A-Level Chemistry: Proton NMR Spectroscopy Analysis | A-Level 化学：质子核磁共振波谱解析

📚 A-Level Chemistry: Proton NMR Spectroscopy Analysis | A-Level 化学：质子核磁共振波谱解析

Proton nuclear magnetic resonance (¹H NMR) spectroscopy is one of the most powerful tools for determining the structure of organic molecules. In CIE A-Level Chemistry, you are expected to interpret a given ¹H NMR spectrum to deduce the environment of hydrogen atoms, their relative numbers, and the neighbouring groups within a molecule.

质子核磁共振波谱（¹H NMR）是确定有机分子结构最强有力的工具之一。在 CIE A-Level 化学考试中，你被要求解读给定的 ¹H NMR 谱图，以推断氢原子的化学环境、相对数目以及分子内的相邻基团。

1. What Is Proton NMR? | 什么是质子核磁共振？

¹H NMR spectroscopy measures the absorption of radiofrequency radiation by hydrogen nuclei placed in a strong magnetic field. The exact frequency absorbed depends on the electronic environment around each proton, allowing chemists to distinguish different types of hydrogen atoms in a molecule.

¹H NMR 波谱测量的是处于强磁场中的氢原子核对射频辐射的吸收。吸收的精确频率取决于每个质子周围的电子环境，这使得化学家能够区分分子中不同类型的氢原子。

The spectrum is a plot of signal intensity (y-axis) against chemical shift (x-axis, in parts per million, ppm). Each distinct set of equivalent protons produces one signal, and the position, area, and splitting of that signal carry structural information.

谱图是以信号强度（纵轴）对化学位移（横轴，单位为百万分之一 ppm）作图。每一组等效的质子产生一个信号，该信号的位置、面积和裂分都携带结构信息。

2. Chemical Shift and Shielding | 化学位移与屏蔽效应

Chemical shift (δ) is the resonance frequency of a proton relative to a standard, measured in ppm. It reflects how shielded or deshielded a proton is by surrounding electrons. More deshielded protons appear at higher δ values (downfield, to the left).

化学位移（δ）是质子相对于标准化合物的共振频率，以 ppm 为单位。它反映了质子被周围电子屏蔽或去屏蔽的程度。去屏蔽越强的质子出现在越高的 δ 值处（低场，谱图左侧）。

Electronegative atoms such as oxygen, nitrogen, or halogens withdraw electron density, deshielding nearby hydrogens and increasing their chemical shift. Conversely, alkyl groups donate electron density slightly, shielding protons and giving lower δ values.

氧、氮或卤素等电负性原子会拉走电子密度，使邻近的氢去屏蔽并增大其化学位移。相反，烷基会轻微地提供电子密度，使质子屏蔽增强，从而给出较低的 δ 值。

The key relationship is that δ is proportional to the local magnetic field experienced by the proton. A shielded proton experiences a slightly lower effective field and absorbs at a lower frequency, giving a smaller δ value.

关键关系在于 δ 与质子所经历的局部磁场成正比。被屏蔽的质子感受到的有效场强略低，因此吸收频率较低，给出的 δ 值较小。

3. TMS: The Internal Standard | TMS：内标物

Chemical shifts are measured relative to tetramethylsilane, Si(CH₃)₄, abbreviated TMS. TMS is assigned a δ value of 0 ppm. It is used because all twelve hydrogen atoms are chemically equivalent, it is inert and volatile, and its signals do not interfere with most organic proton signals.

化学位移是相对于四甲基硅烷 Si(CH₃)₄（缩写为 TMS）测得的。TMS 被规定为 δ = 0 ppm。使用它的原因是其十二个氢全部化学等价、惰性且易挥发，并且它的信号不干扰大多数有机质子的信号。

TMS is added in a small amount directly to the sample. The tiny peak at δ = 0 is a convenient internal reference point. In exam questions, you may be asked why TMS is suitable: answer with “single sharp peak, chemically inert, volatile, low boiling point, non-toxic, and separate from most organic peaks.”

TMS 以少量直接加入样品中。δ = 0 处的小峰是方便的内标参考点。考试中可能会问你为什么选择 TMS：回答应为“单峰、化学惰性、易挥发、沸点低、无毒、且与大多数有机峰分离”。

4. Integration: Peak Area and Proton Count | 积分：峰面积与质子数

The area under each NMR signal is proportional to the number of hydrogen atoms giving that signal. The spectrometer automatically integrates each peak, often shown as a step curve or number above the peak.

