Mass Spectrometry in Structural Identification | A-Level 化学：质谱法在结构鉴定中的应用

📚 Mass Spectrometry in Structural Identification | A-Level 化学：质谱法在结构鉴定中的应用

Mass spectrometry (MS) is one of the most powerful analytical techniques in modern chemistry. It allows chemists to determine the relative molecular mass of a compound, identify atoms and structural fragments, and deduce molecular formulae from isotopic patterns. In this article, we explore how mass spectrometry is used for structural identification, with a particular focus on the CIE A-Level Chemistry syllabus.

质谱法是现代化学中最强大的分析技术之一。它使化学家能够确定化合物的相对分子质量、识别原子和结构碎片，并通过同位素模式推导分子式。在本文中，我们将重点探讨质谱法如何用于结构鉴定，特别聚焦于 CIE A-Level 化学考纲。

1. Principles of Mass Spectrometry | 质谱法的基本原理

A mass spectrometer measures the mass-to-charge ratio (m/z) of gaseous ions. The sample is vaporised and ionised, usually by electron impact (EI) or electrospray ionisation (ESI). The resulting ions are accelerated in an electric field and deflected by a magnetic field according to their m/z values. Ions with smaller m/z are deflected more than those with larger m/z. A detector records the relative abundance of each ion, producing a mass spectrum.

质谱仪测量气态离子的质荷比（m/z）。样品被气化并电离，通常采用电子轰击（EI）或电喷雾电离（ESI）。生成的离子在电场中加速，并在磁场中根据其 m/z 值发生偏转。m/z 较小的离子比 m/z 较大的离子偏转更大。检测器记录每个离子的相对丰度，从而产生质谱图。

2. The Molecular Ion and Relative Molecular Mass | 分子离子与相对分子质量

The molecular ion, M⁺, is formed when a molecule loses one electron. For a molecule with molecular formula CₐHᵦOᵧNδ, the m/z value of M⁺ corresponds to the relative molecular mass (Mᵣ). For example, the mass spectrum of butanone (CH₃COCH₂CH₃) shows a molecular ion peak at m/z = 72, which is the relative molecular mass of butanone.

分子离子 M⁺ 是分子失去一个电子后形成的。对于分子式为 CₐHᵦOᵧNδ 的分子，M⁺ 的 m/z 值对应相对分子质量（Mᵣ）。例如，丁酮（CH₃COCH₂CH₃）的质谱图在 m/z = 72 处显示分子离子峰，这正是丁酮的相对分子质量。

Mᵣ = m/z of M⁺ (for singly charged ions)

Mᵣ = M⁺ 峰的 m/z 值（对于单电荷离子）

In electron impact ionisation, the molecular ion often undergoes fragmentation. The molecular ion peak is usually the peak with the highest m/z in the spectrum, but it may be weak or even absent for some compounds. In such cases, soft ionisation techniques such as electrospray ionisation are used. ESI produces [M + H]⁺ or [M + Na]⁺ ions, and the relative molecular mass is calculated by subtracting the mass of the attached proton or sodium ion.

在电子轰击电离中，分子离子常发生碎裂。分子离子峰通常是谱图中 m/z 最大的峰，但对于某些化合物可能很弱甚至消失。这种情况下，可使用软电离技术如电喷雾电离（ESI）。ESI 产生 [M + H]⁺ 或 [M + Na]⁺ 离子，通过减去所附质子或钠离子的质量即可计算相对分子质量。

3. Fragmentation and Structural Information | 碎裂与结构信息

Fragmentation is a key feature of electron impact mass spectrometry. When a molecular ion fragments, it breaks into smaller ions and neutral fragments. The m/z values of the fragment ions provide clues about the structural units present in the molecule. For example, a fragment at m/z = 29 corresponds to the ethyl cation C₂H₅⁺, while m/z = 43 corresponds to the acylium ion CH₃CO⁺ or propyl cation C₃H₇⁺. These fragment peaks help chemists reconstruct the skeleton of the molecule.

碎裂是电子轰击质谱法的一个关键特征。当分子离子碎裂时，它会分解成较小的离子和中性碎片。碎片离子的 m/z 值提供了分子中存在的结构单元线索。例如，m/z = 29 的碎片对应于乙基阳离子 C₂H₅⁺，而 m/z = 43 对应于乙酰基阳离子 CH₃CO⁺ 或丙基阳离子 C₃H₇⁺。这些碎片峰帮助化学家重建分子的骨架。

Common neutral losses include:

常见的中性丢失包括：

Loss of a methyl radical (•CH₃, 15 units) — indicates a methyl group attached to the structure.

