2025 CCEA A-Level Chemistry 1110 Practice Paper with Answers

Lattice enthalpy of formation becomes more exothermic as the ionic charges increase and the ionic radii decrease, since both raise the charge density and strengthen electrostatic attraction. \( Mg^{2+} \) and \( O^{2-} \) carry twice the charge of \( Na^+ \) and \( Cl^- \) and are also smaller, so the attraction between them is much stronger.

1 mark: A.

\( \Delta G = \Delta H - T\Delta S = -572 - (298 \times (-0.327)) = -572 + 97.5 = -474.5 \approx -475 \text{ kJ mol}^{-1} \).

1 mark: A.

Second order overall with first order in A means the reaction is first order with respect to B. Tripling [A] multiplies rate by 3; doubling [B] multiplies rate by 2. Overall factor \( = 3 \times 2 = 6 \).

1 mark: B.

\( K_c \) depends only on temperature. Increasing pressure shifts the position of equilibrium (favouring the side with fewer gas moles) but does not change the value of \( K_c \) itself, since T is unchanged.

1 mark: C.

A buffer requires a weak acid together with a reservoir of its conjugate base (or vice versa) in comparable amounts. Propanoic acid (weak acid) with sodium propanoate (source of the conjugate base \( CH_3CH_2COO^- \)) satisfies this; HCl is a strong acid, so A and D contain no weak acid/conjugate base pair, and C contains no weak acid at all.

1 mark: B.

C2 of butan-2-ol is bonded to four different groups: \( -OH \), \( -H \), \( -CH_3 \) and \( -CH_2CH_3 \), making it a chiral (asymmetric) carbon. This gives rise to two non-superimposable mirror-image forms — optical isomers (enantiomers).

1 mark: B.

Aldehydes such as butanal are readily oxidised (to carboxylic acids) by acidified dichromate(VI), reducing orange \( Cr_2O_7^{2-} \) to green \( Cr^{3+} \). Ketones such as butan-2-one have no hydrogen on the carbonyl carbon and cannot be oxidised under these conditions, so no colour change occurs.

1 mark: B.

In the liquid and vapour phases, carboxylic acid molecules pair up as dimers, each pair held together by two hydrogen bonds (O-H···O=C on each side). This effectively doubles the intermolecular attraction compared with a single hydrogen bond in alcohols, requiring more energy to separate the molecules and so raising the boiling point.

1 mark: A.

In an amide, the lone pair on nitrogen delocalises particularly effectively into the carbonyl \( \pi \) system (partial C-N double bond character), reducing the electrophilicity of the carbonyl carbon and making \( N \) a poor leaving group. This makes amides the least reactive of the three towards nucleophiles.

1 mark: C.

A Kekulé structure predicts three shorter C=C bonds (about 0.134 nm) alternating with three longer C-C bonds (about 0.154 nm). Experimentally, all six bonds in benzene are the same length, about 0.139 nm, intermediate between the two — consistent with a delocalised \( \pi \) system spread evenly around the ring rather than three localised double bonds.

1 mark: B.

First electron affinity is the enthalpy change when one mole of gaseous 1- ions is formed from one mole of gaseous atoms by gaining one electron per atom, under standard conditions. It is always exothermic (negative), since the incoming electron is attracted to the positive nucleus.

1 mark: correct definition (1 mol gaseous atoms gaining 1 electron each to form 1 mol gaseous 1- ions); 1 mark: negative/exothermic correctly stated. [2]

\( \Delta S_{surroundings} \) is positive for an exothermic reaction. Heat is transferred from the system to the surroundings, increasing the thermal motion of particles in the surroundings and so increasing their entropy (disorder).

1 mark: positive stated; 1 mark: correct explanation linking heat release to increased disorder of surroundings. [2]

\( K_w = [H^+][OH^-] \). At 298 K, \( K_w = 1.00 \times 10^{-14} \text{ mol}^2\text{dm}^{-6} \).

1 mark: correct expression \( K_w=[H^+][OH^-] \); 1 mark: correct value and units at 298 K. [2]

The buffer contains a large reservoir of undissociated \( HCOOH \) molecules. When a small amount of alkali is added, the \( OH^- \) ions are removed by reaction with \( HCOOH \): \( HCOOH + OH^- \rightarrow HCOO^- + H_2O \), so the concentration of \( OH^- \) (and hence the pH) changes only slightly.

1 mark: recognition that HCOOH acts as a reservoir that reacts with added OH-; 1 mark: correct equation showing HCOOH + OH- → HCOO- + H2O. [2]

On each carbon of the \( C=C \) bond, the two attached groups are ranked by the atomic number of the atom directly bonded (higher atomic number = higher priority). If the two higher-priority groups lie on the same side of the double bond, the isomer is designated Z (zusammen); if they lie on opposite sides, it is designated E (entgegen).

1 mark: correct method for ranking priority (higher atomic number directly attached = higher priority); 1 mark: correct distinction that Z is same side and E is opposite side for the two higher-priority groups. [2]

Add 2,4-dinitrophenylhydrazine (2,4-DNP) reagent to the unknown compound. Formation of an orange or yellow precipitate confirms the presence of a carbonyl group (aldehyde or ketone).

1 mark: 2,4-DNP (2,4-dinitrophenylhydrazine) named as the reagent; 1 mark: correct positive observation, orange/yellow precipitate. [2]

\( 2CH_3CH_2CH_2COOH + Mg \rightarrow (CH_3CH_2CH_2COO)_2Mg + H_2 \).

1 mark: correctly balanced equation with correct formulae for both products. [1]

Alkaline hydrolysis of ethyl propanoate gives sodium propanoate, \( CH_3CH_2COONa \), and ethanol, \( C_2H_5OH \).

1 mark: both correct organic products stated (sodium propanoate and ethanol). [1]

This is a nucleophilic addition-elimination (condensation) reaction, giving ethyl propanoate, \( CH_3CH_2COOC_2H_5 \), and propanoic acid, \( CH_3CH_2COOH \), as the two organic products.

1 mark: both correct products named (ethyl propanoate and propanoic acid); reaction type accepted as supporting detail. [1]

The \( -NH_2 \) group is activating (electron-donating by lone pair delocalisation into the ring) and is an ortho/para-director.

1 mark: activating and ortho/para both correctly stated. [1]

The 2-position (ortho) is close to the bulky methyl group, so steric hindrance makes attack there less favourable than at the less hindered 4-position (para), even though both are activated positions.

1 mark: correct reference to steric hindrance near the methyl group disfavouring the ortho (2-) position. [1]

A first-order concentration-time graph has a constant half-life: successive halvings of concentration take the same amount of time, regardless of the starting concentration.

1 mark: correct reference to constant (concentration-independent) half-life. [1]

By Hess's law (Born–Haber cycle): \( \Delta H_f^{\ominus} = \Delta H_{at}(Mg) + IE_1 + IE_2 + 2\Delta H_{at}(Br) + 2EA(Br) + LE \). Sum of known terms \( = 148 + 738 + 1451 + 2(112) + 2(-325) = 148+738+1451+224-650 = 1911 \text{ kJ mol}^{-1} \). So \( LE = \Delta H_f^{\ominus} - 1911 = -524 - 1911 = -2435 \text{ kJ mol}^{-1} \).

1 mark: correct Hess's law cycle set up with all six terms; 1 mark: atomisation and both ionisation energy terms for Mg summed correctly; 1 mark: atomisation and electron affinity terms for 2×Br correctly doubled; 1 mark: correct rearrangement to isolate LE; 1 mark: correct final answer −2435 kJ mol⁻¹ (accept ±5). [5]

Feasibility requires \( \Delta G \le 0 \), i.e. \( \Delta H \le T\Delta S \), so the minimum temperature is when \( \Delta H = T\Delta S \): \( T = \dfrac{\Delta H}{\Delta S} = \dfrac{128000 \text{ J mol}^{-1}}{387 \text{ J K}^{-1}\text{mol}^{-1}} = 330.7 \approx 331 \text{ K} \).

1 mark: condition for feasibility stated (\( \Delta G \le 0 \) or \( \Delta H = T\Delta S \)); 1 mark: \( \Delta H \) correctly converted to J mol⁻¹; 1 mark: correct division method; 1 mark: correct final answer 331 K. [4]

Comparing Exp 1 and 2: [X] doubles (Y constant) and rate quadruples \( (\times4) \), so order with respect to X is 2. Comparing Exp 2 and 3: [Y] doubles (X constant) and rate doubles \( (\times2) \), so order with respect to Y is 1. Rate equation: \( \text{rate} = k[X]^2[Y] \). Using Exp 1: \( k = \dfrac{2.5\times10^{-3}}{(0.050)^2(0.050)} = \dfrac{2.5\times10^{-3}}{1.25\times10^{-4}} = 20 \). Overall order 3, so units of k are \( \text{mol}^{-2}\text{dm}^{6}\text{s}^{-1} \).

