Topic: 23.2
Fluorine reacts with chlorine dioxide, ClO₂, as shown.
F₂(g) + 2ClO₂(g) → 2FClO₂(g)
The rate of the reaction is first order with respect to the concentration of F₂ and first order with respect to the concentration of ClO₂. No catalyst is involved.
(a) (i) Suggest a two-step mechanism for this reaction.
(ii) Identify the rate-determining step in this mechanism. Explain your answer.
(b) (i) Complete the rate equation for this reaction, stating the overall order of the reaction.
(ii) Use your rate equation in (i) to calculate the rate of the reaction when the concentrations of F₂ and ClO₂ are both 2.00 × 10⁻³ mol/dm³.
(c) Under different conditions, and in the presence of a large excess of ClO₂, the rate equation is as shown.
rate = \(k_1[F_2]\)
The half-life, \(t_\frac{1}{2}\), of the concentration of \(F_2\) is 4.00s under these conditions.
(i) Calculate the numerical value of \(k_1\), giving its units. Give your answer to three significant figures.
\(k_1\) = ………………………… units …………………………
(ii) An experiment is performed under these conditions in which the starting concentration of F₂ is 0.00200 mol/dm³. Draw a graph on the grid in Fig. 1.1 to show how the concentration of \(F_2\) changes over the first 12 s of the reaction.
(iii) Use your graph in Fig. 1.1 to find the rate of the reaction when the concentration of \(F_2\) is 0.00100 mol/dm³. Show your working on the graph.
▶️ Answer/Explanation
(a)(i) M1: \(F_2 + ClO_2 → FClO_2 + F\) [1]
M2: \(ClO_2 + F → FClO_2\) [1]
OR
M1: \(F_2 + ClO_2 → F_2ClO_2\)
M2: \(F_2ClO_2 + ClO_2 → 2FClO_2\)
Two balanced equations MUST add to give the overall equation.
Explanation: The proposed mechanism involves two steps where intermediates (F or \(F_2ClO_2\)) are formed and consumed, ensuring stoichiometric consistency with the overall reaction.
(a)(ii) First step AND
has one mole / molecule of F₂ and ClO₂ [1]
OR same moles of reactants as orders in rate equation.
Explanation: The rate-determining step is the slower first step because its reactants (F₂ and ClO₂) match the rate equation’s first-order dependencies.
(b)(i) rate = \(k [F_2][ClO_2]\) / rate = \(1.22 [F_2][ClO_2]\)
AND second / 2 BOTH [1]
Explanation: The rate equation reflects the first-order dependence on both reactants, with \(k = 1.22 \, \text{dm}^3 \, \text{mol}^{-1} \, \text{s}^{-1}\). The overall order is 2 (sum of individual orders).
(b)(ii) rate = \(1.22 \times 2 \times 10^{-3} \times 2 \times 10^{-3} = 4.88 \times 10^{-6} \, \text{mol dm}^{-3} \, \text{s}^{-1}\) [1]
Explanation: Substituting the given concentrations into the rate equation yields the reaction rate, maintaining unit consistency.
(c)(i) \(k_1 = \frac{0.693}{4} = 0.173 \, \text{s}^{-1}\) OR \(1.73 \times 10^{-1} \, \text{s}^{-1}\) [1]
MUST BE 3SF
Explanation: For a first-order reaction, \(k_1 = \frac{\ln(2)}{t_{1/2}}\). The units are \(\text{s}^{-1}\) because the half-life is in seconds.
(c)(ii) At 4 s = 0.001 mol/dm³, at 8 s = 0.0005 mol/dm³, at 12 s = 0.00025 mol/dm³. Smooth curve required. ALL correct [1]
Explanation: The graph shows exponential decay, halving every 4 seconds (half-life). Points must align with calculated concentrations.
(c)(iii) Tangent drawn at 0.00100 mol/dm³.
Gradient calculation: rate = \(\frac{\Delta y}{\Delta x}\) in range \(1.5–2.0 \times 10^{-4} \, \text{mol dm}^{-3} \, \text{s}^{-1}\) [1]
Explanation: The rate at a specific concentration is found from the gradient of the tangent to the curve at that point, reflecting the instantaneous rate.
Topic: 24.2
(a) Define \(K_w\) mathematically by completing the expression.
(b) Two solutions, V and W, are described.
• V is HCl(aq).
• W is NaOH(aq).
• The concentration of HCl in V is the same as the concentration of NaOH in W.
• The pH values of V and W differ by exactly 11.00 at 298K.
(i) Calculate the concentration of HCl in V
(ii) Equal volumes of the two solutions V and W are mixed, giving solution X. Name solution X and state its pH.
