📚 Edexcel A-Level Chemistry: Combined Spectral Analysis (MS, IR and NMR) | 爱德思 A-Level 化学：波谱综合分析（质谱、红外与核磁共振）
In Edexcel A-Level Chemistry, structure determination questions often ask you to combine mass spectrometry, infrared spectroscopy and ¹H NMR evidence. A single spectrum is rarely enough to identify an unknown organic compound; the marks come from linking each piece of evidence to a structural feature.
在爱德思 A-Level 化学中，结构推断题常要求你综合质谱、红外光谱和 ¹H 核磁共振的证据。单一谱图通常不足以鉴定未知有机物；得分点在于把每条证据与某个结构特征联系起来。
This article builds the three techniques into one working method: read the molecular ion first, decide which functional groups are present or absent from the infrared spectrum, then use proton and carbon NMR to assemble the carbon skeleton and count the hydrogen environments.
本文将三种技术整合成一套可操作的解题流程：先读分子离子峰，再根据红外光谱判断哪些官能团存在或不存在，最后用氢谱与碳谱拼出碳骨架并统计氢环境数目。
1. Why Combined Analysis? | 为什么需要综合分析
Each technique answers a different question. Mass spectrometry gives the molecular mass and molecular formula, infrared spectroscopy identifies functional groups, and ¹H NMR shows the number, kind and neighbouring relationships of hydrogen atoms.
每种技术回答不同问题。质谱给出分子质量和分子式，红外光谱识别官能团，而 ¹H 核磁共振显示氢原子的数量、种类和邻接关系。
Using them together is especially important for structural isomers. For example, an ester and a carboxylic acid may have the same molecular formula, but IR and NMR can distinguish them quickly.
把它们结合起来对结构异构体尤其重要。例如，酯和羧酸可能具有相同的分子式，但红外光谱和核磁共振可以迅速区分它们。
ou the formula could be C₃H₆O₂, but it cannot tell you whether the oxygens belong to an acid, an ester or a hydroxy aldehyde. The infrared spectrum answers the functional-group question; the NMR spectra settle the connectivity.
任何单一谱图本身都不具决定性。m/z = 74 处的分子离子峰说明分子式可能是 C₃H₆O₂，但它无法告诉你两个氧原子属于羧酸、酯还是羟基醛。红外光谱回答官能团问题；核磁共振谱最终确定连接方式。
2. The Three Techniques at a Glance | 三大技术分工速览
Before attempting a question, fix in your mind what each method can and cannot deliver. This table is a useful revision summary and also a planning tool for long structured questions.
在做题之前，先在脑中明确每种方法能给什么、不能给什么。下表既是复习小结，也是应对长结构化试题的解题规划工具。
| Technique | Information obtained | Key data |
|---|---|---|
| Mass spectrometry (MS) | Relative molecular mass and, with high resolution, molecular formula; fragmentation clues | m/z of the molecular ion M⁺•; fragment ion m/z values |
| Infrared spectroscopy (IR) | Which bonds and functional groups are present, and which are absent | Wavenumber / cm⁻¹ and whether the absorption is broad or sharp |
| ¹H NMR | Number of proton environments, ratio of protons, neighbouring protons | Chemical shift δ / ppm, integration ratio, splitting pattern |
| ¹³C NMR | Number of carbon environments; DEPT identifies CH₃, CH₂, CH and quaternary carbons | Number of peaks; chemical shift δ / ppm |
A well-written answer uses the language of cause and effect: 'the broad absorption at 2500-3300 cm⁻¹ shows an O-H group hydrogen-bonded as in a carboxylic acid, therefore the compound is an acid rather than an ester.'
高分答案使用因果语言："2500-3300 cm⁻¹ 处的宽吸收说明存在以羧酸形式氢键结合的 O-H 基团，因此该化合物是酸而不是酯。"
3. Mass Spectrometry: The Molecular Ion | 质谱：分子离子峰
The molecular ion is formed when a molecule is ionised and loses one electron, producing a radical cation, written M⁺•. Its m/z value equals the relative molecular mass of the compound.
分子离子是分子被电离并失去一个电子后形成的自由基阳离子，写作 M⁺•。其 m/z 值等于该化合物的相对分子质量。
M + e⁻ → M⁺• + 2e⁻ (m/z of M⁺• = Mr)
The molecular ion is usually the peak at the highest m/z value, although it may be weak if the ion fragments readily, and you must discount small peaks caused by the ¹³C isotope (about 1.1% of the intensity of the M⁺ peak for each carbon atom, giving an M+1 peak).
分子离子峰通常是 m/z 值最大的峰，但如果离子容易碎裂，它可能很弱；同时必须排除由 ¹³C 同位素引起的小峰（每个碳原子约为 M⁺ 峰强度的 1.1%，形成 M+1 峰）。
ble bond. That single deduction already narrows the possibilities enormously before you look at any other spectrum.
