📚 Combined Analytical Techniques (065): Edexcel A-Level Chemistry | 综合结构解析技术（065）：爱德思 A-Level 化学
In Edexcel A-Level Chemistry, the ability to combine mass spectrometry, infrared spectroscopy, 1H NMR and 13C NMR is a core examined skill. This revision guide explains how to use molecular formula, fragmentation patterns, functional group absorptions, chemical shifts, integration and splitting to deduce an organic structure step by step.
在爱德思 A-Level 化学中，综合运用质谱、红外光谱、1H 核磁共振和 13C 核磁共振是核心考点。本复习指南逐步讲解如何利用分子式、碎片离子、官能团吸收、化学位移、积分和裂分推断有机结构。
1. The Combined Analytical Strategy | 综合解析策略
The key idea is that no single technique usually gives the complete structure. You combine molecular formula, mass spectrometry, infrared spectroscopy and NMR data like pieces of a puzzle to identify the exact organic molecule.
核心思想是通常没有单一技术能给出完整结构。你像拼图一样组合分子式、质谱、红外光谱和核磁共振数据，从而确定确切的有机分子。
- Mass spectrometry gives relative molecular mass and fragments.
- 质谱提供相对分子质量和碎片离子信息。
- Infrared spectroscopy identifies key functional groups present or absent.
- 红外光谱判断关键官能团是否存在。
- 1H NMR shows hydrogen environments, relative numbers of protons and neighbouring protons.
- 1H 核磁共振显示氢环境、质子相对数量以及相邻质子。
- 13C NMR shows the number of different carbon environments.
- 13C 核磁共振显示不同碳环境的数目。
2. Mass Spectrometry: Molecular Ion and Fragmentation | 质谱：分子离子与裂解
The molecular ion peak, M+, gives the relative molecular mass. In high-resolution mass spectrometry, the exact m/z value can be used to determine the molecular formula because atomic masses are not whole numbers.
分子离子峰 M+ 给出相对分子质量。在高分辨质谱中，精确的 m/z 值可用于确定分子式，因为原子质量并非整数。
om CH₃⁺. Halogen isotopes produce characteristic M and M+2 peaks for chlorine and bromine.
碎片离子峰也具有诊断意义。例如，m/z = 29 的峰通常来自 C₂H₅⁺ 或 CHO⁺，而 m/z = 15 常见于 CH₃⁺。卤素同位素会使氯和溴出现特征的 M 与 M+2 峰。
| Halogen | Pattern |
| Cl | M : M+2 ≈ 3 : 1 |
| Br | M : M+2 ≈ 1 : 1 |
质谱中氯和溴的同位素峰模式是常见考点，可据此判断分子中是否含有 Cl 或 Br。
3. Infrared Spectroscopy: Key Absorptions | 红外光谱：关键吸收
Infrared spectroscopy identifies the presence of bonds and functional groups. You must quote the absorption range and the bond responsible in exam answers.
红外光谱用于识别化学键和官能团。在考试答案中必须引用吸收范围以及对应的化学键。
| Bond | Wavenumber range / cm⁻¹ |
| O–H in alcohols/phenols | 3200–3550, broad |
| O–H in carboxylic acids | 2500–3300, very broad |
| N–H | 3300–3500 |
| C–H | 2800–3100 |
| C≡N | 2200–2250 |
| C=O | 1650–1750 |
| C=C in alkenes | 1620–1680 |
| C–O | 1000–1300 |
红外光谱中，O–H 的宽峰、C=O 的强吸收以及 C–O 的存在是区分醇、羧酸、酯、醛和酮的重要证据。
4. 1H NMR: Chemical Shift and Reference | 1H 核磁共振：化学位移与参照
In 1H NMR, each chemically distinct hydrogen environment produces a signal at a characteristic chemical shift, δ, measured in ppm relative to TMS.
