Year 13 CIE Chemistry: Case Study Practical Drills | 案例分析实战演练

📚 Year 13 CIE Chemistry: Case Study Practical Drills | 案例分析实战演练

Case study questions in CIE A-Level Chemistry (Paper 4) require you to integrate knowledge from multiple topics – organic synthesis, physical calculations, analytical techniques – into a single, cohesive answer. This article presents eight realistic case studies, each broken down step-by-step, to help you develop the analytical thinking and precise application skills essential for top marks.

在 CIE A-Level 化学考试（卷四）中，案例分析题要求你将多个模块的知识——有机合成、物理化学计算、分析技术——整合成一个连贯的答案。本文提供八个贴近真题的案例分析，每个都按步骤拆解，帮助你培养赢得高分所必需的分析思维与精准应用能力。

1. Introduction to Case Study Approach | 案例分析方法简介

Effective case study answering involves identifying the relevant chemical principles, planning a logical sequence of working, and presenting the solution with clear scientific terminology. Always read the entire question to grasp the context before diving into calculations or mechanism drawing.

高效地解答案例分析需要识别相关的化学原理、规划逻辑清晰的工作流程，并用清晰的科学术语呈现答案。一定要先通读整个题目，把握全局背景，再开始计算或机理推导。

Key skills include converting verbal descriptions into chemical equations, linking experimental data to rate laws or equilibrium expressions, and deducing structural features from spectra. Each case study below simulates a typical CIE scenario and models how to deconstruct it.

关键技能包括将文字描述转化为化学方程式、将实验数据关联到速率定律或平衡表达式、以及从波谱推断结构特征。下面的每个案例都模拟了典型的 CIE 考题情景，并展示了如何拆解题目。

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2. Case Study 1: Multi-step Organic Synthesis | 案例一：多步有机合成

Scenario: Propose a three-step synthesis of 4-nitroaniline starting from benzene. For each step, name the reagents and conditions, and draw the mechanism for Step 1.

情景：请以苯为原料，设计三步合成 4-硝基苯胺。对每一步，命名试剂与条件，并画出第一步的反应机理。

Step 1 identification: Benzene must first be nitrated to give nitrobenzene. Electrophilic substitution is the only pathway for benzene.

步骤 1 识别：苯必须先硝化生成硝基苯。苯环只能发生亲电取代反应。

Reagents: concentrated HNO₃ and concentrated H₂SO₄, heated under reflux at 50–55 °C. The electrophile NO₂⁺ is generated in situ.

试剂：浓 HNO₃ 和浓 H₂SO₄，50–55 °C 下回流加热。亲电试剂 NO₂⁺ 在反应中原位生成。

Mechanism (Step 1): The nitronium ion attacks the delocalised π-electron system of benzene, forming a carbocation intermediate, which then loses a proton to restore aromaticity, yielding nitrobenzene.

机理（步骤 1）：硝鎓离子进攻苯的离域 π 电子体系，形成一个碳正离子中间体，随后失去一个质子恢复芳香性，得到硝基苯。

Step 2: Reduction of nitrobenzene to phenylamine. Use tin and concentrated HCl, heat, followed by neutralisation with NaOH. The functional group –NO₂ is reduced to –NH₂.