每个 NMR 信号下的峰面积与产生该信号的氢原子数目成正比。波谱仪会自动对每个峰进行积分，通常以阶梯曲线或峰上方的数字表示。

Integration ratios rather than absolute numbers are read directly. For example, if a spectrum contains three peaks in the ratio 3:2:1, the molecule may contain 6, 4, or 2 hydrogens in those environments, depending on the molecular formula. You should use the molecular formula to determine the absolute number.

直接读出的是积分比值而非绝对值。例如，如果一张谱图包含三个峰，积分比为 3:2:1，那么该分子在各个环境中可能含有 6、4 或 2 个氢，具体取决于分子式。你需要利用分子式来确定绝对数目。

For a molecule with formula C₄H₁₀O, an NMR spectrum may show three peaks with integration ratio 3:2:1 and total 10 hydrogens. The smallest integer ratio would be multiplied by 1.67? No, you should find the common factor. In this case the actual numbers are 5:3.33:1.67? That is not integer. Usually you choose the smallest integer ratio that sums to the number of H in the formula. If the integration trace gives 6:4:2, then divide by 2 to get 3:2:1. Then compare with formula: total = 6 parts, but the molecule has 10 H, so multiply each ratio by 10/6? That is not neat. Actually if peaks are 3:2:1, total = 6 units; to match 10 H, each unit = 10/6 = 1.67, not integer. So likely the integration trace would show actual counts: e.g., 5:2:3? Let’s keep explanation simple: “Integration values tell you relative numbers, not absolute counts”.

例如，对分子式 C₄H₁₀O，若谱图显示三个峰积分比为 3:2:1，总份数为 6，但分子中总氢数为 10，则不能直接相乘。实际上谱仪给出的积分值已经按比例反映氢数，你需要用分子式中的总氢数去校准。常见的考题会给出积分曲线下的数值，例如 6:4:2 或 3:2:1，而分子式总氢数可以帮助你确定具体是 3 个、2 个、1 个还是 6 个、4 个、2 个。

5. Spin-Spin Splitting: The n+1 Rule | 自旋-自旋裂分：n+1 规则

Adjacent non-equivalent protons can cause a signal to split into multiple peaks. If a proton (or a set of equivalent protons) has n neighbouring protons on an adjacent carbon, its signal appears as n+1 peaks (multiplets).

相邻的非等效质子会使信号裂分为多个峰。如果某个质子（或一组等效质子）在相邻碳上有 n 个邻近质子，其信号就表现为 n+1 个峰（多重峰）。

Common splitting patterns include: n=0 gives a singlet (s), n=1 gives a doublet (d) with intensity ratio 1:1, n=2 gives a triplet (t) with ratio 1:2:1, n=3 gives a quartet (q) with ratio 1:3:3:1.

常见的裂分模式包括：n=0 为单峰（s），n=1 为双重峰（d）强度比 1:1，n=2 为三重峰（t）强度比 1:2:1，n=3 为四重峰（q）强度比 1:3:3:1。

For example, in ethanol, CH₃CH₂OH, the CH₃ protons are coupled to the CH₂ protons (n=2), so the CH₃ signal is a triplet. The CH₂ protons are coupled to the CH₃ protons (n=3), so the CH₂ signal is a quartet. The OH proton is often a singlet because it may exchange rapidly.

例如，在乙醇 CH₃CH₂OH 中，CH₃ 质子与 CH₂ 质子耦合（n=2），因此 CH₃ 信号为三重峰；CH₂ 质子与 CH₃ 质子耦合（n=3），因此 CH₂ 信号为四重峰；OH 质子通常为单峰，因为它可能快速交换。

The splitting pattern therefore reveals the number of hydrogen atoms on adjacent carbon atoms, helping you to assemble the carbon skeleton.

因此，裂分模式揭示了相邻碳上氢原子的数目，帮助你组装碳骨架。

6. Equivalent and Non-Equivalent Protons | 等效与非等效质子

Protons that occupy identical chemical environments are called chemically equivalent. They have the same chemical shift and do not split each other. Non-equivalent protons have different environments and may split one another if within three bonds (usually on adjacent carbons).

