📚 Case Study Analysis in Pre-U CAIE Chemistry: Practical Workout on Aspirin Synthesis and Purity | 案例分析实战演练：阿司匹林合成与纯度测定
In Pre-U CAIE Chemistry, case study questions require you to apply theoretical knowledge to a real-world laboratory scenario. You will be presented with a detailed description of an experiment, including raw data, observations and potential pitfalls, and you must interpret, calculate and critically evaluate. This article walks you through a complete case study centred on the synthesis of aspirin, its purification and subsequent purity analysis. The aim is to build your confidence in tackling similar extended-response tasks that blend organic chemistry, quantitative analysis and practical skills.
在Pre-U CAIE化学考试中，案例分析题要求你将理论知识应用于真实的实验室场景。题目通常会提供详细的实验描述，包含原始数据、观察结果和潜在问题，你需要进行解读、计算和批判性评价。本文将以阿司匹林的合成为核心案例，带你完整经历其合成、提纯和纯度分析的全过程，旨在帮助你建立应对此类综合性试题的信心，这类题目融合了有机化学、定量分析和实验技能。
1. Understanding the Case Study | 理解案例研究
A typical Pre-U case study will begin with a brief background, then outline a student's experimental procedure and present a table of results. Your job is not simply to recall facts but to engage with the chemistry behind each step. You must identify the reaction type, explain the purpose of every technique, perform stoichiometric calculations, assess the purity of the product and suggest improvements. This mirrors the real scientific process and is heavily examined.
典型的Pre-U案例分析会先给出简要背景，接着概述某位学生的实验步骤，并呈现结果数据表。你的任务不是简单复述事实，而是要理解每一步背后的化学原理。你需要辨别反应类型，解释每种技术的用途，完成化学计量计算，评估产品纯度，并提出改进建议。这模拟了真实的科学研究过程，也是考试的重点。
2. The Context: Aspirin (Acetylsalicylic Acid) | 案例背景：阿司匹林（乙酰水杨酸）
Aspirin is the common name for acetylsalicylic acid, an ester formed from salicylic acid and ethanoic anhydride. It is a widely used analgesic and anti-inflammatory drug. In the laboratory, its synthesis is a classic example of an esterification reaction that can be carried out with relative ease, making it a perfect subject for a case study integrating organic preparation, purification and quantitative analysis.
阿司匹林是乙酰水杨酸的俗称，是由水杨酸和乙酸酐形成的酯。它是一种广泛使用的镇痛和抗炎药物。在实验室中，其合成是酯化反应的经典范例，操作相对简便，因此成为将有机制备、提纯和定量分析融为一体的绝佳案例素材。
| Substance | Formula | Molar Mass (g mol⁻¹) |
|---|---|---|
| Salicylic acid | C₇H₆O₃ | 138.12 |
| Ethanoic anhydride | C₄H₆O₃ | 102.09 |
| Acetylsalicylic acid | C₉H₈O₄ | 180.16 |
3. Synthesis: Esterification Reaction | 合成：酯化反应
en salicylic acid and ethanoic anhydride is a nucleophilic acyl substitution. The phenolic – + C₄H₆O₃ → C₉H₈O₄ + CH₃COOH. Note that ethanoic acid is the by-product, not water. This reaction is catalysed by concentrated phosphoric acid or sulfuric acid, which protonates the carbonyl oxygen to make the anhydride more electrophilic.
水杨酸和乙酸酐的反应属于亲核酰基取代。水杨酸的酚羟基进攻乙酸酐的羰基碳，使乙酸根作为离去基团离去。反应的化学方程式为：C₇H₆O₃ + C₄H₆O₃ → C₉H₈O₄ + CH₃COOH。注意副产品是乙酸而不是水。该反应通常由浓磷酸或浓硫酸催化，酸使羰基氧质子化，增强了乙酸酐的亲电性。
In a typical student experiment, 2.00 g of salicylic acid is reacted with 5.00 cm³ of ethanoic anhydride (density 1.08 g cm⁻³) in the presence of 5 drops of concentrated phosphoric acid. The mixture is warmed on a water bath at about 60 °C for 15 minutes. The ethanoic anhydride is used in excess to drive the equilibrium and ensure complete reaction of the more valuable salicylic acid.
