Edexcel A-Level Chemistry: Mastering Combined Spectroscopic Techniques | 爱德思 A-Level 化学：综合光谱分析技术精讲

📚 Edexcel A-Level Chemistry: Mastering Combined Spectroscopic Techniques | 爱德思 A-Level 化学：综合光谱分析技术精讲

In Edexcel A-Level Chemistry, modern analytical techniques are not examined in isolation. You are expected to combine infrared spectroscopy, mass spectrometry, and nuclear magnetic resonance spectroscopy to deduce the structure of an unknown organic compound. This article builds a clear, exam-ready framework for using combined spectroscopic evidence, with worked examples and common pitfalls.

在爱德思 A-Level 化学中，现代分析技术并不是孤立考查的。你需要综合红外光谱、质谱和核磁共振波谱，推断未知有机化合物的结构。本文构建一套清晰、贴近考试的综合性波谱分析框架，并配有实例和常见误区。

1. Why Combined Techniques Matter | 为什么需要综合技术

No single technique gives you the complete structure. IR identifies functional groups, mass spectrometry provides the molecular mass and fragmentation clues, and NMR maps the carbon and hydrogen environments. Combining these pieces is what turns a set of spectra into a reliable structural identification.

没有任何一种技术能单独给出完整结构。红外光谱识别官能团，质谱提供相对分子质量和碎片信息，核磁共振则描绘碳和氢的环境。将这些信息组合起来，才能把一组谱图转化为可靠的结构鉴定。

The exam will often present a molecular formula, an IR spectrum, a mass spectrum, and one or two NMR spectra. Your job is to use every piece of data, not to rely on a single spectrum.

考试通常会给出分子式、红外光谱、质谱以及一张或两张核磁共振谱图。你的任务是使用每一份数据，而不是只依赖某一张谱图。

2. Infrared Spectroscopy: Functional Group Fingerprints | 红外光谱：官能团的指纹区

Infrared spectroscopy detects bond vibrations. Each type of bond absorbs energy at a characteristic wavenumber, so IR is the fastest way to confirm the presence or absence of major functional groups such as O-H, C=O, C=C, and C-H.

红外光谱检测化学键的振动。每种键在特征波数处吸收能量，因此红外是确认 O-H、C=O、C=C、C-H 等主要官能团存在与否的最快方法。

Bond or functional group	Typical wavenumber / cm⁻¹
O-H alcohol, broad	3200-3600
C-H alkane	2850-2950
C=O ester	1735-1750
C=O ketone/aldehyde	1680-1740
C=C alkene	1620-1680

In a worked problem, a strong peak at about 1740 cm⁻¹ often points to an ester or a similar carbonyl compound. A broad peak around 3300 cm⁻¹ usually indicates an alcohol or carboxylic acid O-H.

在解题中，约 1740 cm⁻¹ 的强吸收峰通常指向酯或类似的羰基化合物。约 3300 cm⁻¹ 的宽峰通常表明存在醇或羧酸的 O-H。

3. Mass Spectrometry: Molecular Mass and Fragments | 质谱：相对分子质量与碎片

Mass spectrometry first gives you the molecular ion peak, usually labelled M⁺ or M. This peak corresponds to the relative molecular mass of the intact molecule. High-resolution mass spectrometry can even give the exact molecular formula from accurate mass data.

质谱首先给出分子离子峰，通常标记为 M⁺ 或 M。该峰对应完整分子的相对分子质量。高分辨质谱甚至可以根据精确质量数据给出准确的分子式。

Fragmentation peaks are also useful. For example, an ester such as ethyl ethanoate may produce fragments at m/z = 43, 45, 61, and 73. In A-Level questions, you are usually expected to recognise simple fragments rather than analyse complex rearrangements.

碎片峰也很有用。例如，乙酸乙酯等酯可能在 m/z = 43、45、61、73 处产生碎片。在 A-Level 题目中，通常要求你识别简单碎片，而不是分析复杂的重排。

m/z = relative mass of the ion ÷ charge on the ion

In practice, most ions have a charge of +1, so m/z is numerically equal to the mass of the fragment.

实际上，大多数离子带 +1 电荷，因此 m/z 在数值上等于碎片的质量。

4. ¹H NMR: Chemical Shift and Environment | ¹H 核磁共振：化学位移与化学环境

Proton NMR tells you how many different hydrogen environments exist in the molecule. Each environment produces a signal at a characteristic chemical shift, measured in parts per million. The number of signals in a ¹H NMR spectrum gives the number of non-equivalent proton environments.

