📚 Edexcel A Level Chemistry Topic 19: Combined Analytical Techniques | 爱德思 A Level 化学专题 19：综合分析技术
In Edexcel A Level Chemistry Topic 19, you will learn how to combine evidence from mass spectrometry, infrared spectroscopy, carbon-13 NMR, proton NMR and chromatography to determine the structure of an unknown organic compound. Examiners do not usually ask for one technique in isolation; they give a set of data and expect you to build the molecule piece by piece.
在爱德思 A Level 化学专题 19 中，你将学习如何综合质谱、红外光谱、碳-13 核磁共振、质子核磁共振和色谱的证据，来确定未知有机物的结构。考官通常不会孤立考查单一技术；他们给出多组数据，希望考生逐步拼出分子结构。
1. The Overall Strategy: From Data to Structure | 总体策略：从数据到结构
Start with the mass spectrum to find the molecular ion peak, which gives the relative molecular mass. Then use the molecular formula, if given or obtained from high-resolution mass spectrometry, to calculate the degree of unsaturation. This tells you how many rings or double bonds are present.
先从质谱入手，找到分子离子峰，得到相对分子质量。然后利用给定分子式，或高分辨质谱得到的分子式，计算不饱和度。这能告诉你分子中有多少个环或双键。
Infrared spectroscopy identifies the main functional groups by absorption bands. Carbon-13 NMR tells you how many different carbon environments exist, while proton NMR gives the number of proton environments, their relative numbers and how they are connected through spin-spin splitting.
红外光谱通过吸收峰识别主要官能团。碳-13 核磁共振显示不同碳化学环境的数量，而质子核磁共振给出质子的环境数量、相对数目以及通过自旋-自旋裂分反映的连接方式。
- M⁺ peak → molecular mass
- IR bands → functional groups
- ¹³C NMR peaks → number of unique carbons
- ¹H NMR peaks, integration, splitting → proton arrangement
- M⁺ 峰 → 分子质量
- 红外吸收带 → 官能团
- ¹³C 核磁峰 → 不同碳的数量
- ¹H 核磁峰、积分、裂分 → 质子排列
2. Mass Spectrometry: Molecular Ion and Formula | 质谱：分子离子与分子式
In electron impact mass spectrometry, the molecule loses one electron to form a positive radical cation called the molecular ion, M⁺. The peak at the highest m/z value may be the molecular ion, provided there is no significant trapping of molecular ions or contamination.
在电子轰击质谱中，分子失去一个电子形成带正电的自由基阳离子，称为分子离子，写作 M⁺。最高 m/z 值的峰可能就是分子离子峰，前提是不存在明显的分子离子捕获或杂质。
sses. For example, a compound with M⁺ = 72.0575 may be C₄H₈O rather than C₃H₄O₂ or C₃H₈N₂.
高分辨质谱可将 m/z 测量到小数点后四位，从而根据精确原子质量确定分子式。例如，M⁺ = 72.0575 的化合物可能是 C₄H₈O，而不是 C₃H₄O₂ 或 C₃H₈N₂。
IHD = (2C + 2 + N − H − X) ÷ 2
不饱和度 IHD = (2C + 2 + N − H − X) ÷ 2
onds. A benzene ring contributes 4 degrees of unsaturation: three C=C bonds plus one ring.
用这个公式计算环和双键总数。苯环提供 4 个不饱和度：三个 C=C 双键加一个环。
3. Fragmentation Patterns: Recognising Stable Ions | 碎片化规律：识别稳定离子
Fragmentation peaks help confirm the structure because certain fragments are especially stable. A peak at m/z 43 often indicates C₃H₇⁺ or CH₃CO⁺, while m/z 29 suggests C₂H₅⁺ or CHO⁺.
碎片峰有助于确认结构，因为某些碎片特别稳定。m/z 43 的峰通常表示 C₃H₇⁺ 或 CH₃CO⁺，而 m/z 29 提示 C₂H₅⁺ 或 CHO⁺。
For a ketone such as butanone, alpha cleavage can produce CH₃CO⁺ at m/z 43 and C₂H₅⁺ at m/z 29. Do not try to explain every small peak; focus on the largest fragments that can be linked logically to the proposed structure.
对于丁酮等酮类，α 裂解可以产生 m/z 43 的 CH₃CO⁺ 和 m/z 29 的 C₂H₅⁺。不要试图解释每一个小峰；重点放在能与推测结构合理关联的最大碎片上。
| m/z | Possible ion | 可能离子 |
|---|---|---|
| 15 | CH₃⁺ | CH₃⁺ |
| 29 | C₂H₅⁺ / CHO⁺ | C₂H₅⁺ / CHO⁺ |
| 43 | C₃H₇⁺ / CH₃CO⁺ | C₃H₇⁺ / CH₃CO⁺ |
| 57 | C₄H₉⁺ / C₂H₅CO⁺ | C₄H₉⁺ / C₂H₅CO⁺ |
4. Infrared Spectroscopy: Key Absorption Bands | 红外光谱：关键吸收峰
Infrared spectroscopy detects bond vibrations. A broad, strong absorption around 3200–3550 cm⁻¹ usually indicates an O-H bond in alcohols or carboxylic acids, while a sharp N-H stretch appears around 3300–3500 cm⁻¹ in amines and amides.