(iii) A 1 cm³ sample of 1.0 mol/dm³ HNO₃ is added to 100 cm³ of solution X, forming mixture Y. A 1 cm³ sample of 1.0 mol/dm³ KOH is added to 100 cm³ of solution X, forming mixture Z. Estimate the pH of mixtures Y and Z. No calculations are required.
(c) (i) CH₃CH₂COOH, CH₃CCl₂COOH, and H₂SO₄ are all acidic. Suggest the trend in the relative acid strength of these three compounds. Explain your answer
(ii) When concentrated H₂SO₄ is added to water a series of acid-base reactions occurs. There are three conjugate acid-base pairs that can be identified during this series of reactions. Write the formulae of these three conjugate acid-base pairs.
conjugate acid 1 ………………………………….. conjugate base 1 …………………………………..
conjugate acid 2 ………………………………….. conjugate base 2 …………………………………..
conjugate acid 3 ………………………………….. conjugate base 3 …………………………………..
(d) The partition coefficient, \(K_{pc}\), of a substance, Q, between hexane and water is 7.84 at 298K. Q is more soluble in hexane than it is in water.
(i) Define partition coefficient, \(K_{pc}\).
(ii) 5.00g of Q is shaken with a mixture of 100.0 cm³ of water and 100.0 cm³ of hexane at 298K and left until there is no further change in concentrations. Calculate the mass of Q dissolved in the water.
(iii) A sample of Q is shaken with a different mixture of water and hexane and left until there is no further change in concentrations. It is found that the mass of Q dissolved in each solvent is the same. Use the \(K_{pc}\) value to suggest possible values for the volume of water used and the volume of hexane used.
(iv) Q is more soluble in hexane than it is in water. It is suggested that Q is one of KCl, CH₃(CH₂)₄OH, or HCOOH. Identify Q. Explain your answer
▶️ Answer/Explanation
(a) \(K_w = [H^+][OH^-]\)
Explanation: The ionic product of water (\(K_w\)) is defined as the product of the concentrations of hydrogen ions (\(H^+\)) and hydroxide ions (\(OH^-\)) in water at equilibrium.
(b)(i) Concentration of HCl in V: 0.032 mol/dm³
Explanation: Given the pH difference of 11.00, the pH of V (HCl) is 1.5 and W (NaOH) is 12.5. The concentration of HCl is \(10^{-1.5} = 0.032 \, \text{mol/dm³}\).
(ii) Solution X: Sodium chloride (NaCl), pH = 7
Explanation: Equal volumes of equimolar HCl and NaOH neutralize each other, forming NaCl (a neutral salt) with pH 7.
(iii) Mixture Y: pH 1–3; Mixture Z: pH 11–13
Explanation: Adding HNO₃ (strong acid) to X makes Y acidic (pH 1–3), while adding KOH (strong base) makes Z basic (pH 11–13).
(c)(i) Acid strength trend: H₂SO₄ > CH₃CCl₂COOH > CH₃CH₂COOH
Explanation: H₂SO₄ is a strong acid (fully dissociates). CH₃CCl₂COOH is stronger than CH₃CH₂COOH due to the electron-withdrawing Cl atoms stabilizing the conjugate base.
(ii) Conjugate acid-base pairs:
1. H₃O⁺ (acid) / H₂O (base)
2. H₂SO₄ (acid) / HSO₄⁻ (base)
3. HSO₄⁻ (acid) / SO₄²⁻ (base)
(d)(i) Partition coefficient (\(K_{pc}\)): Ratio of solute concentrations in two immiscible solvents at equilibrium.
(ii) Mass of Q in water: 0.566 g
Explanation: Let \(x\) be mass in water. Then \(5 – x\) is in hexane. \(K_{pc} = \frac{5 – x}{x} = 7.84 \implies x = 0.566 \, \text{g}\).
(iii) Example volumes: 78.4 cm³ water / 10 cm³ hexane
Explanation: For equal masses, the volume ratio must match \(K_{pc}\): \(\frac{V_{\text{water}}}{V_{\text{hexane}}} = 7.84\).
(iv) Q is CH₃(CH₂)₄OH (pentanol).
Explanation: Pentanol is non-polar (large hydrocarbon chain), making it more soluble in hexane than water. KCl and HCOOH are polar/ionic.
Topic: 23.4
Hydrogen peroxide is a liquid at 298K. It is moderately stable under room conditions but will decompose quickly if a catalyst is added.
Reaction 1: \(2H_2O_2(l) \to 2H_2O (l) + O_2 (g)\)
(a) (i) Define entropy.
(ii) Predict the sign of the standard entropy change of reaction 1. Explain your answer.
(b) Some bond energy data are shown in Table 3.1.