例如，m/z = 74 的分子离子峰对应分子式 C₃H₆O₂，不饱和度为 1，因此分子中含有 C=O 或 C=C 双键之一。在查看其他谱图之前，这一条推断就已经大幅缩小了可能性范围。
4. Fragmentation Patterns | 碎片离子与结构线索
Some of the molecular ions break apart in the spectrometer. The charged fragments are detected, and the neutral fragments are not. Because only charged species give peaks, each fragment peak corresponds to a cation, and the difference between the molecular ion and a fragment peak gives the mass of the neutral group lost.
部分分子离子在质谱仪中发生断裂。带电荷的碎片被检测到，中性碎片则不会出峰。由于只有带电物种产生峰，每个碎片峰对应一个阳离子，而分子离子与碎片峰之差即为失去的中性基团的质量。
m/z of fragment = Mr − mass of the neutral fragment lost
Two compounds of formula C₃H₆O₂ illustrate the idea. Propanoic acid, CH₃CH₂COOH, shows M⁺• (M − 17), a peak at m/z = 45 from loss of C₂H₅ (M − 29) and a peak at m/z = 29 for C₂H₅⁺.
两种分子式为 C₃H₆O₂ 的化合物可以说明这一点。丙酸 CH₃CH₂COOH 的 M⁺• 在 m/z = 74，失去 OH 得到 m/z = 57（M − 17）的峰，失去 C₂H₅ 得到 m/z = 45（M − 29）的峰，另有 m/z = 29 的 C₂H₅⁺ 峰。
Methyl ethanoate, CH₃COOCH₃, also has M⁺• mbined with the absence of m/z = 45, is a good indication of an ester rather than an acid.
乙酸甲酯 CH₃COOCH₃ 的 M⁺• 同样在 m/z = 74，但其特征峰为失去 CH₃ 得到的 m/z = 59（M − 15），以及酰基阳离子 CH₃CO⁺ 的 m/z = 43。强 m/z = 43 峰与 m/z = 45 峰的缺失相结合，是酯而非酸的很好证据。
5. Infrared Spectroscopy: Functional Groups | 红外光谱：官能团识别
A covalent bond absorbs infrared radiation at a wavenumber that matches its natural frequency of vibration. Absorptions are reported in wavenumbers, using the unit cm⁻¹, and the region above about 1500 cm⁻¹ is where functional-group absorptions appear.
共价键会吸收与其固有振动频率相匹配的红外辐射。吸收位置用波数表示，单位为 cm⁻¹；约 1500 cm⁻¹ 以上的区域是官能团吸收出现的范围。
| Bond / group | Wavenumber / cm⁻¹ | Appearance |
|---|---|---|
| O-H (alcohol) | 3200-3600 | Broad, strong |
| O-H (carboxylic acid) | 2500-3300 | Very broad, overlaps C-H |
| N-H (amine, amide) | 3300-3500 | Medium, sometimes two peaks |
| C-H | 2850-3100 | Sharp, present in most organics |
| C=O (acid) | 1680-1720 | Strong, sharp |
| C=O (ester) | 1735-1750 | Strong, sharp |
| C=O (aldehyde / ketone) | 1705-1740 | Strong, sharp |
| C=C (alkene) | 1620-1680 | Weak, easily missed |
| C≡N (nitrile) | 2220-2260 | Medium, sharp |
| C-O | 1000-1300 | Strong |
roup, so it cannot be an aldehyde, ketone, carboxylic acid, ester, amide or acyl chloride.
没有证据本身也是证据。如果约 1680 至 1750 cm⁻¹ 之间没有吸收，该化合物就不含 C=O 基团，因此不可能是醛、酮、羧酸、酯、酰胺或酰氯。
6. Reading the Fingerprint Region | 指纹区的读法
The region below about 1500 cm⁻¹, called the fingerprint region, contains many overlapping absorptions caused by complex bending and skeletal vibrations. It is rarely interpreted band by band in an exam.
约 1500 cm⁻¹ 以下的区域称为指纹区，包含许多由复杂弯曲振动和骨架振动引起的重叠吸收。考试中很少逐条解读该区域的吸收带。
Instead, the fingerprint region is used for identification by comparison: the unknown spectrum is matched against a library or a reference spectrum of a known compound. An exact match across the whole fingerprint region is strong evidence that the two compounds are identical.
指纹区主要通过比对来实现鉴定：将未知物谱图与谱库或已知化合物的参考谱图进行匹配。若整个指纹区完全吻合，则强有力地说明两者为同一化合物。
s an alcohol or ester assignment already suggested by the broad O-H or the C=O absorption.
指纹区在考试中最常见的用途，是识别醇和酯在约 1000-1300 cm⁻¹ 处的强 C-O 吸收，用来支持已由宽 O-H 或 C=O 吸收所暗示的醇或酯结论。
7. ¹H NMR: Counting Proton Environments | 氢谱：数氢环境
Protons in identical chemical environments give one signal. Protons are equivalent if they are related by symmetry or by rapid rotation about a single bond, and each set of equivalent protons produces one peak in the spectrum.
