOCCH2CH2COOH. Both acids are soluble in ethoxyethane and in water. Suggest how the value of Kpc of butanedioic acid between ethoxyethane and water compares to the value of Kpc for ethanedioic acid between ethoxyethane and water. Explain your answer.

[2]
5
Figure 1 of 2 for question 5
Figure 2 of 2 for question 5
a

Define standard electrode potential, E⊖, including a description of standard conditions.

[2]
b(i)

Table 5.1 shows the standard electrode potentials for three redox systems. Table 5.1

redox systemelectrode reactionE⊖/V
1Fe3++e−⇌Fe2++0.77
2NO3−+10H++8e−⇌NH4++3H2O+0.87
3Zn2++2e−⇌Zn-0.76

Use the information in Table 5.1 to write equations for two reactions that are feasible. Explain why these reactions are feasible. equation 1 ......................................................................................................................... equation 2 ......................................................................................................................... explanation ........................................................................................................................ ........................................................................................................................................... ...........................................................................................................................................

[3]
b(ii)

Suggest why the feasible reactions in 5(b)(i) may not take place under standard conditions.

[1]
b(iii)

An electrochemical cell is set up to measure E⊖ for redox system 2 in Table 5.1. Complete Figure 5.1 to show a labelled diagram of this electrochemical cell. Include all necessary substances and relevant pieces of apparatus needed to measure E⊖. It is not necessary to state the conditions used.

[3]
c

A different electrochemical cell is set up. The cell reaction for this electrochemical cell is shown. Cr+3Mn3+→3Mn2++Cr3+ΔG⊖=−645.6 kJ mol−1 Calculate the standard cell potential, Ecell⊖, for this cell reaction. Ecell⊖= ...................................................... V

[2]
6
Figure 1 of 6 for question 6
Figure 2 of 6 for question 6
Figure 3 of 6 for question 6
Figure 4 of 6 for question 6
Figure 5 of 6 for question 6
Figure 6 of 6 for question 6
a(i)

Phenol reacts with Br2(aq). State all the expected observations when Br2(aq) is added dropwise to a solution of phenol.

[1]
a(ii)

Draw the structure of the organic product formed when phenol reacts with an excess of Br2(aq).

[1]
b

Explain the relative acidities of phenol and water.

[2]
c(i)

Compound C can undergo both addition and condensation polymerisation. Draw one repeat unit of the addition polymer formed from C.

[1]
c(ii)

Draw two repeat units of the condensation polymer formed from C. The new functional group formed should be displayed.

[2]
d

C can react with LiAlH4. C can also react with an excess of H2(g) with a nickel catalyst. Draw the structure of the organic product formed in each reaction in Figure 6.2.

[2]
e(i)

Compound F can be synthesised from phenol in three steps, as shown in Figure 6.3. Draw the structures of compounds D and E in the boxes in Figure 6.3.

[2]
e(ii)

Suggest reagents and conditions for steps 1 to 3 in Figure 6.3. step 1 ................................................................................................................................ step 2 ................................................................................................................................ step 3 ................................................................................................................................

[3]
7
Figure 1 of 2 for question 7
Figure 2 of 2 for question 7
a(i)

Alanine, CH3CH(NH2)COOH, is an amino acid that contains a chiral centre. Different synthetic routes to produce alanine can lead to the formation of an optically active sample or a racemic mixture. Describe what is meant by optically active.

[1]
a(ii)

Describe what is meant by racemic mixture.

[1]
b(i)

Alanine has an isoelectric point of 6.11. Explain what is meant by isoelectric point.

[1]
b(ii)

Draw the structure of alanine at pH 10.

[1]
c

Alanine molecules can react together to form the tripeptide Ala–Ala–Ala. Draw the structure for this tripeptide. The functional groups formed should be displayed.

[2]
d(i)

Compound J can be synthesised from alanine in three steps, as shown in Figure 7.1. Draw the structures of compounds G and H in the boxes in Figure 7.1.

