- 描述并理解以下术语:(a) 固定相,例如氧化铝(负载在固体载体上);(b) 流动相;极性或非极性溶剂;(c) $R_{\text{f}}$ 值;(d) 溶剂前沿和基线
- 解读 $R_{\text{f}}$ 值
- 从与固定相的相互作用以及在流动相中的相对溶解度方面解释 $R_{\text{f}}$ 值的差异
分析技术
A-Level 化学 · 知识点 37
8:06
分析技术(A2):核磁共振谱
这里有七种不同的分子。每一个都含有五个碳原子、十个氢原子和一个氧原子,并且每一个都包含一个羰基。在质谱中称量它们……
英语旁白 · 英中内嵌字幕
37.1
薄层色谱
教学大纲
来源:Cambridge International 教学大纲
Chromatography 色谱 separates a mixture using two "phases" — one that stays still and one that moves. In thin-layer chromatography 薄层色谱 (TLC):
- the stationary phase 固定相 stays still (for example aluminium oxide on a plate).
- the mobile phase 流动相 moves (a polar or non-polar solvent that travels up the plate).
- the baseline 基线 is the starting line where the spots are placed; the solvent front 溶剂前沿 is the highest level the solvent reaches.
The $R_{\text{f}}$ value compares how far a spot moves with how far the solvent moves:
It is always between 0 and 1. A substance that sticks more strongly to the stationary phase, or is less soluble in the mobile phase, moves less and has a smaller $R_{\text{f}}$.


TLC 途径
跟随一个斑点沿一块板向上,并用 Rf 识别物质。
| English | 中文 | 拼音 |
|---|---|---|
| chromatography/krəʊməˈtɒɡrəfi/ | 色谱 | sè pǔ |
| thin-layer chromatography/θɪn ˈleɪə krəʊməˈtɒɡrəfi/ | 薄层色谱 | báo céng sè pǔ |
| stationary phase/ˈsteɪʃənəri feɪz/ | 固定相 | gù dìng xiāng |
| mobile phase/ˈməʊbaɪl feɪz/ | 流动相 | liú dòng xiāng |
| baseline/ˈbeɪslaɪn/ | 基线 | jī xiàn |
| solvent front/ˈsɒlvənt frʌnt/ | 溶剂前沿 | róng jì qián yán |
37.2
气相/液相色谱
教学大纲
- 描述并理解以下术语:(a) 固定相;涂覆在固体载体上的高沸点非极性液体 (b) 流动相;一种惰性气体 (c) 保留时间
- 根据混合物的百分组成解释气/液色谱图
- 从与固定相相互作用的角度解释保留时间
来源:Cambridge International 教学大纲
In gas/liquid chromatography 气液色谱 (GLC):
- the stationary phase is a high-boiling-point non-polar liquid on a solid support.
- the mobile phase is an unreactive carrier gas.
- the retention time 保留时间 is how long a component takes to pass through.
The area of each peak gives the percentage of that component in the mixture. A component that interacts more with the stationary phase has a longer retention time.

气/液色谱途径
跟随一个挥发性样品通过柱分离到一个色谱图。
| English | 中文 | 拼音 |
|---|---|---|
| gas/liquid chromatography/ɡæs ˈlɪkwɪd krəʊməˈtɒɡrəfi/ | 气液色谱 | qì yè sè pǔ |
| retention time/rɪˈtenʃn taɪm/ | 保留时间 | bǎo liú shí jiān |
37.3
碳-13 核磁共振谱
教学大纲
- 分析并解读简单分子的 碳-13 NMR 谱图,以推断:(a) 所存在碳原子的不同化学环境 (b) 该分子的可能结构
- 预测或解释给定分子的碳-13 NMR 谱图中的峰数
来源:Cambridge International 教学大纲
NMR 核磁共振 (nuclear magnetic resonance) studies how certain nuclei behave in a strong magnetic field.

In carbon-13 NMR, each different chemical environment 化学环境 of carbon gives one peak. So:
- the number of peaks tells you how many different carbon environments there are (equivalent carbons share a peak).
- the position of each peak (its chemical shift) suggests the type of carbon, which helps you deduce possible structures.
碳-13 NMR 实验室
把碳环境匹配到 C-13 NMR 证据。
| English | 中文 | 拼音 |
|---|---|---|
| NMR/ˌen em ˈɑː/ | 核磁共振 | hé cí gòng zhèn |
| chemical environment/ˈkemɪkl enˈvaɪrənmənt/ | 化学环境 | huà xué huán jìng |
37.4
质子(¹H)核磁共振谱
教学大纲
- 分析并解读简单分子的质子 ($^1\text{H}$) NMR 谱图,以推断:(a) 利用 化学位移 值确定存在的不同环境的质子 (b) 根据相对峰面积确定每种类型质子的相对数量 (c) 利用 裂分模式(基于 $n + 1$ 规则,仅限于单峰、双峰、三重峰、四重峰和多重峰)确定与给定质子所连碳原子相邻的碳原子上等价质子的数量 (d) 分子的可能结构
- 预测给定分子中质子的化学位移和裂分模式
- 描述使用 四甲基硅烷 (TMS) 作为化学位移测量标准的原因及方法
- 说明在获取质子 NMR 谱图时使用氘代溶剂(例如 $\text{CDCl}_3$)的必要性
- 描述利用 $\text{D}_2\text{O}$ 通过质子交换来识别 O–H 和 N–H 质子的方法
来源:Cambridge International 教学大纲
Proton NMR looks at the hydrogen atoms (proton 质子 nuclei). From the spectrum you read off:
- chemical environments: protons in different environments appear at different chemical shift 化学位移 values.
- relative numbers: the relative peak areas give the ratio of each type of proton.
- splitting 裂分: a peak is split by the protons on the neighbouring carbon, following the $n+1$ rule — $n$ equivalent neighbours split a peak into $n+1$ lines:
| Neighbours ($n$) | Pattern |
|---|---|
| 0 | singlet 单峰 |
| 1 | doublet 双峰 |
| 2 | triplet 三峰 |
| 3 | quartet 四峰 |
| many | multiplet 多重峰 |