1 mark: order 2 with respect to X correctly deduced; 1 mark: order 1 with respect to Y correctly deduced; 1 mark: correct rate equation rate = k[X]²[Y]; 1 mark: correct value and units of k = 20 mol⁻² dm⁶ s⁻¹ (using any experiment, consistent with candidate's rate equation). [4]

\( K_c = \dfrac{[PCl_3][Cl_2]}{[PCl_5]} = \dfrac{0.40 \times 0.40}{0.30} = \dfrac{0.16}{0.30} = 0.533 \text{ mol dm}^{-3} \) (3 s.f.). Units: \( \dfrac{\text{mol dm}^{-3}\times\text{mol dm}^{-3}}{\text{mol dm}^{-3}} = \text{mol dm}^{-3} \).

1 mark: correct expression for Kc; 1 mark: correct substitution of values; 1 mark: correct value 0.53 (accept 0.533); 1 mark: correct units mol dm⁻³. [4]

\( [H^+] = \sqrt{K_a c} = \sqrt{1.6\times10^{-4} \times 0.20} = \sqrt{3.2\times10^{-5}} = 5.66\times10^{-3} \text{ mol dm}^{-3} \). \( pH = -\log(5.66\times10^{-3}) = 2.25 \).

1 mark: correct approximation \( [H^+]=\sqrt{K_a c} \) used; 1 mark: correct substitution; 1 mark: correct \( [H^+] = 5.66\times10^{-3} \text{ mol dm}^{-3} \); 1 mark: correct final pH = 2.25. [4]

\( pK_a = -\log(1.3\times10^{-5}) = 4.886 \). \( pH = pK_a + \log\dfrac{[\text{salt}]}{[\text{acid}]} = 4.886 + \log\left(\dfrac{0.25}{0.15}\right) = 4.886 + \log(1.667) = 4.886 + 0.222 = 5.11 \).

1 mark: correct pKa calculated from Ka; 1 mark: correct Henderson–Hasselbalch expression used; 1 mark: correct ratio term calculated (log(0.25/0.15) = 0.22); 1 mark: correct final pH = 5.11. [4]

At equivalence, all \( NH_3 \) has been converted to \( NH_4^+ \); total volume \( = 50.0 \text{ cm}^3 \), so \( [NH_4^+] = \dfrac{0.100\times0.0250}{0.0500} = 0.0500 \text{ mol dm}^{-3} \). \( [H^+] = \sqrt{K_a c} = \sqrt{5.6\times10^{-10}\times0.0500} = \sqrt{2.8\times10^{-11}} = 5.29\times10^{-6} \text{ mol dm}^{-3} \). \( pH = -\log(5.29\times10^{-6}) = 5.28 \).

1 mark: correct [NH4+] at equivalence found using diluted total volume; 1 mark: correct application of √(Ka×c) to the NH4+ salt; 1 mark: correct [H+] value; 1 mark: correct final pH = 5.28. [4]

In pure water, \( [H^+] = [OH^-] \), so \( [H^+] = \sqrt{K_w} = \sqrt{5.5\times10^{-14}} = 2.35\times10^{-7} \text{ mol dm}^{-3} \). \( pH = -\log(2.35\times10^{-7}) = 6.63 \). The water is still neutral, since \( [H^+] \) still equals \( [OH^-] \) at this temperature, even though the pH value is less than 7.

1 mark: correct method \( [H^+]=\sqrt{K_w} \); 1 mark: correct [H+] = 2.35×10⁻⁷ mol dm⁻³; 1 mark: correct pH = 6.63; 1 mark: correct conclusion that the water is still neutral because [H+] = [OH-]. [4]

The mechanism is nucleophilic addition. A lone pair on the carbon of the cyanide ion, \( CN^- \), attacks the electron-deficient (\( \delta+ \)) carbonyl carbon of butanone; simultaneously, one pair of electrons from the \( C=O \) \( \pi \) bond moves fully onto the oxygen atom, forming a negatively charged alkoxide (tetrahedral) intermediate ion, \( CH_3CH_2C(O^-)(CH_3)CN \). A lone pair on this \( O^- \) then accepts a proton, \( H^+ \) (from HCN or the solvent), forming the neutral hydroxynitrile product, 2-hydroxy-2-methylbutanenitrile.

1 mark: nucleophilic addition correctly stated as mechanism type; 1 mark: CN⁻ correctly identified as attacking the carbonyl carbon, with correct description of electron pair movement from CN⁻ to C; 1 mark: correct description of the C=O π electrons moving onto O to form the tetrahedral alkoxide intermediate; 1 mark: correct final protonation step forming the hydroxynitrile product. [4]

The lowest common multiple of 6 and 4 electrons is 12, so the reduction half-equation is multiplied by 2 and the oxidation half-equation by 3, then added, cancelling the 12 electrons on each side: \( 2Cr_2O_7^{2-} + 28H^+ + 12e^- \rightarrow 4Cr^{3+} + 14H_2O \) and \( 3CH_3CH_2CH_2CH_2OH + 3H_2O \rightarrow 3CH_3CH_2CH_2COOH + 12H^+ + 12e^- \). Adding and cancelling \( 12H^+ \) and \( 3H_2O \) from each side: \( 2Cr_2O_7^{2-} + 3CH_3CH_2CH_2CH_2OH + 16H^+ \rightarrow 4Cr^{3+} + 3CH_3CH_2CH_2COOH + 11H_2O \).

1 mark: correct multiples chosen (×2 reduction, ×3 oxidation) to balance electrons; 1 mark: half-equations correctly scaled; 1 mark: correct cancellation of electrons, H+ and H2O; 1 mark: correctly balanced overall ionic equation. [4]

Let x = moles of ester formed at equilibrium. Since the reaction shows no change in total moles of species and equal starting amounts of acid and alcohol are used, the mole ratios equal concentration ratios, so \( K_c = \dfrac{x^2}{(1.00-x)^2} = 4.0 \). Taking square roots: \( \dfrac{x}{1.00-x} = 2.0 \), so \( x = 2.0(1.00-x) = 2.0 - 2.0x \), giving \( 3.0x = 2.0 \), \( x = 0.667 \text{ mol} \).

1 mark: correct Kc expression set up in terms of x; 1 mark: correct simplification by taking square roots; 1 mark: correct rearrangement to solve for x; 1 mark: correct final answer, 0.667 mol. [4]

\( K_a = 10^{-pK_a} = 10^{-2.86} = 1.38\times10^{-3} \text{ mol dm}^{-3} \). \( [H^+] = \sqrt{K_a c} = \sqrt{1.38\times10^{-3}\times0.050} = \sqrt{6.9\times10^{-5}} = 8.31\times10^{-3} \text{ mol dm}^{-3} \). \( pH = -\log(8.31\times10^{-3}) = 2.08 \).

1 mark: correct Ka = 1.38×10⁻³ mol dm⁻³ from pKa; 1 mark: correct method [H+]=√(Ka×c); 1 mark: correct [H+] value; 1 mark: correct final pH = 2.08. [4]

Step 1: propanoic acid is treated with sulfur dichloride oxide (thionyl chloride, \( SOCl_2 \)) (or \( PCl_5 \)) at room temperature, forming propanoyl chloride, \( CH_3CH_2COCl \), as the intermediate. Step 2: propanoyl chloride is then reacted with excess concentrated aqueous ammonia at room temperature; the acyl chloride undergoes nucleophilic addition-elimination with \( NH_3 \) to form propanamide, \( CH_3CH_2CONH_2 \), with ammonium chloride as a by-product (from the second equivalent of \( NH_3 \) neutralising the HCl formed).

1 mark: correct reagent for step 1 (SOCl2 or PCl5); 1 mark: correct conditions for step 1 (room temperature); 1 mark: correct intermediate named, propanoyl chloride; 1 mark: correct reagent and conditions for step 2 (excess NH3, room temperature); 1 mark: correct final product named, propanamide. [5]

Anhydrous \( AlCl_3 \) accepts a lone pair from the chlorine of ethanoyl chloride, polarising the \( C-Cl \) bond until it breaks heterolytically, generating the acylium electrophile \( CH_3CO^+ \) and the complex ion \( AlCl_4^- \). A pair of electrons from the delocalised \( \pi \) system of the benzene ring attacks the electron-deficient carbon of \( CH_3CO^+ \), forming a new \( C-C \) bond and a positively charged, non-aromatic arenium (Wheland) intermediate in which the positive charge is delocalised over the remaining ring carbons. A pair of electrons from the \( C-H \) bond on the carbon bearing the acyl group then reforms the aromatic \( \pi \) system, expelling \( H^+ \); this \( H^+ \) reacts with \( AlCl_4^- \), regenerating \( AlCl_3 \) and \( HCl \), so the catalyst is reformed unchanged.

