📚 Edexcel A-Level Chemistry: Combined Analytical Techniques (Mass Spectrometry, IR, NMR) | 组合分析技术（质谱、红外、核磁）
In Edexcel A-Level Chemistry, the modern analytical section requires you to combine mass spectrometry, infrared spectroscopy, and nuclear magnetic resonance spectroscopy to determine organic structures. The key skill is not just identifying one spectrum, but using evidence from all three techniques to build an argument for a specific molecular structure.
在 Edexcel A-Level 化学中，现代分析技术要求你结合质谱、红外光谱和核磁共振光谱来确定有机结构。核心能力不是只识别单一谱图，而是综合三种技术的证据，为某个特定分子结构建立推理。
Examiners rarely ask for a single technique in isolation at the top grades. Instead, they give you a molecular formula, an IR spectrum, an NMR spectrum, and sometimes a mass spectrum, then ask you to justify a structure with reasons. This article walks through each technique in turn, then shows how to weld them together into one coherent answer.
高分题目很少只考单一技术。考卷通常给出分子式、红外谱图、核磁谱图，有时还有质谱图，然后要求你用理由论证某个结构。本文逐一讲解每种技术，再演示如何把它们拼成一个完整的答案。
1. Why Combined Techniques Are Essential | 为什么需要组合分析技术
No single analytical method gives every piece of structural information. Mass spectrometry provides the molecular formula and fragment masses, infrared spectroscopy identifies functional groups, and NMR reveals the carbon and hydrogen skeleton. Combining them allows chemists to confirm the full structure of an unknown compound.
没有一种分析方法能提供全部结构信息。质谱给出分子式和碎片质量，红外光谱识别官能团，核磁共振揭示碳骨架和氢环境。三者结合后，化学家才能确认未知化合物的完整结构。
Think of the three techniques as three witnesses to the same event. Mass spectrometry tells you how heavy the molecule is and how it breaks apart. Infrared spectroscopy tells you which functional groups are present. NMR tells you how the atoms are connected and how many distinct environments exist. A structural claim supported by all three is far stronger than one supported by only one.
可以把这三种技术想成同一事件的三位证人。质谱告诉你分子有多重、如何断裂；红外光谱告诉你存在哪些官能团；核磁共振告诉你原子如何连接、有多少种不同环境。由三者共同支持的结构结论，远比只靠一种技术可靠。
- Mass spectrometry → relative molecular mass, molecular formula, fragmentation clues | 质谱 → 相对分子质量、分子式、碎片线索
- Infrared spectroscopy → functional group identification | 红外光谱 → 官能团识别
- NMR spectroscopy → carbon skeleton and hydrogen environments | 核磁共振 → 碳骨架与氢环境
2. Mass Spectrometry: Finding the Molecular Ion | 质谱：确定分子离子
In a mass spectrometer, the sample is vaporised, ionised (usually by electron impact), accelerated, and separated according to mass-to-charge ratio, m/z. The peak of highest m/z (ignoring small isotope peaks) is the molecular ion peak, written M⁺ or M⁺·, and it gives the relative molecular mass of the compound. For example, a molecular ion peak at m/z = 74 means the molecule has a relative molecular mass of 74.
在质谱仪中，样品先汽化，再电离（通常用电子轰击法），然后加速，并按质荷比 m/z 分离。忽略小的同位素峰后，m/z 最大的峰就是分子离子峰，写作 M⁺ 或 M⁺·，它给出化合物的相对分子质量。例如，m/z = 74 的分子离子峰说明该分子相对分子质量为 74。
M + e⁻ → M⁺· + 2e⁻
This equation shows that the electron impact removes one electron from the molecule, leaving a positively charged radical cation. Because one electron is lost, the molecular ion has an odd electron count and is described as a radical cation.
这个方程说明电子轰击从分子中打掉一个电子，留下一个带正电的自由基阳离子。由于失去一个电子，分子离子具有奇数电子数，被称为自由基阳离子。
Some molecules give very weak molecular ion peaks because they fragment almost completely. Alcohols and heavily branched alkanes are classic examples: the M⁺ peak may be barely visible above the baseline, so you must look carefully and may need to rely on other data.
有些分子的分子离子峰非常弱，因为它们几乎完全碎裂。醇和高度支链化的烷烃就是典型例子：M⁺ 峰可能几乎贴近基线，所以必须仔细观察，必要时依靠其他数据。
3. Fragmentation Patterns and Structural Clues | 碎片模式与结构线索
Once the molecular ion forms, it can break into a positive fragment and a neutral radical. Only the positive fragment is accelerated and detected, so the spectrum shows peaks corresponding to positively charged fragments. The difference between the molecular ion mass and a fragment mass tells you the mass of the group that was lost.