在 1H 核磁共振中，每种化学环境不同的氢都会在相对 TMS 的特征化学位移 δ 处产生信号，单位为 ppm。
- Alkyl R–CH₃, R–CH₂–R: δ 0.5–2.0
- Adjacent to C=O or benzene: δ 2.0–3.0
- Adjacent to O or N: δ 3.0–4.5
- Alkenes and aromatics: δ 5.5–8.0
- Aldehyde R–CHO: δ 9.5–10.0
- Carboxylic acid –COOH: δ 10.0–12.0
TMS, tetramethylsilane, is used as an internal standard because it is inert, volatile, has a single sharp signal and absorbs at a lower frequency than most organic protons.
四甲基硅烷 TMS 用作内标，因为它惰性、易挥发、只有一个尖锐信号，并且其吸收频率低于大多数有机质子。
5. 1H NMR: Integration and Splitting | 1H 核磁共振：积分与裂分
The integration ratio tells you the relative number of protons in each environment. Splitting follows the n+1 rule: a proton signal is split by n equivalent protons on the adjacent carbon.
积分比告诉你每种环境中质子的相对数量。裂分遵循 n+1 规则：某个质子的信号被相邻碳上 n 个等价质子裂分。
| Neighbouring H atoms | Multiplicity |
| 0 | singlet |
| 1 | doublet |
| 2 | triplet |
| 3 | quartet |
For example, the ethyl group CH₃CH₂– typically shows a triplet for CH₃ and a quartet for CH₂ in a 3:2 integration ratio.
例如，乙基 CH₃CH₂– 通常显示 CH₃ 的三重峰和 CH₂ 的四重峰，积分比为 3:2。
6. 13C NMR: Number of Carbon Environments | 13C 核磁共振：碳环境数目
13C NMR gives one peak for each distinct carbon environment. Symmetry reduces the number of peaks: two equivalent carbon atoms in the same environment produce only one signal.
13C 核磁共振为每种不同的碳环境给出一个峰。对称性会减少峰的数量：处于相同环境的两个等价碳原子只产生一个信号。
Chemical shift ranges for 13C NMR are broad but useful: C–C around 5–40 ppm, C–O and C–N around 40–70 ppm, alkene and aromatic carbons around 100–160 ppm, esters and amides around 160–185 ppm, and aldehydes and ketones around 190–220 ppm.
13C 核磁共振的化学位移范围较宽但很有用：C–C 约为 5–40 ppm，C–O 和 C–N 约为 40–70 ppm，烯烃和芳香碳约为 100–160 ppm，酯和酰胺约为 160–185 ppm，醛和酮约为 190–220 ppm。
7. Using Molecular Formula and Degree of Unsaturation | 利用分子式与不饱和度
Before interpreting spectra, calculate the degree of unsaturation, also called DBE or index of hydrogen deficiency. It tells you the total number of rings and π bonds.
在解析谱图之前，先计算不饱和度，也称 DBE 或氢缺失指数。它告诉你环和 π 键的总数。
DBE = (2C + 2 + N − H − X) ÷ 2
DBE of 1 could mean one C=C bond or one ring; a DBE of 4 strongly suggests a benzene ring.
这里 C 是碳原子数，N 是氮原子数，H 是氢原子数，X 是卤素原子数。氧忽略不计。DBE 为 1 表示可能存在一个 C=C 键或一个环；DBE 为 4 则强烈提示苯环。
8. Worked Example: Ethyl Ethanoate | 例题：乙酸乙酯
A compound has molecular formula C₄H₈O₂. Mass spectrometry gives M+ = 88. Infrared shows a strong C=O at 1740 cm⁻¹ and C–O at 1240 cm⁻¹, with no broad O–H absorption.
某化合物分子式为 C₄H₈O₂。质谱给出 M+ = 88。红外在 1740 cm⁻¹ 处有强 C=O 吸收，在 1240 cm⁻¹ 处有 C–O 吸收，且无宽 O–H 吸收。
The 1H NMR shows three signals: δ 1.1, triplet, 3H; δ 2.0, singlet, 3H; δ 4.1, quartet, 2H. The 13C NMR shows four peaks.