[2]
d(ii)

Suggest reagents and conditions for steps 1 to 3 in Figure 7.1. step 1 ................................................................................................................................ step 2 ................................................................................................................................ step 3 ................................................................................................................................

[3]
8
Figure 1 of 3 for question 8
Figure 2 of 3 for question 8
Figure 3 of 3 for question 8
a

Give the full name of the substance used in NMR spectroscopy as the standard for chemical shift measurements.

[1]
b(i)

Compound K is analysed by carbon-13 NMR and proton (1H) NMR spectroscopy in D2O. Predict the number of peaks in the carbon-13 NMR spectrum of K.

[1]
b(ii)

The proton (1H) NMR spectrum of K in D2O shows four peaks, for the proton environments labelled w, x, y and z on the structure of K in Figure 8.2. Complete Table 8.1 to identify the splitting patterns for proton environments w, x, y and z.

[2]
b(iii)

A separate sample of K is dissolved in CDCl3 and analysed using proton (1H) NMR spectroscopy. The proton (1H) NMR spectrum of K dissolved in CDCl3 contains one more peak than the spectrum of K dissolved in D2O. Explain why there is one more peak, and suggest the chemical shift range in which it occurs. chemical shift range δ= ........................................................ ppm

[2]
c

Compound K reacts with an excess of hot NaOH(aq). The products are isolated from the reaction mixture at pH 10. Draw the structures of the two organic products of the complete alkaline hydrolysis of K.

[2]
9
Figure 1 of 10 for question 9
Figure 2 of 10 for question 9
Figure 3 of 10 for question 9
Figure 4 of 10 for question 9
Figure 5 of 10 for question 9
Figure 6 of 10 for question 9
Figure 7 of 10 for question 9
Figure 8 of 10 for question 9
Figure 9 of 10 for question 9
Figure 10 of 10 for question 9
a

Compound P and compound Q are isomers. State the relative basicities of P, Q and ammonia. Explain your answer. .............................. > .............................. > .............................. most basic least basic explanation ............................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...................................................................................................................................................

[4]
b(i)

P can be prepared from methylbenzene in two steps, as shown in Figure 9.2. Suggest reagents and conditions for steps 1 and 2 in Figure 9.2. step 1 ................................................................................................................................ step 2 ................................................................................................................................

[2]
b(ii)

Complete the equation in Figure 9.3 for step 2. Use [H] to represent an atom of hydrogen from the reducing agent.

[1]
c(i)

Figure 9.4 shows a two-step synthesis of azo dye N. Draw the structure of compound M in the box in Figure 9.4.

[1]
c(ii)

Give the reagents and conditions for steps 1 and 2 in Figure 9.4. step 1 ................................................................................................................................ step 2 ................................................................................................................................

[2]

Mark scheme

9701/42 · May/June 2026

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1a(i) · 1 mark

  • 1 mark

    they behave as catalysts AND (they form) coloured compounds / complexes / ions BOTH

1a(ii) · 1 mark

  • 1 mark

    the d and s sub-shells / orbitals are close / similar in energy / electrons in d and s sub-shells / orbitals available for bonding

1b(i) · 1 mark

  • 1 mark

    3d: [↑][↑][↑][ ][ ] 4s: [ ]

1b(ii) · 1 mark

  • 1 mark

    dative (covalent) / coordinate

1c · 1 mark

  • 1 mark

    vacant (d) orbital(s) are energetically accessible or vacant (d) orbital(s) can form dative bonds with ligands or vacant (d) orbital(s) can accept a (lone) pair of electrons to form a dative bond

1d(i) · 1 mark

  • 1 mark

    NaOH / OH– and precipitation / ligand exchange / neutralisation / deprotonation / acid-base BOTH

1d(ii) · 2 marks

  • M1

    [Cr(H2O)3(OH)3] + 6NH3 → [Cr(NH3)6]3+ + 3H2O + 3OH–

  • M2

    ligand exchange / replacement / substitution / displacement

1e · 2 marks

  • M1

    one correct 3D structure of [Cr(H2O)3(OH)3]