质子核磁共振告诉你分子中存在多少种不同的氢环境。每种环境在特征化学位移处产生一个信号，化学位移以百万分率表示。¹H NMR 谱中的信号数等于不等价质子环境的数目。

Typical ranges include alkyl protons at δ 0.5-2.0, protons next to an electronegative atom or ester oxygen at δ 3.5-4.5, aromatic protons at δ 6.5-8.0, and aldehyde protons at δ 9.5-10.0.

典型范围包括：烷基质子在 δ 0.5-2.0，与电负性原子或酯氧相邻的质子在 δ 3.5-4.5，芳香质子在 δ 6.5-8.0，醛基质子在 δ 9.5-10.0。

You do not need to memorise exact shift values for every environment, but you should recognise that a signal near δ 9-10 strongly suggests an aldehyde proton.

你不需要记住每种环境的精确位移值，但应当认识到 δ 9-10 附近的信号强烈提示醛基质子。

5. Integration: Counting Protons | 积分：质子计数

The area under each ¹H NMR signal is proportional to the number of protons in that environment. The integration ratio tells you the relative numbers of protons, not necessarily the absolute numbers. You must scale these ratios using the molecular formula.

每个 ¹H NMR 信号下的面积与该环境中的质子数成正比。积分比只说明质子的相对数量，不一定是绝对数量。你必须结合分子式来放大这些比例。

For example, if the integration ratio is 2 : 3 : 3 and the molecular formula is C₄H₈O₂, then the protons add up to 8, so the actual numbers are exactly 2H, 3H, and 3H. If the total were 16, you would double each value.

例如，如果积分比为 2 : 3 : 3，而分子式为 C₄H₈O₂，那么质子总数为 8，因此实际数目就是 2H、3H 和 3H。如果总数是 16，你就需要把每个值扩大一倍。

6. Splitting Patterns: The n+1 Rule | 分裂模式：n+1 规则

A ¹H NMR signal is split by protons on adjacent carbon atoms. The number of peaks is given by the n+1 rule, where n is the number of equivalent neighbouring protons. Thus a proton with two adjacent equivalent protons appears as a triplet.

¹H NMR 信号会被相邻碳原子上的质子裂分。峰的数目由 n+1 规则给出，其中 n 是相邻等价质子的数目。因此，一个具有两个相邻等价质子的氢原子会出现三重峰。

Splitting patterns include singlet, doublet, triplet, quartet, and sometimes multiplet. In structure deduction, an ethyl group -CH₂CH₃ is a classic pattern: the CH₂ is a quartet, and the CH₃ is a triplet.

分裂模式包括单峰、二重峰、三重峰、四重峰，有时还有多重峰。在结构推断中，乙基 -CH₂CH₃ 是经典模式：CH₂ 为四重峰，CH₃ 为三重峰。

Note that protons on an oxygen or nitrogen atom may not split nearby signals, and their signals may be broad. Edexcel questions usually avoid ambiguous exchangeable protons unless they are clearly signalled.

注意，氧或氮原子上的质子可能不会裂分相邻信号，而且它们自身的信号可能较宽。爱德思题目通常会避免模棱两可的可交换质子，除非有明确提示。

7. ¹³C NMR: Carbon Environments | ¹³C 核磁共振：碳环境

Carbon-13 NMR shows each non-equivalent carbon environment as a separate signal. Because carbon environments are often easier to count than proton environments, ¹³C NMR is a powerful check on symmetry and functional groups.

碳-13 核磁共振将每种不等价碳环境显示为独立的信号。由于碳环境通常比氢环境更容易计数，¹³C NMR 是检查分子对称性和官能团的有力工具。

For example, ethanoic acid CH₃COOH has two carbon environments and therefore two ¹³C signals. Methylbenzene C₆H₅CH₃ has five carbon environments because of symmetry in the benzene ring.

例如，乙酸 CH₃COOH 有两种碳环境，因此有两个 ¹³C 信号。甲苯 C₆H₅CH₃ 由于苯环的对称性而有五种碳环境。

Carbonyl carbons usually appear in a distinct region from approximately δ 160 to 220, making them easy to spot.

羰基碳通常出现在大约 δ 160 至 220 的独立区域内，因此很容易识别。

8. A Five-Step Problem-Solving Framework | 五步解题框架

Use this repeatable sequence when you face a combined spectra question. First, calculate the degree of unsaturation from the molecular formula. Second, use IR to identify the main functional groups. Third, use the mass spectrum to confirm the molecular mass and look for fragment clues.