在一个典型的学生实验中，将2.00 g水杨酸与5.00 cm³乙酸酐（密度1.08 g cm⁻³）混合，并加入5滴浓磷酸。混合物在约60 °C的水浴中加热15分钟。乙酸酐过量使用，以推动平衡并确保价值更高的水杨酸完全反应。
4. Reaction Conditions and Yield | 反应条件与产率
rategy to maximise the yield of aspirin based on the limiting reagent, salicylic acid. Let' ol. Since the stoichiometry is 1:1, salicylic acid is clearly the limiting reactant. Warming the mixture increases the rate of reaction, but temperatures above 80 °C should be avoided to prevent decomposition of the product and unwanted side reactions such as hydrolysis of the aspirin back to salicylic acid.
使用过量乙酸酐是基于限制试剂水杨酸来提高阿司匹林产率的常用策略。我们来验证：水杨酸的摩尔数 = 2.00 g / 138.12 g mol⁻¹ ≈ 0.01448 mol。乙酸酐的摩尔数 = (5.00 cm³ × 1.08 g cm⁻³) / 102.09 g mol⁻¹ ≈ 0.0529 mol。由于化学计量比为1:1，水杨酸显然是限制反应物。加热可以提高反应速率，但应避免温度超过80 °C，以防止产物分解和副反应，例如阿司匹林水解回水杨酸。
5. Purification: Recrystallisation | 提纯：重结晶
After cooling, the crude aspirin precipitates and is filtered under suction. The solid is impure and must be recrystallised. The student dissolves the crude product in the minimum volume of hot ethanol, then adds warm water dropwise until cloudiness persists. Upon cooling in an ice bath, pure aspirin crystals form. The choice of solvent mixture (ethanol‑water) exploits the fact that aspirin is more soluble in hot ethanol but poorly soluble in cold water. This technique removes unreacted salicylic acid and other impurities that remain dissolved in the mother liquor.
反应液冷却后，粗阿司匹林析出，经抽滤收集。该固体不纯，必须进行重结晶。学生用最少量热乙醇溶解粗产物，然后逐滴加入温水直至溶液出现持续浑浊。在冰浴中冷却后，纯阿司匹林结晶析出。选择乙醇‑水混合溶剂是利用阿司匹林在热乙醇中溶解度大而在冷水中的溶解度差的特性。此技术可以除去未反应的水杨酸和其他仍溶解在母液中的杂质。
rded as 1.58 g. The percentage yield is then calculated: (1.58 g / theoretical mass) × 100% mass = 0.01448 mol × 180.16 g mol⁻¹ = 2.61 g. Thus, percentage yield = (1.58 / 2.61) × 100% ≈ 60.5%.
纯化后的晶体经减压过滤收集，用少量冰冷水洗涤并干燥。记录干燥阿司匹林的质量为1.58 g。然后计算产率：(1.58 g / 理论产量) × 100%。理论产量 = 0.01448 mol × 180.16 g mol⁻¹ = 2.61 g。因此，产率 = (1.58 / 2.61) × 100% ≈ 60.5%。
6. Purity Analysis: Melting Point Determination | 纯度分析：熔点测定
A sharp melting point range is indicative of a pure substance. The student determines the melting point of the recrystallised aspirin using a melting point apparatus and obtains a range of 134–136 °C. The literature value for pure aspirin is 136 °C. The slight depression and broadening suggest a small amount of impurity, possibly salicylic acid or moisture. A mixed melting point determination with an authentic sample of aspirin would confirm identity and purity more rigorously.