丢失甲基自由基（•CH₃，15 个质量单位）— 表明结构中存在甲基。
Loss of a water molecule (H₂O, 18 units) — indicates the presence of a hydroxyl group (-OH).

丢失水分子（H₂O，18 个质量单位）— 表明存在羟基（-OH）。
Loss of a carbon monoxide molecule (CO, 28 units) — typical of ketones, aldehydes or phenols.

丢失一氧化碳分子（CO，28 个质量单位）— 常见于酮、醛或酚类化合物。
Loss of a hydrogen chloride molecule (HCl, 36 units) — indicates a chloroalkane.

丢失氯化氢分子（HCl，36 个质量单位）— 表明是氯代烷烃。
Loss of a bromine atom (Br, 79 or 81 units) — indicates a bromoalkane.

丢失溴原子（Br，79 或 81 个质量单位）— 表明是溴代烷烃。

4. Isotopic Peaks: Chlorine and Bromine | 同位素峰：氯和溴

Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. Chlorine has two naturally occurring isotopes: ³⁵Cl (75%) and ³⁷Cl (25%). Bromine has ⁷⁹Br (51%) and ⁸¹Br (49%). When a molecule contains chlorine or bromine, the mass spectrum shows characteristic isotopic peaks.

同位素是质子数相同但中子数不同的同种元素的原子。氯有两种天然同位素：³⁵Cl（75%）和 ³⁷Cl（25%）。溴有 ⁷⁹Br（51%）和 ⁸¹Br（49%）。当分子中含有氯或溴时，质谱图会显示特征的同位素峰。

For a compound containing one chlorine atom, the M⁺ and [M+2]⁺ peaks appear in a 3:1 ratio. For a compound containing one bromine atom, the M⁺ and [M+2]⁺ peaks appear in a 1:1 ratio. This is because the relative abundance of ³⁵Cl:³⁷Cl is 3:1, while that of ⁷⁹Br:⁸¹Br is approximately 1:1.

对于含一个氯原子的化合物，M⁺ 峰和 [M+2]⁺ 峰以 3:1 的比例出现。对于含一个溴原子的化合物，M⁺ 峰和 [M+2]⁺ 峰以 1:1 的比例出现。这是因为 ³⁵Cl:³⁷Cl 的相对丰度为 3:1，而 ⁷⁹Br:⁸¹Br 的相对丰度约为 1:1。

One Cl atom: M⁺ : [M+2]⁺ = 3 : 1

一个 Cl 原子：M⁺ : [M+2]⁺ = 3 : 1

One Br atom: M⁺ : [M+2]⁺ ≈ 1 : 1

一个 Br 原子：M⁺ : [M+2]⁺ ≈ 1 : 1

If a molecule contains two chlorine atoms, the M⁺, [M+2]⁺ and [M+4]⁺ peaks appear approximately in a 9:6:1 ratio. This ratio can be predicted using the binomial expansion of (a + b)ⁿ, where a is the abundance of the lighter isotope, b is the abundance of the heavier isotope, and n is the number of chlorine atoms.

如果分子含有两个氯原子，M⁺、[M+2]⁺ 和 [M+4]⁺ 峰以约 9:6:1 的比例出现。该比例可通过二项式展开 (a + b)ⁿ 预测，其中 a 是较轻同位素的丰度，b 是较重同位素的丰度，n 是氯原子数目。

5. The Nitrogen Rule | 氮规则

The nitrogen rule is a useful guideline for determining the molecular formula. For a neutral molecule containing carbon, hydrogen, oxygen, nitrogen and halogens, the relative molecular mass is even if the number of nitrogen atoms is even (including zero), and odd if the number of nitrogen atoms is odd. Therefore, if a molecular ion has an odd m/z value, the compound must contain an odd number of nitrogen atoms. Conversely, an even m/z value indicates an even number of nitrogen atoms (including zero).