红外光谱检测键的振动。3200–3550 cm⁻¹ 附近的宽而强的吸收通常表示醇或羧酸中的 O-H 键，而胺和酰胺中 N-H 的伸缩振动在 3300–3500 cm⁻¹ 附近出现尖锐吸收。
up C=O is one of the most diagnostic bands. Esters and carboxylic acids absorb around 1700–1750 cm⁻¹, while aldehydes and ketones absorb around 1680–1745 cm⁻¹. A C≡N nitrile group gives a sharp, medium peak near 2220–2260 cm⁻¹.
羰基 C=O 是最具诊断意义的吸收之一。酯和羧酸约在 1700–1750 cm⁻¹ 吸收，醛和酮约在 1680–1745 cm⁻¹ 吸收。腈基 C≡N 在 2220–2260 cm⁻¹ 附近出现尖锐的中等强度峰。
| Bond | Wavenumber / cm⁻¹ | Note |
|---|---|---|
| O-H alcohol | 3200–3550 | broad |
| O-H acid | 2500–3300 | very broad |
| N-H | 3300–3500 | sharp |
| C≡N | 2220–2260 | sharp |
| C=O ester/acid | 1700–1750 | strong |
| C=O aldehyde/ketone | 1680–1745 | strong |
5. Carbon-13 NMR: Counting Unique Carbon Environments | 碳-13 核磁共振：统计不同碳环境
A ¹³C NMR spectrum records one peak for each unique carbon environment, not for every carbon atom if symmetry makes carbons equivalent. For example, methylbenzene gives six peaks: four from the benzene ring, one from the methyl carbon and one from the ring carbon attached to CH₃.
碳-13 核磁共振谱为每一种不同的碳环境记录一个峰；如果对称性使某些碳等价，就不会每个碳原子都出峰。例如，甲苯给出六个峰：四个来自苯环，一个来自甲基碳，一个来自与 CH₃ 相连的环碳。
mical shifts in ¹³C NMR are reported in ppm from TMS. Typically, alkyl carbons appear at 0–50 ppm, carbons attached to oxygen at 50–90 ppm, C=C carbons at 110–160 ppm, and carbonyl carbons above 160 ppm.
碳-13 核磁共振化学位移以 TMS 为基准，单位为 ppm。通常烷基碳出现在 0–50 ppm，与氧相连的碳在 50–90 ppm，C=C 碳在 110–160 ppm，羰基碳在 160 ppm 以上。
- 0–50 ppm: C-C
- 50–90 ppm: C-O
- 110–160 ppm: C=C
- 160–185 ppm: C=O ester/acid/amide
- 190–220 ppm: C=O aldehyde/ketone
6. Proton NMR: Chemical Shift and Integration | 质子核磁共振：化学位移与积分
A ¹H NMR spectrum records one signal for each unique proton environment. The area under a signal, called the integration trace, is proportional to the number of protons in that environment. If the integration ratio is 3:2:1, the actual numbers could be 3, 2 and 1, or 6, 4 and 2.
质子核磁共振谱为每一种不同的质子环境记录一个信号。信号下方的面积称为积分曲线，与该环境中的质子数成正比。如果积分比为 3:2:1，实际质子数可能是 3、2 和 1，也可能是 6、4 和 2。
Chemical shifts help identify the type of proton. Alkyl protons attached to saturated carbon appear at 0.5–2.0 ppm, protons on carbon bonded to an electronegative atom such as O or Cl appear at 3.0–4.5 ppm, alkene protons at 4.5–6.0 ppm, and aromatic protons at 6.5–8.5 ppm.
化学位移有助于识别质子类型。与饱和碳相连的烷基质子出现在 0.5–2.0 ppm，与氧或氯等电负性原子相连的碳上的质子在 3.0–4.5 ppm，烯烃质子在 4.5–6.0 ppm，芳香质子在 6.5–8.5 ppm。
Aldehyde protons are very distinctive at 9.4–10.0 ppm, while carboxylic acid protons often appear very broadly at 10.0–13.0 ppm. Alcohol O-H protons can appear anywhere from 0.5–5.0 ppm and can be identified by their disappearance after D₂O shake.
醛基质子在 9.4–10.0 ppm 非常特征，羧酸质子在 10.0–13.0 ppm 常出现很宽的信号。醇的 O-H 质子可出现在 0.5–5.0 ppm，可通过 D₂O 振荡后信号消失来识别。
7. Spin-Spin Splitting: The n + 1 Rule | 自旋-自旋裂分：n + 1 规则
A proton signal is split by non-equivalent protons on adjacent carbon atoms. If a proton has n equivalent neighbouring protons, its signal is split into n + 1 peaks. Equivalent protons do not split each other.
