NMR Spectroscopy Exam Essentials | IB CIE 化学：核磁共振考点精讲

📚 NMR Spectroscopy Exam Essentials | IB CIE 化学：核磁共振考点精讲

Nuclear Magnetic Resonance (NMR) spectroscopy is one of the most powerful analytical tools in modern chemistry, providing detailed information about the carbon–hydrogen framework of organic molecules. For IB and CIE chemistry students, mastering NMR means understanding both the underlying physical principles and the practical interpretation of spectra. This article distils the essential concepts – from nuclear spin and chemical shift to spin–spin splitting and integration – and shows you how to tackle typical exam questions with confidence.

核磁共振波谱是现代化学中最为强大的分析工具之一，能够提供有机分子碳氢骨架的详细信息。对于 IB 和 CIE 化学学生而言，掌握 NMR 意味着既要理解其背后的物理原理，也要学会实际解析谱图。本文凝练了核自旋、化学位移、自旋–自旋分裂以及积分等核心概念，并指导你如何自信地应对典型考题。

1. Fundamentals of NMR: Nuclear Spin and Resonance | NMR 的基本原理：核自旋与共振

NMR spectroscopy relies on the behaviour of certain atomic nuclei in an external magnetic field. Nuclei with an odd mass number (such as ¹H and ¹³C) possess a property called nuclear spin, which generates a tiny magnetic moment. When placed in a strong magnetic field (B₀), these nuclei can align either with or against the field, creating two distinct energy states. The energy difference between these states lies in the radiofrequency region of the electromagnetic spectrum. Irradiation with a short radiofrequency pulse matching this energy gap causes nuclei to flip from the lower to the higher energy state – a phenomenon called resonance. The subsequent relaxation generates a signal that is recorded as a free induction decay and converted into an NMR spectrum.

核磁共振波谱依赖于某些原子核在外磁场中的行为。质量数为奇数的核（如 ¹H 和 ¹³C）具有一种称为核自旋的性质，从而产生微小的磁矩。当这类核被置于强磁场（B₀）中时，它们可以顺着或逆着磁场方向排列，形成两个不同的能级。这两个能级之间的能量差落在电磁波谱的射频区。用与能隙匹配的短射频脉冲照射样品，会使原子核从低能级跃迁到高能级——这种现象称为共振。随后的弛豫过程产生的信号被记录为自由感应衰减信号，并转化为 NMR 谱图。

For the exam, you must recall that NMR-active nuclei require an odd number of protons or neutrons, or both. ¹²C, with six protons and six neutrons, is not NMR active, which is why ¹³C (6 protons, 7 neutrons) is used instead. In proton NMR (¹H NMR), the hydrogen nuclei in different chemical environments absorb at slightly different frequencies, giving us structural information.

考试中需要牢记，具有 NMR 活性的核要求质子数或中子数为奇数，或两者皆为奇数。¹²C 有 6 个质子和 6 个中子，没有 NMR 活性，因此改用 ¹³C（6 个质子，7 个中子）。在质子核磁共振（¹H NMR）中，处于不同化学环境的氢核吸收频率略有差异，从而为我们提供结构信息。

2. Chemical Shift (δ) | 化学位移 (δ)

The exact resonance frequency of a nucleus depends on its local electronic environment. Electrons surrounding a nucleus shield it from the full effect of the applied magnetic field. Nuclei in electron-rich environments experience greater shielding and resonate at a lower frequency (upfield); those attached to electronegative atoms or adjacent to π systems are deshielded and resonate at a higher frequency (downfield). The chemical shift, symbolised by δ and measured in parts per million (ppm), is defined as the difference in resonance frequency relative to a reference compound, tetramethylsilane (TMS), divided by the operating frequency of the spectrometer.