  • M2

    correct 3D structure of the other isomer

1f(i) · 2 marks

  • M1

    [Fe(H2O)6]3+ / Fe3+ weakens O–H bonds / polarises O–H bonds

  • M2

    [Fe(H2O)6]3+ → [Fe(H2O)5OH]2+ + H+

1f(ii) · 1 mark

  • 1 mark

    Fe3+ has lower pH / Fe2+ has a higher pH and Fe3+ has a higher charge / smaller radii / higher charge density (so easier to hydrolyse) BOTH ora

1g · 2 marks

  • M1

    (Ag+) d-subshell is full / complete or d10 or d-orbitals are full

  • M2

    no d-d(*) transition or no d electrons promoted / excited or cannot absorb light in the visible spectrum

1h · 5 marks

  • M1

    any two of: • moles MnO4– = 0.025 × 0.02180 = 5.45 × 10–4 • moles C2O42– = 2.5 × 5.45 × 10–4 = 1.3625 × 10–3 ecf (moles MnO4– × 2.5) • moles OH– = 0.150 × 0.0136 = 2.04 × 10–3 • moles H2C2O4 = ½ × 2.04 × 10–3 = 1.02 × 10–3 ecf (moles OH– x ½)

  • M2

    all four of the above mole calculations correct

  • M3

    moles Na2C2O4 = 1.3625 × 10–3 – 1.02 × 10–3 = 3.425 × 10–4 ecf (in 25 cm3 A)

  • M4

    moles Na2C2O4 = 3.425 × 10–4 × 20 = 6.85 × 10–3 ecf (in 500 cm3 A) AND mass Na2C2O4 = 6.85 × 10–3 × 134 = 0.9179 g

  • M5

    % = 0.9179 / 5.00 × 100 = 18.4 / 18.358 min 2sf ecf

2a · 1 mark

  • 1 mark

    BOTH arrows (ARROW 1 from O– to C–O bond; ARROW 2 from C–O bond to the other O–)

2b · 2 marks

  • 1 mark

    PbCO3(most) CaCO3 CuCO3 (least)

  • 1 mark

    (stability increases) as (ionic) radii increases / size of M2+ increases / or Pb2+ largest radii ora AND less polarisation / distortion of carbonate ion / anion / CO3(2)– or less weakening of CO bond / C-O / C=O

2c · 3 marks

  • M1

    ΔHlatt and ΔHhyd (both) become less exothermic / less negative

  • M2

    ΔHhyd changes more / changes faster / dominant factor / becomes less exothermic by a larger extent ora

  • M3

    ΔHsol becomes less exothermic / less negative or ΔHsol becomes more endothermic / more positive

2d(i) · 2 marks

  • M1

    (Ksp for Mg(OH)2) = [Mg2+][OH–]2

  • M2

    units mol3 dm–9 ecf from M1

2d(ii) · 1 mark

  • 1 mark

    [H+] = 10–10.35 = 4.47 × 10–11 AND [OH–] = 1 × 10–14 ÷ 4.47 × 10–11 = 2.24 × 10–4 min 2sf

2d(iii) · 1 mark

  • 1 mark

    Ksp = (0.5 × 2.24 × 10–4) × (2.24 × 10–4)2 = 5.61(2) / 5.62 × 10–12 u / c min 2sf ecf for 2(d)(i) and 2(d)(ii)

3a(i) · 4 marks

  • M1

    Exp 1 & 2 [CH3OH] × 1.5 rate × 1.5 or [CH3OH] × 2 / 3 rate × 2 / 3 and so first order to [CH3OH] u / c

  • M2

    Exp 3 & 4 [H+] × 1.33 rate × 1.33 or [H+] × 3 / 4 rate × 3 / 4 and so first order to [H+] u / c

  • M3

    Exp 2 & 3 [CH3CHO] × 0.5, [CH3OH] × 2 and [H+] × 3 rate × 3 and so first order to [CH3CHO] ecf

  • M4

    Rate = k[CH3OH][H+][CH3CHO] u / c ecf or rate = 1.25 [CH3OH][H+][CH3CHO]

