Edexcel A-Level Chemistry: Combined Analytical Techniques | 爱德思 A-Level 化学：综合分析技术

📚 Edexcel A-Level Chemistry: Combined Analytical Techniques | 爱德思 A-Level 化学：综合分析技术

Organic structure determination in Edexcel A-Level Chemistry, especially in the Modern Analytical Techniques II topic, almost always requires you to combine mass spectrometry, infrared spectroscopy and NMR data rather than rely on one spectrum. This article gives you a reliable deduction strategy, the key data ranges you must know, a full worked example and the common exam pitfalls that cost marks.

在爱德思 A-Level 化学，尤其是“现代分析技术 II”这一专题中，确定有机结构几乎总是需要综合质谱、红外光谱与核磁共振波谱数据，而不是依赖单一谱图。本文提供可靠的推断策略、必须掌握的关键数据范围、一个完整的例题解析，以及最常见的失分陷阱。

1. Overview of Combined Analytical Techniques | 综合分析技术总览

Combined analytical techniques use complementary evidence. Mass spectrometry gives the relative molecular mass and fragmentation patterns; infrared spectroscopy identifies functional groups; ¹H NMR reveals hydrogen environments, their ratios and splitting; ¹³C NMR counts non-equivalent carbon environments. When used together, these techniques can determine the full structure of an unknown organic compound.

综合分析技术利用互补的证据。质谱提供相对分子质量和碎片模式；红外光谱鉴别官能团；¹H NMR 显示氢环境、氢原子比例与裂分；¹³C NMR 统计非等效碳环境。综合使用这些技术，可以确定未知有机化合物的完整结构。

annot tell them apart because they all show a C=O and a C-O absorption. Only the NMR data, which counts environments, can distinguish ethanoate from propanoate or from the branched isomer.

单一技术之所以永远不够，原因在于同分异构现象。四种不同的酯可以共享分子式 C₄H₈O₂，仅凭红外光谱无法区分它们，因为它们都显示 C=O 与 C-O 吸收。只有能够统计环境数目的核磁共振数据，才能把乙酸酯、丙酸酯或支链异构体区分开来。

In exam questions, you will normally be given just enough data from all four techniques. Your job is to use each piece of evidence in a logical order rather than guessing from a single spectrum.

在考试题目中，你通常会得到来自这四种技术的刚好足够的数据。你的任务是按逻辑顺序使用每一条证据，而不是仅凭单一谱图猜测。

2. Mass Spectrometry: Finding the Molecular Formula | 质谱：确定分子式

The highest m/z peak in a simple molecular ion region is usually the molecular ion peak, written as M⁺. The m/z value of M⁺ gives the relative molecular mass Mᵣ. In Edexcel A-Level questions, you often combine this with percentage composition data: use the percentages to find the empirical formula, then divide Mᵣ by the empirical formula mass to find the molecular formula.

在简单分子离子区中，最高的 m/z 峰通常就是分子离子峰，记作 M⁺。M⁺ 的 m/z 值给出相对分子质量 Mᵣ。在爱德思 A-Level 题目中，你常需要把它与百分比组成数据结合：先用百分含量求出实验式，再用 Mᵣ 除以实验式质量求出分子式。

n = Mᵣ ÷ (empirical formula mass); molecular formula = (empirical formula)ₙ

Example: an unknown contains 40.0% C, 6.7% H and 53.3% al formula is CH₂O with mass 30. Then n = 60 ÷ 30 = 2, so the molecular formula is C₂H₄O₂.

例如：某未知物质量组成为 40.0% C、6.7% H、53.3% O，其 M⁺ 峰在 m/z = 60。用各百分含量除以相对原子质量得到 C : H : O = 1 : 2 : 1，实验式为 CH₂O，式量为 30。于是 n = 60 ÷ 30 = 2，分子式为 C₂H₄O₂。

You should also recognise three further features of mass spectra that examiners like to test. First, high-resolution mass spectrometry measures m/z to several decimal places, which allows the molecular formula to be determined directly, because the exact masses of the isotopes of C, H, N and O differ slightly. Second, the M+1 peak arises from the presence of carbon-13 and grows in relative height as the number of carbon atoms increases, so it can be used to estimate the number of carbons in the molecule. Third, an odd Mᵣ value strongly suggests an odd number of nitrogen atoms (the nitrogen rule), because nitrogen has an even relative atomic mass but forms three bonds.