面对综合谱图题时，使用这个可重复的步骤。第一，根据分子式计算不饱和度。第二，用红外光谱识别主要官能团。第三，用质谱确认相对分子质量并寻找碎片线索。

Fourth, use ¹H NMR and ¹³C NMR to map hydrogen and carbon environments. Fifth, assemble the fragments into a complete structure and check that every spectrum is consistent with your answer.

第四，利用 ¹H NMR 和 ¹³C NMR 标出氢和碳环境。第五，将碎片组装成完整结构，并检查每个谱图是否与你的答案一致。

DBE = (2C + 2 + N − H − X) ÷ 2

Here C is the number of carbons, N is nitrogens, H is hydrogens, and X is halogens. Each degree of unsaturation represents a ring or a π bond.

其中 C 是碳原子数，N 是氮原子数，H 是氢原子数，X 是卤素原子数。每个不饱和度代表一个环或一个 π 键。

9. Worked Example: Deducing Ethyl Ethanoate from Spectra | 实例：从谱图推断乙酸乙酯

Consider a compound with molecular formula C₄H₈O₂. Its IR spectrum shows a strong absorption at 1740 cm⁻¹. The mass spectrum gives a molecular ion at m/z = 88. The ¹H NMR spectrum shows three signals: δ 1.25 triplet, δ 2.05 singlet, and δ 4.12 quartet.

考虑一个分子式为 C₄H₈O₂ 的化合物。其红外光谱在 1740 cm⁻¹ 处有强吸收。质谱给出 m/z = 88 的分子离子。¹H NMR 谱显示三个信号：δ 1.25 三重峰、δ 2.05 单峰和 δ 4.12 四重峰。

The degree of unsaturation is (2×4+2−8)÷2 = 1, so one π bond is likely a C=O. The IR peak at 1740 cm⁻¹ suggests an ester, not a simple ketone or aldehyde. Integration gives 3H, 3H, and 2H, which totals 8H and matches the formula.

不饱和度为 (2×4+2−8)÷2 = 1，因此一个 π 键很可能是 C=O。红外 1740 cm⁻¹ 的峰提示为酯，而不是简单的酮或醛。积分给出 3H、3H、2H，总计 8H，与分子式一致。

The triplet at δ 1.25 and quartet at δ 4.12 form an ethyl group -OCH₂CH₃. The singlet at δ 2.05 is a methyl group attached to a carbonyl, CH₃CO-. Putting these together gives CH₃COOCH₂CH₃, ethyl ethanoate.

δ 1.25 的三重峰和 δ 4.12 的四重峰构成乙基 -OCH₂CH₃。δ 2.05 的单峰是与羰基相连的甲基 CH₃CO-。将这些片段连接起来得到 CH₃COOCH₂CH₃，即乙酸乙酯。

10. Exam Pitfalls and Key Takeaway Checklist | 考试常见误区与核心清单

A common mistake is to read spectra in isolation. For example, seeing a C=O peak does not automatically mean a ketone; the exact wavenumber and NMR shift can distinguish an ester from a ketone or an acid.

一个常见误区是孤立地解读谱图。例如，看到 C=O 峰并不自动意味着酮；精确的波数和 NMR 位移可以区分离酯、酮或酸。

Another pitfall is forgetting to check integration totals against the molecular formula. Always scale the relative integration ratio to the actual number of protons. A ratio of 1 : 2 : 3 does not mean one, two, and three protons unless the formula confirms it.

另一个误区是忘记将积分总和与分子式核对。始终要把相对积分比换算成实际质子数。1 : 2 : 3 的比例并不一定代表一个、两个和三个质子，除非分子式确认如此。

Always calculate DBE first.

始终先计算不饱和度。
Use IR for functional groups, MS for Mr, ¹H NMR for hydrogen environments, and ¹³C NMR for carbon environments.

用红外识别官能团，用质谱确定相对分子质量，用 ¹H NMR 分析氢环境，用 ¹³C NMR 分析碳环境。
Check your final structure against every spectrum before writing the answer.

在写下答案之前，将最终结构与每张谱图进行核对。

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Edexcel A-Level Chemistry: Mastering Combined Spectroscopic Techniques | 爱德思 A-Level 化学：综合光谱分析技术精讲