Nuclear Magnetic Resonance (NMR) in IGCSE WJEC Chemistry | IGCSE WJEC 化学：核磁共振考点精讲

📚 Nuclear Magnetic Resonance (NMR) in IGCSE WJEC Chemistry | IGCSE WJEC 化学：核磁共振考点精讲

Nuclear magnetic resonance (NMR) spectroscopy is a powerful analytical technique that reveals the structure of organic molecules by studying the magnetic properties of certain atomic nuclei. For IGCSE WJEC Chemistry, understanding the basics of proton (¹H) NMR allows you to identify the number and type of hydrogen environments in a molecule, which is essential for deducing molecular structure.

核磁共振（NMR）波谱是一种强大的分析技术，通过研究某些原子核的磁性质来揭示有机分子的结构。对于 IGCSE WJEC 化学，了解质子（¹H）核磁共振的基础知识可以帮助你确定分子中氢环境的数量和类型，这对于推断分子结构至关重要。

1. What is Nuclear Magnetic Resonance? | 什么是核磁共振？

Nuclear magnetic resonance relies on the fact that nuclei of certain atoms, such as hydrogen-1 (¹H), behave like tiny magnets because they have a property called spin. When placed in a strong external magnetic field, these nuclei can align either with or against the field. The energy difference between these two orientations lies in the radio wave region of the electromagnetic spectrum.

核磁共振依据的是某些原子核（如氢-1 ¹H）由于具有自旋特性而表现得像微小磁体这一事实。当置于强外部磁场中时，这些原子核可以顺着或逆着磁场方向排列。这两种取向之间的能量差位于电磁波谱的射频波段。

By applying a pulse of radio waves at exactly the right frequency, nuclei in the lower energy state can ‘flip’ to the higher energy state – this is resonance. The exact frequency absorbed depends on the local chemical environment of each nucleus, which is what makes NMR so useful for structure determination.

通过施加恰好合适频率的射频脉冲，低能态的原子核可以“翻转”到高能态——这就是共振。所吸收的精确频率取决于每个原子核所处的局部化学环境，这正是 NMR 在结构测定中如此有用的原因。

2. The NMR Spectrometer and Sample Preparation | 核磁共振波谱仪与样品制备

An NMR spectrometer consists of a very strong, stable magnet (often superconducting), a radio frequency transmitter and receiver, and a computer to process the signals. The sample is dissolved in a suitable solvent, typically one that contains no hydrogen atoms itself, such as deuterated chloroform (CDCl₃), so that the solvent does not interfere with the proton NMR spectrum.

核磁共振波谱仪由一块非常强且稳定的磁体（通常是超导磁体）、射频发射器和接收器，以及用于处理信号的计算机组成。样品溶解在合适的溶剂中，通常是不含氢原子的溶剂，例如氘代氯仿（CDCl₃），这样溶剂就不会干扰质子 NMR 谱图。

A small amount of tetramethylsilane (TMS, (CH₃)₄Si) is added to the sample as an internal reference standard. All chemical shifts are measured relative to the signal from TMS, which is assigned a value of exactly 0 parts per million (ppm).

样品中加入少量四甲基硅烷（TMS，(CH₃)₄Si）作为内标参考物。所有化学位移都是相对于 TMS 的信号来测量的，TMS 的化学位移被精确地定为 0 ppm（百万分之一）。

3. Chemical Shift (δ) – The Key to Identifying Proton Environments | 化学位移 (δ) – 识别质子环境的关键

Not all hydrogen nuclei absorb radio waves at the same frequency because electrons surrounding each nucleus shield it from the external magnetic field to different extents. The chemical shift, given the symbol δ (delta), is the measure of this difference, expressed in parts per million (ppm). The more deshielded a proton is (e.g. near electronegative atoms), the higher its δ value and the further ‘downfield’ it appears on the spectrum.

并非所有氢原子核都以相同的频率吸收射频波，因为每个原子核周围的电子对其屏蔽外部磁场的程度不同。化学位移，用符号 δ（delta）表示，就是这种差异的量度，以百万分之一（ppm）为单位。质子越去屏蔽（例如靠近电负性原子），其 δ 值越大，在谱图中越出现在“低场”位置。

In IGCSE questions, you may be provided with a table of typical chemical shift ranges. A simplified version is shown below.

在 IGCSE 考题中，你可能会看到一个典型化学位移范围的数据表。以下是一个简化版本。

Proton Environment	δ (ppm)
R—CH₃ (alkyl)	0.9 – 1.7
R—CH₂—R’	1.2 – 1.4
CH₃—C=O (adjacent to carbonyl)	2.0 – 2.5
R—O—CH₃ (ether / alcohol next to O)	3.3 – 4.0
R—OH (alcohol —OH)	1.0 – 5.5 (variable, often broad)
R—COOH (carboxylic acid —OH)	10.0 – 13.0
C₆H₅—H (aromatic)	6.5 – 8.5
R—CHO (aldehyde H—C=O)	9.0 – 10.0

Remember that δ values are not required to be memorised for IGCSE, but you must be able to use a data sheet to identify proton environments.

