📚 Nuclear Magnetic Resonance Spectroscopy | 核磁共振波谱
Nuclear magnetic resonance (NMR) spectroscopy is one of the most powerful analytical techniques available to chemists, allowing the detailed structural elucidation of organic molecules by probing the magnetic environments of atomic nuclei. This A-Level AQA Chemistry guide provides a complete and rigorous treatment of NMR spectroscopy, from the underlying physics to spectral interpretation and exam-style applications.
核磁共振波谱法是化学家可用的最强有力分析技术之一，它通过探测原子核所处的磁环境来揭示有机分子的精细结构。本 A-Level AQA 化学指南对 NMR 波谱进行完整且严谨的阐述，涵盖基础物理原理、谱图解析以及考试型应用。
1. Fundamental Principles of NMR | NMR 的基本原理
Certain nuclei, such as ¹H (protons) and ¹³C, possess a property called spin, which generates a tiny magnetic moment. When placed in an external magnetic field, these nuclei align either with the field (lower energy, α-state) or against it (higher energy, β-state). The energy difference between these two spin states depends on the strength of the applied magnetic field and the gyromagnetic ratio of the nucleus.
某些原子核，如 ¹H（质子）和 ¹³C，具有一种称为自旋的性质，会产生微小的磁矩。当置于外磁场中时，这些原子核要么顺着磁场排列（较低能态，α 态），要么逆着磁场排列（较高能态，β 态）。这两种自旋态之间的能量差取决于外加磁场强度以及原子核的旋磁比。
The frequency of radiation required to flip a nucleus between spin states is given by the resonance condition:
使原子核在两个自旋态之间翻转所需辐射频率由共振条件给出：
ΔE = hν = (γhB₀)/(2π)
where γ is the gyromagnetic ratio, B₀ is the applied magnetic field strength, and h is Planck’s constant. In a typical NMR instrument operating at 400 MHz for ¹H, the applied field is approximately 9.4 tesla.
其中 γ 为旋磁比，B₀ 为外加磁场强度，h 为普朗克常数。在以 400 MHz 频率运行、用于检测 ¹H 的典型 NMR 仪器中，所施加的磁场约为 9.4 特斯拉。
2. Chemical Shift and the δ Scale | 化学位移与 δ 标度
The precise resonance frequency of a nucleus depends on its electronic environment. Electrons surrounding a nucleus circulate in the applied magnetic field, generating a small opposing field that shields the nucleus. This shielding effect reduces the effective field experienced by the nucleus, causing it to resonate at a slightly lower frequency. The variation in resonance frequency due to electronic shielding is called the chemical shift.
原子核的精确共振频率取决于其电子环境。原子核周围的电子在外加磁场中循环运动，产生一个微小的反向磁场，对该原子核起到屏蔽作用。这种屏蔽效应降低了原子核感受到的有效磁场，使其在略低的频率下共振。由电子屏蔽引起的共振频率变化称为化学位移。
Chemical shift δ is measured in parts per million (ppm) relative to a reference compound, typically tetramethylsilane (TMS). The chemical shift is calculated using the formula:
化学位移 δ 以百万分之一（ppm）为单位，相对于参考化合物（通常是四甲基硅烷，TMS）进行测量。化学位移按以下公式计算：
δ = (ν_sample − ν_TMS) / ν_TMS × 10⁶
Since δ is a ratio of frequency differences to the operating frequency, it is independent of the instrument’s field strength, allowing spectra collected on different machines to be compared directly. Tetramethylsilane is chosen as the reference because its twelve equivalent protons are highly shielded, producing a single sharp peak at δ = 0, which conveniently falls outside the typical range of most organic protons.
由于 δ 是频率差与工作频率之比，它与仪器的磁场强度无关，因此不同仪器上采集的谱图可以直接比较。选择四甲基硅烷作为参考化合物，是因为其十二个等效质子受到高度屏蔽，在 δ = 0 处产生单一尖锐峰，恰好在大多数有机质子的典型范围之外。
3. Equivalent Protons and Integration | 等效质子与积分
Protons that occupy identical chemical environments in a molecule are called chemically equivalent protons. These protons experience the same shielding and therefore resonate at the same frequency, producing a single signal. For example, all three protons in a methyl group (−CH₃) are equivalent to each other, but they are not equivalent to protons in a neighbouring methylene group (−CH₂−).