1 mark: correct generation of the CH3CO+ electrophile and AlCl4⁻ by AlCl3 polarising/breaking the C–Cl bond; 1 mark: correct description of ring π electrons attacking the acylium carbon, forming a new C–C bond; 1 mark: correct description of the arenium (Wheland) intermediate with delocalised positive charge; 1 mark: correct loss of H+ to reform the aromatic ring; 1 mark: correct regeneration of AlCl3 catalyst from AlCl4⁻ and H+. [5]

\( pK_a = -\log(1.74\times10^{-5}) = 4.76 \). Using \( pH = pK_a + \log\dfrac{[\text{salt}]}{[\text{acid}]} \): \( 4.50 = 4.76 + \log\dfrac{[\text{salt}]}{[\text{acid}]} \), so \( \log\dfrac{[\text{salt}]}{[\text{acid}]} = -0.26 \), giving \( \dfrac{[\text{salt}]}{[\text{acid}]} = 0.55 \). Choosing \( [CH_3COOH] = 0.200 \text{ mol dm}^{-3} \) gives \( [CH_3COONa] = 0.110 \text{ mol dm}^{-3} \). For 250 cm³: moles \( CH_3COOH = 0.0500 \text{ mol} \), mass \( = 0.0500\times60.0 = 3.00 \text{ g} \); moles \( CH_3COONa = 0.0275 \text{ mol} \), mass \( = 0.0275\times82.0 = 2.26 \text{ g} \). These masses are accurately weighed on a balance, dissolved in distilled water in a 250 cm³ volumetric flask, and made up to the mark with thorough mixing. The buffer contains a large reservoir of both undissociated \( CH_3COOH \) and \( CH_3COO^- \) ions. On adding a small amount of acid, most of the extra \( H^+ \) is removed by \( CH_3COO^- + H^+ \rightarrow CH_3COOH \); on adding a small amount of alkali, the \( OH^- \) is removed by \( CH_3COOH + OH^- \rightarrow CH_3COO^- + H_2O \). In both cases, the ratio \( \dfrac{[\text{salt}]}{[\text{acid}]} \) changes only slightly, so the pH changes only slightly.

Band A (5–6 marks): correct pKa and target ratio calculation, correct (or clearly consistent) masses of acid and salt for 250 cm³, a clear preparative method (weighing, dissolving, volumetric flask to the mark), and a full, correct explanation of buffering action against both added acid and added alkali using appropriate equations; fluent, accurate use of specialist terms. Band B (3–4 marks): correct or near-correct calculation and reasonable preparative method, but explanation of buffering action incomplete (e.g. only one direction covered) or with minor errors. Band C (1–2 marks): basic/fragmented answer, e.g. states a buffer resists pH change without a correct calculation or without a coherent method. Indicative content: pKa = 4.76; ratio [salt]/[acid] = 0.55; masses ≈3.00 g acid and ≈2.26 g salt (or consistent values) in a 250 cm³ volumetric flask; reservoir of CH3COOH and CH3COO⁻; CH3COO⁻ + H⁺ → CH3COOH; CH3COOH + OH⁻ → CH3COO⁻ + H2O. [6]

Ethanoic acid is first converted to the more reactive acyl chloride, ethanoyl chloride, by heating gently with sulfur dichloride oxide (thionyl chloride, \( SOCl_2 \)); the by-products, \( SO_2 \) and HCl, are gaseous and escape, leaving ethanoyl chloride, which is collected by simple distillation. This ethanoyl chloride is then added slowly, with cooling (e.g. in an ice bath, since the reaction is exothermic and vigorous), to an excess of phenylamine dissolved in a suitable solvent; nucleophilic addition-elimination occurs, with the nitrogen lone pair of phenylamine attacking the carbonyl carbon, displacing chloride and forming N-phenylethanamide, with excess phenylamine mopping up the HCl by-product as phenylammonium chloride. The crude solid product is filtered off using a Büchner funnel under reduced pressure, then purified by recrystallisation: it is dissolved in a minimum volume of hot suitable solvent (e.g. hot water or aqueous ethanol), the hot solution is filtered to remove insoluble impurities, and allowed to cool slowly so that pure crystals grow while soluble impurities remain in solution; the crystals are filtered again, washed with a small volume of cold solvent, and dried, for example in a desiccator or low-temperature oven. The purity and identity of the product can be checked by measuring its melting point and comparing to the known sharp literature value.

Band A (5–6 marks): a full, coherent, correctly sequenced synthetic and purification route covering: conversion of ethanoic acid to ethanoyl chloride using SOCl2 with correct reasoning; reaction of the acyl chloride with excess phenylamine (with cooling, given the vigorous reaction) to form N-phenylethanamide; filtration of the crude solid; recrystallisation method (dissolve in minimum hot solvent, hot filter, cool to crystallise, filter, wash, dry); and a check of purity by melting point; fluent, accurate use of specialist terms. Band B (3–4 marks): most key steps present (acyl chloride formation, reaction with amine, some purification) but with some detail missing, e.g. no clear recrystallisation method or no purity check. Band C (1–2 marks): only a fragmented or partial method, e.g. mentions reacting the two compounds without a viable synthetic route or without any purification step. Indicative content: SOCl2 converts acid to acyl chloride; react acyl chloride with excess phenylamine, cooling; filter crude solid; recrystallise from minimum hot solvent; hot filter; cool to crystallise; filter, wash, dry; check melting point. [6]

All ions are accelerated to the same kinetic energy, so lighter ions (or those with greater charge, i.e. smaller m/z) travel faster and reach the detector sooner. Time of flight increases with increasing mass-to-charge ratio, which is the basis of separation in a TOF instrument.

1 mark: A.

Ethanol has three chemically distinct proton environments: the \( CH_3 \) protons, the \( CH_2 \) protons, and the \( OH \) proton, so three signals are seen at low resolution.

1 mark: B.

A primary standard must NOT be volatile, since loss of the solid through evaporation/sublimation during weighing or storage would introduce a systematic error into the calculated mass and hence the concentration of solution prepared.

1 mark: C.

\( R_f = \dfrac{\text{distance moved by the spot (component)}}{\text{distance moved by the solvent front}} \), both measured from the origin (baseline).

1 mark: B.

In excess concentrated HCl, chloride ions act as ligands and displace water, forming the tetrahedral complex \( [CuCl_4]^{2-} \): \( [Cu(H_2O)_6]^{2+} + 4Cl^- \rightleftharpoons [CuCl_4]^{2-} + 6H_2O \). This is a ligand exchange with a change in coordination number from 6 (octahedral) to 4 (tetrahedral), and the colour changes from pale blue to yellow/green.

1 mark: A.

Zinc is oxidised (anode) and \( Cu^{2+} \) is reduced (cathode). \( E_{cell} = E^{\ominus}_{cathode} - E^{\ominus}_{anode} = 0.34 - (-0.76) = +1.10 \text{ V} \).

1 mark: C.

The lone pair on the nitrogen of phenylamine overlaps with, and becomes partly delocalised into, the aromatic \( \pi \) system. This makes the lone pair less available to bond to (accept a proton from) \( H^+ \), so phenylamine is a much weaker base than ethylamine, whose nitrogen lone pair is fully localised and available.

1 mark: B.

\( LiAlH_4 \) is a strong reducing agent. It reduces the carbonyl group of the amide, ultimately converting \( C=O \) to \( CH_2 \) and forming the primary amine \( RCH_2NH_2 \).

1 mark: C.

At the isoelectric point, the amino group is fully protonated (\( -NH_3^+ \)) and the carboxyl group is fully deprotonated (\( -COO^- \)); these charges cancel, giving a zwitterion with zero overall (net) charge.

1 mark: C.

Two different monomers, each with two reactive functional groups, join with the loss of a small molecule (water) at each new amide (peptide) linkage formed. This is condensation polymerisation.

1 mark: B.

Chromium has the anomalous configuration \( [Ar]3d^5 4s^1 \). Forming \( Cr^{3+} \) removes the single 4s electron first, then two 3d electrons, giving \( [Ar]3d^3 \). By Hund's rule, all three 3d electrons occupy separate orbitals with parallel spin, so there are 3 unpaired electrons.