Practical points
- tetramethylsilane 四甲基硅烷 (TMS) is the standard, set at a chemical shift of $0$.
- a deuterated solvent 氘代溶剂 (such as $\text{CDCl}_3$) is used so that the solvent itself gives no proton signal.
- shaking the sample with $\text{D}_2\text{O}$ makes the O–H and N–H peaks disappear (their hydrogen is swapped for deuterium), which identifies those protons.
Worked example. A compound $\text{C}_4\text{H}_8\text{O}_2$ gives three proton NMR peaks: a triplet at $\delta\ 1.2$ (area 3), a quartet at $\delta\ 4.1$ (area 2), and a singlet at $\delta\ 2.0$ (area 3). Deduce the structure. Read the areas first, then the splitting. Areas $3:2:3$ give three proton environments holding 3, 2 and 3 hydrogens. Apply the $n+1$ rule in reverse: a triplet (area 3) is a $\text{CH}_3$ with 2 neighbours, and a quartet (area 2) is a $\text{CH}_2$ with 3 neighbours - each splitting the other, which is the classic ethyl group, $\text{CH}_3\text{CH}_2-$. That $\text{CH}_2$ lies far downfield at $\delta\ 4.1$, so it is attached to an oxygen. The remaining singlet (area 3) is a $\text{CH}_3$ with no neighbours at $\delta\ 2.0$, so it sits next to the C=O. Together: $\text{CH}_3\text{COOCH}_2\text{CH}_3$, ethyl ethanoate. A triplet-and-quartet pair is almost always an ethyl group - spot it first and the rest follows.
Common mistakes.
- Counting the protons in a group to predict its splitting. The splitting comes from the $n$ equivalent protons on the neighbouring carbon: a $\text{CH}_3$ next to a $\text{CH}_2$ is a triplet, not a quartet.
- Splitting an O–H peak. An O–H proton usually appears as a singlet and does not split its neighbours, because it exchanges quickly; the $\text{D}_2\text{O}$ shake removes it.
- Reading peak heights as the proton ratio. The ratio comes from the peak areas (integration).
- Counting every carbon as a separate $^{13}\text{C}$ peak. Carbons related by symmetry share a peak: propanone has two carbon environments, not three.
- Saying TMS is used because it is "a reference". Say why it suits: all 12 of its protons are in one environment, giving a single strong peak, and it is inert and volatile.
- Using an ordinary $\text{CHCl}_3$ solvent. Its own proton would show on the spectrum; use a deuterated solvent such as $\text{CDCl}_3$.
- Calling the spot nearest the baseline in TLC the least polar. With a polar stationary phase (aluminium oxide or silica), the more polar compound sticks more and moves less, giving a smaller $R_f$.
质子 NMR 实验室
把质子证据匹配到化学环境和邻居。
| English | 中文 | 拼音 |
|---|---|---|
| proton/ˈprəʊtɒn/ | 质子 | zhì zi |
| chemical shift/ˈkemɪkl ʃɪft/ | 化学位移 | huà xué wèi yí |
| splitting/ˈsplɪtɪŋ/ | 裂分 | liè fēn |
| singlet/ˈsɪŋɡlɪt/ | 单峰 | dān fēng |
| doublet/ˈdʌblət/ | 双峰 | shuāng fēng |
| triplet/ˈtrɪplɪt/ | 三峰 | sān fēng |
| quartet/kwɔːˈtet/ | 四峰 | sì fēng |
| multiplet/ˈmʌltɪplɪt/ | 多重峰 | duō zhòng fēng |
| tetramethylsilane/ˈtetrəmiːθɪlsɪleɪn/ | 四甲基硅烷 | sì jiǎ jī guī wán |
| deuterated solvent/ˈdjuːtəreɪtɪd ˈsɒlvənt/ | 氘代溶剂 | dāo dài róng jì |
37.4
考试技巧
- In $^{13}\text{C}$ NMR the number of peaks equals the number of carbon environments — use symmetry to count them.
- In $^1\text{H}$ NMR use chemical shift (data booklet), integration (ratio of H's) and splitting (the $n+1$ rule): a triplet + quartet means an ethyl group.
- TMS is the reference ($\delta = 0$); a $\text{D}_2\text{O}$ shake removes O-H and N-H peaks.
- Combine IR, mass spectrum and NMR to deduce a structure, stating which evidence gives which feature.
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