步骤 2：硝基苯还原为苯胺。使用锡和浓盐酸，加热，随后用 NaOH 中和。官能团 –NO₂ 被还原为 –NH₂。

Step 3: The amino group is strongly activating and ortho/para-directing. Nitration of phenylamine would give a mixture of 2- and 4-nitro products, but the direct reaction is problematic because the amine can be protonated. Instead, a protection strategy is often used, but for simplicity we can state that controlled mononitration of acetanilide (from acetylation) followed by hydrolysis gives the para isomer. However, the question asks for a three-step route from benzene: (1) nitration, (2) reduction, (3) nitration of phenylamine? That would give 2- and 4- isomers but 4-nitroaniline can be separated. In CIE context, students are expected to recognise that to obtain 4-nitroaniline, one can nitrate phenylamine in acidic solution; the –NH₃⁺ group is meta-directing? Actually, aniline in nitration mixture forms anilinium ion which is meta-directing, so the major product would be 3-nitroaniline. The safe route is: nitration of benzene to nitrobenzene; reduction to phenylamine; acetylation to acetanilide; nitration (mostly para); hydrolysis to 4-nitroaniline. That’s four steps, but the case study could be adapted. Alternatively, we can use a synthesis of 4-nitrophenol from benzene: nitration, then nucleophilic substitution of chlorine, etc. Let’s adjust the scenario to: “Propose a synthesis of 4-nitrophenol from benzene” to avoid protection complications. That is a classic three-step route: nitration of benzene -> nitrobenzene; then chlorination to 1-chloro-4-nitrobenzene (need to consider directive effects: nitro group is meta-directing, so direct chlorination of nitrobenzene would give 3-chloronitrobenzene, not 4-). So that fails. Better choose a feasible route: from benzene to 4-nitrobenzoic acid via Friedel-Crafts alkylation then oxidation then nitration? Hmm.

To keep things CIE-friendly and accurate, let’s use a simple case: synthesis of 4-bromonitrobenzene from benzene? Direct bromination of benzene gives bromobenzene, then nitration gives 4-nitrobromobenzene because Br is ortho/para-directing. That’s two steps. A three-step could be: benzene to bromobenzene (Br₂, FeBr₃), then nitration to 4-nitrobromobenzene (HNO₃/H₂SO₄, 50°C). A third step: conversion of -Br to -OH via nucleophilic substitution (NaOH, high temp) to give 4-nitrophenol. That works: bromobenzene can undergo nucleophilic aromatic substitution under forcing conditions (NaOH, 300°C) but that’s not typical A-level. Actually, CIE covers nucleophilic substitution of chlorobenzene requiring high temp and pressure. So we can do: benzene -> chlorobenzene (Cl₂, AlCl₃) -> nitration to 4-nitrochlorobenzene -> hydrolysis to 4-nitrophenol (NaOH, high pressure, then acidification). This is feasible and well-known. So I’ll modify Case Study 1 to that route. Will rewrite accordingly.

Case Study 1 scenario: Propose a three-step synthesis of 4-nitrophenol from benzene. Include reagents and conditions for each step, and the mechanism for the first step.

This is correct. Let’s implement.

Step 1: Chlorination of benzene. Reagents: Cl₂, anhydrous AlCl₃ (halogen carrier). Mechanism: electrophilic substitution with Cl⁺ electrophile generated.

Step 2: Nitration of chlorobenzene. Reagents: conc HNO₃, conc H₂SO₄, 50-55°C. The Cl is ortho/para-directing, giving mainly 4-nitrochlorobenzene.

Step 3: Hydrolysis of 4-nitrochlorobenzene. Reagents: NaOH(aq), high temperature and pressure (e.g. 300°C, 200 atm), followed by acidification with dilute HCl. The -Cl is replaced by -OH.

Thus accurate. I’ll present that.

So for Case Study 1: Multi-step Organic Synthesis. I need to pair English and Chinese paragraphs. I’ll describe the problem, then each step with conditions and mechanism detail.

Will do.

Step 1: Electrophilic chlorination of benzene using chlorine gas and anhydrous AlCl₃ as a Lewis acid catalyst. The electrophile Cl⁺ is generated via Cl₂ + AlCl₃ → Cl⁺ + AlCl₄⁻. Attack by benzene π-electron cloud leads to chlorobenzene.

步骤1：使用氯气和无水 AlCl₃ 作为路易斯酸催化剂，对苯进行亲电氯化。亲电试剂 Cl⁺ 经由 Cl₂ + AlCl₃ → Cl⁺ + AlCl₄⁻ 生成。苯的 π 电子云进攻，得到氯苯。

Step 2: Nitration of chlorobenzene. The chlorine atom is ortho/para-directing due to its +M effect. Under standard nitration conditions (conc HNO₃/conc H₂SO₄, 50-55°C), the nitro group enters mainly the para position to give 4-nitrochlorobenzene.