1 mark: correct configuration [Ar]3d³; 1 mark: correct number of unpaired electrons, 3. [2]

A ligand is a molecule or ion with at least one lone pair of electrons that forms a dative (coordinate) bond to a central metal ion; in \( [Cu(H_2O)_6]^{2+} \), each \( H_2O \) molecule is a ligand. The coordination number is the number of dative bonds formed by ligands to the central ion; in \( [Cu(H_2O)_6]^{2+} \) this is 6.

1 mark: correct definition of ligand with reference to a lone pair/dative bond, illustrated by H2O; 1 mark: correct definition of coordination number, illustrated as 6. [2]

If only distilled water remains inside the pipette, it would dilute the solution subsequently drawn up, reducing its concentration and introducing a systematic error into the volume/amount actually delivered. Rinsing with a small volume of the solution to be measured removes this residual water/any other solution present, ensuring the concentration of the measured sample is unchanged.

1 mark: recognition that residual water/other solution would dilute the sample; 1 mark: correct conclusion that rinsing with the solution itself avoids this dilution error. [2]

A burette has a fine-bore tap that allows the reagent to be added slowly, drop by drop near the end point, and a finely graduated scale that can be read to \( \pm0.05 \text{ cm}^3 \). A measuring cylinder cannot deliver such small, controlled volumes or be read with comparable precision, so it would give an inaccurate and imprecise end point.

1 mark: correct reference to controlled, dropwise addition near the end point; 1 mark: correct reference to the greater precision/accuracy of burette readings compared with a measuring cylinder. [2]

1-Bromopropane is heated with a large excess of concentrated ammonia dissolved in ethanol, in a sealed tube (under pressure). A large excess of \( NH_3 \) is used so that an incoming molecule of 1-bromopropane is far more likely to collide with and react with \( NH_3 \) than with the primary amine product, minimising further substitution to secondary/tertiary amines and quaternary ammonium salts.

1 mark: correct reagent and conditions (excess ethanolic ammonia, sealed tube/heated under pressure); 1 mark: correct explanation that excess NH3 minimises further substitution of the amine product. [2]

Propanamide is heated under reflux with dilute hydrochloric acid. This hydrolyses the amide to propanoic acid, \( CH_3CH_2COOH \) (with the nitrogen released as the ammonium salt, \( NH_4Cl \)).

1 mark: correct reagent and conditions (dilute HCl, reflux); 1 mark: correct organic product named, propanoic acid. [2]

A peptide bond (an amide linkage, \( -CONH- \)) forms between the carboxyl group of one amino acid and the amine group of another, with loss of water.

1 mark: peptide bond (or amide bond). [1]

Standard electrode potentials are measured at 298 K, with all ion concentrations at 1 mol dm⁻³ and any gases at 100 kPa (1 atm) pressure, relative to the standard hydrogen electrode (0.00 V).

1 mark: all three conditions correctly stated (298 K; 1 mol dm⁻³; 100 kPa), or equivalent detail. [1]

\( D_2O \) exchanges rapidly with labile O-H and N-H protons in the sample; the signal(s) due to these protons disappear (or greatly reduce in size) when the spectrum is re-run, allowing them to be distinguished from C-H protons, which do not exchange.

1 mark: correct explanation that O-H/N-H protons exchange with D2O and their signal disappears. [1]

The M+1 peak arises from the small natural abundance (about 1.1%) of the \( ^{13}C \) isotope; molecules containing one \( ^{13}C \) atom instead of \( ^{12}C \) have a relative mass one unit higher, giving a small M+1 peak.

1 mark: correct reference to the natural abundance of ¹³C. [1]

Successive phases allow safety (toxicity, side effects) and effectiveness to be assessed on progressively larger and more diverse groups, minimising the risk of unforeseen harm before the drug is used by the general population.

1 mark: valid reason given, e.g. establishing safety/efficacy/correct dosage before wider release. [1]

Condensation polymers contain hydrolysable functional groups (e.g. ester or amide linkages) in the backbone, which can be broken down by water or enzymes. Addition polymers have a continuous, non-polar carbon-carbon backbone with no such susceptible linkages, so they resist hydrolysis and persist in the environment.

1 mark: correct reference to hydrolysable ester/amide linkages in condensation polymers versus an inert C-C backbone in addition polymers. [1]

This is cis-trans (geometric) isomerism. In the cis isomer, the two chloride ligands occupy adjacent positions on the octahedron, at 90° to each other, with the four ammonia ligands occupying the remaining four positions. In the trans isomer, the two chloride ligands occupy opposite positions, at 180° to each other, directly across the central cobalt ion from one another, with the four ammonia ligands in the remaining square-planar arrangement around the equator.

1 mark: cis-trans (geometric) isomerism correctly named; 1 mark: correct description of the cis arrangement (Cl ligands adjacent, 90°); 1 mark: correct description of the trans arrangement (Cl ligands opposite, 180°); 1 mark: correct reference to the remaining NH3 ligands occupying the other positions in each case. [4]

\( K_{stab} = \dfrac{[[Cu(NH_3)_4(H_2O)_2]^{2+}]}{[[Cu(H_2O)_6]^{2+}][NH_3]^4} \) (water, as solvent, is omitted from the expression). A very large value of \( K_{stab} \) means the equilibrium position lies almost entirely to the right, so at equilibrium the concentration of the ammine complex is very much greater than that of the aqua complex; this shows that the ammine complex, \( [Cu(NH_3)_4(H_2O)_2]^{2+} \), is considerably more stable (thermodynamically favoured) than the original aqua complex.

1 mark: correct Kstab expression with correct powers; 1 mark: water correctly omitted (or noted as constant/solvent); 1 mark: correct statement that equilibrium lies far to the right for large Kstab; 1 mark: correct conclusion that the ammine complex is much more stable than the aqua complex. [4]

(i) With only a small quantity of ammonia, once some primary amine (propan-1-amine) has formed, it competes with the remaining \( NH_3 \) for reaction with unreacted 1-bromopropane; since the amine is also nucleophilic, further substitution occurs, giving a mixture of primary, secondary and tertiary amines and, eventually, a quaternary ammonium salt. (ii) With a large excess of ammonia, at any point in the reaction there is a huge excess of \( NH_3 \) molecules compared with amine product molecules, so 1-bromopropane molecules are far more likely to collide with and react with \( NH_3 \) than with the amine; this strongly favours the primary amine, propan-1-amine, as the major product.

1 mark: correct identification of a mixture of amines/quaternary salt for (i); 1 mark: correct reasoning that further substitution competes with excess NH3 for (i); 1 mark: correct identification of predominantly primary amine for (ii); 1 mark: correct reasoning based on large excess of NH3 out-competing the amine product for (ii). [4]

Nucleophilic addition-elimination occurs between the nitrogen lone pair of methylamine and the carbonyl carbon of propanoyl chloride, displacing chloride. The secondary amide formed is N-methylpropanamide, structural formula \( CH_3CH_2CONHCH_3 \). A second molecule of methylamine reacts with the HCl by-product to form methylammonium chloride, \( CH_3NH_3^+Cl^- \).

1 mark: correct structural formula CH3CH2CONHCH3; 1 mark: correct systematic name N-methylpropanamide; 1 mark: correct identification of nucleophilic addition-elimination as the mechanism type; 1 mark: correct second product named, methylammonium chloride. [4]

The carboxyl group of glycine can condense with the amine group of alanine, losing water, to give the dipeptide glycylalanine: \( H_2NCH_2CONHCH(CH_3)COOH \). (Equally valid: alanylglycine, \( H_2NCH(CH_3)CONHCH_2COOH \), formed the other way round.) In either structure, the peptide bond is the \( -CO-NH- \) linkage joining the two amino acid residues.

1 mark: correct structural formula of a valid dipeptide (either Gly-Ala or Ala-Gly); 1 mark: correct loss of one water molecule reflected in the structure; 1 mark: peptide (amide) bond correctly identified as -CONH-; 1 mark: correct location of the peptide bond shown between the two residues. [4]

Compound Q is ethyl ethanoate, \( CH_3COOCH_2CH_3 \). The singlet at \( \delta \) 2.0 (3H) is the \( CH_3CO- \) group, with no adjacent protons, so it is unsplit. The quartet at \( \delta \) 4.1 (2H) is the \( -OCH_2- \) group; it is deshielded by the neighbouring oxygen (high shift) and split into a quartet by the adjacent \( CH_3 \) group (n+1 = 4, from 3 neighbouring protons). The triplet at \( \delta \) 1.3 (3H) is the terminal \( CH_3 \) group, split into a triplet by the adjacent \( CH_2 \) group (n+1 = 3, from 2 neighbouring protons); the integration ratio 3:3:2 and molecular formula \( C_4H_8O_2 \) both confirm the structure.