敏锐的熔点范围是纯物质的标志。学生用熔点测定仪测定重结晶阿司匹林的熔点，获得的范围为134–136 °C。纯阿司匹林的文献值为136 °C。轻微的降低和熔程变宽表明存在少量杂质，可能是水杨酸或水分。与阿司匹林标准品进行混合熔点测定将能更严格地确认其身份和纯度。
7. Titrimetric Analysis: Back Titration | 滴定分析：返滴定
: C₉H₈O₄ + 2NaOH → C₇H₄O₃Na₂ + CH₃COONa + 2H₂O. The excess NaOH is then titrated against standard hydrochloric acid. Procedure: 0.50 g of the prepared aspirin is dissolved in 25.0 cm³ of 1.00 mol dm⁻³ NaOH (excess) and heated gently to complete hydrolysis. The solution is transferred to a 250.0 cm³ volumetric flask and made up to the mark. A 25.0 cm³ aliquot is titrated against 0.100 mol dm⁻³ HCl using phenolphthalein indicator. The titration value is 22.50 cm³ of HCl.
为了定量分析纯度，学生进行了返滴定。加入已知过量的氢氧化钠水解脱去阿司匹林的乙酰基。反应为：C₉H₈O₄ + 2NaOH → C₇H₄O₃Na₂ + CH₃COONa + 2H₂O。过量的NaOH随后用标准盐酸滴定。步骤：将0.50 g制备的阿司匹林溶于25.0 cm³ 1.00 mol dm⁻³ NaOH（过量）中，并微热至完全水解。溶液转移至250.0 cm³容量瓶并定容。移取25.0 cm³该溶液，以酚酞为指示剂，用0.100 mol dm⁻³ HCl滴定，消耗HCl 22.50 cm³。
8. Calculations Walkthrough | 计算步骤详解
= 0.0250 mol. This NaOH is consumed by both the aspirin hydrolysis and the HCl titration.
首先，求出最初加入的NaOH总摩尔数：25.0 cm³ × 1.00 mol dm⁻³ = 0.0250 mol。这些NaOH被阿司匹林水解和HCl滴定共同消耗。
(25.0 in 250.0), the total excess NaOH in the whole solution = 0.00225 mol × 10 = 0.0225 mol.
滴定25.0 cm³等分试样消耗的HCl摩尔数：22.50 cm³ × 0.100 mol dm⁻³ = 0.00225 mol。由于这是原始水解液的1/10稀释（25.0定容至250.0），全量溶液中过量的NaOH = 0.00225 mol × 10 = 0.0225 mol。
Thus, NaOH that reacted with aspirin = total NaOH – excess NaOH = 0.0250 mol – 0.0225 mol = 0.00250 mol.
因此，与阿司匹林反应的NaOH = 总NaOH – 过量NaOH = 0.0250 mol – 0.0225 mol = 0.00250 mol。
1 mol aspirin reacts with 2 mol NaOH. So moles of aspirin = 0.00250 mol / 2 = 0.00125 mol.
从水解方程可知，1 mol阿司匹林与2 mol NaOH反应。所以阿司匹林的摩尔数 = 0.00250 mol / 2 = 0.00125 mol。
.50 g sample = 0.00125 mol × 180.16 g mol⁻¹ = 0.2252 g. Purity = (0.2252 g / 0.50 g) × 100% = 45.0% g result, indicating significant impurity or incomplete drying, or perhaps student error).
0.50 g样品中阿司匹林的质量 = 0.00125 mol × 180.16 g mol⁻¹ = 0.2252 g。纯度 = (0.2252 g / 0.50 g) × 100% = 45.0%（这个结果不理想，表明杂质较多或干燥不充分，也可能是学生操作失误）。
Compare with the yield calculation earlier: the 1.58 g of recrystallised aspirin was assumed largely pure, but here the titration on 0.50 g of that same batch shows only 45% purity. This discrepancy invites critical evaluation – perhaps the recrystallised product still contained a lot of unreacted salicylic acid, or the sample for titration was not dry, or the back titration procedure had systematic errors.