步骤2：氯苯的硝化。氯原子由于 +M 效应是邻/对位定位基。在标准硝化条件下（浓 HNO₃/浓 H₂SO₄，50-55°C），硝基主要进入对位，生成 4-硝基氯苯。

Step 3: Nucleophilic aromatic substitution. Heating 4-nitrochlorobenzene with aqueous NaOH under high pressure replaces –Cl with –OH, forming sodium 4-nitrophenoxide. Subsequent acidification with dilute HCl liberates the phenol product, 4-nitrophenol.

步骤3：亲核芳香取代。在高压力下用 NaOH 水溶液加热 4-硝基氯苯，–Cl 被 –OH 取代，生成 4-硝基苯酚钠。随后用稀盐酸酸化，游离出酚产物 4-硝基苯酚。

Mechanism for Step 1: In the electrophilic substitution, the π-electrons of benzene attack the Cl⁺ electrophile, forming a sigma-complex (arenium ion) with a positive charge delocalised over the ring. A chloride ion AlCl₄⁻ abstracts a proton, restoring aromaticity and releasing HCl.

步骤1 机理：亲电取代中，苯的 π 电子进攻 Cl⁺ 亲电试剂，形成一个正电荷离域在环上的 σ-络合物（芳正离子）。一个氯离子 AlCl₄⁻ 夺取质子，恢复芳香性并释放 HCl。

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3. Case Study 2: Redox Titration and Percentage Purity | 案例二：氧化还原滴定与纯度计算

Scenario: A 1.50 g sample of impure iron(II) sulfate, FeSO₄·7H₂O, is dissolved in dilute sulfuric acid and titrated with 0.0200 mol dm⁻³ KMnO₄ solution. 24.50 cm³ of the KMnO₄ solution is required to reach the endpoint. Calculate the percentage purity of the iron(II) sulfate crystals.

情景：将 1.50 g 不纯的硫酸亚铁晶体 FeSO₄·7H₂O 溶于稀硫酸，用 0.0200 mol dm⁻³ KMnO₄ 溶液滴定，到达终点时消耗 24.50 cm³。计算该硫酸亚铁晶体的百分纯度。

The balanced redox equation (acidified): MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.

配平的氧化还原方程式（酸性条件下）：MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O。

Moles of MnO₄⁻ used = c × V = 0.0200 × 0.02450 = 4.90 × 10⁻⁴ mol.

所用 MnO₄⁻ 的物质的量 = c × V = 0.0200 × 0.02450 = 4.90 × 10⁻⁴ mol。

From stoichiometry, 1 mol MnO₄⁻ reacts with 5 mol Fe²⁺, so moles of Fe²⁺ = 5 × 4.90 × 10⁻⁴ = 2.45 × 10⁻³ mol.

根据化学计量，1 mol MnO₄⁻ 与 5 mol Fe²⁺ 反应，所以 Fe²⁺ 物质的量 = 5 × 4.90 × 10⁻⁴ = 2.45 × 10⁻³ mol。

Molar mass of FeSO₄·7H₂O = (55.8 + 32.1 + 64.0) + 7×18.0 = 278.0 g mol⁻¹. Mass of pure FeSO₄·7H₂O in sample = 2.45 × 10⁻³ × 278.0 = 0.681 g.

FeSO₄·7H₂O 的摩尔质量 = (55.8 + 32.1 + 64.0) + 7×18.0 = 278.0 g mol⁻¹。样品中纯物质质量 = 2.45 × 10⁻³ × 278.0 = 0.681 g。

Percentage purity = (0.681 / 1.50) × 100% = 45.4%.

百分纯度 = (0.681 / 1.50) × 100% = 45.4%。

This low purity indicates significant inert impurities. Always check that Fe²⁺ is fully oxidised by the acidified permanganate and that no air oxidation occurs before titration.