质子信号会被相邻碳上的非等价质子裂分。如果某质子有 n 个等价的相邻质子，其信号裂分为 n + 1 重峰。等价质子之间不裂分。
For example, in ethanol CH₃CH₂OH, the CH₃ protons are adjacent to two CH₂ protons, so the CH₃ signal is a triplet. The CH₂ protons are adjacent to three CH₃ protons, so the CH₂ signal is a quartet. The OH proton is usually a singlet or broad peak.
例如，在乙醇 CH₃CH₂OH 中，CH₃ 质子的相邻碳上有两个 CH₂ 质子，因此 CH₃ 信号为三重峰。CH₂ 质子的相邻碳上有三个 CH₃ 质子，因此 CH₂ 信号为四重峰。OH 质子通常为单峰或宽峰。
Common splitting patterns include: singlet for no adjacent protons, doublet for one neighbour, triplet for two neighbours, quartet for three neighbours, and septet for six neighbours. Use integration to decide the number of protons producing each signal.
常见裂分模式包括：无相邻质子时为单峰，一个相邻质子时为二重峰，两个相邻质子时为三重峰，三个相邻质子时为四重峰，六个相邻质子时为七重峰。用积分判断每个信号对应的质子数。
8. Chromatography and Purity | 色谱与纯度
Thin-layer chromatography and gas chromatography are used to check purity and separate components. In TLC, the retention factor Rf is calculated as the distance moved by the spot divided by the distance moved by the solvent front. A pure compound gives a single well-defined spot under UV or after staining.
薄层色谱和气相色谱用于检验纯度并分离组分。在薄层色谱中，保留因子 Rf 等于组分斑点移动距离除以溶剂前沿移动距离。纯化合物在紫外灯下或显色后给出单一清晰的斑点。
In gas chromatography, each component has a characteristic retention time. The relative peak areas on a gas chromatogram give the relative amounts of the components in a mixture. This can be used to monitor the extent of a reaction or identify impurities.
在气相色谱中，每个组分都有特征的保留时间。气相色谱图上的相对峰面积表示混合物中各组分的相对含量。这可用于监测反应程度或识别杂质。
9. Combining the Evidence: A Worked Strategy | 组合证据：解题策略
ak at 1735 cm⁻¹, four ¹³C NMR peaks, and ¹H NMR signals at δ 1.2 (3H, triplet), δ 2.1 (2H, quartet), δ 3.7 (3H, singlet). The M⁺ value and carbonyl peak suggest a methyl ester.
假设某化合物 M⁺ = 88，红外光谱在 1735 cm⁻¹ 有吸收，碳-13 核磁共振有四个峰，质子核磁共振信号为 δ 1.2（3H，三重峰）、δ 2.1（2H，四重峰）、δ 3.7（3H，单峰）。M⁺ 值和羰基峰提示为甲酯。
. Assembling these gives CH₃CH₂COOCH₃, methyl propanoate, with formula C₄H₈O₂ and M⁺ = 88.
三重峰和四重峰表明存在连接在羰基上的乙基 CH₃CH₂-。3.7 ppm 处 3H 的单峰提示 OCH₃ 基团。组合这些信息得到 CH₃CH₂COOCH₃，即丙酸甲酯，分子式为 C₄H₈O₂，M⁺ = 88。
Always check your proposed structure against every piece of data: molecular formula, IHD, fragment ions, IR bands, number of ¹³C signals, number of ¹H signals, integration and splitting. If any one piece contradicts, revise the structure.
一定要将推测结构与所有数据核对：分子式、不饱和度、碎片离子、红外吸收、¹³C 信号数、¹H 信号数、积分和裂分。只要有任何一项矛盾，就要修改结构。
10. Common Pitfalls and Exam Tips | 常见失分点与考试技巧
A common mistake is to ignore symmetry. For example, in propane CH₃CH₂CH₃, there are only two carbon environments because the two CH₃ groups are equivalent; a ¹³C NMR spectrum therefore shows two peaks, not three.
常见错误是忽略对称性。例如，丙烷 CH₃CH₂CH₃ 只有两种碳环境，因为两个 CH₃ 基团等价；因此碳-13 核磁共振谱只显示两个峰，而不是三个。
Another pitfall is forgetting to use integration in proton NMR. If two signals have integration ratio 6:1, the signal for 6 protons could be two equivalent CH₃ groups, not a single CH₆ group. Also remember that O-H and N-H protons may appear broad and can exchange with D₂O.
另一个易错点是忘记在质子核磁共振中使用积分。如果两个信号的积分比为 6:1，6 个质子的信号可能是两个等价的 CH₃ 基团，而不是一个 CH₆ 基团。还要记住 O-H 和 N-H 质子可能呈宽峰，并能与 D₂O 发生交换。
When explaining splitting, quote the number of adjacent protons explicitly: 'The CH₂ signal is a quartet because it is adjacent to three equivalent CH₃ protons.' This earns the marking point much more reliably than just saying 'quartet'.
在解释裂分时，要明确指出相邻质子数：“CH₂ 信号是四重峰，因为它与三个等价的 CH₃ 质子相邻。”这比只说“四重峰”更能确保得分。
IR → functional groups | ¹³C → carbon count | ¹H → proton map | MS → mass & formula
红外
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