1 mark: correct structure identified as ethyl ethanoate; 1 mark: correct assignment of the δ 2.0 singlet to CH3CO- (no adjacent H); 1 mark: correct assignment of the δ 4.1 quartet to OCH2 (split by adjacent CH3, deshielded by O); 1 mark: correct assignment of the δ 1.3 triplet to the terminal CH3 (split by adjacent CH2). [4]

There are three distinct proton environments. The terminal \( CH_3 \) protons are adjacent to the central \( CH_2 \) (2 protons), so by the n+1 rule they appear as a triplet. The central \( CH_2 \) protons are adjacent to both the \( CH_3 \) (3 protons) and the \( CH_2Cl \) (2 protons), a total of 5 neighbouring protons, so they appear as a sextet (n+1 = 6). The \( CH_2Cl \) protons are adjacent to the central \( CH_2 \) (2 protons), so they appear as a triplet. In total, 3 signals are expected.

1 mark: CH3 correctly predicted as a triplet; 1 mark: central CH2 correctly predicted as a sextet, with correct reasoning (5 neighbouring H); 1 mark: CH2Cl correctly predicted as a triplet; 1 mark: correct total of 3 signals stated. [4]

With molecular formula \( C_3H_6O_2 \) and \( M_r = 74 \), the ester is ethyl methanoate, \( HCOOC_2H_5 \). Cleavage of the \( O-C \) bond, with loss of the \( HCOO \) fragment, produces the ethyl cation, \( C_2H_5^+ \), which has \( m/z = 29 \) and accounts for this fragment peak.

1 mark: correct molecular formula matched to ethyl methanoate; 1 mark: correct structure HCOOC2H5 given; 1 mark: correct fragment ion identified as C2H5+; 1 mark: correct reasoning linking m/z = 29 to loss of the HCOO group. [4]

The half-cell with the more negative (lower) electrode potential, \( I_2/I^- \), is written on the left (oxidation, anode); the half-cell with the more positive potential, \( Fe^{3+}/Fe^{2+} \), is written on the right (reduction, cathode): \( Pt(s)\,|\,I^-(aq),I_2(aq)\,||\,Fe^{3+}(aq),Fe^{2+}(aq)\,|\,Pt(s) \). \( E_{cell} = E^{\ominus}(Fe^{3+}/Fe^{2+}) - E^{\ominus}(I_2/I^-) = 0.77 - 0.54 = +0.23 \text{ V} \).

1 mark: I2/I- half-cell correctly placed on the left with inert Pt electrode shown; 1 mark: Fe3+/Fe2+ half-cell correctly placed on the right; 1 mark: correct single/double line notation used; 1 mark: correct EMF value, +0.23 V. [4]

Autocatalysis occurs when a product of a reaction itself acts as a catalyst for that same reaction. Here, \( Mn^{2+} \) ions, formed as \( MnO_4^- \) is reduced, catalyse the further reaction between \( MnO_4^- \) and \( C_2O_4^{2-} \) (via intermediate manganese oxidation states). At the very start of the reaction there is no \( Mn^{2+} \) present, so the reaction proceeds only by the slow, uncatalysed pathway between two negatively charged ions (which repel each other). As the reaction proceeds, \( Mn^{2+} \) accumulates and increasingly catalyses the reaction, so the rate rises. Eventually, as the concentrations of \( MnO_4^- \) and \( C_2O_4^{2-} \) fall as they are used up, the rate decreases again despite the catalyst being present, since rate also depends on reactant concentration.

1 mark: correct definition of autocatalysis (a product catalyses its own formation); 1 mark: Mn2+ correctly identified as the autocatalyst, produced during the reaction; 1 mark: correct explanation of the slow initial rate (no catalyst present yet, and/or repulsion between two anions); 1 mark: correct explanation of the rate increasing as Mn2+ builds up, then decreasing as reactants are depleted. [4]

Moles HCl \( = 0.150 \times 0.0300 = 4.50\times10^{-3} \text{ mol} \). Since \( NaOH + HCl \rightarrow NaCl + H_2O \) is a 1:1 reaction, moles NaOH in the 25.0 cm³ portion \( = 4.50\times10^{-3} \text{ mol} \). Scaling up to the full 250 cm³ (×10): moles NaOH \( = 0.0450 \text{ mol} \). Mass NaOH \( = 0.0450 \times 40.0 = 1.80 \text{ g} \). Percentage purity \( = \dfrac{1.80}{2.00}\times100 = 90.0\% \).

1 mark: correct moles HCl calculated; 1 mark: correct 1:1 mole ratio applied to find moles NaOH in the aliquot; 1 mark: correct scaling ×10 and mass NaOH calculated; 1 mark: correct final percentage purity, 90.0%. [4]

Moles \( MnO_4^- = 0.0200 \times 0.01840 = 3.68\times10^{-4} \text{ mol} \). By the 1:5 ratio, moles \( Fe^{2+} = 5 \times 3.68\times10^{-4} = 1.84\times10^{-3} \text{ mol} \), present in 25.0 cm³ (0.0250 dm³). Concentration \( = \dfrac{1.84\times10^{-3}}{0.0250} = 0.0736 \text{ mol dm}^{-3} \).

1 mark: correct moles MnO4- calculated; 1 mark: correct 1:5 ratio applied to find moles Fe2+; 1 mark: correct conversion of 25.0 cm³ to dm³; 1 mark: correct final concentration, 0.0736 mol dm⁻³. [4]

Moles \( S_2O_3^{2-} = 0.100 \times 0.02130 = 2.13\times10^{-3} \text{ mol} \). By the 2:1 ratio in the thiosulfate equation, moles \( I_2 = \tfrac{1}{2}(2.13\times10^{-3}) = 1.065\times10^{-3} \text{ mol} \). By the 2:1 ratio of \( Cu^{2+} \) to \( I_2 \) in the first equation, moles \( Cu^{2+} = 2\times1.065\times10^{-3} = 2.13\times10^{-3} \text{ mol} \), present in 25.0 cm³ (0.0250 dm³). Concentration \( = \dfrac{2.13\times10^{-3}}{0.0250} = 0.0852 \text{ mol dm}^{-3} \).

1 mark: correct moles S2O3²⁻ calculated; 1 mark: correct moles I2 found using the 2:1 ratio; 1 mark: correct moles Cu2+ found using the 2:1 ratio from the first equation; 1 mark: correct final concentration, 0.0852 mol dm⁻³. [4]

The dichromate half-equation involves 6 electrons, so the iron half-equation must be multiplied by 6 to balance electrons: \( 6Fe^{2+} \rightarrow 6Fe^{3+} + 6e^- \). Adding this to the dichromate half-equation and cancelling the 6 electrons on each side gives: \( Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \rightarrow 2Cr^{3+} + 7H_2O + 6Fe^{3+} \).

1 mark: correct recognition that the Fe half-equation must be multiplied by 6; 1 mark: correctly scaled iron half-equation; 1 mark: correct cancellation of the 6 electrons; 1 mark: correctly balanced overall ionic equation. [4]

\( E_{cell} = E^{\ominus}(MnO_4^-/Mn^{2+}) - E^{\ominus}(Br_2/Br^-) = 1.51 - 1.07 = +0.44 \text{ V} \). Since this is positive, the reaction is thermodynamically feasible: acidified \( MnO_4^- \) can oxidise \( Br^- \) to \( Br_2 \).

1 mark: correct EMF calculation method; 1 mark: correct value, +0.44 V; 1 mark: correct conclusion that the reaction is feasible, with reasoning based on the positive EMF. [3]

\( Ag^+/Ag \) has the more positive electrode potential, so it is reduced (cathode): \( Ag^+ + e^- \rightarrow Ag \). \( Fe^{3+}/Fe^{2+} \) is oxidised (anode): \( Fe^{2+} \rightarrow Fe^{3+} + e^- \). Adding these (electrons already balanced, 1:1): \( Ag^+ + Fe^{2+} \rightarrow Ag + Fe^{3+} \). \( E_{cell} = E^{\ominus}(Ag^+/Ag) - E^{\ominus}(Fe^{3+}/Fe^{2+}) = 0.80 - 0.77 = +0.03 \text{ V} \).