在分子中占据相同化学环境的质子称为化学等效质子。这些质子受到相同的屏蔽作用，因而在相同频率处共振，产生单一信号。例如，甲基（−CH₃）中的三个质子彼此等效，但它们与相邻亚甲基（−CH₂−）中的质子并不等效。
Integration of an NMR signal gives the relative number of protons responsible for that signal. The area under each peak is proportional to the number of equivalent protons producing it. For instance, in the ¹H NMR spectrum of ethanol (CH₃CH₂OH), the integrals of the three signals corresponding to CH₃, CH₂, and OH appear in the ratio 3 : 2 : 1. This information is essential for assigning signals to specific proton environments within a molecule.
NMR 信号的积分给出产生该信号的质子的相对数目。每个峰下的面积与产生该信号的等效质子数目成正比。例如，在乙醇（CH₃CH₂OH）的 ¹H NMR 谱中，对应于 CH₃、CH₂ 和 OH 的三个信号的积分比率为 3 : 2 : 1。该信息对于将信号归属到分子中特定质子环境至关重要。
4. Factors Affecting Chemical Shift | 影响化学位移的因素
Several structural factors influence the chemical shift of a proton. The most important is electronegativity. Electronegative atoms such as oxygen, nitrogen, and halogens withdraw electron density from nearby protons, thereby deshielding them. Deshielded protons experience a larger effective field and resonate at higher δ values (further downfield).
若干结构因素影响质子的化学位移。最重要的是电负性。氧、氮和卤素等电负性原子从附近质子处吸引电子密度，从而使其去屏蔽。去屏蔽的质子感受到更大的有效磁场，在更高的 δ 值处共振（更靠低场）。
A general guide to ¹H chemical shift ranges is provided below:
以下是 ¹H 化学位移范围的一般指南：
| Proton environment | Chemical shift δ / ppm |
| Alkane (R−CH₃) | 0.8 – 1.0 |
| R−CH₂−R | 1.2 – 1.4 |
| R−CH₂−Cl | 3.2 – 3.6 |
| R−CH₂−O | 3.3 – 4.0 |
| Aromatic (Ar−H) | 6.5 – 8.0 |
| Aldehyde (R−CHO) | 9.4 – 10.0 |
| Carboxylic acid (R−COOH) | 10 – 13 |
The deshielding effect of electronegative atoms decreases rapidly with distance, so protons on carbons more than three bonds away from a substituent are generally little affected.
电负性原子的去屏蔽效应随距离增大而迅速减弱，因此与取代基相距三个键以上的碳上的质子通常受到很小的影响。
5. Spin-Spin Splitting (n+1 Rule) | 自旋-自旋裂分（n+1 规则）
Protons that are non-equivalent and separated by three bonds or fewer interact magnetically through the bonding electrons, a phenomenon known as spin-spin coupling. This interaction causes each signal to split into multiple peaks. The number of peaks into which a signal is split follows the n+1 rule, where n is the number of equivalent protons on adjacent carbon atoms.
非等效且相隔三个键或更少的质子通过成键电子发生磁相互作用，这一现象称为自旋-自旋耦合。这种相互作用使每个信号分裂为多重峰。一个信号分裂为峰的数目遵循 n+1 规则，其中 n 为相邻碳原子上等效质子的数目。
For example, consider the ¹H NMR spectrum of ethanol. The CH₃ protons are adjacent to a CH₂ group containing two equivalent protons, so the CH₃ signal is split into a triplet (n+1 = 3). Conversely, the CH₂ protons are adjacent to a CH₃ group with three equivalent protons, so the CH₂ signal is split into a quartet (n+1 = 4). The relative peak intensities within a multiplet follow Pascal’s triangle:
例如，考虑乙醇的 ¹H NMR 谱。CH₃ 质子与含有两个等效质子的 CH₂ 基团相邻，因此 CH₃ 信号裂分为三重峰（n+1 = 3）。相反，CH₂ 质子与具有三个等效质子的 CH₃ 基团相邻，因此 CH₂ 信号裂分为四重峰（n+1 = 4）。多重峰内各峰的相对强度遵循杨辉三角：
singlet 1 : 1
doublet 1 : 2 : 1
triplet 1 : 3 : 3 : 1
quartet 1 : 4 : 6 : 4 : 1
It is important to note that protons exchanging rapidly, such as OH and NH protons, often do not show splitting due to chemical exchange broadening. Additionally, equivalent protons do not split each other.