原子核的精确共振频率取决于其所处的局部电子环境。核外的电子会对外加磁场产生屏蔽作用。处于富电子环境中的核受到较大的屏蔽，共振频率较低（高场）；而与电负性原子相连或邻近 π 体系的核则被去屏蔽，共振频率较高（低场）。化学位移用 δ 表示，单位为百万分之一（ppm），定义为样品共振频率与参比化合物四甲基硅烷（TMS）的频率差除以谱仪的工作频率。

You must be able to use a chemical shift data table provided in the exam to identify the types of protons or carbon atoms in a molecule. Typical ¹H NMR chemical shift ranges: alkyl protons (R–CH₃) 0.7–1.2 ppm; protons adjacent to a carbonyl (CH₃–C=O) 2.0–2.5 ppm; protons next to an electronegative atom such as oxygen (R–O–CH₃) 3.3–4.0 ppm; aromatic protons 6.5–8.0 ppm; aldehyde protons (R–CHO) 9.5–10.0 ppm; and carboxylic acid protons (R–COOH) 10.0–12.0 ppm, often broad and variable. ¹³C NMR chemical shifts span roughly 0–220 ppm, with carbonyl carbons appearing above 160 ppm, aromatic carbons 110–160 ppm, and alkane carbons 0–50 ppm.

你必须能够利用考试提供的数据表来鉴别分子中质子或碳原子的类型。典型的 ¹H NMR 化学位移范围：烷基质子（R–CH₃）0.7–1.2 ppm；与羰基相邻的质子（CH₃–C=O）2.0–2.5 ppm；与电负性原子（如氧）相邻的质子（R–O–CH₃）3.3–4.0 ppm；芳香族质子 6.5–8.0 ppm；醛基质子（R–CHO）9.5–10.0 ppm；羧酸质子（R–COOH）10.0–12.0 ppm，通常为宽峰且可变。¹³C NMR 化学位移范围约为 0–220 ppm，羰基碳出现在 160 ppm 以上，芳香碳 110–160 ppm，烷基碳 0–50 ppm。

δ = (ν_sample − ν_TMS) ÷ ν_spectrometer × 10⁶ ppm

3. Proton NMR: Number of Signals and Equivalent Protons | 质子 NMR：信号数目与等价质子

The number of distinct signals in a ¹H NMR spectrum tells us how many chemically non-equivalent proton environments exist in the molecule. Chemically equivalent protons are those that are in identical chemical environments, usually related by symmetry or free rotation. For example, all three protons of a methyl group are generally equivalent, as are the two methyl groups in propane bonded to the central carbon. In a symmetrical molecule like 1,4-dimethylbenzene (p-xylene), the four aromatic protons are equivalent, giving only one aromatic signal.

¹H NMR 谱中不同信号峰的数目表明分子中存在多少种化学上不等价的质子环境。化学等价质子是指处于完全相同化学环境中的质子，通常与对称性或自由旋转有关。例如，甲基的三个质子通常是等价的；丙烷中连接在中心碳上的两个甲基也是等价的。在对称分子如 1,4-二甲基苯（对二甲苯）中，四个芳香质子等价，仅产生一个芳香信号。

A common exam question asks you to predict the number of ¹H NMR signals for a given structure. Consider propan-2-ol: the two methyl groups (CH₃–) are equivalent by symmetry, giving one signal; the CH proton gives a second signal; the OH proton gives a third signal. Thus three signals are observed. Beware of hydrogen bonding: OH and NH protons may sometimes be broad and exchangeable, but they still count as a distinct signal.

常见的考题是要求预测给定结构中的 ¹H NMR 信号数目。以丙-2-醇为例：两个甲基（CH₃–）因对称性而等价，给出一个信号；CH 质子给出第二个信号；OH 质子给出第三个信号。因此可观察到三个信号。需注意氢键的影响：OH 和 NH 质子有时呈宽峰并可交换，但仍算作一个独立的信号。

4. Integration and Relative Peak Areas | 积分与相对峰面积

The area under each ¹H NMR signal is directly proportional to the number of protons responsible for that signal. The spectrometer automatically integrates the peaks, and the integration trace is displayed either as a step curve or as numerical values beneath the peaks. The ratio of peak areas reveals the ratio of chemically distinct protons. For example, in ethyl ethanoate (CH₃COOCH₂CH₃), the three-proton singlet of the CH₃–C=O group, the two-proton quartet of the –OCH₂– group, and the three-proton triplet of the –CH₃ group appear in an integration ratio of 3 : 2 : 3.