1 mark: correct identification of Ag+/Ag as the reduction (cathode) half-reaction; 1 mark: correct identification of Fe2+/Fe3+ as the oxidation (anode) half-reaction; 1 mark: correctly combined overall ionic equation; 1 mark: correct EMF value, +0.03 V. [4]

Percentage of N \( = \dfrac{14}{89}\times100 = 15.7\% \) (3 s.f.).

1 mark: correct Mr (89) used as the denominator; 1 mark: correct method, mass N ÷ Mr × 100; 1 mark: correct final answer, 15.7%. [3]

\( [OH^-] = \sqrt{K_b c} = \sqrt{4.4\times10^{-4}\times0.10} = \sqrt{4.4\times10^{-5}} = 6.63\times10^{-3} \text{ mol dm}^{-3} \). \( pOH = -\log(6.63\times10^{-3}) = 2.18 \). \( pH = 14.00 - pOH = 14.00 - 2.18 = 11.82 \).

1 mark: correct method [OH-]=√(Kb×c); 1 mark: correct [OH-] value; 1 mark: correct pOH = 2.18; 1 mark: correct final pH = 11.82 (allow follow-through). [4]

In heterogeneous catalysis, the catalyst is in a different phase (solid) from the reactants (gases). Gas molecules of CO and NO diffuse to, and are adsorbed onto, active sites on the surface of the solid Pt/Pd/Rh catalyst, forming weak bonds to metal atoms at the surface. This adsorption weakens the bonds within the reactant molecules and holds them close together in a favourable orientation, providing an alternative reaction pathway of lower activation energy than the uncatalysed gas-phase reaction. Reaction occurs between the adsorbed species on the surface, forming \( CO_2 \) and \( N_2 \) (from the overall reaction \( 2CO + 2NO \rightarrow 2CO_2 + N_2 \)). Once formed, these product molecules are more weakly bonded to the surface than the original reactants and desorb (leave the surface), freeing the active site so that further reactant molecules can be adsorbed and the cycle repeated. The catalyst itself is typically spread as a fine coating over a honeycomb support of large surface area, to maximise the number of active sites available and so the rate of reaction.

Band A (5–6 marks): a full, coherent explanation covering adsorption of both reactant gases onto active sites, weakening of bonds/favourable orientation lowering activation energy, reaction occurring on the surface, desorption of products freeing the active site, and reference to the large surface area of the honeycomb support maximising active sites; fluent, accurate use of specialist terms. Band B (3–4 marks): most of the key stages present (adsorption, reaction, desorption) but explanation of how this lowers activation energy or the role of surface area is less developed. Band C (1–2 marks): only a basic/fragmented account, e.g. states the catalyst provides a surface for reaction without a clear description of adsorption and desorption. Indicative content: adsorption of CO and NO onto active sites; weakening of bonds/correct orientation; lower activation energy pathway; surface reaction to form CO2 and N2; desorption of products; active site freed for further reactants; large surface area (honeycomb) maximises rate. [6]

An accurately weighed mass of the impure ethanedioic acid crystals is dissolved in distilled water and transferred quantitatively, using a funnel and rinsings, into a 250 cm³ volumetric flask, which is then made up to the graduation mark and mixed thoroughly. A 25.0 cm³ aliquot of this solution is pipetted into a conical flask, a few drops of phenolphthalein indicator are added, and the solution is titrated against a standardised sodium hydroxide solution of known concentration added from a burette, swirling constantly, until the indicator just turns from colourless to permanent pale pink. The titration is repeated until at least two concordant titres (within 0.10 cm³ of each other) are obtained, and a mean titre is calculated from these. Since ethanedioic acid is diprotic (\( (COOH)_2 + 2NaOH \rightarrow (COONa)_2 + 2H_2O \)), the moles of NaOH used are calculated from the mean titre and its concentration, and moles of ethanedioic acid in the aliquot are found using the 1:2 mole ratio. This is scaled up by a factor of 10 (250/25) to find the total moles of ethanedioic acid in the original 250 cm³, which is converted to a mass using its molar mass; the percentage purity is then found by dividing this mass by the mass of impure sample originally weighed out, and multiplying by 100.

Band A (5–6 marks): a full, coherent, correctly sequenced method covering accurate weighing and dissolving; quantitative transfer to a 250 cm³ volumetric flask made up to the mark; pipetting a 25.0 cm³ aliquot; titration against standard NaOH with phenolphthalein to a correct colourless-to-pink end point; repeating to concordance and taking a mean titre; and a clear, correct calculation outline using the 1:2 mole ratio, scaling by 10, and converting to mass and % purity; fluent, accurate use of specialist terms. Band B (3–4 marks): most steps present but some detail (e.g. the diprotic 1:2 ratio, the scaling factor, or concordance) missing or method less clearly sequenced. Band C (1–2 marks): only a fragmented or partial method, e.g. mentions titration against NaOH but no clear volumetric-flask/dilution step or no coherent calculation method. Indicative content: dissolve weighed sample; 250 cm³ volumetric flask; 25.0 cm³ aliquots; phenolphthalein; titrate vs standard NaOH; colourless to pink end point; concordant titres/mean; 1:2 mole ratio (diprotic acid); scale by 10; convert to mass and % purity. [6]

(a) On dropwise addition of NaOH, a pale blue gelatinous precipitate of copper(II) hydroxide forms immediately; in excess NaOH, the precipitate remains, insoluble and unchanged. (b) On dropwise addition of ammonia, the same pale blue precipitate of \( Cu(OH)_2 \) forms; in excess ammonia, the precipitate dissolves to give a deep (royal) blue solution, due to formation of the complex ion \( [Cu(NH_3)_4(H_2O)_2]^{2+} \). (c) A cream/off-white precipitate of copper(I) iodide forms, and the solution above it turns brown/yellow-brown due to iodine liberated in solution. (d) \( Cu^{2+}(aq) + 2OH^-(aq) \rightarrow Cu(OH)_2(s) \). (e) The cation present in X is \( Cu^{2+} \).

(a) [3]: 1 mark pale blue precipitate formed with dropwise NaOH; 1 mark precipitate described as gelatinous; 1 mark precipitate insoluble/unchanged in excess NaOH. (b) [3]: 1 mark pale blue precipitate formed with dropwise ammonia; 1 mark precipitate dissolves in excess ammonia; 1 mark correct description of the resulting deep/royal blue solution. (c) [2]: 1 mark cream/white precipitate observed; 1 mark brown/yellow-brown colouration of the solution (iodine) also noted. (d) [2]: 1 mark correct species; 1 mark correctly balanced ionic equation with state symbols. (e) [2]: 1 mark Cu²⁺ stated; 1 mark consistent with all observations given. [12]

(a) Titre 1 \( = 23.55-0.05=23.50 \text{ cm}^3 \); Titre 2 \( = 23.60-0.05=23.55 \text{ cm}^3 \); Titre 3 \( = 23.65-0.10=23.55 \text{ cm}^3 \). (b) The rough titre (24.70) is not concordant and is discarded. Titres 1, 2 and 3 (23.50, 23.55, 23.55) all lie within 0.10 cm³ of each other and are concordant. Mean \( = \dfrac{23.50+23.55+23.55}{3} = 23.53 \text{ cm}^3 \). (c) Moles \( MnO_4^- = 0.0500 \times 0.02353 = 1.18\times10^{-3} \text{ mol} \). By the 5:2 ratio, moles \( C_2O_4^{2-} = \dfrac{5}{2}\times1.18\times10^{-3} = 2.94\times10^{-3} \text{ mol} \), present in the 25.0 cm³ (0.0250 dm³) aliquot. Concentration \( = \dfrac{2.94\times10^{-3}}{0.0250} = 0.118 \text{ mol dm}^{-3} \). The solution is warmed because the uncatalysed reaction between \( MnO_4^- \) and \( C_2O_4^{2-} \), two negatively charged ions that repel each other, is otherwise very slow at room temperature; warming increases the rate to a practical speed for titration (the reaction is also autocatalysed by the \( Mn^{2+} \) product as it forms).

