📚 Combined Science 192: NMR Spectroscopy | 综合科学192：核磁共振波谱
Nuclear magnetic resonance (NMR) spectroscopy is one of the most powerful analytical tools in modern chemistry. For Edexcel A-Level Chemistry, NMR appears as a core technique used to determine the structure of organic molecules by revealing the chemical environments of hydrogen-1 and carbon-13 nuclei.
核磁共振波谱是现代化学中最强大的分析工具之一。在 Edexcel A-Level 化学中，NMR 是一种核心分析技术，它通过揭示氢-1 和碳-13 原子核的化学环境来确定有机分子的结构。
1. What NMR Reveals | 核磁共振揭示什么
NMR spectroscopy tells us how many different proton or carbon environments exist in a molecule, how many atoms occupy each environment, and how adjacent proton groups are connected. It cannot directly measure molecular mass, so it is often used alongside mass spectrometry and infrared spectroscopy.
核磁共振波谱可以告诉我们在一个分子中存在多少种不同的质子或碳环境、每种环境中有多少个原子，以及相邻的质子基团如何连接。它不能直接测量分子质量，因此通常与质谱和红外光谱一起使用。
In organic structure determination, NMR answers three key questions: how many non-equivalent protons or carbons are present, in what ratio, and what is the carbon skeleton connectivity.
在有机结构确定中，NMR 回答三个关键问题：存在多少种非等价的质子或碳、它们的比例是多少，以及碳骨架的连接方式是什么。
2. Magnetic Spin and Resonance | 磁自旋与共振
Certain nuclei, including ¹H and ¹³C, possess nuclear spin. When placed in a strong external magnetic field B₀, their spins align either with the field (lower energy, α-state) or against it (higher energy, β-state). The energy difference ΔE is proportional to the field strength.
某些原子核，包括 ¹H 和 ¹³C，具有核自旋。当置于强外部磁场 B₀ 中时，它们的自旋要么与磁场平行（低能级，α 态），要么与磁场反平行（高能级，β 态）。能量差 ΔE 与磁场强度成正比。
Radiofrequency radiation supplies exactly ΔE, causing nuclei to flip from the lower to the higher spin state. This absorption is called resonance and is detected as an NMR signal.
射频辐射恰好提供 ΔE 的能量，使原子核从低自旋态翻转到高自旋态。这种吸收称为共振，并被检测为 NMR 信号。
3. Chemical Shift δ | 化学位移 δ
Electrons around a nucleus shield it from the external magnetic field. Nuclei in different chemical environments experience slightly different effective magnetic fields, so they resonate at different frequencies. This is recorded as the chemical shift, symbol δ, in parts per million (ppm).
原子核周围的电子会对外部磁场产生屏蔽作用。不同化学环境中的原子核感受到的有效磁场略有不同，因此它们的共振频率也不同。这被记录为化学位移，符号为 δ，单位为百万分之一（ppm）。
δ = (νₛₐₘₚₗₑ − ν_TMS) ÷ ν_operating × 10⁶ ppm
Where νₛₐₘₚₗₑ is the resonance frequency of the sample nucleus, ν_TMS is the reference frequency, and ν_operating is the operating frequency of the spectrometer. Using ppm ensures δ values are independent of magnetic field strength.
其中 νₛₐₘₚₗₑ 是样品原子核的共振频率，ν_TMS 是参比频率，ν_operating 是波谱仪的运行频率。使用 ppm 可以确保 δ 值与磁场强度无关。
4. Tetramethylsilane Standard | 四甲基硅烷标准
Tetramethylsilane, Si(CH₃)₄, abbreviated TMS, is the standard reference for both ¹H and ¹³C NMR. It is assigned δ = 0 ppm.