分子离子形成后，会断裂成一个正碎片和一个中性自由基。只有正碎片被加速和检测，因此谱图中出现的峰对应带正电的碎片。分子离子质量与某个碎片质量之差，就是失去基团的质量。
M⁺ → fragment⁺ + radical·
Common neutral species lost include a methyl radical (·CH₃, mass 15), a water molecule (H₂O, mass 18), and an ethyl radical (·C₂H₅, mass 29). Detecting these losses helps you guess which functional group is present.
常见的脱去中性物种包括甲基自由基（·CH₃，质量 15）、水分子（H₂O，质量 18）和乙基自由基（·C₂H₅，质量 29）。识别这些丢失有助于推断存在哪些官能团。
- A peak at m/z = 29 often indicates CH₃CH₂⁺ or CHO⁺ | m/z = 29 的峰常指示 CH₃CH₂⁺ 或 CHO⁺
- A peak at m/z = 45 often indicates CH₃CH₂O⁺ from an alcohol or ether | m/z = 45 的峰常指示来自醇或醚的 CH₃CH₂O⁺
- A peak at m/z = 77 indicates the C₆H₅⁺ phenyl cation from an aromatic ring | m/z = 77 指示来自芳环的苯基阳离子 C₆H₅⁺
- A strong peak at m/z = 43 often indicates CH₃CO⁺ from a methyl ketone | m/z = 43 的强峰常指示来自甲基酮的 CH₃CO⁺
The most intense peak in the spectrum is called the base peak and is assigned an abundance of 100%. It does not have to be the molecular ion peak; in fact, for many compounds the base peak is a stable fragment such as a tertiary carbocation.
谱图中最强的峰称为基峰，其丰度定为 100%。基峰不一定是分子离子峰；事实上，对许多化合物来说，基峰是稳定碎片，例如叔碳正离子。
4. Isotope Peaks: M+1 and M+2 | 同位素峰：M+1 与 M+2
Because carbon-13 occurs naturally at about 1.1% abundance, every carbon-containing compound shows a small M+1 peak one unit above the molecular ion. This peak is not a fragment; it is the molecular ion containing one ¹³C atom instead of ¹²C.
由于碳-13 的天然丰度约为 1.1%，任何含碳化合物都会在分子离子上方一个质量单位处出现一个小的 M+1 峰。这个峰不是碎片峰，而是含有一个 ¹³C 原子（代替 ¹²C）的分子离子。
The M+2 peak becomes important when chlorine or bromine is present. Chlorine has two isotopes, ³⁵Cl and ³⁷Cl, in roughly a 3:1 ratio, so a chlorine-containing compound shows an M and M+2 pair of peaks with heights approximately 3:1. Bromine has ⁷⁹Br and ⁸¹Br in roughly a 1:1 ratio, so a bromine-containing compound shows an M and M+2 pair of almost equal height.
当分子中含有氯或溴时，M+2 峰变得非常重要。氯有两种同位素 ³⁵Cl 与 ³⁷Cl，比例约为 3:1，因此含氯化合物会出现高度比约为 3:1 的 M 和 M+2 峰对。溴有 ⁷⁹Br 与 ⁸¹Br，比例约为 1:1，因此含溴化合物会出现高度几乎相等的 M 和 M+2 峰对。
| Element | 元素 | Isotope pattern | 同位素峰型 | Typical clue | 典型线索 |
| Carbon | 碳 | Small M+1 | 小的 M+1 峰 | Confirms carbon present | 确认含碳 |
| Chlorine | 氯 | M : M+2 ≈ 3 : 1 | One Cl atom | 含一个氯原子 |
| Bromine | 溴 | M : M+2 ≈ 1 : 1 | One Br atom | 含一个溴原子 |
M : M+2 : M+4 ≈ 9 : 6 : 1, which is a distinctive fingerprint that examiners love to test.
如果含有两个氯原子，峰型会变成 M : M+2 : M+4 ≈ 9 : 6 : 1，这是考官喜欢考查的独特指纹特征。
5. High-Resolution Mass Spectrometry and Molecular Formula | 高分辨质谱与分子式
High-resolution mass spectrometry measures m/z to several decimal places. Because isotopes have slightly different exact masses, you can use the precise mass of the molecular ion to determine the molecular formula directly.