(a) [3]: 1 mark each for Titre 1 = 23.50, Titre 2 = 23.55, Titre 3 = 23.55 cm³. (b) [3]: 1 mark for correctly excluding the rough titre; 1 mark for correctly identifying titres 1–3 as concordant (within 0.10 cm³); 1 mark for correct mean titre 23.53 cm³ (allow follow-through from candidate's own titre values). (c) [5]: 1 mark for moles MnO4⁻ correctly calculated; 1 mark for correct 5:2 ratio applied to find moles C2O4²⁻; 1 mark for correct division by aliquot volume in dm³; 1 mark for correct final concentration, 0.118 mol dm⁻³ (accept follow-through); 1 mark for a valid reason for warming (increases the rate of an otherwise slow reaction between two anions/reaches a practical rate for titration). [11]

(a) Effervescence (bubbles of a colourless gas, \( CO_2 \)) is observed with one liquid, propanoic acid, since it is acidic enough to react with the carbonate; the other two liquids (propanal and propanone) show no reaction, as neither is sufficiently acidic. This identifies propanoic acid. (b) The remaining two liquids are warmed separately with acidified potassium dichromate(VI). One liquid (propanal, an aldehyde) is oxidised: the solution changes colour from orange to green, and a rise in temperature of the reaction mixture is observed/measured, since oxidation is exothermic. The other liquid (propanone, a ketone) shows no colour change and no significant temperature rise, since ketones cannot be oxidised under these conditions. (c) The liquid identified as the aldehyde (propanal) is warmed gently with Tollens' reagent (ammoniacal silver nitrate); a positive result is a silver mirror forming on the inside of the test tube (alternatively, Fehling's solution warmed gently gives a brick-red precipitate). (d) L, M and N are propanoic acid, propanal and propanone (matched to the candidate's own test results).

(a) [3]: 1 mark sodium carbonate test applied to all three samples; 1 mark correct positive observation for the acid (effervescence/CO2 gas); 1 mark propanoic acid correctly identified, with the other two giving no reaction. (b) [3]: 1 mark correct reagent/conditions (acidified potassium dichromate(VI), warmed); 1 mark correct colour change (orange to green) and temperature rise noted for the aldehyde; 1 mark correct no-reaction/no significant temperature change noted for the ketone. (c) [3]: 1 mark correct confirmatory reagent (Tollens'/Fehling's); 1 mark correct conditions (gentle warming); 1 mark correct positive observation (silver mirror, or brick-red precipitate). (d) [2]: 1 mark all three correctly identified; 1 mark identifications consistent with the observations recorded in (a)-(c). [11]

The alcohol, excess acidified potassium dichromate(VI) solution and anti-bumping granules are placed in a round-bottomed flask fitted with a Liebig condenser held vertically above the flask (water entering at the bottom of the condenser jacket and leaving at the top, running countercurrent to the vapour), forming a reflux set-up; the flask is heated (e.g. on a heating mantle or water bath) for an extended period to ensure the alcohol is fully oxidised through to the carboxylic acid. The condenser continuously cools and condenses any vapours (alcohol, intermediate aldehyde, water) rising from the boiling mixture, returning them as liquid to the flask; this prevents loss of volatile reactant and intermediate product to the atmosphere, allowing the reaction to be heated continuously for as long as needed without loss of material, ensuring complete oxidation to the carboxylic acid.

1 mark: round-bottomed flask with Liebig condenser held vertically, described correctly; 1 mark: correct heating method mentioned (e.g. heating mantle/water bath) with anti-bumping granules; 1 mark: correct reference to extended heating time needed for full oxidation to the acid (rather than stopping at the aldehyde); 1 mark: correct explanation that the condenser condenses vapours, returning them to the flask; 1 mark: correct conclusion that this prevents loss of volatile alcohol/intermediate, allowing continuous heating. [5]

The reaction mixture containing phenylamine is placed in a flask, and steam is passed into the mixture from a steam generator (or the mixture itself is heated with water, generating its own steam). The phenylamine, which is immiscible with water and reasonably volatile, co-distils with the steam at a temperature below 100 °C; the resulting vapour mixture passes into a condenser, where it condenses, and the distillate (a mixture of water and phenylamine) is collected. Because phenylamine and water are immiscible, the distillate separates into two layers, and the phenylamine layer is separated using a separating funnel, then dried and purified by further distillation. Steam distillation is preferred to simple heating because it allows phenylamine to be distilled over at a temperature well below its own boiling point (184 °C), avoiding prolonged heating at high temperature which could cause the amine to decompose or oxidise/discolour in air.

1 mark: correct description of steam being passed into/generated within the reaction mixture; 1 mark: correct reference to phenylamine co-distilling with steam below 100 °C due to immiscibility/volatility; 1 mark: correct description of condensation and collection of the distillate; 1 mark: correct description of separating the immiscible layers (e.g. using a separating funnel); 1 mark: correct explanation that steam distillation avoids prolonged heating at high temperature, preventing decomposition. [5]

A Büchner funnel is fitted with a circle of filter paper (moistened first with a little solvent to seat it) and placed in a Büchner flask, which is connected via thick-walled rubber tubing to a vacuum pump or water aspirator. The reaction mixture is poured into the funnel; the reduced pressure created below the filter paper pulls the liquid (filtrate) through into the flask much faster than gravity filtration, while the solid product is retained on the filter paper. The solid is washed in situ with a small volume of ice-cold solvent (to remove soluble impurities while minimising loss of product through dissolving) and air is drawn through it under vacuum to partially dry it, before it is transferred for further drying (e.g. in a desiccator or low-temperature oven).

1 mark: correct set-up of Büchner funnel with filter paper, connected to a Büchner flask and vacuum source; 1 mark: correct explanation that reduced pressure speeds up filtration compared with gravity filtration; 1 mark: correct description of washing the solid with a small volume of cold solvent; 1 mark: correct description of drying the solid product. [4]

A small amount of the dry solid product is packed into a fine capillary tube, which is placed in an electric melting point apparatus (or oil bath with thermometer) alongside a thermometer, and the sample is heated slowly, noting the temperature at which melting begins and is complete. If the observed melting point is sharp (melts over a narrow range, typically less than 1-2 °C) and matches the known literature melting point of the expected product, this confirms both that the product is likely the correct compound and that it is reasonably pure. If the sample melts over a broad range and/or at a lower temperature than the literature value, this indicates the presence of impurities, since impurities generally lower and broaden the melting point of a solid.

1 mark: correct method described (capillary tube, melting point apparatus, thermometer, slow heating); 1 mark: correct comparison to literature value to confirm identity; 1 mark: correct statement that a sharp melting point indicates purity; 1 mark: correct statement that a lowered/broadened melting point indicates impurity. [4]

Propan-1-ol is reacted with ethanoic anhydride (or, more vigorously, ethanoyl chloride), with a few drops of concentrated sulfuric acid as catalyst (if using the anhydride), at room temperature or with gentle warming under reflux; nucleophilic addition-elimination occurs, forming propyl ethanoate. The crude product mixture is then purified: it is shaken with aqueous sodium hydrogencarbonate in a separating funnel to remove unreacted acid/acid by-product (releasing \( CO_2 \) gas, so the tap is opened periodically to release pressure), and the upper ester layer is run off/separated from the aqueous layer. The ester layer is then dried using an anhydrous drying agent such as anhydrous magnesium sulfate, filtered to remove the solid drying agent, and finally purified by simple distillation, collecting the fraction that boils at the known boiling point of propyl ethanoate.

1 mark: correct reagent (ethanoic anhydride or ethanoyl chloride) and catalyst/conditions; 1 mark: correct washing step with sodium hydrogencarbonate to remove acidic impurities, with pressure release noted; 1 mark: correct separation of the ester layer using a separating funnel; 1 mark: correct drying step with an anhydrous drying agent, followed by filtration; 1 mark: correct final purification by distillation, collecting at the correct boiling point. [5]

Phenylamine is dissolved in excess dilute hydrochloric acid and cooled in an ice bath to below 5 °C; a cold solution of sodium nitrite is then added slowly, generating nitrous acid in situ, which reacts with the phenylamine to form the diazonium salt, benzenediazonium chloride, \( C_6H_5N_2^+Cl^- \). The temperature must be kept below 5 °C throughout, since the diazonium salt is unstable and decomposes (releasing \( N_2 \) gas and forming phenol) if the temperature rises. This cold diazonium salt solution is then added to a cold, alkaline solution of phenol (phenol dissolved in aqueous sodium hydroxide); the diazonium ion acts as an electrophile and undergoes electrophilic substitution onto the strongly activated phenol ring (predominantly at the position para to the \( -OH \) group), forming an intensely coloured orange/yellow azo compound, which precipitates from solution.