需要注意，快速交换的质子，如 OH 和 NH 质子，由于化学交换展宽通常不显示裂分。此外，等效质子彼此不产生裂分。
6. Solvent Selection and Deuteration | 溶剂选择与氘代
NMR spectra are typically recorded in solution, and the solvent must be chosen carefully to avoid overwhelming signals from the sample. Protonated solvents contain abundant ¹H nuclei that would produce intense, interfering peaks. Therefore, deuterated solvents such as CDCl₃, D₂O, and (CD₃)₂SO are used instead.
NMR 谱通常在溶液中进行记录，必须谨慎选择溶剂以避免来自样品信号的干扰。含氢溶剂含有大量 ¹H 核，会产生强烈且造成干扰的峰。因此，使用氘代溶剂如 CDCl₃、D₂O 和 (CD₃)₂SO。
Deuterium (²H) has a different gyromagnetic ratio from ¹H and resonates at a very different frequency, so it does not appear in the ¹H NMR spectrum. However, residual protons in the deuterated solvent (e.g., CHCl₃ in CDCl₃) still appear as small impurity peaks. In CDCl₃, this residual CHCl₃ peak appears as a singlet at δ = 7.26 ppm, which is well known to practitioners.
氘（²H）具有与 ¹H 不同的旋磁比，在非常不同的频率处共振，因此不会出现在 ¹H NMR 谱中。然而，氘代溶剂中残留的质子（例如 CDCl₃ 中的 CHCl₃）仍会作为小的杂质峰出现。在 CDCl₃ 中，残留 CHCl₃ 峰在 δ = 7.26 ppm 处显示为单峰，这是从业者熟知的。
Exchangeable protons such as O−H and N−H can be identified by adding D₂O to the sample. These protons exchange with deuterium from D₂O, causing their signals to disappear from the ¹H NMR spectrum. This technique is invaluable for confirming the presence of alcohol, phenol, amine, and carboxylic acid functional groups.
可交换质子如 O−H 和 N−H 可通过向样品中加入 D₂O 来识别。这些质子与 D₂O 中的氘发生交换，导致其信号从 ¹H NMR 谱中消失。该技术对于确认醇、酚、胺和羧酸官能团的存在非常宝贵。
7. Chemical Exchange and OH/NH Protons | 化学交换与 OH/NH 质子
Hydroxyl and amino protons behave differently from C−H protons in NMR spectroscopy. Because they participate in rapid chemical exchange with other molecules of the same or similar species (especially in protic solvents or in the presence of trace acid or base), their signals are often broad singlets that do not exhibit coupling to adjacent protons.
羟基和氨基质子在 NMR 波谱中的行为与 C−H 质子不同。由于它们与相同或相似种类的其他分子（尤其是在质子性溶剂中或存在痕量酸或碱时）发生快速化学交换，它们的信号通常是宽单峰，不显示与相邻质子的耦合。
The chemical shift of an OH proton is also concentration- and temperature-dependent, because hydrogen bonding alters the degree of shielding. In dilute solutions or at higher temperatures, hydrogen bonding is reduced, and the OH proton typically appears at a lower δ value. This variability means that OH signals can be assigned with confidence only after performing a D₂O shake experiment, in which the signal disappears as the OH proton is replaced by deuterium.
OH 质子的化学位移还与浓度和温度相关，因为氢键改变屏蔽程度。在稀溶液或较高温度下，氢键减弱，OH 质子通常出现在较低 δ 值处。这种可变性意味着 OH 信号只有在进行 D₂O 振荡实验后才能可靠归属，在该实验中，OH 质子被氘取代后信号消失。
8. Interpretation of ¹H NMR Spectra | ¹H NMR 谱图的解析
The systematic interpretation of a ¹H NMR spectrum involves four key steps. First, count the number of signals, which indicates the number of distinct proton environments. Second, examine the chemical shift of each signal to deduce the electronic environment, such as whether the proton is attached to an sp³ carbon, an aromatic ring, or a carbonyl-adjacent carbon. Third, analyse the splitting pattern to determine the number of protons on adjacent carbons. Fourth, use the integration ratio to establish the relative number of protons in each environment.