高分辨质谱能把 m/z 测量到小数点后好几位。由于各同位素精确质量略有不同，你可以用分子离子的精确质量直接确定分子式。
C = 12.0000 H = 1.0079 O = 15.9949 N = 14.0031
For example, both C₃H₈O and C₂H₄O₂ have a nominal relative molecular mass of 60. Their exact masses differ: C₃H₈O gives 60.0575, while C₂H₄O₂ gives 60.0211. A measured value of 60.021 therefore points to C₂H₄O₂, an acid or ester rather than an alcohol or ether.
例如，C₃H₈O 和 C₂H₄O₂ 的名义相对分子质量都是 60，但精确质量不同：C₃H₈O 为 60.0575，C₂H₄O₂ 为 60.0211。若测得 60.021，就指向 C₂H₄O₂，即酸或酯，而不是醇或醚。
ated formula CₙH₂ₙ₊₂. A formula with two fewer hydrogens than saturated suggests one C=C bond or one ring.
得到候选分子式后，要检查不饱和度。相对于饱和通式 CₙH₂ₙ₊₂，每增加一个环或一个双键，氢原子数就减少 2。氢数比饱和式少 2，说明存在一个 C=C 键或一个环。
6. Infrared Spectroscopy: Identifying Functional Groups | 红外光谱：识别官能团
Infrared spectroscopy works because covalent bonds absorb infrared radiation at frequencies that match their vibrational modes. Different bonds vibrate at different frequencies, so the position of an absorption band tells you which bonds are present.
红外光谱的原理是：共价键会吸收与其振动模式频率相匹配的红外辐射。不同化学键的振动频率不同，因此吸收带的位置可以告诉你存在哪些化学键。
Two vibrational modes matter most at A-Level: stretching, where the bond length changes rhythmically, and bending, where the bond angle changes. Stretching absorptions usually appear at higher wavenumbers than bending absorptions.
A-Level 阶段最重要的两种振动模式是：伸缩振动（键长周期性变化）和弯曲振动（键角变化）。伸缩振动的吸收通常出现在比弯曲振动更高的波数位置。
Infrared spectra are plotted with wavenumber in cm⁻¹ decreasing from left to right, typically from 4000 cm⁻¹ to 400 cm⁻¹. The region above about 1500 cm⁻¹ is called the functional group region, because characteristic absorptions for common groups appear there.
红外谱图的横轴波数以 cm⁻¹ 表示，从左到右递减，通常从 4000 cm⁻¹ 到 400 cm⁻¹。约 1500 cm⁻¹ 以上的区域称为官能团区，常见基团的特征吸收都出现在这里。
7. Key IR Absorptions and the Fingerprint Region | 关键红外吸收与指纹区
You should be able to quote approximate wavenumber ranges for the functional groups in the Edexcel specification. Examiners accept a range rather than an exact number, so learn the intervals rather than single values.
你需要能说出 Edexcel 考纲中官能团的大致波数范围。考官接受一个区间而非精确数值，所以要记住区间而不是单一数字。
| Bond | 化学键 | Wavenumber / cm⁻¹ | 波数 | Appearance | 峰形 |
| O-H (alcohol) | O-H（醇） | 3230-3550 | Broad | 宽峰 |
| O-H (carboxylic acid) | O-H（羧酸） | 2500-3300 | Very broad | 非常宽 |
| N-H (amine) | N-H（胺） | 3300-3500 | Sharp, often two bands | 尖锐，常为两条 |
| C-H | C-H | 2850-3100 | Sharp | 尖锐 |
| C≡N | C≡N | 2220-2260 | Sharp, medium | 尖锐，中等 |
| C=O (aldehyde, ketone) | C=O（醛、酮） | 1680-1750 | Strong, sharp | 强而尖锐 |
| C=O (ester) | C=O（酯） | 1735-1750 | Strong, plus C-O at 1000-1300 |
| C=C | C=C | 1620-1680 | Weak to medium | 弱到中等 |
The region below about 1500 cm⁻¹ is the fingerprint region. It contains many overlapping bands from bending and skeletal vibrations, and it is unique to each compound. Matching a sample spectrum against a reference spectrum in this region is strong evidence of identity, even though individual bands are hard to assign.
约 1500 cm⁻¹ 以下的区域是指纹区，包含许多来自弯曲振动和骨架振动的重叠谱带，对每种化合物都是独一无二的。把样品谱图与该区内的标准谱图比对，是证明同一性的有力证据，尽管单个谱带很难归属。
If instead the broad O-H appears at 3230-3550 cm⁻¹ with no C=O band, you have an alcohol.