1 mark: correct reagents for diazotisation (NaNO2 and dilute HCl); 1 mark: correct low-temperature condition (below 5 °C, ice bath) correctly justified by instability of the diazonium salt; 1 mark: correct product named, benzenediazonium chloride; 1 mark: correct coupling conditions (cold, alkaline solution of phenol); 1 mark: correct description of electrophilic substitution forming a coloured azo dye precipitate. [5]

The crude solid is dissolved in the minimum volume of hot (near-boiling) water needed to fully dissolve it, since benzamide is far more soluble in hot water than in cold. The hot solution is filtered quickly through a preheated (fluted) filter funnel to remove any insoluble impurities while the solution is still hot, preventing premature crystallisation in the funnel. The filtrate is then left to cool slowly to room temperature (and further in an ice bath), allowing pure benzamide crystals to form slowly, while soluble impurities remain dissolved in the mother liquor at their lower concentration. The crystals are collected by filtration (e.g. Büchner filtration), washed with a small volume of ice-cold water, and dried. The purity of the final product can be confirmed by determining its melting point: a sharp melting point matching the literature value for benzamide indicates a pure sample, whereas a lowered, broadened melting point would indicate remaining impurity.

1 mark: correct method (dissolve in minimum hot water, hot filtration); 1 mark: correct description of slow cooling to grow pure crystals while impurities remain in solution; 1 mark: correct final filtration, washing with cold solvent, and drying; 1 mark: correct method for confirming purity (sharp melting point matching literature value). [4]

A small, concentrated spot of the amino acid mixture is applied near one corner of a square sheet of chromatography paper, using a capillary tube, and allowed to dry. The paper is placed in a tank containing the first solvent (below the level of the spot) and left to develop until the solvent front nears the top; the paper is removed, dried, then rotated through 90° and placed in a second, different solvent to develop in the perpendicular direction, spreading out amino acids that overlapped after the first run. Since amino acids are colourless, the developed, dried chromatogram is sprayed with ninhydrin solution and gently heated (or left to develop over time); ninhydrin reacts with amino acids to form purple/blue coloured spots, making their positions visible. The \( R_f \) value of each spot (distance moved by the spot ÷ distance moved by the solvent front, for each solvent direction) is calculated and compared with the known \( R_f \) values of reference amino acids run under the same conditions, allowing each amino acid in the original mixture to be identified.

1 mark: correct description of spotting the mixture and developing in two perpendicular solvent directions; 1 mark: correct explanation of why two directions are used (separating spots that overlap in one solvent); 1 mark: correct role of ninhydrin (forms coloured spots with colourless amino acids); 1 mark: correct use of Rf values (calculated and compared with reference values) to identify each amino acid. [4]

Moles NaOH \( = 0.100 \times 0.0200 = 2.00\times10^{-3} \text{ mol} \), equal to moles HA in the 25.0 cm³ portion (1:1 acid-base reaction). Scaling up to the full 250 cm³ (×10): total moles HA \( = 0.0200 \text{ mol} \). Molar mass \( = \dfrac{1.20}{0.0200} = 60.0 \text{ g mol}^{-1} \). The empirical formula mass of \( CH_2O \) is 30; since \( 60.0 \div 30 = 2 \), the molecular formula is \( C_2H_4O_2 \), i.e. \( CH_3COOH \), ethanoic acid.

1 mark: correct moles NaOH = moles HA in the aliquot; 1 mark: correct scaling ×10 to total moles in 250 cm³; 1 mark: correct molar mass calculation, 60.0 g mol⁻¹; 1 mark: correct molecular formula, C2H4O2 (n=2 correctly found); 1 mark: correctly identified as ethanoic acid. [5]

Initial moles HCl \( = 0.400 \times 0.0250 = 0.0100 \text{ mol} \). Moles NaOH used (= moles unreacted HCl) \( = 0.200 \times 0.0350 = 7.00\times10^{-3} \text{ mol} \). Moles HCl reacted with \( CaCO_3 \): \( 0.0100 - 7.00\times10^{-3} = 3.00\times10^{-3} \text{ mol} \). From \( CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2 \), moles \( CaCO_3 = \dfrac{3.00\times10^{-3}}{2} = 1.50\times10^{-3} \text{ mol} \). Mass \( CaCO_3 = 1.50\times10^{-3}\times100 = 0.150 \text{ g} \). Percentage purity \( = \dfrac{0.150}{0.200}\times100 = 75.0\% \).

1 mark: correct initial moles HCl; 1 mark: correct moles unreacted HCl from moles NaOH; 1 mark: correct moles HCl reacted with CaCO3 (by subtraction); 1 mark: correct 2:1 ratio applied to find moles/mass of CaCO3; 1 mark: correct final percentage, 75.0% (allow follow-through). [5]

Moles \( MnO_4^- = 0.0200 \times 0.02400 = 4.80\times10^{-4} \text{ mol} \). By the 5:2 ratio, moles \( H_2O_2 = \dfrac{5}{2}\times4.80\times10^{-4} = 1.20\times10^{-3} \text{ mol} \), in the 25.0 cm³ diluted portion. Scaling to the full 250 cm³ diluted solution (×10): moles \( H_2O_2 = 0.0120 \text{ mol} \), which was originally present in 10.0 cm³ (0.0100 dm³) of undiluted solution. Concentration \( = \dfrac{0.0120}{0.0100} = 1.20 \text{ mol dm}^{-3} \). In g dm⁻³: \( 1.20\times34 = 40.8 \text{ g dm}^{-3} \).

1 mark: correct moles MnO4- calculated; 1 mark: correct moles H2O2 found using the 5:2 ratio; 1 mark: correct scaling ×10 to the full 250 cm³ diluted volume; 1 mark: correct concentration of the original undiluted solution in mol dm⁻³, 1.20 mol dm⁻³; 1 mark: correct conversion to 40.8 g dm⁻³. [5]

Initial moles NaOH \( = 0.400 \times 0.0250 = 0.0100 \text{ mol} \). Moles HCl used (= moles unreacted NaOH) \( = 0.200 \times 0.02150 = 4.30\times10^{-3} \text{ mol} \). Moles NaOH reacted with aspirin: \( 0.0100 - 4.30\times10^{-3} = 5.70\times10^{-3} \text{ mol} \). By the 1:2 ratio, moles aspirin \( = \dfrac{5.70\times10^{-3}}{2} = 2.85\times10^{-3} \text{ mol} \). Mass aspirin \( = 2.85\times10^{-3}\times180 = 0.513 \text{ g} \). Percentage \( = \dfrac{0.513}{0.600}\times100 = 85.5\% \).

1 mark: correct initial moles NaOH; 1 mark: correct moles unreacted NaOH from moles HCl; 1 mark: correct moles NaOH reacted with aspirin (by subtraction); 1 mark: correct 1:2 ratio applied to find mass of aspirin; 1 mark: correct final percentage, 85.5% (allow follow-through). [5]

Initial moles HCl \( = 0.500 \times 0.100 = 0.0500 \text{ mol} \). Moles NaOH used (= moles unreacted HCl) \( = 0.250 \times 0.03200 = 8.00\times10^{-3} \text{ mol} \). Moles HCl reacted with ethylamine (1:1): \( 0.0500 - 8.00\times10^{-3} = 0.0420 \text{ mol} \), equal to moles ethylamine. Mass ethylamine \( = 0.0420\times45 = 1.89 \text{ g} \). Percentage \( = \dfrac{1.89}{2.00}\times100 = 94.5\% \).

1 mark: correct initial moles HCl; 1 mark: correct moles unreacted HCl from moles NaOH; 1 mark: correct moles ethylamine (1:1 with reacted HCl); 1 mark: correct mass of ethylamine; 1 mark: correct final percentage, 94.5% (allow follow-through). [5]

(a) Initial moles NaOH \( = 0.600 \times 0.100 = 0.0600 \text{ mol} \). Moles HCl used (= moles unreacted NaOH) \( = 0.300 \times 0.04200 = 0.0126 \text{ mol} \). Moles NaOH that reacted with the ester \( = 0.0600 - 0.0126 = 0.0474 \text{ mol} \). (b) By the 1:1 ratio, moles ester \( = 0.0474 \text{ mol} \); mass \( = 0.0474\times88 = 4.17 \text{ g} \). Percentage purity \( = \dfrac{4.17}{5.00}\times100 = 83.4\% \). A back-titration is needed because the ester hydrolysis is a slow reaction that must be heated under reflux to completion; there is no sharp, observable end point (such as an indicator colour change) that can be monitored during this reaction, so an excess of NaOH is added to ensure complete hydrolysis, and the unreacted NaOH remaining afterwards is then determined by a normal, direct titration against standard acid.

(a) [3]: 1 mark correct initial moles NaOH; 1 mark correct moles unreacted NaOH from moles HCl; 1 mark correct moles NaOH reacted with the ester (by subtraction). (b) [3]: 1 mark correct mass of ester and percentage purity, 83.4% (allow follow-through); 1 mark correct explanation that the hydrolysis reaction has no observable/direct end point during reflux; 1 mark correct explanation that excess NaOH ensures complete reaction, with the excess then determined by direct titration. [6]