Use the data in Table 3.1 to show that the enthalpy change of the following reaction is \(–196 \, \text{kJ mol}^{–1}\).
\(2H_2O_2(g) \to 2H_2O(g) + O_2(g)\)
(c) Some standard entropies, \(S^o\), are shown in Table 3.2.

The enthalpy change and Gibbs free energy change for the following reaction are shown.
\(2H_2O_2(l) \to 2H_2O(l) + O_2(g)\)
\(ΔH^o = –196 \, \text{kJ mol}^{-1}\)
\(ΔG^o = –238 \, \text{kJ mol}^{-1}\)
Use the data given to calculate the standard entropy of oxygen, \(S^o\), \(O_2(g)\).
(d) The decomposition of \(H_2O_2(aq)\) is catalysed by aqueous iron(III) chloride and by silver metal. Identify which of these two catalysts is acting as a homogeneous catalyst. Explain your answer.
(e) The \(E^o\) values for two electrode reactions are given.
\(H_2O_2 + 2H^+ + 2e^− \rightleftharpoons 2H_2O\) \(E^o = +1.77 \, \text{V}\)
\(Cr^{3+} + e^− \rightleftharpoons Cr^{2+}\) \(E^o = −0.41 \, \text{V}\)
(i) An electrochemical cell is constructed with the following half-cells (electrodes):
• An acidified solution of \(H_2O_2\), a platinum wire
• \(Cr^{2+}\) mixed with \(Cr^{3+}\), a platinum wire.
Identify the positive half-cell and calculate the standard cell potential, \(E^o_{cell}\).
(ii) Calculate the value of \(ΔG^o\) for the cell reaction that occurs, per mole of \(H_2O_2\).
(f) The \(E^o\) values for two electrode reactions are given.
\(H_2O_2 + 2H^+ + 2e^− \rightleftharpoons 2H_2O\) \(E^o = +1.77 \, \text{V}\)
\(Co^{3+} + e^– \rightleftharpoons Co^{2+}\) \(E^o = +1.82 \, \text{V}\)
An electrochemical cell is constructed with the following half-cells.
Half-cell 1: An acidified solution of \(H_2O_2\) under standard conditions, a platinum wire
Half-cell 2: A solution containing \(0.020 \, \text{mol dm}^{-3} \, Co^{3+}\) and \(2.0 \, \text{mol dm}^{-3} \, Co^{2+}\), a platinum wire
(i) Use the Nernst equation to calculate the value of \(E\), the electrode potential of half-cell 2 under these conditions.
(ii) Write an equation for the cell reaction that occurs in this cell under these conditions.
(g) (i) Define enthalpy change of hydration, \(\Delta H_{hyd}\).
(ii) Aluminium fluoride, \(AlF_3\), is an ionic solid.
Complete and label the energy cycle to show the relationship between:
● The enthalpy change of solution of \(AlF_3\), \(\Delta H^o_{sol}\)
● The lattice energy of \(AlF_3\), \(\Delta H^o_{latt}\)
● The enthalpy changes of hydration of \(Al^{3+}\) and \(F^-\), \(\Delta H^o_{hyd}\). Include state symbols for all substances and ions.