纯度较低表明存在大量惰性杂质。永远确保 Fe²⁺ 被酸性高锰酸钾完全氧化，并在滴定前不发生空气氧化。

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4. Case Study 3: Kinetics and Rate Equations | 案例三：动力学与速率方程

Scenario: The reaction 2NO(g) + O₂(g) → 2NO₂(g) was studied by measuring initial rates. Data: Experiment 1: [NO]₀ = 0.010, [O₂]₀ = 0.010, rate = 2.5 × 10⁻⁵ mol dm⁻³ s⁻¹; Expt 2: [NO]₀ = 0.010, [O₂]₀ = 0.020, rate = 5.0 × 10⁻⁵; Expt 3: [NO]₀ = 0.030, [O₂]₀ = 0.020, rate = 4.5 × 10⁻⁴. Determine the rate equation, rate constant, and its units.

情景：反应 2NO(g) + O₂(g) → 2NO₂(g) 通过测定初始速率进行研究。数据：实验 1：[NO]₀ = 0.010，[O₂]₀ = 0.010，速率 = 2.5 × 10⁻⁵ mol dm⁻³ s⁻¹；实验 2：[NO]₀ = 0.010，[O₂]₀ = 0.020，速率 = 5.0 × 10⁻⁵；实验 3：[NO]₀ = 0.030，[O₂]₀ = 0.020，速率 = 4.5 × 10⁻⁴。求速率方程、速率常数及其单位。

Comparing Expt 1 and 2: [NO] constant, [O₂] doubles → rate doubles. Rate ∝ [O₂]¹.

比较实验 1 和 2：[NO] 不变，[O₂] 加倍 → 速率加倍。速率 ∝ [O₂]¹。

Comparing Expt 2 and 3: [O₂] constant, [NO] triples (0.010 to 0.030) → rate increases from 5.0 × 10⁻⁵ to 4.5 × 10⁻⁴, a factor of 9 (3²). Rate ∝ [NO]².

比较实验 2 和 3：[O₂] 不变，[NO] 增至三倍 (0.010 → 0.030) → 速率由 5.0 × 10⁻⁵ 提高到 4.5 × 10⁻⁴，是 9 倍 (3²)。速率 ∝ [NO]²。

Rate equation: rate = k[NO]²[O₂]. Overall order = 3.

速率方程：速率 = k[NO]²[O₂]。总反应级数 = 3。

Using Expt 1: k = rate / ([NO]²[O₂]) = 2.5×10⁻⁵ / ((0.010)²×0.010) = 2.5×10⁻⁵ / 1.0×10⁻⁶ = 25 dm⁶ mol⁻² s⁻¹.

代入实验 1：k = 2.5×10⁻⁵ / ((0.010)²×0.010) = 25 dm⁶ mol⁻² s⁻¹。单位：dm⁶ mol⁻² s⁻¹。

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5. Case Study 4: Thermodynamics – Born-Haber Cycle | 案例四：热力学——玻恩-哈伯循环

Scenario: Construct a Born-Haber cycle for potassium chloride (KCl) and use the following data (in kJ mol⁻¹) to calculate the lattice enthalpy: ΔH_f°(KCl) = -437; ΔH_at°(K) = +89; first ionisation energy of K = +418; ΔH_at°(Cl₂) = +244; electron affinity of Cl = -349.

情景：构建氯化钾 (KCl) 的玻恩-哈伯循环，并使用以下数据 (kJ mol⁻¹) 计算其晶格焓：ΔH_f°(KCl) = -437; ΔH_at°(K) = +89; K 的第一电离能 = +418; ΔH_at°(Cl₂) = +244; Cl 的电子亲和能 = -349。

The cycle: K(s) → K(g) [ΔH_at]; K(g) → K⁺(g) + e⁻ [I.E.]; ½Cl₂(g) → Cl(g) [½ΔH_at(Cl₂) = +122]; Cl(g) + e⁻ → Cl⁻(g) [E.A.]; K⁺(g) + Cl⁻(g) → KCl(s) [lattice enthalpy ΔH_L]. Sum of steps = ΔH_f°.