你还应识别质谱中另外三个常考特征。第一，高分辨质谱把 m/z 测到小数点后几位，由于 C、H、N、O 的同位素精确质量略有差异，可以直接确定分子式。第二，M+1 峰来自碳-13 的存在，其相对高度随碳原子数增加而增大，因此可用来估算分子中碳原子数目。第三，Mᵣ 为奇数强烈提示氮原子数为奇数（氮规则），因为氮的相对原子质量为偶数，却形成三根键。

Chlorine and bromine produce recognisable isotope patterns: a chlorine-containing compound shows an M and M+2 pair with approximately a 3 : 1 ratio, whereas bromine gives an approximately 1 : 1 ratio. Do not confuse the molecular ion peak with the base peak, which is simply the most intense peak in the spectrum and may be a fragment rather than M⁺.

氯和溴会产生特征性同位素峰形：含氯化合物出现 M 与 M+2 一对峰，比例约为 3 : 1；含溴化合物则约为 1 : 1。不要混淆分子离子峰与基峰，基峰只是谱图中强度最大的峰，它可能是碎片峰而非 M⁺。

3. Fragmentation Patterns and Structural Clues | 碎片模式与结构线索

In the ionisation chamber, the molecular ion is a radical cation with excess energy. It can break into a smaller cation and a radical, and only the positively charged fragment is detected. Each fragment gives an m/z value that corresponds to a possible combination of atoms lost from the molecular ion.

在离子化室中，分子离子是一个带有多余能量的自由基阳离子。它可以断裂成一个较小的阳离子和一个自由基，而只有带正电的碎片会被检测到。每个碎片给出一个 m/z 值，对应于从分子离子中失去的某种原子组合。

stent with an ethyl group lost from an acid such as propanoic acid. If a peak appears at m/z = 43 from an Mᵣ of 74, then 31 units are lost, suggesting CH₃O and therefore a methyl ester.

广告 / Sponsored

牛津高阶英汉双解词典 第10版 商务印书馆

¥168

📱 去京东购买

京东精选

一个实用的技巧是用 Mᵣ 减去碎片 m/z。若 Mᵣ = 74 且 m/z = 45 处出现强峰，则失去 29 个质量单位，与丙酸等酸失去乙基相符。若 Mᵣ = 74 而 m/z = 43 处出现峰，则失去 31 个单位，提示失去 CH₃O，因此可能是甲酯。

Remember that the molecular ion peak is not always strong. Alcohols, branched alkanes and compounds with weak C-C bonds fragment so readily that M⁺ may be a small peak, so always look for the highest m/z peak in the region rather than the tallest peak overall.

请记住，分子离子峰不一定很强。醇、支链烷烃以及含弱 C-C 键的化合物很容易断裂，M⁺ 可能只是一个很小的峰，因此应寻找该区域中 m/z 最大的峰，而不是整张谱图中最高的峰。

4. Infrared Spectroscopy: Identifying Functional Groups | 红外光谱：鉴别官能团

Infrared spectroscopy works because covalent bonds absorb infrared radiation at frequencies that match their natural vibrational frequencies. A bond absorbs at a wavenumber that depends on the bond strength and on the masses of the atoms involved: stronger bonds and lighter atoms absorb at higher wavenumbers.

红外光谱的原理是共价键吸收与其自然振动频率相匹配的红外辐射。化学键吸收的波数取决于键的强度与所连原子的质量：键越强、原子越轻，吸收波数越高。

ethene are weak absorbers while polar bonds such as C=O give intense, unmistakable peaks.

只有会改变分子偶极矩的振动才会产生吸收，这就是为什么乙烯中对称的 C=C 键吸收很弱，而 C=O 这样的极性键则给出强烈而明确的峰。

For Edexcel, the key skill is not to interpret the whole spectrum but to read specific absorptions and translate them into functional groups. A single spectrum usually tells you which functional groups are present and, just as importantly, which are absent.