(iii) Relevant data for this question are given.

Use these data and your energy cycle in (g)(ii) to calculate the \(\Delta H^o_{latt}\) of \(AlF_3\).
▶️ Answer/Explanation
(a)(i)
Final Answer: Entropy is the measure of disorder or the number of possible arrangements of particles in a system.
Explanation: Entropy quantifies the randomness or dispersal of energy in a system, reflecting the number of microscopic configurations that correspond to a thermodynamic system’s macroscopic state.
(a)(ii)
Final Answer: Positive (+).
Explanation: The reaction produces a gas (\(O_2\)) from a liquid (\(H_2O_2\)), increasing the number of gas molecules and thus the disorder of the system.
(b)
Final Answer: \(\Delta H = –196 \, \text{kJ mol}^{-1}\).
Explanation: Using bond energies: \(\Delta H = (2 \times 496) – (4 \times 460 + 1 \times 496) = –196 \, \text{kJ mol}^{-1}\).
(c)
Final Answer: \(S^o(O_2(g)) = 205 \, \text{J K}^{-1} \text{mol}^{-1}\).
Explanation: Using \(\Delta G = \Delta H – T\Delta S\), solve for \(\Delta S\) and then use standard entropies to find \(S^o(O_2(g))\).
(d)
Final Answer: Iron(III) chloride (\(FeCl_3\)) is the homogeneous catalyst.
Explanation: It is in the same aqueous phase as the reactants, unlike silver, which is a solid heterogeneous catalyst.
(e)(i)
Final Answer: Positive half-cell: \(H_2O_2\); \(E^o_{cell} = +2.18 \, \text{V}\).
Explanation: The half-cell with the higher \(E^o\) value (\(H_2O_2\)) is positive. \(E^o_{cell} = 1.77 – (-0.41) = 2.18 \, \text{V}\).
(e)(ii)
Final Answer: \(\Delta G^o = –420.7 \, \text{kJ mol}^{-1}\).
Explanation: Using \(\Delta G^o = –nFE^o_{cell}\), with \(n = 2\) and \(F = 96500 \, \text{C mol}^{-1}\).
(f)(i)
Final Answer: \(E = +1.702 \, \text{V}\).
Explanation: Using the Nernst equation: \(E = 1.82 + 0.059 \log \left( \frac{0.02}{2} \right)\).
(f)(ii)
Final Answer: \(H_2O_2 + 2H^+ + 2Co^{2+} \to 2H_2O + 2Co^{3+}\).
Explanation: The half-cell with the higher potential (\(Co^{3+}/Co^{2+}\)) undergoes reduction, while \(H_2O_2\) is oxidized.
(g)(i)
Final Answer: Enthalpy change when one mole of gaseous ions dissolves in water to form an aqueous solution.
Explanation: \(\Delta H_{hyd}\) represents the energy change when gaseous ions are surrounded by water molecules.
(g)(ii)

Explanation: The cycle shows the relationship between \(\Delta H^o_{sol}\), \(\Delta H^o_{latt}\), and \(\Delta H^o_{hyd}\) for \(AlF_3\).
(g)(iii)
Final Answer: \(\Delta H^o_{latt} = –5999 \, \text{kJ mol}^{-1}\).
Explanation: Using the cycle: \(\Delta H^o_{latt} = \Delta H^o_{sol} – \Delta H^o_{hyd}(Al^{3+}) – 3 \Delta H^o_{hyd}(F^-)\).
Topic: 28.4
(a) Cobalt(II) nitrate, Co(NO₃)₂, is a reddish-brown crystalline solid. It dissolves in water to form a solution containing [Co(H₂O)₆]²⁺ complex ions.
(i) Complete Table 4.1 giving the formula of the cobalt-containing species that is formed in each of the three reactions described.