氮规则是确定分子式的有用指南。对于含碳、氢、氧、氮和卤素的中性分子，如果氮原子数为偶数（包括零），相对分子质量为偶数；如果氮原子数为奇数，相对分子质量为奇数。因此，如果分子离子的 m/z 值为奇数，化合物必定含有奇数个氮原子。反之，偶数 m/z 值表明氮原子数为偶数（包括零）。

For example, propanamide (CH₃CH₂CONH₂) has one nitrogen atom and a relative molecular mass of 73 (odd). In contrast, ethyl acetate (CH₃COOCH₂CH₃) has no nitrogen and a relative molecular mass of 88 (even).

例如，丙酰胺（CH₃CH₂CONH₂）含有一个氮原子，相对分子质量为 73（奇数）。相比之下，乙酸乙酯（CH₃COOCH₂CH₃）不含氮，相对分子质量为 88（偶数）。

6. The M+1 Peak and Carbon-13 | M+1 峰与碳-13

Carbon-13 (¹³C) has a natural abundance of approximately 1.1% relative to carbon-12 (¹²C). For a molecule containing n carbon atoms, the probability of having exactly one ¹³C atom is approximately n × 1.1% of the molecular ion peak. The [M+1]⁺ peak intensity can therefore be used to estimate the number of carbon atoms in the molecule.

碳-13（¹³C）相对于碳-12（¹²C）的自然丰度约为 1.1%。对于含有 n 个碳原子的分子，恰好含有一个 ¹³C 原子的概率约为分子离子峰强度的 n × 1.1%。因此，[M+1]⁺ 峰的强度可用于估算分子中的碳原子数目。

Relative intensity of [M+1]⁺ ≈ n × 1.1% of M⁺

[M+1]⁺ 的相对强度 ≈ M⁺ 的 n × 1.1%

For example, benzene (C₆H₆) has six carbon atoms. The [M+1]⁺ peak should be approximately 6 × 1.1% = 6.6% of the M⁺ peak intensity. This calculation assists chemists in confirming the number of carbon atoms in an unknown compound.

例如，苯（C₆H₆）有六个碳原子。[M+1]⁺ 峰应约为 M⁺ 峰强度的 6 × 1.1% = 6.6%。这一计算帮助化学家确认未知化合物中的碳原子数目。

7. High-Resolution Mass Spectrometry | 高分辨质谱法

In A-Level chemistry, relative atomic masses are usually rounded to whole numbers. However, high-resolution mass spectrometry can measure m/z values to four or more decimal places. This allows chemists to distinguish between ions with the same nominal mass but different exact masses.

在 A-Level 化学中，相对原子质量通常四舍五入为整数。然而，高分辨质谱法可将 m/z 值测量到小数点后四位或更多位。这使得化学家能够区分具有相同名义质量但精确质量不同的离子。

For example, both CO⁺ (m/z = 27.9949) and C₂H₄⁺ (m/z = 28.0313) have a nominal mass of 28, but their exact masses are different. High-resolution mass spectrometry can distinguish these two ions, enabling the unambiguous determination of the molecular formula. The exact mass of an ion is calculated by summing the exact masses of all atoms in the ion.

例如，CO⁺（m/z = 27.9949）和 C₂H₄⁺（m/z = 28.0313）的名义质量都是 28，但精确质量不同。高分辨质谱法能区分这两种离子，从而明确确定分子式。离子的精确质量通过累加离子中所有原子的精确质量来计算。

Ion	Nominal mass	Exact mass
CO⁺	28	27.9949
C₂H₄⁺	28	28.0313
N₂⁺	28	28.0061

The table above shows three different ions with the same nominal mass of 28 but different exact masses. In CIE A-Level examinations, you may be asked to use exact masses to identify the molecular formula of an unknown compound.

上表显示了三种名义质量均为 28 但精确质量不同的离子。在 CIE A-Level 考试中，你可能会被要求使用精确质量来鉴定未知化合物的分子式。

8. Determining Molecular Formula: A Step-by-Step Approach | 确定分子式：分步方法

To determine the molecular formula from a mass spectrum, follow these steps:

根据质谱图确定分子式，请遵循以下步骤：

Identify the molecular ion peak (M⁺) and record its m/z value.

识别分子离子峰（M⁺）并记录其 m/z 值。
Check whether the M⁺ peak is odd or even to apply the nitrogen rule.

检查 M⁺ 峰是奇数还是偶数，以应用氮规则。
Measure the relative intensities of [M+1]⁺ and M⁺ to estimate the number of carbon atoms.