四甲基硅烷，Si(CH₃)₄，缩写为 TMS，是 ¹H 和 ¹³C NMR 的标准参比物。它的 δ 值被指定为 0 ppm。
TMS is chemically unreactive, non-toxic, volatile, and easily removed from the sample. It gives one sharp singlet because all twelve equivalent protons and all four equivalent carbons appear together in highly shielded environments.
TMS 化学性质不活泼、无毒、易挥发，并且容易从样品中除去。它能产生一个尖锐的单峰，因为十二个等价的质子和四个等价的碳都出现在高度屏蔽的同一环境中。
5. Proton NMR Environments | 质子 NMR 环境
The number of signals in a ¹H NMR spectrum equals the number of non-equivalent proton environments in the molecule. Equivalent protons are those related by symmetry or rapid rotation, such as the three protons in a methyl group on a carbon with no chiral centre.
¹H NMR 谱中信号的数量等于分子中非等价质子环境的数量。等价质子是由对称性或快速旋转关联的质子，例如没有手性中心的碳上的甲基中三个质子。
For example, ethanal CH₃CHO has two proton environments: the methyl protons CH₃ and the aldehyde proton CHO. Its ¹H NMR spectrum therefore shows two signals.
例如，乙醛 CH₃CHO 有两种质子环境：甲基质子 CH₃ 和醛基质子 CHO。因此它的 ¹H NMR 谱显示两个信号。
6. Splitting Patterns and the n+1 Rule | 裂分规律与 n+1 规则
A proton signal is split into multiplets by spin-spin coupling with non-equivalent protons on adjacent carbon atoms. The number of lines is given by the n+1 rule, where n is the number of equivalent protons on the neighbouring carbon atom.
质子信号会因与相邻碳原子上非等价质子的自旋-自旋耦合而裂分为多重峰。谱线数量由 n+1 规则给出，其中 n 是相邻碳原子上等价质子的数量。
- n = 0 → singlet, 1 line
- n = 1 → doublet, 2 lines
- n = 2 → triplet, 3 lines
- n = 3 → quartet, 4 lines
中文对应：
- n = 0 → 单峰，1 条线
- n = 1 → 双峰，2 条线
- n = 2 → 三重峰，3 条线
- n = 3 → 四重峰，4 条线
The intensity ratios of lines in a multiplet follow Pascal’s triangle: a doublet is 1:1, a triplet is 1:2:1, and a quartet is 1:3:3:1.
多重峰中各谱线的强度比遵循帕斯卡三角：双峰为 1:1，三重峰为 1:2:1，四重峰为 1:3:3:1。
7. Integration Traces | 积分曲线
The area under each proton NMR signal is proportional to the number of protons that produce that signal. Modern spectrometers draw an integration trace, and the step heights give the relative number of protons in each environment.
每个质子 NMR 信号下方的面积与产生该信号的质子数成正比。现代波谱仪会绘制积分曲线，阶梯高度给出各环境中质子的相对数量。
For example, in ethanol CH₃CH₂OH, the integration ratio for the CH₃, CH₂, and OH protons is 3:2:1, provided the OH proton is not exchanged. Integration is essential for proposing the correct molecular formula fragment.
例如，在乙醇 CH₃CH₂OH 中，CH₃、CH₂ 和 OH 质子的积分比为 3:2:1，前提是 OH 质子未被交换。积分对于提出正确的分子式片段至关重要。
8. Carbon-13 NMR | 碳-13 核磁共振
¹³C NMR produces one signal per non-equivalent carbon atom. Because ¹³C is only about 1.1% abundant, coupling between adjacent ¹³C nuclei is rare, and the spectrum is usually recorded with proton decoupling, so all signals appear as singlets.
¹³C NMR 中每个非等价碳原子产生一个信号。由于 ¹³C 的丰度仅约 1.1%，相邻 ¹³C 核之间的耦合很少见，并且谱图通常在质子去耦条件下记录，因此所有信号都表现为单峰。
The ¹³C chemical shift range is much wider than ¹H, typically δ 0–220 ppm. Carbonyl carbons appear at δ 160–220 ppm, alkene and aromatic carbons at δ 100–150 ppm, and alkyl carbons at δ 0–50 ppm.