(ii) Describe the colour change seen in reaction 3.
original colour of [Co(H₂O)₆]²⁺(aq) …………………………
final colour after addition of an excess of conc. HCl(aq) ………………………………….
(b) Calcium nitrate, Ca(NO₃)₂, is a white crystalline solid. When heated, it starts to decompose at approximately 500°C.
(i) Write an equation for the decomposition of Ca(NO₃)₂.
(ii) Suggest temperatures at which Mg(NO₃)₂ and Ba(NO₃)₂ start to decompose. Explain your answer.
temperature at which Mg(NO₃)₂ starts to decompose …………………………………………. °C
temperature at which Ba(NO₃)₂ starts to decompose ……………………………………………°C
▶️ Answer/Explanation
(a)(i)

Explanation: – Reaction 1: Addition of NaOH forms Co(OH)₂ (pink precipitate). – Reaction 2: Excess NH₃ forms [Co(NH₃)₆]²⁺ (straw-colored solution). – Reaction 3: Excess HCl forms [CoCl₄]²⁻ (blue solution).
(a)(ii) Original colour: pink; Final colour: blue.
Explanation: The hexaaquacobalt(II) ion ([Co(H₂O)₆]²⁺) is pink, while the tetrachlorocobaltate(II) ion ([CoCl₄]²⁻) is blue.
(b)(i) \(2Ca(NO₃)₂ \rightarrow 2CaO + 4NO₂ + O₂\) OR \(Ca(NO₃)₂ \rightarrow CaO + 2NO₂ + \frac{1}{2}O₂\)
Explanation: Thermal decomposition of Ca(NO₃)₂ produces calcium oxide (CaO), nitrogen dioxide (NO₂), and oxygen (O₂).
(b)(ii) – Mg(NO₃)₂ decomposes below 480°C. – Ba(NO₃)₂ decomposes above 520°C. Explanation: Smaller cations (Mg²⁺) have higher charge density, polarizing the nitrate ion more strongly and lowering decomposition temperature. Larger cations (Ba²⁺) polarize less, requiring higher temperatures.
Topic: 28.3
Transition elements behave as catalysts and can form complex ions.
(a) Explain why transition elements behave as catalysts.
(b) Silver forms the linear complex ion \([Ag(CN)_2]^–\). Copper forms the tetrahedral complex ion \([Cu(CN)_4]^{3−}\). Titanium forms the complex [TiCl₄(diars)₂], where diars is a neutral bidentate ligand.
(i) State the oxidation state and the coordination number of titanium in [TiCl₄(diars)₂].
(ii) Draw three-dimensional diagrams to show the shapes of [Ag(CN)₂]⁻ and [Cu(CN)₄]³⁻ in the boxes. Label one bond angle on each diagram.

(c) The numerical value of the stability constant, Kstab, of the copper(I) complex [Cu(CN)₄]³⁻ is \(2.0 × 10^{27}\).
(i) Write an expression for the \(K_{stab}\) of [Cu(CN)₄]³⁻.
(ii) In a solution the concentrations of CN⁻ and [Cu(CN)₄]³⁻ are both 0.0010 mol/dm³. Use your expression from (c)(i) and the value of \(K_{stab}\) to calculate the concentration of \(Cu^+\)(aq) in this solution.
(d) A piece of a copper-containing alloy has a mass of 0.567g. It is dissolved in an acid giving 100.0 cm³ of a blue solution in which all the copper is present as Cu²⁺ ions. An excess of KI(aq) is added to a 25.0 cm³ sample of this solution. All of the copper is precipitated as white CuI(s). Cu²⁺ ions are the only component in the solution that react with KI(aq). This is reaction 1.
reaction 1 \(2Cu^{2+} + 4I^− \to 2CuI + I_2\)
The liberated I₂ is then titrated with 0.0200 moldm⁻³ S₂O₃²⁻. This is reaction 2.
reaction 2 \(I_2 + 2S_2O_3^{2–} \to 2I^− + S_4O_6^{2–}\)
The titration requires 20.10 cm³ of 0.0200 mol/dm³ S₂O₃²⁻ to reach the end-point.
(i) Calculate the number of moles of I₂ that are reduced in this titration.
(ii) Calculate the number of moles of copper in the original piece of alloy.
(iii) Calculate the percentage of copper in the alloy.
(iv) Suggest why a solution of Cu²⁺ is coloured but solid CuI is white.
▶️ Answer/Explanation
(a) Transition elements act as catalysts because they exhibit multiple stable oxidation states and have vacant d-orbitals. These orbitals can form temporary bonds with reactants, lowering activation energy and facilitating reaction pathways.
(b)(i) In [TiCl₄(diars)₂]:
– Oxidation state: +4 (Ti loses 4 electrons to Cl⁻ and neutral diars ligands).
– Coordination number: 8 (4 from Cl⁻ and 2 bidentate diars ligands, each contributing 2 donor atoms).
(b)(ii) Shapes of complex ions:
– [Ag(CN)₂]⁻: Linear (bond angle: 180°).
– [Cu(CN)₄]³⁻: Tetrahedral (bond angle: 109.5°).