1H 核磁共振显示三个信号：δ 1.1，三重峰，3H；δ 2.0，单峰，3H；δ 4.1，四重峰，2H。13C 核磁共振显示四个峰。
bond and no ring. The triplet at δ 1.1 and quartet at δ 4.1 indicate an ethyl group CH₃CH₂– δ 2.0 is an isolated CH₃ next to C=O. Combining these gives CH₃COOCH₂CH₃, ethyl ethanoate.
DBE = (2×4 + 2 − 8) ÷ 2 = 1，因此有一个 C=O 键且无环。δ 1.1 的三重峰和 δ 4.1 的四重峰表明存在一个连接氧的乙基 CH₃CH₂–。δ 2.0 的单峰是与 C=O 相邻的孤立 CH₃。综合得出 CH₃COOCH₂CH₃，即乙酸乙酯。
9. Distinguishing Isomers: Propanoic Acid vs Methyl Ethanoate | 区分同分异构体：丙酸与乙酸甲酯
Both propanoic acid and methyl ethanoate have molecular formula C₃H₆O₂, so mass spectrometry alone cannot distinguish them. Infrared and NMR are needed.
丙酸和乙酸甲酯的分子式都是 C₃H₆O₂，因此仅凭质谱无法区分。需要红外和核磁共振。
Propanoic acid shows a very broad O–H absorption from 2500 to 3300 cm⁻¹ and a C=O at about 1710 cm⁻¹. Its 1H NMR includes a carboxylic acid proton near δ 10–12, which disappears after D₂O shaking.
丙酸在 2500 至 3300 cm⁻¹ 有非常宽的 O–H 吸收，并在约 1710 cm⁻¹ 有 C=O 吸收。其 1H 核磁共振在 δ 10–12 附近有羧酸质子信号，加 D₂O 振摇后该信号消失。
Methyl ethanoate has no O–H absorption. It shows a C=O at about 1740 cm⁻¹ and a C–O at about 1220 cm⁻¹. Its 1H NMR has two singlets: one for CH₃CO at δ 2.0 and one for OCH₃ at δ 3.7.
乙酸甲酯没有 O–H 吸收。它在约 1740 cm⁻¹ 有 C=O 吸收，在约 1220 cm⁻¹ 有 C–O 吸收。其 1H 核磁共振有两个单峰：一个 CH₃CO 位于 δ 2.0，另一个 OCH₃ 位于 δ 3.7。
10. Common Pitfalls and Exam Technique | 常见错误与应试技巧
Many students lose marks by forgetting to quote the bond and wavenumber range together in IR answers, or by missing symmetry effects in 13C NMR. Always account for every signal and its integration before proposing a structure.
许多学生在红外答题时忘记同时引用化学键和波数范围，或者在 13C 核磁共振中忽略对称性影响。提出结构之前，一定要解释每一个信号及其积分。
- For IR, write 'O–H stretch at 3200–3550 cm⁻¹' rather than just 'O–H'.
- 红外要写 '3200–3550 cm⁻¹ 的 O–H 伸缩振动'，而不是只写 'O–H'。
- For NMR splitting, count only non-equivalent neighbouring protons.
- 核磁共振裂分时，只数相邻碳上非等价的质子。
- Check that your proposed structure has the exact molecular formula and DBE.
- 检查你提出的结构是否具有精确的分子式和不饱和度。
- Use D₂O exchange to confirm OH or NH protons if the question mentions it.
- 如果题目提到 D₂O 交换，可用其确认 OH 或 NH 质子。
A clear written sequence: molecular formula → DBE → MS → IR → 13C NMR → 1H NMR → proposed structure, is the most reliable way to gain full marks.
清晰的书写顺序：分子式 → 不饱和度 → 质谱 → 红外 → 13C 核磁共振 → 1H 核磁共振 → 提出结构，是获得满分的最可靠方法。
Published by TutorHao | Chemistry Revision Series | aleveler.com
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