测量 [M+1]⁺ 与 M⁺ 的相对强度，估算碳原子数目。
Examine the [M+2]⁺ peak. A 3:1 M⁺:[M+2]⁺ ratio indicates one Cl atom; a 1:1 ratio indicates one Br atom.

检查 [M+2]⁺ 峰。M⁺:[M+2]⁺ = 3:1 表明含一个 Cl 原子；1:1 表明含一个 Br 原子。
Use exact masses (if provided) to calculate the molecular formula from the accurate Mᵣ value.

使用精确质量（如果提供）从准确的 Mᵣ 值计算分子式。
Consider fragment peaks to confirm structural features.

考虑碎片峰以确认结构特征。

For example, consider a compound with M⁺ at m/z = 46 (even, no nitrogen). The [M+1]⁺ intensity is about 2.2% of M⁺, suggesting approximately 2 carbon atoms (2 × 1.1% = 2.2%). The remaining mass is 46 − 24 = 22. Possible formula: C₂H₆O (Mᵣ = 46). This could be ethanol or methoxymethane. The presence of a fragment at m/z = 31 (CH₂OH⁺) would confirm ethanol, while a fragment at m/z = 15 (CH₃⁺) alone would be less conclusive.

例如，考虑一个 M⁺ 在 m/z = 46（偶数，不含氮）的化合物。[M+1]⁺ 强度约为 M⁺ 的 2.2%，表明约有 2 个碳原子（2 × 1.1% = 2.2%）。剩余质量为 46 − 24 = 22。可能的分子式：C₂H₆O（Mᵣ = 46）。这可能是乙醇或甲氧基甲烷。m/z = 31（CH₂OH⁺）处的碎片峰可确认为乙醇，而仅有 m/z = 15（CH₃⁺）的碎片峰则不足以确定。

9. Fragmentation Patterns of Common Functional Groups | 常见官能团的碎裂模式

Carboxylic acids typically lose a hydroxyl radical (•OH, 17 units) or show a fragment at m/z = 45 (COOH⁺). Aliphatic aldehydes and ketones often lose an alkyl radical, producing acylium ions RCO⁺. For example, propanone (CH₃COCH₃) gives a strong fragment at m/z = 43 (CH₃CO⁺). Primary alcohols often show loss of H₂O (18 units) and a fragment at m/z = 31 (CH₂OH⁺).

羧酸通常会失去羟基自由基（•OH，17 个质量单位）或在 m/z = 45（COOH⁺）处显示碎片峰。脂肪醛和酮通常会丢失一个烷基自由基，产生酰基阳离子 RCO⁺。例如，丙酮（CH₃COCH₃）在 m/z = 43（CH₃CO⁺）处产生强碎片峰。伯醇通常显示失去 H₂O（18 个质量单位）并在 m/z = 31（CH₂OH⁺）处产生碎片峰。

Aromatic compounds often show a stable fragment at m/z = 77, corresponding to the phenyl cation C₆H₅⁺. Esters exhibit characteristic fragmentation involving cleavage adjacent to the carbonyl group, often producing an acylium ion RCO⁺ or an alkoxy ion RO⁺. An understanding of these typical fragmentation patterns is essential for interpreting mass spectra.

芳香族化合物通常在 m/z = 77 处显示稳定碎片峰，对应于苯基阳离子 C₆H₅⁺。酯类表现出特征碎裂，涉及羰基相邻键的断裂，通常产生酰基离子 RCO⁺ 或烷氧基离子 RO⁺。理解这些典型碎裂模式对于解读质谱图至关重要。

10. Worked Example: Identification of an Unknown Compound | 实例解析：未知化合物的鉴定

An unknown organic compound has a mass spectrum with a molecular ion peak at m/z = 88 and an [M+2]⁺ peak with approximately 0.8% of the intensity of the M⁺ peak. The [M+1]⁺ peak is about 4.4% of M⁺. No nitrogen is present. A prominent fragment is observed at m/z = 43.