占据完全相同化学环境的质子称为化学等效质子。它们具有相同的化学位移，并且彼此不裂分。非等效质子具有不同的环境，如果在三根键以内（通常在相邻碳上）则可能相互裂分。

In alkanes such as propane, CH₃CH₂CH₃, the two end CH₃ groups are equivalent to each other, and the central CH₂ is different. So propane shows two NMR signals: a triplet (for CH₃, n = 2 adjacent H) and a sextet? Wait actually CH₂ has six adjacent H across two CH₃ groups, so n=6, giving a septet (n+1=7). But in many exam contexts, symmetry simplifies. Let’s check: propane has two equivalent methyl groups, total 6 H on two carbons, both adjacent to CH₂, so CH₂ sees 6 equivalent methyl protons? Are the methyl protons equivalent to each other? Yes, the two CH₃ are equivalent, and all 6 H are equivalent? In propane, the two methyl groups are equivalent, and within each methyl the three H are equivalent by rotation. So all 6 methyl H are chemically equivalent? They are equivalent to each other overall by symmetry? They are on different carbons but symmetry makes them equivalent. So the CH₂ signal is split into 7 peaks (septet). The CH₃ signal is coupled to 2 equivalent H from CH₂, so doublet? Actually n=2? For CH₃, adjacent CH₂ has 2 hydrogen, so n=2, triplet. So propane: CH₃ triplet, CH₂ septet. This is advanced but could be mentioned. However CIE A-Level usually focuses on simple molecules like ethanol, propanone, ethyl ethanoate. Avoid overly complex examples unless needed. We’ll mention symmetry: internal mirror plane can create equivalence.

在丙烷 CH₃CH₂CH₃ 这类烷烃中，两端的 CH₃ 基团彼此等效，中心的 CH₂ 不同。因此丙烷显示两个 NMR 信号：CH₃ 为三重峰（相邻 2 个 H），CH₂ 为七重峰（相邻 6 个等效甲基 H）。

Homonuclear coupling only occurs between non-equivalent protons. Equivalent protons do not split one another because their magnetic states are averaged. Therefore you need to identify equivalent sets before predicting splitting.

同核耦合仅发生在非等效质子之间。等效质子不会互相裂分，因为它们的磁态被平均。因此，在预测裂分之前你需要先识别等效组。

7. Common Chemical Shift Ranges | 常见化学位移区间

Type of Proton	Chemical Shift δ / ppm
Alkane R–CH₃	0.9–1.0
Alkane R–CH₂–R	1.2–1.4
Alkane R₃CH	1.5–2.0
R–CO–CH₃ (methyl ketone)	2.1–2.6
R–O–CH₃ (ether)	3.2–4.0
–CH₂–OH or –CH₂–O–	3.3–4.5
–OH (alcohol)	1.0–5.0 (variable)
–COOH (carboxylic acid)	10–12
–CHO (aldehyde)	9–10
Aromatic C₆H₅–	6.5–8.0

You should memorise the approximate ranges for aldehyde, carboxylic acid, aromatic, alcohol/ether, and alkyl protons. In CIE questions, a table of chemical shifts is often provided, but for quick answers you should know the extremes: –COOH ~10–12, –CHO ~9–10, aromatic ~6.5–8, O–CH₃ ~3.5–4, CH₃–CO ~2.1.

你应该熟记醛、羧酸、芳香、醇/醚和烷基质子的近似范围。CIE 题目中通常会提供化学位移表，但为了快速作答，你应记住极端值：–COOH 约 10–12，–CHO 约 9–10，芳香约 6.5–8，O–CH₃ 约 3.5–4，CH₃–CO 约 2.1。

Remember that exchangeable protons like –OH and –NH can appear at variable positions depending on concentration, temperature, and solvent. Their signals may also be broad singlets.

记住像 –OH 和 –NH 这样的可交换质子会因浓度、温度和溶剂不同而出现在不同位置。它们的信号也常常是宽单峰。

8. Solvents and Deuterium | 溶剂与氘代

NMR samples are usually dissolved in a deuterated solvent such as CDCl₃ (deuterated chloroform) or D₂O (heavy water). Deuterium (²H) has a different magnetic resonance from protium (¹H), so it does not produce signals in the ¹H spectrum.

