📚 Year 13 OCR Chemistry: Case Study Practice | Year 13 OCR 化学：案例分析实战演练
In A Level Chemistry, the ability to apply knowledge across different topics to solve unfamiliar problems is essential for top marks. This article presents a series of integrated case studies that mirror the style of OCR synoptic questions, combining organic synthesis, spectroscopic analysis, mechanistic reasoning, and practical skills. Each case will guide you through a systematic approach to deconstruct complex scenarios, just as you would in an exam.
在 A Level 化学中，将不同专题的知识整合起来解决陌生问题的能力是获得高分的关键。本文通过一系列综合案例分析，模拟 OCR 的综合性考题风格，融合有机合成、波谱分析、机理推断和实验技能。每个案例都会引导你一步步系统地解构复杂情境，就像你在考试中需要做的那样。
1. Introduction to Integrated Case Studies | 综合案例分析引言
OCR examination papers often present a scenario where multiple strands of the specification are woven together. You might be given spectral data for an intermediate in a multi-step synthesis and asked to deduce its structure, then outline a mechanism for its formation, and finally suggest a purification method. Developing a structured problem-solving routine is crucial.
OCR 试卷经常将考纲中的多个分支融合到一个情境中。比如，给你一个多步合成中间体的波谱数据，要求推断结构，然后写出其形成机理，最后建议纯化方法。培养一套结构化的解题流程至关重要。
2. Case 1: Identifying an Unknown Compound | 案例一：鉴定未知化合物
An organic liquid (X) with molecular formula C₄H₈O₂ gives a positive result with 2,4-dinitrophenylhydrazine but does not react with Tollens’ reagent. Its IR spectrum shows a strong absorption at 1735 cm⁻¹ and a broad peak around 3000 cm⁻¹. The ¹H NMR spectrum has three signals: a singlet at δ 2.1 (3H), a triplet at δ 1.3 (3H), and a quartet at δ 4.1 (2H). We will work through the deduction step by step.
一种分子式为 C₄H₈O₂ 的有机液体 (X) 与 2,4-二硝基苯肼反应呈阳性，但不与托伦斯试剂反应。其红外光谱在 1735 cm⁻¹ 处有强吸收，在 3000 cm⁻¹ 附近有宽峰。¹H 核磁共振谱有三个信号：δ 2.1 处单峰 (3H)、δ 1.3 处三重峰 (3H) 和 δ 4.1 处四重峰 (2H)。我们将逐步推演。
3. Using IR and Mass Spectrometry | 运用红外与质谱分析
The IR peak at 1735 cm⁻¹ is characteristic of a C=O stretch in an ester or a saturated ketone/aldehyde, but since broad O–H absorption around 3000 cm⁻¹ is present (overlapping C–H), a carboxylic acid is likely. The broadness suggests hydrogen bonding. The molecular formula C₄H₈O₂ with one C=O and one O–H group accounts for all atoms, leaving a C₃H₇ fragment. The mass spectrum would show a molecular ion peak at m/z = 88.
红外 1735 cm⁻¹ 的吸收峰是酯类或饱和酮/醛中 C=O 伸缩振动的特征，但 3000 cm⁻¹ 附近的宽峰（与 C–H 重叠）表明存在 O–H，很可能是羧酸。峰形较宽说明有氢键。分子式 C₄H₈O₂，含一个 C=O 和一个 O–H 基团，剩余 C₃H₇ 片段。质谱中会显示 m/z = 88 的分子离子峰。
The positive 2,4-DNP test confirms a carbonyl group, while a negative Tollens’ test rules out an aldehyde. This supports the carboxylic acid hypothesis, since carboxylic acids do not give a positive Tollens’ test. The carbonyl is part of a carboxyl group, not an aldehyde.
2,4-二硝基苯肼呈阳性证实有羰基，而托伦斯试验阴性排除醛。这支持了羧酸的推测，因为羧酸不会与托伦斯试剂反应。羰基属于羧基，而非醛基。
4. Interpreting ¹H NMR Data | 解读 ¹H 核磁共振谱
The NMR signals provide the final pieces: a singlet at δ 2.1 ppm (3H) indicates a methyl group adjacent to a carbonyl (CH₃C=O), deshielded by the C=O. A triplet at δ 1.3 (3H) and a quartet at δ 4.1 (2H) together suggest an ethyl group (–CH₂CH₃) where the CH₂ is deshielded, likely attached to an electronegative oxygen. The coupling pattern (triplet + quartet) indicates a CH₂CH₃ spin system, typical of an ethyl ester or an ethoxy group.