3a(ii) · 1 mark

  • 1 mark

    mol–2dm6s–1 u / c ecf from 3(a)(i)

3a(iii) · 1 mark

  • 1 mark

    initial rate = 0.0195 u / c ecf from 3(a)(i) min 2sf

3b · 2 marks

  • 1 mark

    step 1: protonated propanone H3C—C(=+OH)—CH3

  • 1 mark

    step 3: bromo-propanone H3C—C(=O)—CH2Br + HBr

3c · 2 marks

  • 1 mark

    (reactants) adsorb (to surface of catalyst)

  • 1 mark

    bonds (in reactant) weaken AND (reaction occurs and the) product(s) are desorbed / reaction occurs and substances are desorbed

4a · 1 mark

  • 1 mark

    (the) ratio of the concentrations of a solute / substance / compound between two (immiscible) solvents / two liquids (at) equilibrium

4b · 3 marks

  • M1

    Solution 1: moles of (COOH)2 in ether = 0.00165 (subsumes moles of NaOH = 0.0033) OR Solution 2: mass of (COOH)2 in ether = 0.1485 g

  • M2

    Solution 1: moles of (COOH)2 in water = 0.0300 – 0.00165 = 0.02835 ecf OR Solution 2: mass of (COOH)2 in water = 2.70 – 0.1485 = 2.5515 g ecf

  • M3

    Kpc = 0.00165 / 0.02835 = 0.058(2) min 2sf ecf from bullets 3 & 4 only

4c · 2 marks

  • 1 mark

    Kpc would be higher / larger (if Kpc < 1 in 4(b)) (for butanedioic acid) (ecf from 4(b)) ora

  • 1 mark

    butanedioic acid is less polar / more non-polar ora AND so less soluble in water / more soluble in ethoxyethane ora (or dissolves less in water / dissolves more in ethoxyethane)

5a · 2 marks

  • M1

    emf / potential difference / voltage and between / compared to and a standard hydrogen electrode and another electrode

  • M2

    under standard conditions of 1 atm / 101 kPa / 100 kPa, 298 K / 25 °C, (all solutions being) 1M / 1 mol dm–3 u / c

5b(i) · 3 marks

  • M1

    any one or two of the three equations correct

  • M2

    all three equations correct: • Zn + 2Fe3+ → Zn2+ + 2Fe2+ • NO3– + 10H+ + 8Fe2+ → NH4+ + 3H2O + 8Fe3+ • NO3– + 10H+ + 4Zn → NH4+ + 3H2O + 4Zn2+

  • M3

    Ecell is positive (to be feasible) or Eo of Fe3+ / Fe2+ and NO3– / NH4+ is more positive or Eo of redox system 1 and 2 are more positive ora u / c

5b(ii) · 1 mark

  • 1 mark

    high Ea for the reaction OWTTE

5b(iii) · 3 marks

  • 1 mark

    any 3 of the 9 marking features correct

  • 1 mark

    any 6 of the 9 marking features correct

  • 1 mark

    all 9 marking features correct

5c · 2 marks

  • M1

    ΔGo = -nEcellF AND –645.6 × 1000 = –3 × E × 96 500

  • M2

    Ecell = (+)2.23 (V) min 2sf

6a(i) · 1 mark

  • 1 mark

    white ppt. and (solution) decolourises / turns colourless / orange colour disappears

6a(ii) · 1 mark

  • 1 mark

    structure of 2,4,6-tribromophenol

6b · 2 marks

  • M1

    lone pair on oxygen / O delocalised / overlaps and into the ring / π system / delocalised system or p-orbital on oxygen / O delocalised / overlaps and with ring / π system / delocalised system

  • M2

    weakens O—H and phenol is more acidic or (phenol forms) anion / conjugate base / phenoxide ion is more stable and phenol is more acidic