一种未知有机化合物的质谱图显示分子离子峰在 m/z = 88，[M+2]⁺ 峰的强度约为 M⁺ 的 0.8%。[M+1]⁺ 峰约为 M⁺ 的 4.4%。不含氮。在 m/z = 43 处观察到显著碎片峰。

First, estimate the number of carbon atoms: [M+1]⁺ / M⁺ ≈ 0.044. Dividing by 0.011 gives approximately 4 carbon atoms (4 × 1.1% = 4.4%). The mass of 4 carbon atoms is 48. The remaining mass is 88 − 48 = 40. Since the compound has no nitrogen, we need to distribute the remaining mass among hydrogen and oxygen atoms. Try C₄H₈O₂ (Mᵣ = 4×12 + 8×1 + 2×16 = 88). This formula fits perfectly.

首先估算碳原子数：[M+1]⁺ / M⁺ ≈ 0.044。除以 0.011 得到约 4 个碳原子（4 × 1.1% = 4.4%）。4 个碳原子的质量为 48。剩余质量为 88 − 48 = 40。由于化合物不含氮，我们需要在氢和氧之间分配剩余质量。尝试 C₄H₈O₂（Mᵣ = 4×12 + 8×1 + 2×16 = 88）。该分子式完全吻合。

The fragment at m/z = 43 is consistent with CH₃CO⁺ (mass 43) or C₃H₇⁺. The molecular formula C₄H₈O₂ could correspond to an ester such as ethyl ethanoate (CH₃COOCH₂CH₃) or a carboxylic acid such as butanoic acid (CH₃CH₂CH₂COOH). Butanoic acid typically shows a fragment at m/z = 45 (COOH⁺) and loss of OH (17), while ethyl ethanoate shows a strong fragment at m/z = 43 (CH₃CO⁺). The presence of a strong m/z = 43 peak suggests ethyl ethanoate. The fragmentation CH₃COOC₂H₅ → CH₃CO⁺ + OC₂H₅• is highly favourable.

m/z = 43 处的碎片与 CH₃CO⁺（质量 43）或 C₃H₇⁺ 一致。分子式 C₄H₈O₂ 可能对应酯类如乙酸乙酯（CH₃COOCH₂CH₃）或羧酸如丁酸（CH₃CH₂CH₂COOH）。丁酸通常在 m/z = 45（COOH⁺）处显示碎片峰并失去 OH（17 个单位），而乙酸乙酯在 m/z = 43（CH₃CO⁺）处显示强碎片峰。m/z = 43 强峰的存在提示乙酸乙酯。碎裂 CH₃COOC₂H₅ → CH₃CO⁺ + OC₂H₅• 是非常有利的。

11. Common Pitfalls and Examination Tips | 常见陷阱与考试技巧

In CIE A-Level examinations, students often confuse the molecular ion peak with the base peak. The base peak is the most intense peak, set to 100% relative abundance, and corresponds to the most stable fragment ion — it is not always the molecular ion. Another common error is assuming that the M⁺ peak is always visible; for some compounds, the M⁺ peak may be very weak or absent under electron impact ionisation.

在 CIE A-Level 考试中，学生经常混淆分子离子峰和基峰。基峰是最强的峰，其相对丰度设为 100%，对应最稳定的碎片离子——它不总是分子离子。另一个常见错误是假设 M⁺ 峰总是可见；对于某些化合物，在电子轰击电离下 M⁺ 峰可能非常弱或根本不存在。

When using the nitrogen rule, remember that it applies to neutral molecules. If the spectrum was obtained by ESI, the observed m/z corresponds to [M + H]⁺, and you must subtract 1 before applying the nitrogen rule. Always check whether the question provides high-resolution data; if so, use exact masses rather than nominal masses.

使用氮规则时，请记住它适用于中性分子。如果谱图通过 ESI 获得，观察到的 m/z 对应 [M + H]⁺，在应用氮规则前必须减去 1。始终检查题目是否提供高分辨数据；如果提供，应使用精确质量而非名义质量。

Finally, for this question type, always cross-check the molecular formula with the degree of unsaturation, also called the index of hydrogen deficiency (IHD). For C₄H₈O₂, IHD = (2×4 + 2 − 8) / 2 = 1, indicating one double bond or ring. This is consistent with a carbonyl group or an ester linkage.

最后，对于此类题型，务必用不饱和度（也称为缺氢指数，IHD）交叉验证分子式。对于 C₄H₈O₂，IHD = (2×4 + 2 − 8) / 2 = 1，表明有一个双键或环。这与羰基或酯键一致。

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Mass Spectrometry in Structural Identification | A-Level 化学：质谱法在结构鉴定中的应用