核磁共振给出最后线索：δ 2.1 ppm 处的单峰 (3H) 表明与羰基相邻的甲基 (CH₃C=O)，受 C=O 去屏蔽影响。δ 1.3 的三重峰 (3H) 和 δ 4.1 的四重峰 (2H) 共同暗示一个乙基 (–CH₂CH₃)，其中 CH₂ 被去屏蔽，可能连接在电负性氧上。三重峰+四重峰的耦合模式表明 CH₂CH₃ 自旋体系，这正是乙酯或乙氧基的特征。
Putting everything together, the compound must be propanoic acid, CH₃CH₂COOH. However, check the NMR: propanoic acid would show the CH₂ as a quartet (δ ~2.3–2.4) and CH₃ as a triplet (δ ~1.1), with a broad O–H around δ 10–12. Our signals are shifted: the CH₂ is at δ 4.1 – too far downfield for propanoic acid. That suggests the CH₂ is attached to an oxygen, so we have an ester, not an acid. The singlet at δ 2.1 is an acetyl methyl (CH₃COO–). Thus the structure is ethyl ethanoate, CH₃COOCH₂CH₃. Indeed, ethyl ethanoate (C₄H₈O₂) has a carbonyl in an ester (1735 cm⁻¹), no broad O–H (the 3000 cm⁻¹ broad peak is just C–H, not O–H – re-evaluation: the broad peak at 3000 cm⁻¹ is likely the C–H stretches overlapping, not truly an O–H; esters lack O–H). Correct interpretation: the IR shows ester C=O, and no O–H; the NMR confirms an ethoxy group. The earlier O–H assumption was wrong. The broadness around 3000 cm⁻¹ is typical of C–H stretching, not O–H. So compound X is ethyl ethanoate.
综合来看，化合物最初可能被认为是丙酸，但核磁位移不符：如果是丙酸，CH₂ 的化学位移应在 δ 2.3–2.4 左右，而此处是 δ 4.1，表明 CH₂ 直接连在氧上。因此是酯类。δ 2.1 是乙酰基的甲基 (CH₃COO–)。所以结构为乙酸乙酯 CH₃COOCH₂CH₃。红外中 3000 cm⁻¹ 附近并非 O–H 宽峰，而是 C–H 伸缩振动叠加，酯没有 O–H。因此化合物 X 鉴定为乙酸乙酯。
5. Chemical Verification and Derivatisation | 化学验证与衍生化
To confirm, we could hydrolyse the ester with dilute acid or base. Acid hydrolysis yields ethanoic acid and ethanol; the products can be identified by smell or by further tests. Alkaline hydrolysis produces the carboxylate salt and ethanol. Distilling off ethanol and testing with acidified dichromate(VI) would give a green colour, confirming a primary or secondary alcohol. The remaining solution after acidification would smell of vinegar.
为确证结构，可用稀酸或稀碱水解该酯。酸性水解得到乙酸和乙醇；产物可通过气味或进一步检验鉴别。碱性水解得到羧酸盐和乙醇。蒸出乙醇后用酸化重铬酸钾(VI)检验，出现绿色可确认伯醇或仲醇。剩余溶液酸化后会有醋味。
6. Case 2: Designing a Multi-Step Synthesis | 案例二：多步合成路线设计
Consider the synthesis of 4-nitrophenyl ethanoate from benzene. This requires introducing an ester group and a nitro group onto the aromatic ring. Since esterification of phenol with ethanoic acid is possible, we could first make phenol, then esterify, then nitrate. However, the directing effects must be considered.