¹H NMR 谱中每个信号峰的面积与该信号所对应的质子数目成正比。谱仪自动对峰进行积分，积分线可以显示为阶梯曲线或峰下方的数值。峰面积之比反映了化学性质不同的质子数目之比。例如，在乙酸乙酯（CH₃COOCH₂CH₃）中，CH₃–C=O 的三质子单峰、–OCH₂– 的二质子四重峰以及 –CH₃ 的三质子三重峰的积分比为 3 : 2 : 3。

When solving exam problems, always normalise the integration values to the simplest whole-number ratio. If you obtain ratios like 1.5 : 1, multiply by 2 to get 3 : 2, which likely corresponds to one CH₃ and one CH₂ group. Integration is a powerful tool for deducing the molecular formula fragment by fragment.

在解题时，务必将积分值归一化为最简整数比。如果得到类似于 1.5 : 1 的比例，乘以 2 即可变为 3 : 2，这可能对应一个 CH₃ 和一个 CH₂ 基团。积分是逐步推导分子式的有力工具。

5. Spin–Spin Splitting and the n+1 Rule | 自旋–自旋分裂与 n+1 规则

Spin–spin coupling arises from the interaction between non-equivalent protons on adjacent carbon atoms. A proton (or a set of equivalent protons) senses the small magnetic fields of neighbouring protons, which can be aligned with or against the external field. This splits the signal of the observed proton into multiple lines. The multiplicity (splitting pattern) follows the n+1 rule: a proton with n equivalent neighbouring protons on adjacent carbon(s) is split into n+1 peaks. Equivalent protons do not couple with each other.

自旋–自旋耦合源于相邻碳上非等价质子之间的相互作用。一个质子（或一组等价质子）会感受到邻近质子产生的微小磁场，这些磁场可以与外磁场同向或反向，从而使被观察质子的信号分裂为多重谱线。裂分峰形遵守 n+1 规则：若某个质子有 n 个处于邻位碳上的等价质子，则其信号将分裂为 n+1 个峰。等价质子彼此之间不发生耦合。

A classic example is the ethyl group (–CH₂CH₃). The CH₃ protons have two neighbouring protons on the adjacent CH₂ group, so they appear as a triplet (n=2, n+1=3). The CH₂ protons are coupled to three neighbouring CH₃ protons, so they appear as a quartet (n=3, n+1=4). The intensities of the lines in a multiplet follow Pascal’s triangle: doublet 1:1, triplet 1:2:1, quartet 1:3:3:1, etc.

典型的例子是乙基 (–CH₂CH₃) 。CH₃ 的质子有邻位 CH₂ 基团上的两个质子，因此裂分为三重峰（n=2，n+1=3）。CH₂ 的质子则与三个邻位 CH₃质子耦合，因此裂分为四重峰（n=3，n+1=4）。多重峰中各谱线的强度遵循帕斯卡三角形：双峰为 1:1，三重峰为 1:2:1，四重峰为 1:3:3:1，依此类推。

Protons bonded to oxygen or nitrogen (OH, NH) are often broad and may not show coupling to adjacent protons, especially in protic solvents. In exam questions, you are usually expected to consider OH as a singlet unless splitting is explicitly indicated.

与氧或氮相连的质子（OH、NH）通常表现为宽峰，并且可能不会与邻位质子发生耦合，特别是在质子性溶剂中。在考试题中，除非题目明确给出了裂分信息，否则通常将 OH 视为单峰。

6. Carbon-13 NMR Spectroscopy | ¹³C NMR 波谱

¹³C NMR is complementary to ¹H NMR because it directly probes the carbon skeleton. Due to the low natural abundance of ¹³C (about 1.1 %), ¹³C spectra are recorded using signal averaging, and coupling between adjacent ¹³C nuclei is rarely observed. More importantly, the standard ¹³C spectrum is proton-decoupled: all ¹³C–¹H couplings are removed, so each chemically distinct carbon gives a single sharp peak. The number of peaks therefore equals the number of non-equivalent carbon environments.