NMR 样品通常溶解在氘代溶剂中，如 CDCl₃（氘代氯仿）或 D₂O（重水）。氘（²H）与氕（¹H）具有不同的磁共振，因此不会在 ¹H 谱中产生信号。

If an alcohol, amine, or carboxylic acid is present, dissolving in D₂O leads to deuterium exchange: –OH becomes –OD and disappears from the ¹H NMR spectrum. This is a useful technique for confirming the presence of labile hydrogens.

如果存在醇、胺或羧酸，在 D₂O 中溶解会导致氘交换：–OH 变为 –OD 并从 ¹H NMR 谱中消失。这是确认活泼氢存在的有用技术。

In CIE practical questions, you may be asked why D₂O is used: it removes interfering solvent peaks and verifies exchangeable protons by their disappearance.

在 CIE 实验类问题中，你可能会被问到为什么使用 D₂O：它可以去除溶剂峰干扰，并通过活泼氢信号的消失来验证其存在。

9. Interpreting an NMR Spectrum: Step by Step | 解读 NMR 谱图：分步指南

When faced with a molecule’s ¹H NMR spectrum, follow these steps:

面对一张分子的 ¹H NMR 谱图时，请按以下步骤进行：

Step 1: Count the number of signals. Each signal corresponds to one chemically equivalent set of protons.

第一步：数出信号数量。每个信号对应一组化学等效质子。
Step 2: Measure the integration ratio of each signal. This tells you the relative number of H in each set.

第二步：测量每个信号的积分比。这告诉你每组中氢的相对数量。
Step 3: Determine the chemical shift of each signal. Correlate with functional groups (e.g., δ 9–10 suggests aldehyde).

第三步：确定每个信号的化学位移。与官能团关联（例如 δ 9–10 提示醛）。
Step 4: Examine the splitting pattern of each signal. Use the n+1 rule to deduce the number of H on adjacent carbon atoms.

第四步：观察每个信号的裂分模式。利用 n+1 规则推断相邻碳上的氢数。
Step 5: Combine the information to propose a structural fragment, then fit the fragments to the molecular formula and degree of unsaturation.

第五步：综合信息提出结构片段，然后根据分子式和不饱和度来拼接这些片段。

Always check that the total number of hydrogens in your proposed structure equals the molecular formula and that the predicted NMR spectrum matches the observed number of signals, integrations, and splittings.

始终检查你提出的结构中氢的总数是否等于分子式，并确保预测的 NMR 谱与观测到的信号数、积分和裂分一致。

10. Worked Example: Ethyl Ethanoate | 实例分析：乙酸乙酯

Consider the molecule ethyl ethanoate, CH₃COOCH₂CH₃. Its molecular formula is C₄H₈O₂. The ¹H NMR spectrum shows three signals in an integration ratio of 3:2:3.

考虑乙酸乙酯分子 CH₃COOCH₂CH₃。其分子式为 C₄H₈O₂。其 ¹H NMR 谱显示三个信号，积分比为 3:2:3。

The signal at δ ≈ 2.0 is a singlet integrating to 3H: this is the CH₃ group attached to the carbonyl (CH₃–CO). No adjacent H on that carbonyl carbon, hence a singlet.

δ ≈ 2.0 处的信号是单峰，积分为 3H：这是与羰基相连的 CH₃ 基团（CH₃–CO）。由于羰基碳上没有相邻 H，因此为单峰。

The signal at δ ≈ 4.1 is a quartet integrating to 2H: this is the –OCH₂– group. The four peaks come from coupling to the three CH₃ protons on the ethyl group (n=3, so 4 peaks).

δ ≈ 4.1 处的信号是四重峰，积分为 2H：这是 –OCH₂– 基团。四个峰来自与乙基上三个 CH₃ 质子的耦合（n=3，所以 4 个峰）。

The signal at δ ≈ 1.2 is a triplet integrating to 3H: this is the terminal CH₃ of the ethyl group. The three peaks come from coupling to the two –OCH₂– protons (n=2, so 3 peaks).

δ ≈ 1.2 处的信号是三重峰，积分为 3H：这是乙基末端的 CH₃。三个峰来自与两个 –OCH₂– 质子的耦合（n=2，所以 3 个峰）。

Thus the spectrum immediately distinguishes the acetyl methyl (singlet, δ 2.0) from the ethyl methyl (triplet, δ 1.2), proving the structure.