请记住，IGCSE 不要求记忆 δ 值，但你必须能够使用数据表来识别质子环境。

4. Equivalent Protons – Why One Peak May Represent Several H Atoms | 等性质子 – 为什么一个峰可能代表多个氢原子

Protons that are in identical chemical environments are called equivalent protons. They experience exactly the same electron shielding and therefore give a single NMR signal. For example, in ethane (CH₃—CH₃), all six protons are equivalent because rotation about the C—C bond makes the two methyl groups indistinguishable; thus only one peak is observed.

处于完全相同化学环境中的质子称为等性质子。它们经历完全相同的电子屏蔽，因此只产生一个 NMR 信号。例如，在乙烷 (CH₃—CH₃) 中，全部六个质子是等价的，因为围绕 C—C 键的旋转使两个甲基难以区分；因此只观察到一个峰。

In contrast, in ethanol (CH₃—CH₂—OH), there are three distinct sets of equivalent protons: the CH₃ protons, the CH₂ protons, and the OH proton. A low-resolution NMR spectrum of ethanol would therefore show three peaks, with relative intensities (areas) in the ratio 3:2:1.

相反，在乙醇 (CH₃—CH₂—OH) 中，有三组不同的等性质子：CH₃ 质子、CH₂ 质子和 OH 质子。因此，乙醇的低分辨率 NMR 谱图将显示三个峰，相对强度（面积）比为 3:2:1。

5. Integration – What Peak Areas Tell You | 积分 – 峰面积告诉你什么

The area under each NMR peak is directly proportional to the number of protons that give rise to that signal. This is called integration. Modern spectrometers automatically integrate the peaks and display the relative areas either as printed numbers or as an integral trace (a stepped curve whose heights correspond to the number of protons).

每个 NMR 峰下的面积与产生该信号的质子数成正比。这称为积分。现代波谱仪会自动积分峰，并将相对面积显示为打印数字或积分轨迹（一种阶梯曲线，其高度对应质子数）。

For example, if a spectrum shows three peaks with integration values 3:2:1, this means the molecule contains three types of proton environment in that numerical ratio. In combination with chemical shift data, integration helps you assemble the molecular fragments.

例如，如果一个谱图显示三个峰，积分值为 3:2:1，这意味着分子包含三种质子环境，其数量比即为此比例。结合化学位移数据，积分帮助你组合分子片段。

6. Low-Resolution vs High-Resolution NMR | 低分辨率与高分辨率 NMR

IGCSE WJEC often distinguishes between low-resolution and high-resolution proton NMR. In a low-resolution spectrum, each set of chemically equivalent protons appears as a single peak – the fine structure caused by coupling is not resolved. The spectrum simply tells you about the number of proton environments and their relative amounts.

IGCSE WJEC 经常区分低分辨率和高分辨率质子 NMR。在低分辨率谱图中，每组化学等价的质子表现为一个单峰——由偶合引起的精细结构未被分辨出来。谱图只告诉你质子环境的数量及其相对数量。

In a high-resolution spectrum, peaks are split into multiplets due to spin-spin coupling with protons on adjacent carbon atoms. This splitting provides additional information about the number of neighbouring protons, following the n+1 rule.

在高分辨率谱图中，由于与相邻碳原子上的质子发生自旋-自旋偶合，峰分裂为多重峰。这种裂分提供了关于相邻质子数量的额外信息，遵循 n+1 规则。

7. Spin-Spin Coupling and the n+1 Rule | 自旋-自旋偶合与 n+1 规则

Spin-spin coupling arises because the magnetic field experienced by a proton is influenced by the spin states of non-equivalent protons on adjacent (usually neighbouring) carbon atoms. The signal for a proton with n equivalent neighbouring protons is split into n+1 lines. This is the n+1 rule. Coupling does not occur between equivalent protons (they give a singlet) or between protons separated by more than three bonds (generally).

自旋-自旋偶合的产生是因为质子所经历的磁场受到相邻（通常是邻接）碳原子上非等性质自旋态的影响。具有 n 个等价邻位质子的质子信号会分裂成 n+1 条谱线。这就是 n+1 规则。等性质子之间不产生偶合（它们给出单峰），相隔超过三个键的质子之间通常也不偶合。

Common splitting patterns include: singlet (s), doublet (d), triplet (t), quartet (q), and multiplet. For example, in a CH₃—CH₂— group, the CH₂ protons (adjacent to CH₃ with 3 protons) appear as a quartet (3+1), while the CH₃ protons (adjacent to CH₂ with 2 protons) appear as a triplet (2+1).