¹³C NMR 与 ¹H NMR 互为补充，因为它直接探测碳骨架。由于 ¹³C 的自然丰度很低（约 1.1 %），¹³C 谱需要通过信号累加来记录，而且相邻 ¹³C 核之间的耦合极少被观察到。更重要的是，标准的 ¹³C 谱为质子去耦谱：所有 ¹³C–¹H 耦合被消除，因此每一种化学性质不同的碳都只出一个尖锐的单峰。因此，峰的数量就等于分子中非等价碳环境的数目。

You should be able to predict the number of ¹³C NMR signals for a molecule by identifying symmetry-equivalent carbons. For example, benzene has six identical CH carbons, giving one signal. Toluene (C₆H₅CH₃) has five different aromatic carbon environments and one methyl carbon, giving six signals in total. For butan-2-one (CH₃COCH₂CH₃), the four carbon atoms are all in different environments, so four ¹³C signals are observed.

你应能通过识别分子的对称等价碳来预测 ¹³C NMR 的信号数目。例如，苯有六个相同的 CH 碳，因此只有一个信号。甲苯（C₆H₅CH₃）有五种不同的芳香碳环境和一个甲基碳，一共六个信号。对于丁-2-酮（CH₃COCH₂CH₃），四个碳原子均处于不同的化学环境，因此可观察到四个 ¹³C 信号。

A typical ¹³C chemical shift data table is used exactly like the proton version. The key ranges to remember are: C=O (aldehydes and ketones) 190–220 ppm; C=O (esters, acids, amides) 160–185 ppm; aromatic carbons 110–160 ppm; C–O (alcohols, ethers) 50–90 ppm; alkane carbons 0–50 ppm.

¹³C 化学位移数据表的使用方法与质子谱完全相同。需要记住的关键范围是：C=O（醛、酮）190–220 ppm；C=O（酯、酸、酰胺）160–185 ppm；芳香碳 110–160 ppm；C–O（醇、醚）50–90 ppm；烷基碳 0–50 ppm。

7. Interpreting NMR Spectra: A Systematic Approach | NMR 谱图解析的系统方法

When faced with an unknown NMR spectrum in the exam, adopt a logical step-by-step strategy. First, examine the molecular formula if provided; calculate the degree of unsaturation to anticipate possible rings or π bonds. Second, in ¹H NMR, count the number of signals to determine the number of proton environments. Third, look at the integration trace to obtain the relative numbers of protons in each environment. Fourth, analyse the splitting pattern of each signal using the n+1 rule to deduce the number of adjacent protons. Fifth, use the chemical shift values to assign each signal to a particular type of proton (alkyl, alkoxy, carbonyl-adjacent, aromatic, etc.). Finally, piece together the structural fragments, bearing in mind that a proton with n neighbours must be connected to a carbon bearing those n protons.

考试中遇到未知物的 NMR 谱图时，应采用一套合乎逻辑的步骤。首先，如果给出了分子式，要计算其不饱和度，以预估可能存在的环或 π 键。其次，在 ¹H NMR 中，数出峰的数目，以确定质子环境的种类。第三，观察积分线，得出各环境中质子的相对数目。第四，利用 n+1 规则分析每个信号的裂分情况，推断邻位质子的数目。第五，利用化学位移数值将各信号归入特定的质子类型（烷基、烷氧基、羰基邻位、芳环等）。最后，将各结构片段拼合起来，要记住：一个拥有 n 个邻位质子的质子，必然连接在一个带有这 n 个质子的碳上。

For example, a compound with molecular formula C₄H₈O₂ gives three ¹H NMR signals: a triplet at δ 1.2 (3H), a quartet at δ 4.1 (2H), and a singlet at δ 2.0 (3H). The triplet and quartet pattern with integration 3:2 suggests an ethyl group –CH₂CH₃ attached to an electronegative atom (the CH₂ is at 4.1 ppm, indicating an O–CH₂– fragment). The singlet at 2.0 ppm (3H) is a methyl adjacent to a carbonyl (CH₃–C=O). Joining the two fragments gives the ester ethyl ethanoate, CH₃COOCH₂CH₃.