因此谱图能立刻区分乙酰甲基（单峰，δ 2.0）和乙基甲基（三重峰，δ 1.2），从而证明结构。

11. Worked Example: Propanal | 实例分析：丙醛

Propanal, C₂H₅CHO, is an aldehyde with structure CH₃CH₂CHO. Its ¹H NMR spectrum should show three signals with integration ratios 3:2:1.

丙醛 C₂H₅CHO，结构为 CH₃CH₂CHO。其 ¹H NMR 谱应显示三个信号，积分比为 3:2:1。

The aldehyde proton (–CHO) appears at δ ≈ 9.7 as a triplet. It couples to the two H on the adjacent CH₂ (n=2), hence a triplet. This confirms the presence of a –CH₂–CHO group.

醛基质子（–CHO）出现在 δ ≈ 9.7，为三重峰。它与相邻 CH₂ 上的两个 H 耦合（n=2），因此为三重峰。这证实了 –CH₂–CHO 基团的存在。

The CH₂ group between CH₃ and CHO appears at δ ≈ 2.4 as a quartet. It couples to the three CH₃ protons and also to the one aldehyde proton. Strictly, it is a doublet of quartets due to two different neighbouring groups, but in A-Level you can treat the combined coupling as more than 4 peaks? Wait: the CH₂ has adjacent CH₃ (3 equivalent H) and adjacent aldehyde H (1 H). These are non-equivalent neighbours with different coupling constants, leading to a complex multiplet. However, some simplified courses approximate as a “quartet” because only the CH₃ coupling is considered? Actually that is incorrect. Let’s discuss carefully: ethyl group attached to aldehyde: CH₃–CH₂–CHO. The CH₂ sees 3 H from CH₃ and 1 H from CHO, so n=4? Not exactly; because the neighbouring H are not all equivalent, the simple n+1 rule doesn’t directly apply. The signal is a complex multiplet with up to 8 lines, but often in A-Level they may simplify and say it is a quartet? No, they might avoid this. Better not use propanal if complexity arises. Use a simpler molecule like methyl ethanoate? But we already used ethyl ethanoate. Let’s choose another example: butan-2-one? Or 1-bromopropane? Let’s use 1-bromopropane, CH₃CH₂CH₂Br. It has three signals: CH₃ (triplet, δ 1.0), CH₂Br (triplet, δ 3.4), and central CH₂ (sextet? Actually central CH₂ has 5 neighbouring H: 3 from CH₃ and 2 from CH₂Br, but not equivalent? They are non-equivalent? CH₃ and CH₂Br protons have different chemical shifts; the central CH₂ is coupled to both sets with different coupling constants, so it would show a complex multiplet, often approximated as a sextet if the coupling constants are similar. In A-Level, they may accept a sextet (n=5). That is a bit advanced. Simpler: use propan-2-ol? CH₃CH(OH)CH₃: two signals? Actually CH₃ doublet, CH quintet, OH singlet. That’s still complex. Let’s use ethoxyethane (diethyl ether), (CH₃CH₂)₂O: only two signals: CH₃ triplet and CH₂ quartet. That’s simple. Or propanone: CH₃COCH₃: one signal singlet. But we need an example with functional groups. Let’s use methyl propanate? Hmm.

We can keep “Worked Example: Ethyl Ethanoate” as section 10 and add another “Worked Example: Propanone” or “Ethanol”. For ethanol: CH₃CH₂OH: three signals: CH₃ triplet (δ ~1.2), CH₂ quartet (δ ~3.7), OH singlet (δ ~1-5 depending). This is classic. Let’s use ethanol as second example. It’s simple and demonstrates –OH.

Let’s rewrite Section 11: Worked Example: Ethanol. Propanal is problematic. Yes, switch to ethanol.

Section content: Ethanol CH₃CH₂OH. Three signals: CH₃ triplet (δ 1.2), CH₂ quartet (δ 3.7), OH singlet (δ ~2.5). Why OH singlet? Because OH does not couple to CH₂ significantly due to rapid exchange. Integration ratio 3:2:1.

We’ll write that.

Also need mention splitting of OH can be variable.

Let’s craft accordingly.

Need ensure no unnecessary content, but complete.

Need final note “Published by TutorHao | Chemistry Revision Series | aleveler.com”. Use exactly.

Let’s proceed to write full HTML. Ensure all paragraphs paired. Need include “TITLE: …” first line exactly.

Let’s write.TITLE: A-Level Chemistry: Proton NMR Spectroscopy Analysis | A-Level 化学：质子核磁共振波谱解析