考虑由苯合成 4-硝基苯基乙酸酯。这需要往苯环上引入酯基和硝基。由于苯酚可与乙酸酯化，可以先制备苯酚，再酯化，最后硝化。但必须考虑定位效应。
7. Retrosynthetic Analysis and Directing Groups | 逆合成分析与定位基团
The target molecule has an ester group at position 1 and a nitro group at position 4. The ester group is meta-directing and deactivating, while the nitro group is strongly meta-directing and deactivating. If we nitrate first, we get nitrobenzene; subsequent Friedel-Crafts acylation or alkylation is impossible because nitrobenzene is too deactivated. So the sequence must start with a group that is ortho/para-directing and activating, then convert it to the desired substituent.
目标分子的 1 号位是酯基，4 号位是硝基。酯基是间位定位致钝基团，硝基也是强间位定位致钝基团。若先硝化得到硝基苯，后续傅克酰基化或烷基化均无法进行，因为硝基苯太钝化。因此次序必须从邻对位定位且活化苯环的基团开始，再将其转化为所需取代基。
A classic route is: benzene → nitrobenzene → phenylamine → phenol → phenyl ethanoate → 4-nitrophenyl ethanoate. However, phenylamine is highly activating and ortho/para-directing; direct nitration would give multiple substitution. Instead, we can protect the amino group by converting to an amide, or better, start with phenol synthesis via cumene process, but that is beyond scope. A cleaner OCR-level route: benzene → bromobenzene → phenol (by nucleophilic substitution with NaOH at high temperature and pressure, then acidification). Then esterify phenol to phenyl ethanoate. The phenyl ethanoate has an ortho/para-directing ester group? Actually, the ester group attached to the ring through oxygen is a moderate activator and ortho/para-directing due to the lone pair on oxygen. Wait: in phenyl ethanoate, the acetyl group is on the oxygen; the ring is attached to O, which is electron-donating by resonance (+M). So the –OCOCH₃ group is ortho/para-directing and moderately activating. Therefore, nitration of phenyl ethanoate will give a mixture of 2- and 4-nitrophenyl ethanoates. The 4-isomer can be separated by recrystallisation or chromatography.
经典路线：苯 → 硝基苯 → 苯胺 → 苯酚 → 乙酸苯酯 → 4-硝基苯基乙酸酯。但苯胺高度活化且邻对位定位，直接硝化会多取代。可先将氨基保护成酰胺，或者采用更优路线：苯 → 溴苯 → 苯酚（高温高压下与 NaOH 发生亲核取代，再酸化）。然后酯化得到乙酸苯酯。苯环通过氧连接乙酰基，氧的孤对电子通过共振给电子（+M），因此 –OCOCH₃ 是邻对位定位基，中度活化。硝化乙酸苯酯会得到邻位和对位硝化产物混合物。对位异构体可通过重结晶或色谱分离。
8. Stepwise Reagents and Conditions | 分步试剂与条件
Here is the full synthesis with OCR-acceptable reagents:
以下是符合 OCR 要求的完整合成方案：
| Step | Reaction | Reagents/Conditions |
| 1 | Benzene to bromobenzene | Br₂, FeBr₃, room temp., electrophilic substitution |
| 2 | Bromobenzene to phenol | NaOH(aq), 300°C, high pressure, then H⁺(aq) |
| 3 | Phenol to phenyl ethanoate | Ethanoic anhydride, conc. H₂SO₄ (catalytic), warm |
| 4 | Phenyl ethanoate to 4-nitrophenyl ethanoate | Conc. HNO₃, conc. H₂SO₄, < 55°C; separate isomers |
9. Mechanisms in the Synthesis | 合成中的反应机理
You must be able to draw the mechanism for each key step. For the bromination of benzene, the electrophile is Br⁺ generated by FeBr₃ + Br₂ → Br⁺[FeBr₄]⁻. The mechanism involves attack by the π system, formation of a Wheland intermediate, and loss of H⁺.
你必须能画出每个关键步骤的机理。苯的溴化反应中，亲电试剂为 Br⁺，由 FeBr₃ 与 Br₂ 生成 Br⁺[FeBr₄]⁻。机理包括 π 电子进攻、形成韦兰德中间体、最后失去 H⁺。
For the nucleophilic substitution of bromobenzene to phenol, this proceeds via an addition-elimination mechanism under severe conditions because the C–Br bond is strong and the benzene ring is deactivated. The nucleophile HO⁻ attacks the carbon bearing bromine, a Meisenheimer complex forms, then Br⁻ leaves to regenerate the aromatic system.