例如，一个分子式为 C₄H₈O₂ 的化合物在 ¹H NMR 谱中给出三个信号：δ 1.2 的三重峰（3H）、δ 4.1 的四重峰（2H）和 δ 2.0 的单峰（3H）。积分比为 3:2 的三重峰和四重峰组合提示存在一个乙基 –CH₂CH₃，且连接在电负性原子上（CH₂ 出现在 4.1 ppm 表明存在 O–CH₂– 片段）。δ 2.0 处的单峰（3H）则是与羰基相邻的甲基（CH₃–C=O）。将这两个片段拼接起来就得到酯乙酸乙酯，CH₃COOCH₂CH₃。

In ¹³C NMR, the approach is similar but simplified by the absence of splitting. Count the number of distinct carbon environments, use chemical shifts to assign them, and then verify consistency with the ¹H NMR and the molecular formula.

在 ¹³C NMR 中，解析方法类似，但因无裂分而更为简单。数出不同碳环境的数目，利用化学位移指认它们，然后验证其是否与 ¹H NMR 及分子式相一致。

8. Advanced Considerations: Exchangeable Protons and Solvent Peaks | 进阶考量：可交换质子与溶剂峰

Labile protons such as OH, NH, and SH can undergo rapid exchange with deuterium in D₂O. When a few drops of D₂O are added to an NMR sample and the spectrum is re-run, the signal from exchangeable protons disappears (or is greatly reduced). This D₂O shake test is often used to identify OH or NH signals. The exam might present two spectra – one before and one after D₂O exchange – and ask you to identify the exchangeable proton.

活泼质子，如 OH、NH 和 SH，可与 D₂O 中的氘发生快速交换。向 NMR 样品中加入几滴 D₂O 后重测谱图，可交换质子的信号便会消失（或显著减弱）。这种 D₂O 交换实验常用来鉴定 OH 或 NH 信号。考试中可能会给出两张谱图——一张是交换前，一张是交换后——要求你找出可交换的质子。

Solvent peaks also appear in NMR spectra. Deuterated solvents such as CDCl₃ are used to avoid an overwhelming signal from the solvent itself, but residual non-deuterated solvent molecules give characteristic peaks. In CDCl₃, a small singlet is observed at about δ 7.26 in ¹H NMR. You need to be able to recognise and ignore these solvent signals during interpretation. Common ¹H NMR solvent peaks: CDCl₃ 7.26 ppm; D₂O 4.79 ppm; CD₃OD 3.31 ppm; DMSO-d₆ 2.50 ppm.

NMR 谱图中还会出现溶剂峰。使用氘代溶剂（如 CDCl₃）是为了避免溶剂本身产生强烈的信号，但残留的非氘代溶剂分子仍会给出特征峰。在 CDCl₃ 中，¹H NMR 谱的约 δ 7.26 处会出现一个小的单峰。你在解析谱图时需要能识别并无视这些溶剂峰。常见的 ¹H NMR 溶剂峰：CDCl₃ 7.26 ppm；D₂O 4.79 ppm；CD₃OD 3.31 ppm；DMSO-d₆ 2.50 ppm。

In ¹³C NMR, CDCl₃ shows a characteristic triplet centred at 77.0 ppm (from the coupling between ¹³C and deuterium, I = 1). This signal should not be mistaken for a carbon of the analyte.

在 ¹³C NMR 中，CDCl₃ 显示出一个位于 77.0 ppm 的特征三重峰（由 ¹³C 与氘的耦合所致，氘的自旋 I = 1）。这个信号不可误认为是分析物的碳信号。

9. High-Resolution vs. Low-Resolution NMR | 高分辨与低分辨 NMR

The IB and CIE syllabuses often distinguish between low-resolution and high-resolution ¹H NMR. Low-resolution (or low-res) NMR shows the number of proton environments, their chemical shifts, and integration, but spin–spin coupling is not resolved; each signal appears as a single peak. High-resolution (or high-res) NMR shows full splitting patterns, providing connectivity information as described by the n+1 rule. Exam questions may present a low-resolution spectrum together with integration data and expect you to deduce the structure, or they may give a high-resolution spectrum and ask you to interpret the splitting patterns.