对爱德思考试而言，关键技能不是解读整张谱图，而是读取特定吸收并把它翻译成官能团。一张谱图通常能告诉你存在哪些官能团，同样重要的是，也告诉你哪些官能团不存在。

5. Interpreting Infrared Spectra: Key Ranges and the Fingerprint Region | 红外谱图解读：关键范围与指纹区

The region above about 1500 cm⁻¹ is called the functional group region, and this is where you should focus your attention. The region below 1500 cm⁻¹ is the fingerprint region, which contains many overlapping absorptions and is unique to each compound; it is useful for identification by comparison with a database, but it is not usually used for deduction in exams.

约 1500 cm⁻¹ 以上的区域称为官能团区，这是你应该重点关注的地方。1500 cm⁻¹ 以下的区域是指纹区，包含许多重叠吸收，对每种化合物都是独一无二的；它适合通过与数据库比对来进行鉴定，但在考试中通常不用于结构推断。

c acid; a C=O near 1700 cm⁻¹ with no broad O-H and with an aldehyde C-H pair near 2750 cm⁻¹ indicates an aldehyde; a C=O near 1735-1750 cm⁻¹ with strong C-O absorptions and no O-H indicates an ester.

鉴别官能团靠的是吸收的组合，而不是单个峰。宽 O-H 带加上 1700 cm⁻¹ 附近的强 C=O 表明羧酸；1700 cm⁻¹ 附近有 C=O，没有宽 O-H，但在 2750 cm⁻¹ 附近有一对醛 C-H 峰，表明醛；1735-1750 cm⁻¹ 附近有 C=O，配合强 C-O 吸收且无 O-H，表明酯。

it is not an aldehyde, ketone, acid, ester or amide. If there is a C=O but no C-O absorption, an ester is ruled out. Always state the wavenumber and the bond responsible, for example: an absorption at about 3300 cm⁻¹ due to the O-H bond.

两个“否定性”观察也能得分。如果 1680-1750 cm⁻¹ 区域没有吸收，则该化合物不是羰基化合物，因此不是醛、酮、酸、酯或酰胺。如果有 C=O 却看不到 C-O 吸收，则可排除酯。作答时始终要写出波数与对应化学键，例如：在约 3300 cm⁻¹ 处有一个由 O-H 键引起的吸收。

6. ¹H NMR: Environments, Integration and Splitting | ¹H NMR：氢环境、积分与裂分

h environment is split by its neighbours. The positions of the peaks are measured relative to tetramethylsilane, TMS, which is assigned δ = 0 ppm.

质子核磁共振告诉你三件事：有多少种不同的氢环境、每种环境中有多少个氢原子、以及每种环境被相邻氢如何裂分。峰的位置相对于四甲基硅烷（TMS）测量，TMS 被定为 δ = 0 ppm。

TMS is used as the standard because it gives a single sharp peak, it is chemically inert and non-toxic, it is easily removed because it is volatile, and its twelve equivalent hydrogens give a strong signal.

使用 TMS 作为标准，是因为它给出单一尖峰、化学惰性且无毒、易挥发因而容易除去，并且它的十二个等效氢给出强信号。

Hydrogen environment	Chemical shift δ / ppm
TMS standard	0
R-CH₃	0.7-1.2
R-CH₂-R	1.2-1.6
H-C-O / H-C-Cl	3.1-4.2
Alkene H (C=C-H)	4.5-6.0
Aromatic H	style="border:1px solid # 广告 / Sponsored Pearson Edexcel International AS/A Level 培生爱德思 Alevel 考试 Ch… 化学 备考教辅 · 正版现货 · 与本文内容相关 查看详情 / View on Taobao参考价 ¥59.2 更多咨询请联系16621398022（同微信） 备考资讯每日更新，关注不错过 考纲变化 · 科目分析 · 学习方法 在微信里长按二维码识别关注 或截图后用「扫一扫 → 相册」选择本图 公众号：tutorhao