系统解析 ¹H NMR 谱涉及四个关键步骤。首先，数出信号的数目，它指示不同质子环境的数目。其次，检查每个信号的化学位移以推断电子环境，例如质子是连接在 sp³ 碳、芳环还是与羰基相邻的碳上。第三，分析裂分模式以确定相邻碳上质子的数目。第四，利用积分比率确定每个环境中质子的相对数目。
Consider the ¹H NMR spectrum of propanal (CH₃CH₂CHO). Three signals are expected. The aldehyde proton (−CHO) appears as a triplet at approximately δ 9.8 ppm because it couples to the two equivalent CH₂ protons. The CH₂ protons, being adjacent to both the electron-withdrawing CHO group and the CH₃ group, appear as a multiplet at approximately δ 2.4 ppm. The CH₃ protons appear as a triplet at approximately δ 1.1 ppm. The integrals are in the ratio 1 : 2 : 3.
考虑丙醛（CH₃CH₂CHO）的 ¹H NMR 谱。预期存在三个信号。醛基质子（−CHO）在约 δ 9.8 ppm 处显示为三重峰，因为它与两个等效 CH₂ 质子耦合。CH₂ 质子与吸电子的 CHO 基团和 CH₃ 基团均相邻，在约 δ 2.4 ppm 处显示为多重峰。CH₃ 质子在约 δ 1.1 ppm 处显示为三重峰。积分比率为 1 : 2 : 3。
9. ¹³C NMR Spectroscopy | ¹³C NMR 波谱法
Carbon-13 nuclei also possess spin ½ and are NMR-active, but their natural abundance is only 1.1%. This low abundance, combined with the smaller gyromagnetic ratio of ¹³C compared to ¹H, makes ¹³C NMR signals far weaker and requires many more scans to achieve a good signal-to-noise ratio. However, ¹³C NMR provides crucial information about the carbon skeleton of a molecule.
碳-13 原子核也具有自旋 ½，具有 NMR 活性，但其天然丰度仅为 1.1%。低丰度加之 ¹³C 的旋磁比小于 ¹H，使得 ¹³C NMR 信号弱得多，需要更多次扫描才能获得良好的信噪比。然而，¹³C NMR 提供关于分子碳骨架的关键信息。
In routine ¹³C NMR spectra, proton decoupling is employed so that each unique carbon environment appears as a single sharp line, with no splitting from attached protons. The chemical shift range for ¹³C is much wider than for ¹H, typically spanning 0 – 220 ppm. This large dispersion makes ¹³C spectra especially useful for distinguishing between carbon environments that would be difficult to resolve in ¹H spectra.
在常规 ¹³C NMR 谱中，采用质子去耦，使每个独特的碳环境以单一尖锐谱线出现，不显示与所连接质子的裂分。¹³C 的化学位移范围远宽于 ¹H，通常跨越 0 – 220 ppm。这种大色散使 ¹³C 谱特别有助于区分在 ¹H 谱中难以分辨的碳环境。
| Carbon environment | Chemical shift δ / ppm |
| Alkane (C−C) | 5 – 50 |
| Amine (C−N) | 40 – 65 |
| Alcohol/Ether (C−O) | 50 – 90 |
| Alkene (C=C) | 100 – 165 |
| Aromatic (Ar−C) | 110 – 160 |
| Carboxylic acid/Amide (C=O) | 160 – 185 |
| Aldehyde/Ketone (C=O) | 190 – 220 |
10. Combining NMR with Other Techniques | NMR 与其他技术的结合
NMR spectroscopy is most powerful when combined with other analytical techniques, particularly mass spectrometry (MS) and infrared (IR) spectroscopy. Mass spectrometry provides the molecular mass and molecular formula from the molecular ion and isotopic abundance data, while IR spectroscopy identifies key functional groups through characteristic absorption frequencies.
NMR 波谱与质谱（MS）和红外（IR）波谱等其他分析技术结合时最为强大。质谱通过分子离子和同位素丰度数据提供分子质量和分子式，而红外波谱通过特征吸收频率鉴别关键官能团。
For example, a compound with the molecular formula C₃H₆O₂ from mass spectrometry, showing a strong IR absorption at 1710 cm⁻¹ (indicating a carbonyl group), can be further analysed by ¹H NMR. If the spectrum shows a singlet at δ 3.7 ppm (integrating to 3H) and a singlet at δ 2.0 ppm (integrating to 3H), the compound is methyl ethanoate (CH₃COOCH₃). The absence of splitting and the downfield position of the OCH₃ singlet are diagnostic.