IB 和 CIE 考纲通常会区分低分辨和高分辨 ¹H NMR。低分辨（低磁场）NMR 显示质子环境的数目、化学位移以及积分，但自旋–自旋耦合未被分辨出来；每个信号为单一峰。高分辨（高磁场）NMR 则显示出完整的裂分模式，提供由 n+1 规则描述的连接信息。考题可能给出一张低分辨谱图及其积分数据，要求你推导结构；或者给出一张高分辨谱图，让你解释裂分模式。

Understanding this distinction helps you anticipate the level of detail you will need to extract from the spectrum. For a low-resolution spectrum, you only know how many types of protons exist and their relative numbers, but you lose the adjacency clues. This makes logical deduction from the molecular formula and chemical shift data even more critical.

理解这一区分有助于你预判需从谱图中提取何种级别的细节信息。对于低分辨谱图，你仅仅知道存在几种类型的质子以及它们的相对数目，却失去了邻接信息。这使得基于分子式和化学位移数据的逻辑推导变得更加关键。

10. Common Exam Pitfalls and How to Avoid Them | 常见考试误区与避免方法

Many students lose marks by misidentifying the number of equivalent proton environments. Always look for symmetry elements – a plane of symmetry or a centre of symmetry – within the molecule. Rotational equivalence also matters: methyl groups attached to a single carbon are often equivalent due to rapid rotation, unless restricted (for example, by a bulky substituent at low temperature). Another common error is to assume that an OH proton always couples to adjacent CH protons. In practice, OH signals are often broad singlets without splitting, unless in a very dry, aprotic solvent. Stick to the simplifying assumption often used in exam mark schemes: treat OH as a singlet unless splitting is explicitly shown.

许多学生因错判等价质子环境的数目而失分。要善于寻找分子的对称元素——对称面或对称中心。旋转等效也很重要：连接在同一个碳上的甲基通常因快速旋转而等价，除非受到限制（例如在低温下有大位阻取代基）。另一个常见错误是假设 OH 质子总是与邻位 CH 质子发生耦合。实际上，OH 信号往往是宽的单峰，没有裂分，除非在非常干燥的非质子溶剂中。应遵循考试评分标准常用的简化假设：只要没有明确显示裂分，就将 OH 视为单峰。

Misreading integration ratios is another trap. Always normalise the given numbers to the smallest integer set. If a spectrum shows integrals of 12, 8, and 4 mm, the ratio is 3:2:1. Failing to do this leads to an incorrect molecular formula and impossible fragments. Also, do not forget the solvent peak: a singlet at 7.26 ppm in CDCl₃ is the solvent, not part of your compound.

误读积分比也是一大陷阱。务必将给出的数值归一化为最小整数集。如果谱图显示的积分分别为 12、8 和 4 mm，其比值为 3:2:1。未做归一化将导致错误的分子式和不可能的结构片段。另外，别忘了溶剂峰：CDCl₃ 谱中位于 7.26 ppm 的单峰是溶剂，不属于你的化合物。

Finally, practise constructing the unknown from the fragments. Write down the inferred groups (e.g., CH₃–CH₂–, –O–CH₂–CH₃, CH₃–C=O) and then try to link them logically. Check that every carbon obeys the octet rule and that the combined fragments fit the molecular formula exactly.