2500 至 3300 cm⁻¹ 的宽 O-H 峰与约 1700 cm⁻¹ 的强 C=O 峰同时出现，是羧酸的经典特征。若宽 O-H 峰出现在 3230-3550 cm⁻¹ 而没有 C=O 峰，那就是醇。
8. Carbon-13 NMR: Counting Carbon Environments | 碳-13 核磁：碳环境数目
Carbon-13 NMR tells you how many different carbon environments exist in a molecule. Each distinct environment gives one peak, so the number of peaks equals the number of non-equivalent carbon atoms. The chemical shift of each peak indicates the type of carbon it belongs to.
碳-13 核磁告诉你分子中有多少种不同的碳环境。每一种不同环境给出一个峰，因此峰的数目等于非等价碳原子的数目。每个峰的化学位移则指示该碳的类型。
Chemical shifts are measured relative to tetramethylsilane, TMS, which is assigned a value of 0 ppm. TMS is used because it is chemically inert, volatile, soluble in organic solvents, and gives a single sharp peak well away from most sample signals.
化学位移以四甲基硅烷（TMS）为基准，规定其值为 0 ppm。选用 TMS 是因为它化学惰性、易挥发、可溶于有机溶剂，并且给出一个远离大多数样品信号的单一尖峰。
| Carbon environment | 碳环境 | Shift / ppm | 位移 |
| R-CH₃ (alkane) | R-CH₃（烷烃） | 5-30 |
| C-O (alcohol, ether) | C-O（醇、醚） | 50-70 |
| C=C (alkene) | C=C（烯） | 100-150 |
| Aromatic C | 芳环碳 | 110-160 |
| C=O (ester, acid) | C=O（酯、酸） | 160-185 |
| C=O (aldehyde, ketone) | C=O（醛、酮） | 190-220 |
Symmetry is the key trap here. In butane, CH₃CH₂CH₂CH₃, there are four carbon atoms but only two peaks, because the two terminal carbons are equivalent and the two central carbons are equivalent. Always check for a plane of symmetry before counting peaks.
对称性是这里的最大陷阱。在丁烷 CH₃CH₂CH₂CH₃ 中虽然有四个碳原子，但只有两个峰，因为两个端基碳等价、两个中间碳等价。数峰之前一定要先检查对称面。
9. Proton NMR: Chemical Shift and Integration | 氢核磁：化学位移与积分
Proton NMR, often written ¹H NMR, gives three separate pieces of information: chemical shift, integration, and splitting. Together they let you reconstruct the hydrogen framework of the molecule.
氢核磁（常写作 ¹H NMR）提供三类独立信息：化学位移、积分和裂分。三者结合，就能重建分子的氢骨架。
The chemical shift, measured in ppm downfield from TMS, tells you the electronic environment of each proton. Protons near electronegative atoms or in anisotropic regions such as aromatic rings are deshielded and appear at higher ppm.
化学位移以 ppm 为单位，从 TMS 向低场方向计量，它告诉你每个质子的电子环境。靠近电负性原子、或处于芳环等各向异性区域的质子受到去屏蔽，出现在更高的 ppm 处。
| Proton environment | 氢环境 | Shift / ppm | 位移 |
| R-CH₃ | R-CH₃ | 0.7-1.2 |
| R-CH₂-R | R-CH₂-R | 1.2-1.6 |
| R₃CH | R₃CH | 1.5-1.9 |
| R-O-H (alcohol) | R-O-H（醇） | 1.0-5.0 (variable) |
| C=C-H (alkene) | C=C-H（烯） | 4.5-6.0 |
| Aromatic H | 芳环氢 | 6.5-8.0 |
| R-CHO | R-CHO | 9.0-10.0 |
| R-COOH | R-COOH | 10.0-13.0 |
Integration gives the relative number of protons responsible for each peak, shown as a ratio or as a number printed on the spectrum. If the ratios are 3 : 2 : 1, the molecule has three protons in one environment, two in another, and one in a third, or any whole-number multiple of that ratio.
积分给出每个峰对应的相对质子数，可以表示为比例，也可以在谱图上直接标数字。如果积分比为 3 : 2 : 1，说明分子中某一种环境有 3 个氢，另一种有 2 个，第三种有 1 个，或者是这个比例的整数倍。
10. Splitting, the n+1 Rule, and D₂O Exchange | 裂分、n+1 规则与 D₂O 交换
uring protons is split into n+1 lines, giving a singlet, doublet, triplet, or quartet for n = 0, 1, 2, 3 respectively.
质子峰会被相邻碳原子上的非等价质子裂分。一般规则是：若某质子有 n 个等价的邻近质子，其峰裂分为 n+1 条线，当 n = 0
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