6c(i) · 1 mark

  • 1 mark

    repeat unit with 1,2-substituted phenol and —COOH group

6c(ii) · 2 marks

  • M1

    ester linkage shown between phenol & COOH

  • M2

    rest of the structure correct

6d · 2 marks

  • M1

    structure of 2-(3-hydroxyprop-1-en-1-yl)phenol

  • M2

    structure of 2-(3-hydroxypropyl)phenol / 3-(2-hydroxyphenyl)propanoic acid

6e(i) · 2 marks

  • M1

    correct structure of D

  • M2

    correct structure of E

6e(ii) · 3 marks

  • 1 mark

    step 1: CH3Cl (and AlCl3) CON (aq), CON other reagents

  • 1 mark

    step 2: hot / reflux / heat / boil / T ⩾ 60 °C and (alkaline / acidified / neutral) MnO4– / KMnO4

  • 1 mark

    step 3: CH3COCl CON (aq) CON any other reagent e.g. cH2SO4

7a(i) · 1 mark

  • 1 mark

    (a substance able to) rotate plane polarised light

7a(ii) · 1 mark

  • 1 mark

    (a mixture containing) equal amounts / proportions / concentrations and of optical isomers / enantiomers / stereoisomers

7b(i) · 1 mark

  • 1 mark

    pH at (which a molecule / amino acid) has no overall charge / is neutral / is a zwitterion

7b(ii) · 1 mark

  • 1 mark

    CH3CH(NH2)COO–

7c · 2 marks

  • M1

    two peptide linkage shown (C=O needs to be displayed but allow NH not displayed)

  • M2

    rest of the structure correct DEP on M1 (except for specific ecfs in guidance)

7d(i) · 2 marks

  • M1

    structure of G (CH3CONHCH(CH3)COOH)

  • M2

    structure of H (CH3CONHCH(CH3)COCl)

7d(ii) · 3 marks

  • M1

    step 1: CH3COCl CON (aq), CON additional reagent e.g. AlCl3, cH2SO4

  • M2

    step 2: PCl3 + heat or SOCl2 or PCl5 CON (aq)

  • M3

    step 3: C6H6 and AlCl3 CON (aq)

8a · 1 mark

  • 1 mark

    tetramethylsilane

8b(i) · 1 mark

  • 1 mark

    5 / five

8b(ii) · 2 marks

  • 1 mark

    any two of w = doublet, x = multiplet, y = triplet, z = triplet

  • 1 mark

    all four correct: w = doublet, x = multiplet, y = triplet, z = triplet

8b(iii) · 2 marks

  • M1

    δ 5 –12

  • M2

    (N-H peak appears and) no proton / hydrogen exchanges / replaces / substitutes with deuterium / D / D+ ora

8c · 2 marks

  • M1

    structure of sodium 3-methyl-2-oxobutanoate / 3-methyl-2-oxobutanoate ion

  • M2

    structure of 3-aminopropan-1-ol H2NCH2CH2CH2OH

9a · 4 marks

  • M1

    order of basicity: Q > ammonia > P

  • M2

    (basicity linked to) ability of lone pair / p-orbital to accept / donate to / coordinate with a proton / H+

  • M3

    any one of: • lone pair / p-orbital from N (in P) and delocalised / overlaps with (π-)ring / benzene • electron donating / +I and R / alkyl / CH2 grp (in Q) • increases electron density on N or N is more electron rich or N LP is more available (in Q) ora

  • M4

    all three explanation points correct

9b(i) · 2 marks

  • M1

    step 1: HNO3 and H2SO4 and conc. / concentrated

  • M2

    step 2: Sn and conc. HCl

9b(ii) · 1 mark

  • 1 mark

    6 [H] and 2 H2O

9c(i) · 1 mark

  • 1 mark

    structure of diazonium salt (4-methylbenzenediazonium ion)

9c(ii) · 2 marks

  • M1

    step 1: NaNO2 with HCl or HNO2 and ⩽10 °C

  • M2

    step 2: alkaline / NaOH and sodium 4-hydroxybenzenesulfonate / 4-hydroxybenzenesulfonic acid

Questions

  1. 1
    Question 120 marks
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  2. 2
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