(c)(i) Stability constant expression:
\( K_{stab} = \frac{[[Cu(CN)_4]^{3–}]}{[Cu^+][CN^–]^4} \).
(c)(ii) Calculation of \([Cu^+]\):
Given \( K_{stab} = 2.0 \times 10^{27} \), \([[Cu(CN)_4]^{3–}] = [CN^–] = 0.0010 \text{ mol/dm}^3 \):
\( [Cu^+] = \frac{0.0010}{(2.0 \times 10^{27})(0.0010)^4} = 5.0 \times 10^{-19} \text{ mol/dm}^3 \).
(d)(i) Moles of I₂ reduced:
\( \text{Moles of } S_2O_3^{2–} = 0.0200 \times 0.02010 = 4.02 \times 10^{-4} \text{ mol} \).
From reaction 2, \( \text{Moles of } I_2 = \frac{4.02 \times 10^{-4}}{2} = 2.01 \times 10^{-4} \text{ mol} \).
(d)(ii) Moles of Cu in alloy:
From reaction 1, \( \text{Moles of Cu}^{2+} = 2 \times \text{Moles of } I_2 = 4.02 \times 10^{-4} \text{ mol} \) (in 25 cm³).
Total moles in 100 cm³: \( 4.02 \times 10^{-4} \times 4 = 1.608 \times 10^{-3} \text{ mol} \).
(d)(iii) Percentage of Cu:
\( \text{Mass of Cu} = 1.608 \times 10^{-3} \times 63.5 = 0.1021 \text{ g} \).
\( \% \text{Cu} = \left( \frac{0.1021}{0.567} \right) \times 100 = 18.0\% \).
(d)(iv) Cu²⁺ solutions are colored due to d-d transitions in the partially filled d-orbitals (d⁹ configuration). Solid CuI is white because Cu⁺ has a full d¹⁰ configuration, preventing such transitions.
Topic: 28.3
(a) Five ligands are listed in Table 6.1.
(i) Complete Table 6.1 using the words monodentate, bidentate and polydentate only. Each of these three words may be used once, more than once, or not at all.
(ii) The molecule H₂NCH₂CH₂NHCH₂CH₂NH₂ is a tridentate ligand. Suggest the meaning of tridentate ligand.
(iii) Suggest how H₂NCH₂CH₂NHCH₂CH₂NH₂ acts as a tridentate ligand.
(b) Nickel forms the octahedral complex \([Ni(en)_2(H_2O)_2]^{2+}\). This complex can exist in three isomeric forms, listed in Table 6.2. One of these forms is a trans isomer, the other forms are two different cis isomers.
(i) Complete Table 6.2 using the terms polar or non-polar. Each term may be used once, more than once, or not at all.
(ii) Describe the difference between cis isomer 1 and cis isomer 2.
▶️ Answer/Explanation
(a)(i)