最后，要多练习如何从片段构建未知物。写下推测出的基团（如 CH₃–CH₂–、–O–CH₂–CH₃、CH₃–C=O），然后尝试合理地连接它们。检查每个碳都满足八隅律，并且组合后的片段必须完全符合给出的分子式。

11. Connecting NMR to Infrared Spectroscopy | NMR 与红外光谱的关联

IB and CIE exams frequently pair NMR data with infrared (IR) spectroscopy. IR provides functional group information: a strong C=O stretch at around 1700–1750 cm⁻¹ confirms a carbonyl; a broad O–H stretch at 2500–3300 cm⁻¹ suggests a carboxylic acid; an O–H in alcohols appears at 3200–3600 cm⁻¹; C–O stretches occur at 1000–1300 cm⁻¹. When solving combined spectroscopy problems, first use IR to identify the major functional groups, then use NMR to assemble the complete structure.

IB 和 CIE 考试经常将 NMR 数据与红外光谱 (IR) 结合起来考查。IR 提供官能团信息：约 1700–1750 cm⁻¹ 处强的 C=O 伸缩振动峰确认羰基的存在；2500–3300 cm⁻¹ 处的宽 O–H 伸缩峰提示羧酸；醇的 O–H 出现在 3200–3600 cm⁻¹；C–O 伸缩振动位于 1000–1300 cm⁻¹。在解答综合波谱题时，应首先利用 IR 确定主要的官能团，然后借助 NMR组装出完整的结构。

For example, an IR spectrum with a sharp peak at 1715 cm⁻¹ and a broad peak centred at 3000 cm⁻¹, together with a proton NMR showing a singlet at δ 11.5 (1H, exchangeable), a quartet at δ 2.4 (2H), and a triplet at δ 1.1 (3H), immediately suggests a carboxylic acid with an ethyl group – propanoic acid, CH₃CH₂COOH. The combined use of IR and NMR is a very common examination technique.

例如，一张 IR 谱图在 1715 cm⁻¹ 处有尖峰，在约 3000 cm⁻¹ 处有宽峰，同时质子 NMR 显示 δ 11.5 处的单峰（1H，可交换）、δ 2.4 的四重峰（2H）和 δ 1.1 的三重峰（3H），立即提示这是一个含有乙基的羧酸——丙酸，CH₃CH₂COOH。IR 与 NMR 的联合使用是非常常见的考查方式。

12. Summary and Final Revision Checklist | 总结与最终复习清单

To master NMR for your IB or CIE chemistry exam, ensure you can do the following: define nuclear magnetic resonance and explain the origin of chemical shift; predict the number of ¹H and ¹³C NMR signals for a given structure by identifying chemically equivalent nuclei; interpret integration traces and deduce relative numbers of protons; apply the n+1 rule to predict or analyse splitting patterns; use chemical shift data tables to assign signals to specific chemical environments; recognise common solvent peaks and the behaviour of exchangeable protons; integrate NMR information with IR and mass spectrometry data to elucidate unknown structures; and finally, practise with past paper questions under timed conditions.

想要在 IB 或 CIE 化学考试中掌握 NMR，请确保能完成以下任务：定义核磁共振并解释化学位移的起源；通过识别化学等价核，预测给定结构中的 ¹H 和 ¹³C NMR 信号数目；解读积分线并推导质子的相对数目；应用 n+1 规则预测或分析裂分模式；使用化学位移数据表将信号归入特定的化学环境；识别常见的溶剂峰以及可交换质子的行为；将 NMR 信息与红外、质谱数据相结合以阐明未知结构；最后，在限时条件下多做往年试题的练习。

Keep a concise summary table of key ¹H and ¹³C chemical shifts and coupling patterns close at hand during revision. Understanding the underlying concepts will make you far more confident than relying solely on rote memorisation. Remember: every NMR spectrum tells the story of the molecule’s structure – your job is to read that story carefully.

在复习时，手边要保留一份简洁的 ¹H 和 ¹³C 关键化学位移及耦合模式汇总表。理解背后的概念会比单纯依靠死记硬背让你自信得多。请记住：每一张 NMR 谱图都在讲述分子结构的故事——你的任务就是仔细读懂这个故事。

Published by TutorHao | Chemistry Revision Series | aleveler.com

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NMR Spectroscopy Exam Essentials | IB CIE 化学：核磁共振 考点精讲