将此与之前的产率计算对比：重结晶得到的1.58 g阿司匹林被认为基本纯净，但此处同一批产品中取0.50 g滴定却显示纯度只有45%。这种差异要求我们进行批判性评价——也许重结晶产物中仍含有大量未反应的水杨酸，或者用于滴定的样品未干燥，或者返滴定步骤存在系统误差。
9. Sources of Error and Improvements | 误差来源与改进
Common errors in this case study include: incomplete acetylation due to insufficient heating or too short reaction time; loss of product during recrystallisation if too much solvent is used; inclusion of salicylic acid impurity which also reacts with NaOH in the back titration, leading to an overestimation of NaOH consumed and thus an underestimation of aspirin purity; failure to dry the crystals properly; and inaccurately reading the burette. Improvements could involve using a longer reflux time, ensuring crystallisation is done slowly, washing with ice-cold solvent only, and using a more precise analytical technique such as colorimetry with FeCl₃ to test for residual salicylic acid (which gives a purple complex).
本案例研究中常见的误差包括：因加热不足或反应时间过短导致乙酰化不完全；重结晶时溶剂用量过多导致产物损失；产品中混有水杨酸杂质，它同样会在返滴定中与NaOH反应，导致消耗NaOH的量被高估，从而低估阿司匹林纯度；晶体干燥不充分；以及滴定管读数不准确等。改进措施可以包括：延长回流时间，确保结晶过程缓慢进行，只用冰冷溶剂洗涤，以及采用更精确的分析技术，如用FeCl₃比色法检测残留水杨酸（水杨酸会与铁离子形成紫色配合物）。
10. Linking to Exam Questions | 联系考试题型
In the Pre-U exam, you may be asked to calculate percentage yield, deduce the structure of the product from spectral data, explain the mechanism of esterification, or evaluate the suitability of the purification method. You might also be given titration data and asked to determine purity, often requiring you to construct a logical sequence of calculations. Always show clear working and pay attention to significant figures. Questions frequently ask 'Suggest why the yield is less than 100%' or 'Explain how the impurity affects the melting point and the titration result'.
在Pre-U考试中，你可能需要计算产率，根据光谱数据推导产物结构，解释酯化反应机理，或评价提纯方法的适用性。也可能给出滴定数据并要求计算纯度，这往往需要你构建清晰的逻辑计算步骤。务必展示清晰的推导过程并注意有效数字。题目常会问“说明为何产率低于100%”或“解释杂质如何影响熔点和滴定结果”。
11. Practice Task for Students | 学生实战演练
reacted salicylic acid. Salicylic acid also reacts with NaOH in a 1:2 mole ratio (phenolic – salicylic acid and aspirin in the 0.50 g sample is represented by x g of aspirin and (0.50 – ation and solve for x. (Hint: 2× moles of aspirin + 2× moles of salicylic acid = 0.00250).
现在，请尝试以下扩展题目：假设阿司匹林样品中的杂质全部是未反应的水杨酸。已知水杨酸与NaOH也按1:2摩尔比反应（酚羟基和羧酸基团均参与反应）。若0.50 g样品中含有x g阿司匹林和(0.50 – x) g水杨酸，请利用消耗的NaOH摩尔数（0.00250 mol）建立方程并求解x。（提示：2×阿司匹林摩尔数 + 2×水杨酸摩尔数 = 0.00250）
12. Conclusion | 结语
This case study illustrates how a single practical scenario can integrate multiple areas of the Pre-U syllabus: organic synthesis, stoichiometry, analytical techniques, error analysis and data interpretation. Mastering such exercises will not only prepare you for the examination but also sharpen your scientific thinking. Always approach case studies as a detective would – question every number, every observation, and connect them back to the underlying chemistry.
本案例展示了如何通过一个简单的实验情境将Pre-U课程大纲的多个领域融会贯通：有机合成、化学计量、分析技术、误差分析和数据解读。掌握这类练习不仅能为你备考助力，更能磨砺你的科学思维。请始终以侦探般的眼光对待案例分析——质疑每一个数字、每一条观察，并将其与背后的化学原理联系起来。
Published by TutorHao | Chemistry Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图