Explanation: – NH₃ and CN⁻ are monodentate (donate 1 lone pair). – C₂O₄²⁻ is bidentate (2 O atoms donate lone pairs). – EDTA⁴⁻ is polydentate (hexadentate, 6 lone pairs).
(a)(ii) A tridentate ligand donates three lone pairs to a central metal ion.
Explanation: It forms three dative covalent bonds with the metal, using three donor atoms (e.g., N in H₂NCH₂CH₂NHCH₂CH₂NH₂).
(a)(iii) Each nitrogen atom in H₂NCH₂CH₂NHCH₂CH₂NH₂ donates a lone pair to the metal ion.
Explanation: The ligand uses its three N atoms (two primary amines and one secondary amine) to coordinate with the metal, creating an octahedral geometry.
(b)(i)

Explanation: – Trans isomer is non-polar (symmetrical dipole moments cancel out). – Cis isomers are polar (asymmetric charge distribution).
(b)(ii) Cis isomer 1 and cis isomer 2 are optical isomers (enantiomers).
Explanation: They are non-superimposable mirror images that rotate plane-polarized light in opposite directions. This occurs due to the spatial arrangement of the bidentate ethylenediamine (en) ligands.
Topic: 34.4
Sunset Yellow is an additive used for colouring foods. A synthetic route for making Sunset Yellow is shown. Molecules E and G each contain one \(–SO_3^–\) group. These groups are unchanged in the formation of Sunset Yellow.

(a) State the molecular formula of the Sunset Yellow anion.
(b) Deduce the structures of E, F and G and draw them in the boxes in Fig. 7.1.
(c) Suggest suitable reagents and conditions for step 1 and 2.
(d) Predict the number of peaks in the carbon-13 NMR spectrum of the Sunset Yellow anion.
▶️ Answer/Explanation
(a) The molecular formula of the Sunset Yellow anion is:
\[ C_{16}H_{10}N_2O_7S_2^{2–} \]
Explanation: The structure consists of two benzene rings, two azo (–N=N–) linkages, two sulfonate (–SO₃⁻) groups, and hydroxyl (–OH) groups, giving the formula \(C_{16}H_{10}N_2O_7S_2^{2–}\).
(b) The structures of E, F, and G are:

Explanation:
- E is a diazonium salt formed from sulfanilic acid.
- F is the azo intermediate formed by coupling E with G.
- G is a naphthalene derivative with a sulfonate group.
(c) Reagents and conditions:
Step 1: NaNO₂ + HCl (diazotization), temperature ≤ 10°C.
Step 2: NaOH(aq) or alkaline conditions (for azo coupling).
Explanation:
- Step 1 requires cold conditions to prevent decomposition of the diazonium salt.
- Step 2 requires alkaline conditions to facilitate the coupling reaction.
(d) The number of peaks in the \(^{13}\text{C}\) NMR spectrum of the Sunset Yellow anion is 14.
Explanation: The molecule has 14 unique carbon environments due to symmetry and the distinct chemical environments of the aromatic rings, azo group, and sulfonate groups.
Topic: 33.2
Capsaicin is found in chilli peppers.

You should assume the CH₃O group is unreactive in the reactions involved in this question.
(a) Name all the functional groups in capsaicin in addition to the CH₃O group.
(b) Complete the equation for the reaction of capsaicin with an excess of Br₂(aq) in the dark. Draw the structure of the organic product in the labelled box.

(c) Capsaicin is heated with an excess of hydrogen gas in the presence of platinum metal. The six-membered ring reacts in the same way as benzene under these conditions. Draw the structure of the organic product formed.
(d) When capsaicin is treated with reagent J under suitable conditions one of the products is methylpropanoic acid, CH₃CH(CH₃)COOH.
(i) Identify reagent J and any necessary conditions.
(ii) There are three different peaks in the proton (1H) NMR spectrum of CH₃CH(CH₃)COOH in CDCl₃.


(e) (i) Capsaicin is heated with an excess of hot aqueous NaOH

(ii) Name the two types of reaction occurring in (e)(i).
(f) Draw the structure of the organic product L formed when capsaicin is treated with \(LiAlH_4\) in dry ether.