¹³C 化学位移范围比 ¹H 宽得多，通常为 δ 0–220 ppm。羰基碳出现在 δ 160–220 ppm，烯烃和芳香碳出现在 δ 100–150 ppm，烷基碳出现在 δ 0–50 ppm。
9. Interpreting a Spectrum | 解析谱图
To identify an unknown from NMR data, follow a systematic method. First calculate the degree of unsaturation from the molecular formula using the expression IHD = (2C + 2 + N − H − X)/2, where C is carbon, N nitrogen, H hydrogen, and X halogen.
要根据 NMR 数据鉴定一个未知物，应遵循系统方法。首先由分子式计算不饱和度，表达式为 IHD = (2C + 2 + N − H − X)/2，其中 C 为碳，N 为氮，H 为氢，X 为卤素。
Then list the NMR peaks by δ value, note their integration ratios, and analyse their splitting patterns. Combine the fragments into a structure that accounts for all signals, chemical shifts, and the molecular formula.
然后按 δ 值列出 NMR 峰，记录它们的积分比，并分析裂分模式。将各片段组合成一个能够解释所有信号、化学位移和分子式的结构。
For example, a compound C₃H₆O with a ¹H NMR spectrum showing a 1H singlet at δ 9.8, a 3H singlet at δ 2.1, and a 2H quartet overlapping with a 3H triplet can be identified as propanal CH₃CH₂CHO by matching the aldehyde proton, methyl group, and ethyl chain.
例如，化合物 C₃H₆O 的 ¹H NMR 谱显示 δ 9.8 处有 1H 单峰、δ 2.1 处有 3H 单峰，以及一个 2H 四重峰与一个 3H 三重峰重叠，通过与醛基质子、甲基和乙基链匹配，可鉴定为丙醛 CH₃CH₂CHO。
10. Solvents and Proton Exchange | 溶剂与质子交换
NMR samples are dissolved in solvents that do not contain ordinary hydrogen, such as deuterated chloroform CDCl₃ or deuterated water D₂O. Deuterium has an even mass number and does not produce a signal in the ¹H NMR range.
NMR 样品溶解在不含普通氢的溶剂中，例如氘代氯仿 CDCl₃ 或重水 D₂O。氘的质量数为偶数，在 ¹H NMR 范围内不产生信号。
Labile protons such as OH and NH can undergo rapid exchange with solvent protons. These peaks are often broad and may disappear when D₂O is added, confirming the presence of an exchangeable proton.
OH 和 NH 等活泼质子能与溶剂质子发生快速交换。这些峰通常较宽，加入 D₂O 后可能会消失，从而证实可交换质子的存在。
11. Common Exam Pitfalls | 常见考试失分点
Students often forget that symmetry reduces the number of NMR signals. For example, 1,4-dimethylbenzene has only two ¹H environments and two ¹³C environments despite having eight hydrogen atoms and eight carbon atoms.
学生经常忘记对称性会减少 NMR 信号的数量。例如，1,4-二甲基苯虽然含有八个氢原子和八个碳原子，但仅有两种 ¹H 环境和两种 ¹³C 环境。
Another common error is applying the n+1 rule to equivalent neighbouring protons or across non-adjacent carbon atoms. Coupling usually only occurs over three bonds, and equivalent protons do not split each other.
另一个常见错误是把 n+1 规则套用于等价的相邻质子或不相邻碳原子上的质子。耦合通常只发生在相隔三根键的原子之间，等价质子之间不会相互裂分。
When calculating integration ratios, always simplify the step heights to the smallest whole numbers. For example, a 6:4:2 ratio must be reported as 3:2:1.
在计算积分比时，始终将阶梯高度化简为最小整数。例如，6:4:2 的比值必须写成 3:2:1。
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