循环：K(s) → K(g) [ΔH_at]; K(g) → K⁺(g) + e⁻ [I.E.]; ½Cl₂(g) → Cl(g) [½ΔH_at(Cl₂) = +122]; Cl(g) + e⁻ → Cl⁻(g) [E.A.]; K⁺(g) + Cl⁻(g) → KCl(s) [晶格焓 ΔH_L]。各项之和等于 ΔH_f°。

By Hess’ law: ΔH_f° = ΔH_at(K) + I.E.(K) + ½ΔH_at(Cl₂) + E.A.(Cl) + ΔH_L. Plugging in: -437 = 89 + 418 + 122 + (-349) + ΔH_L → -437 = 280 + ΔH_L → ΔH_L = -717 kJ mol⁻¹.

根据盖斯定律：ΔH_f° = ΔH_at(K) + I.E.(K) + ½ΔH_at(Cl₂) + E.A.(Cl) + ΔH_L。代入：-437 = 89 + 418 + 122 + (-349) + ΔH_L → ΔH_L = -717 kJ mol⁻¹。

The highly exothermic lattice enthalpy reflects strong electrostatic attraction between K⁺ and Cl⁻. Always add the electron affinity as a negative number; sign errors are common.

高度放热的晶格焓反映了 K⁺ 与 Cl⁻ 之间强静电吸引力。务必以负数代入电子亲和能；符号错误是常见失分点。

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6. Case Study 5: Electrochemical Cells | 案例五：电化学电池

Scenario: A cell is constructed with Cu²⁺/Cu half-cell (1.00 mol dm⁻³) connected to a Fe³⁺/Fe²⁺ half-cell (0.100 mol dm⁻³ Fe³⁺, 0.500 mol dm⁻³ Fe²⁺) via a salt bridge. Standard electrode potentials: E°(Cu²⁺/Cu) = +0.34 V, E°(Fe³⁺/Fe²⁺) = +0.77 V. Calculate the EMF of the cell at 298 K and write the overall cell reaction.

情景：由一个含 1.00 mol dm⁻³ Cu²⁺ 的 Cu²⁺/Cu 半电池和一个含 0.100 mol dm⁻³ Fe³⁺、0.500 mol dm⁻³ Fe²⁺ 的 Fe³⁺/Fe²⁺ 半电池经盐桥连接构成电池。标准电极电势：E°(Cu²⁺/Cu) = +0.34 V，E°(Fe³⁺/Fe²⁺) = +0.77 V。计算 298 K 下电池的电动势，并写出电池总反应。

Cell notation: Pt | Fe²⁺, Fe³⁺ || Cu²⁺ | Cu. The right-hand side (cathode) has higher E°, so reduction occurs there: Cu²⁺ + 2e⁻ → Cu. Left side: Fe²⁺ → Fe³⁺ + e⁻ (oxidation).

电池表示式：Pt | Fe²⁺, Fe³⁺ || Cu²⁺ | Cu。右侧（阴极）标准电极势更高，故发生还原：Cu²⁺ + 2e⁻ → Cu；左侧：Fe²⁺ → Fe³⁺ + e⁻（氧化）。

Standard EMF, E°_cell = 0.34 – 0.77 = -0.43 V, but spontaneous reaction is Fe³⁺/Fe²⁺ reducing Cu²⁺, so we reverse cell polarity: actual spontaneous cell will have Fe³⁺ reduced and Cu oxidised because E°(Fe³⁺/Fe²⁺) > E°(Cu²⁺/Cu). So cathode is Fe³⁺/Fe²⁺, anode is Cu/Cu²⁺. E°_cell = 0.77 – 0.34 = 0.43 V. We use Nernst equation for non-standard conditions of Fe³⁺/Fe²⁺ half-cell.

标准电动势 E°_cell = 0.77 – 0.34 = 0.43 V。由于 [Fe³⁺] 和 [Fe²⁺] 非标准，需用能斯特方程计算半电池电势。Fe³⁺/Fe²⁺ 半电池：E = E° + (0.0592/n) log ([Fe³⁺]/[Fe²⁺]) ; n=1. E = 0.77 + 0.0592 log(0.100/0.500) = 0.77 + 0.0592 × (-0.699) ≈ 0.77 – 0.0414 = 0.7286 V.

Cu²⁺

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