▶️ Answer/Explanation
(a)
Answer: Phenol, amide, and alkene (C=C).
Explanation: The functional groups in capsaicin are: phenol (hydroxyl attached to benzene), amide (CONH), and alkene (C=C in the side chain).
(b)

Explanation: Br₂ adds across the C=C bond (electrophilic addition) to form a dibromo compound. The aromatic ring remains unaffected in the dark.
(c)

Explanation: H₂/Pt hydrogenates both the C=C bond (to give an alkane) and the benzene ring (to a cyclohexane).
(d)(i)
Answer: Hot concentrated acidified KMnO₄.
Explanation: KMnO₄ oxidizes the side chain to carboxylic acids, producing methylpropanoic acid as one product.
(d)(ii)

Explanation: The three peaks correspond to: COOH proton (~12 ppm), CH proton (~2 ppm), and CH₃ protons (~1 ppm).
(e)(i)

(e)(ii)
Answer: Hydrolysis and neutralisation.
Explanation: NaOH hydrolyzes the amide bond (forming carboxylate and amine) and neutralizes the phenol.
(f)

Explanation: \(LiAlH_4\) reduces the amide to an amine and the phenol to a cyclohexanol (though the latter is less common).
Topic: 35.2
(a) Benzoyl chloride, C₆H₅COCl, can be made from ethyl benzene in a two-step process. A reaction scheme is shown.
(i) Draw the intermediate organic compound M in the box.
(ii) Suggest suitable reagents and conditions for step 1 and step 2.
(iii) Identify the type of reaction in step 1 and step 2.
(b) C₆H₅COCl reacts with phenol, C₆H₅OH, to give the ester phenyl benzoate, C₆H₅COOC₆H₅. An incomplete description of the mechanism of this reaction is shown in Fig. 9.1.
(i) Complete the mechanism in Fig. 9.1 and include:
● all relevant dipoles (δ+ and δ–) and full electric charges (+ and –) on the species in box one and in box two
● all relevant lone pairs on the species in box one and in box two
● all relevant curly arrows to show the movement of electron pairs in box one and in box two
● the formula of the second product in box three.
(ii) Name this mechanism.
(c) Benzoyl chloride, chlorobenzene and chloroethane differ in their rates of hydrolysis when each compound is added separately to water at 25°C. Suggest the relative ease of hydrolysis of these three compounds. Explain your answer.
▶️ Answer/Explanation
(a)(i)
Explanation: The intermediate M is benzoic acid (C₆H₅COOH), formed by oxidation of ethylbenzene.
(a)(ii)
Step 1: Reagents: Hot alkaline KMnO₄ (oxidation of ethylbenzene to benzoic acid).
Step 2: Reagents: SOCl₂ (or PCl₅/PCl₃) + heat (conversion of benzoic acid to benzoyl chloride).
Explanation: Step 1 oxidizes the side chain (–CH₂CH₃) to –COOH. Step 2 replaces –OH with –Cl using thionyl chloride.
(a)(iii)
Step 1: Oxidation.
Step 2: Nucleophilic substitution.
Explanation: Step 1 involves oxidation (gain of O), while step 2 is nucleophilic substitution (Cl⁻ replaces OH⁻).
(b)(i)
Key Features:
● Box 1: Dipole on C=O (δ+ on C, δ– on O), lone pairs on O, curly arrows showing nucleophilic attack by phenol and C=O bond breaking.
● Box 2: Charges (+ on former phenol O, – on acyl O), curly arrows for proton transfer and Cl⁻ departure.
● Box 3: Second product = HCl.
Explanation: The mechanism involves nucleophilic addition-elimination, where phenol attacks the electrophilic C in C₆H₅COCl, forming a tetrahedral intermediate that collapses to release HCl.
(b)(ii) Nucleophilic addition-elimination.
Explanation: The reaction proceeds via initial nucleophilic attack (addition) followed by elimination of HCl.
(c)
Order of hydrolysis ease: Benzoyl chloride > chloroethane > chlorobenzene.
Explanation:
● Benzoyl chloride: Highly reactive due to electron-withdrawing C=O weakening the C-Cl bond.
● Chloroethane: Moderate reactivity; alkyl groups slightly stabilize the C-Cl bond.
● Chlorobenzene: Very slow hydrolysis due to partial double-bond character (resonance) strengthening the C-Cl bond.