处于相同化学环境的氢原子只给出一个信号。若氢原子之间通过对称性或单键的快速旋转而相互等价，则它们等价；每一组等价的氢在谱图中产生一个峰。
Ethanol, CH₃CH₂OH, has three environments: the CH₃ protons, the CH₂ protons and the O-H proton. Butan-1-ol has five, while butan-2-ol also has five, yet the splitting patterns differ, which is exactly how the two isomers are distinguished.
乙醇 CH₃CH₂OH 有三种环境：CH₃ 上的氢、CH₂ 上的氢和 O-H 上的氢。丁-1-醇有五种环境，丁-2-醇也有五种，但它们的裂分方式不同，这正是区分两个异构体的关键。
Watch out for symmetry. In methyl ethanoate, CH₃COOCH₃, the two CH₃ groups are not equivalent, because one is attached to a carbonyl carbon and the other to an oxygen; the spectrum therefore shows two singlets rather than one.
注意对称性。在乙酸甲酯 CH₃COOCH₃ 中，两个 CH₃ 并不等价，因为一个连在羰基碳上，另一个连在氧上，因此谱图显示两个单峰而不是一个。
8. Chemical Shift, Integration and Splitting | 化学位移、积分与裂分
systems shift the signal downfield to higher δ values, measured relative to TMS at δ = 0.
峰的位置即化学位移 δ（单位 ppm），取决于氢原子所处的磁环境。电负性原子和邻近的 π 体系使信号向低场移动，δ 值增大；位移均相对于 δ = 0 的 TMS 测定。
| Proton environment | Typical δ / ppm |
|---|---|
| TMS reference | 0 |
| Alkyl C-H | 0.7-1.6 |
| C-H next to C=O | 2.1-2.6 |
| C-H next to O (alcohol, ether, ester) | 3.3-4.3 |
| Alkene C-H | 4.5-6.0 |
| Aromatic C-H | 6.5-8.0 |
| Aldehyde C-H | 9.0-10.0 |
| Carboxylic acid O-H | 9.0-13.0 |
Integration gives the relative number of protons in each environment; the ratios must be reduced to whole numbers, so an integration of 1:2:3 could mean 3:6:9. Splitting follows the n + 1 rule: a proton with n equivalent protons on adjacent carbon atoms is split into n + 1 peaks.
积分给出每种环境下氢原子的相对数目；比例必须化为整数比，因此 1:2:3 的积分也可能表示 3:6:9。裂分遵循 n + 1 规则：若某氢原子的相邻碳上有 n 个等价氢，则其信号裂分为 n + 1 个峰。
singlet = 0 neighbours, doublet = 1, triplet = 2, quartet = 3
Alcohol O-H protons are an exception: they are often broad, may be absent, and are removed by shaking with D₂O. If a peak disappears after a D₂O shake, it was an O-H or N-H proton.
醇的 O-H 氢是例外：它通常较宽、有时不出峰，并可通过加入 D₂O 振荡而消失。若某峰在 D₂O 振荡后消失，说明它是 O-H 或 N-H 氢。
9. ¹³C NMR and DEPT | 碳-13 核磁与 DEPT
Carbon-13 NMR shows one peak for each distinct carbon environment, so the number of peaks reveals how much symmetry the molecule has. In Edexcel questions it is most often used to confirm an assignment suggested by the proton spectrum.
碳-13 核磁对每一种不同的碳环境给出一个峰，因此峰的数目揭示了分子的对称程度。在爱德思试题中，它多用于印证氢谱所提示的结构归属。
ks, but at different shifts, with the ester carbonyl carbon near δ = 170 and the OCH₃ carbon near δ = 52.
丙酸 CH₃CH₂COOH 给出三个 ¹³C 峰，与三种氢环境一致。乙酸甲酯同样给出三个碳峰，但位移不同，其中酯羰基碳约在 δ = 170，OCH₃ 碳约在 δ = 52。
DEPT spectra add another layer: DEPT-90 shows only CH carbons, while DEPT-135 shows CH and CH₃ carbons pointing up and CH₂ carbons pointing down, with quaternary carbons absent altogether. This is powerful evidence for a proposed skeleton.
DEPT 谱提供额外信息：DEPT-90 只显示 CH 碳，而 DEPT-135 中 CH 和 CH₃ 碳向上、CH₂ 碳向下，季碳则完全不出峰。这对验证所提出的碳骨架非常有力。
10. Worked Example: Ester or Carboxylic Acid | 实例：酯还是羧酸
Two unknown compounds, A and B, both have molecular formula C₃H₆O₂ and both show M⁺• at m/z = 74. Use the IR and ¹H NMR data to identify each one.
两种未知物 A
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