例如，质谱给出的分子式为 C₃H₆O₂、红外在 1710 cm⁻¹ 处有强吸收（表明羰基）的化合物，可通过 ¹H NMR 进一步分析。如果谱图在 δ 3.7 ppm 处显示单峰（积分为 3H），在 δ 2.0 ppm 处显示单峰（积分为 3H），则该化合物为乙酸甲酯（CH₃COOCH₃）。无裂分以及 OCH₃ 单峰的低场位置具有诊断意义。
11. Exam-Style Problem Solving | 考试型问题解答
A common A-level question asks students to deduce the structure of an unknown compound from its molecular formula and spectral data. Consider a compound with molecular formula C₄H₈O₂ showing the following ¹H NMR data: a triplet at δ 1.2 ppm (3H), a singlet at δ 2.0 ppm (3H), and a quartet at δ 4.1 ppm (2H).
常见的 A-level 问题要求学生从分子式和波谱数据推断未知化合物的结构。考虑分子式为 C₄H₈O₂ 的化合物显示以下 ¹H NMR 数据：δ 1.2 ppm 处三重峰（3H），δ 2.0 ppm 处单峰（3H），以及 δ 4.1 ppm 处四重峰（2H）。
The quartet at δ 4.1 ppm indicates a CH₂ group attached to an electronegative oxygen atom (typical of an ester O−CH₂), while the triplet at δ 1.2 ppm shows a CH₃ group adjacent to a CH₂ group. The singlet at δ 2.0 ppm is consistent with CH₃ attached to a carbonyl group (CH₃CO−). Thus, the structure is ethyl ethanoate: CH₃COOCH₂CH₃. The degree of unsaturation from the formula C₄H₈O₂ is one, consistent with the single C=O bond in the ester.
δ 4.1 ppm 处的四重峰表明 CH₂ 基团连接在电负性氧原子（典型的酯 O−CH₂）上，而 δ 1.2 ppm 处的三重峰显示与 CH₂ 基团相邻的 CH₃ 基团。δ 2.0 ppm 处的单峰与连接在羰基上的 CH₃（CH₃CO−）一致。因此，该结构为乙酸乙酯：CH₃COOCH₂CH₃。由分子式 C₄H₈O₂ 计算的不饱和度为 1，与酯中的单个 C=O 键一致。
12. Limitations and Common Pitfalls | 局限性与常见错误
Several limitations of NMR spectroscopy should be noted. The technique requires relatively pure samples and is less sensitive than mass spectrometry. Overlapping signals in complex molecules or in ¹H spectra with narrow chemical shift ranges can complicate interpretation. Additionally, compounds with no hydrogen atoms, such as CCl₄, produce no ¹H NMR spectrum, and paramagnetic impurities can cause severe line broadening.
应注意到 NMR 波谱的几个局限性。该技术要求相对纯净的样品，灵敏度低于质谱。复杂分子中信号重叠或 ¹H 谱化学位移范围狭窄可能使解析复杂化。此外，不含氢原子的化合物如 CCl₄ 不产生 ¹H NMR 谱，顺磁性杂质可导致严重谱线展宽。
Common exam pitfalls include forgetting that OH and NH signals are often broad singlets without splitting; assuming that the n+1 rule applies to protons separated by more than three bonds (it does not); and incorrectly identifying non-equivalent protons in symmetrical molecules. For example, in 2,2-dimethylpropane (C(CH₃)₄), all twelve protons are equivalent and produce only one signal despite the large number of hydrogen atoms. Careful attention to molecular symmetry is essential for accurate spectral interpretation.
常见考试错误包括：忘记 OH 和 NH 信号通常是宽单峰且不裂分；假设 n+1 规则适用于相隔超过三个键的质子（实际不适用）；以及在对称分子中错误识别非等效质子。例如，在 2,2-二甲基丙烷（C(CH₃)₄）中，全部十二个质子等效，尽管氢原子数目众多，却只产生一个信号。密切注意分子对称性对于准确波谱解析至关重要。
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