常见的裂分模式包括：单峰 (s)、双峰 (d)、三重峰 (t)、四重峰 (q) 和多重峰。例如，在 CH₃—CH₂— 基团中，CH₂ 质子（与含 3 个质子的 CH₃ 相邻）表现为四重峰 (3+1)，而 CH₃ 质子（与含 2 个质子的 CH₂ 相邻）表现为三重峰 (2+1)。

8. Interpreting Simple ¹H NMR Spectra Step by Step | 逐步解析简单 ¹H NMR 谱图

When faced with an NMR spectrum in an IGCSE WJEC exam, follow a systematic approach.

在 IGCSE WJEC 考试中面对 NMR 谱图时，请遵循系统的方法。

1. Determine the number of distinct signals (peaks). This tells you the number of different proton environments in the molecule.

1. 确定不同信号（峰）的个数。这告诉你分子中不同质子环境的数量。

2. Use the integration values (or the heights of integral steps) to find the ratio of protons in each environment.

2. 利用积分值（或积分台阶的高度）找出每种环境中质子的比例。

3. For each signal, note its chemical shift δ and use a data table to deduce the type of proton environment (e.g., alkyl, aromatic, aldehyde, O—H).

3. 对于每个信号，记下其化学位移 δ，并使用数据表推断质子环境的类型（例如，烷基、芳香族、醛基、O—H）。

4. If high-resolution splitting is shown, examine the multiplicity (singlet, doublet, etc.) to work out the number of protons on the adjacent carbon atom, applying the n+1 rule.

4. 如果显示了高分辨裂分，检查多重性（单峰、双峰等），应用 n+1 规则算出相邻碳原子上的质子数。

5. Piece together the fragments to suggest a molecular structure that is consistent with all the data, including the molecular formula if given.

5. 将所有片段组合起来，提出一个与所有数据（如果给定了分子式也包括在内）一致的结构。

9. The Role of TMS and Deuterated Solvents | TMS 和氘代溶剂的作用

Tetramethylsilane, TMS ((CH₃)₄Si), is chosen as the reference compound because all 12 of its protons are chemically equivalent, producing a single, sharp, strong signal. Moreover, its electrons shield the protons exceptionally well, so it gives a signal at a very low δ value (0 ppm), well away from most organic proton signals. TMS is also chemically inert, volatile (and hence easily removed after analysis), and soluble in most organic solvents.

四甲基硅烷（TMS，(CH₃)₄Si）被选作参考化合物，因为它的 12 个质子全部化学等价，产生单一、尖锐、强的信号。此外，它的电子对质子屏蔽极好，因此它给出的信号 δ 值非常低（0 ppm），远离大多数有机质子的信号。TMS 还具有化学惰性、易挥发（因此分析后容易去除）以及可溶于大多数有机溶剂的优点。

Deuterated solvents such as CDCl₃ are used because deuterium (²H) nuclei have a magnetic moment that is very different from ¹H, so they do not produce signals in the ¹H NMR range. This ensures only the sample’s protons are observed.

使用氘代溶剂如 CDCl₃ 是因为氘核 (²H) 的磁矩与 ¹H 差异很大，因此不会在 ¹H NMR 范围内产生信号。这确保只观察到样品本身的质子。

10. Common Exam Pitfalls and Application Tips | 常见考试陷阱与应用技巧

One frequent mistake is confusing the integration ratio with the actual number of protons. Always remember that the integration gives only the simplest whole-number ratio. If a ratio is 1:2:3, the actual numbers could be 1, 2, 3 or 2, 4, 6, etc., depending on the molecular formula. Cross-check with the total number of hydrogens available.

一个常见错误是混淆积分比与实际质子数。始终记住，积分仅给出最简整数比。如果比值为 1:2:3，实际的数目可能是 1、2、3 或 2、4、6 等，取决于分子式。需要用可用的氢总数进行交叉核对。

Another pitfall is overlooking the possibility of broad signals from —OH or —NH protons. These can appear as slightly broadened singlets due to hydrogen bonding and variable exchange rates; their position can vary and they may even disappear if the solvent contains D₂O (which causes H–D exchange).

另一个陷阱是忽视来自 —OH 或 —NH 质子的宽峰可能性。由于氢键和可变的交换速率，这些峰可能表现为略微展宽的单峰；它们的位置可能变化，如果溶剂中含有 D₂O（会引起 H–D 交换），这些峰甚至可能消失。

When tackling a spectrum, label each peak with its δ value, multiplicity, and integration. Then construct a table. This will help you avoid missing connections and give a clear answer.

处理谱图时，给每个峰标注其 δ 值、多重性和积分。然后构建一个表格。这将帮助你避免遗漏关联并使答案更清晰。

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Nuclear Magnetic Resonance (NMR) in IGCSE WJEC Chemistry | IGCSE WJEC 化学：核磁共振 考点精讲