溴苯转化为苯酚的亲核取代是在苛刻条件下经历加成-消除机理，因为 C–Br 键较强且苯环被钝化。亲核试剂 HO⁻ 进攻连接溴的碳，形成迈森海默络合物，然后 Br⁻ 离去恢复芳香体系。
Esterification of phenol uses ethanoic anhydride rather than ethanoic acid for better yield; the mechanism is nucleophilic addition-elimination. The oxygen of phenol attacks the carbonyl carbon of the anhydride, pushing electrons to form a tetrahedral intermediate, which then collapses, losing ethanoate ion and producing the ester.
苯酚的酯化使用乙酸酐而非乙酸以获得更高产率；机理为亲核加成-消除。酚氧进攻酐的羰基碳，电子转移形成四面体中间体，随后该中间体崩塌，丢失乙酸根离子，生成酯。
Nitration of phenyl ethanoate follows the standard electrophilic substitution: nitronium ion NO₂⁺ (generated from HNO₃ + H₂SO₄) attacks the ring preferentially at the para position due to steric and electronic factors.
乙酸苯酯的硝化遵循标准的亲电取代机理：硝酰正离子 NO₂⁺（由 HNO₃ 与 H₂SO₄ 产生）优先进攻对位，受空间和电子效应影响。
10. Spectroscopic Monitoring of the Synthesis | 合成过程中的波谱监控
At each stage, you could use IR and NMR to check the product. For instance, after step 2, the O–H stretch of phenol appears around 3200–3600 cm⁻¹, broad, and the C–Br stretch (500–600 cm⁻¹) disappears. After esterification, a sharp C=O appears at ~1760 cm⁻¹. After nitration, two strong NO₂ absorptions at ~1520 and ~1350 cm⁻¹ appear. In the ¹H NMR, the aromatic region will show different splitting patterns: phenyl ethanoate has a complex multiplet; after nitration, the para-substituted ring gives two doublets (AA’XX’ system).
每一步都可用红外和核磁检验产物。例如，第二步后，酚的 O–H 伸缩振动出现在 3200–3600 cm⁻¹（宽峰），而 C–Br 峰（500–600 cm⁻¹）消失。酯化后，~1760 cm⁻¹ 出现尖锐的 C=O 峰。硝化后，~1520 和 ~1350 cm⁻¹ 出现两个强硝基吸收峰。在 ¹H NMR 中，芳香区呈现不同的裂分方式：乙酸苯酯为复杂的多重峰，硝化后对位取代的苯环显示两组双峰（AA’XX’ 体系）。
11. Purity and Yield Considerations | 纯度与产率考量
The nitration of phenyl ethanoate produces both 2- and 4-isomers, typically in a ratio of about 1:2 due to steric hindrance at the ortho position. To obtain pure 4-nitrophenyl ethanoate, you can exploit differences in boiling point or solubility. Recrystallisation from ethanol/water is effective, as the para isomer is less soluble. Purity can be assessed by thin-layer chromatography or melting point determination (literature value for 4-nitrophenyl ethanoate is around 81–82 °C).
乙酸苯酯硝化同时生成邻和对位异构体，由于邻位的位阻效应，比例约为 1:2。要获得纯 4-硝基苯基乙酸酯，可利用沸点或溶解度的差异。用乙醇/水重结晶十分有效，因对位异构体溶解度较小。纯度检测可用薄层色谱或熔点测定（4-硝基苯基乙酸酯的文献熔点约 81–82 °C）。
12. Preparing for Exam Case Studies | 考试案例分析备考策略
When tackling an OCR case study question, always scan the entire question first, noting the given data. Start with the parts you find easiest, building confidence. For structure elucidation, systematically list what each spectrum tells you and only then combine the clues. For synthesis, consider directing effects, required reagents, and potential side reactions. Practice with past papers under timed conditions, and write mechanisms clearly using curly arrows.
在解答 OCR 案例分析题时，务必先通览全题，标出所给数据。从你最有把握的部分开始，建立信心。结构解析时，要系统列出每种波谱提供的信息，然后再综合线索。设计合成路线时，要考虑定位效应、所需试剂和可能的副反应。定时练习历年真题，用弯箭头清晰绘制机理。
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