Combined Analytical Techniques: IR, Mass Spectrometry and NMR | 综合分析技术：红外光谱、质谱与核磁共振

📚 Combined Analytical Techniques: IR, Mass Spectrometry and NMR | 综合分析技术：红外光谱、质谱与核磁共振

In Edexcel A-Level Chemistry, determining the structure of an unknown organic compound is a core skill. No single spectroscopic technique can always provide the full picture. Mass spectrometry gives the relative molecular mass and fragmentation clues, infrared spectroscopy identifies functional groups, and NMR spectroscopy reveals the carbon-hydrogen framework. By combining these three techniques, chemists can piece together the exact molecular structure, just like solving a puzzle.

在 Edexcel A-Level 化学中，确定未知有机化合物的结构是一项核心技能。没有一种波谱技术总能提供完整信息。质谱给出相对分子质量和碎片线索，红外光谱识别官能团，核磁共振波谱揭示碳氢骨架。通过综合运用这三种技术，化学家可以像拼图一样拼出确切的分子结构。

1. The Role of Combined Techniques in Structure Determination | 综合分析技术在结构测定中的作用

Combined analytical techniques are essential because each method provides only part of the structural story. A mass spectrum can tell you the molecular mass, but not how the atoms are connected. An IR spectrum can confirm a carbonyl group, but not whether it is an aldehyde or a ketone. NMR can reveal the number and types of hydrogen environments, but without the molecular formula from MS, you might still be guessing. Using all three together lets you build a consistent structural formula.

综合分析技术之所以必不可少，是因为每种方法只提供结构信息的一部分。质谱可以告诉你分子质量，但无法说明原子如何连接。红外光谱可以确认羰基，但不能区分醛或酮。核磁共振可以揭示氢环境的数目和类型，但如果没有质谱给出的分子式，你仍然可能在猜测。将三者结合使用，才能构建出一致的结构式。

In exam questions, you will usually be given two or three spectra for the same unknown compound. Your job is to extract the evidence systematically. Start with the molecular mass, then identify functional groups, and finally map out the carbon-hydrogen skeleton using NMR. This logical approach reduces errors and earns method marks.

在考试题中，通常会给出同一未知化合物的两到三张谱图。你的任务是系统地提取证据。先从分子质量入手，然后识别官能团，最后利用 NMR 描绘碳氢骨架。这种逻辑方法可以减少错误并获得步骤分。

2. Mass Spectrometry: Principles and Molecular Ion | 质谱：原理与分子离子

In a mass spectrometer, molecules are ionised, usually by electron impact, and then accelerated through a magnetic field. The ions are separated according to their mass-to-charge ratio (m/z). The molecular ion peak, often labelled M⁺, corresponds to the whole molecule after losing one electron. Its m/z value gives the relative molecular mass (Mr) of the compound.

在质谱仪中，分子通常通过电子轰击被电离，然后在磁场中加速。离子根据其质荷比 (m/z) 进行分离。分子离子峰，通常标记为 M⁺，对应失去一个电子后的整个分子。其 m/z 值给出了化合物的相对分子质量 (Mr)。

High-resolution mass spectrometry (HRMS) can measure m/z values to several decimal places, allowing the molecular formula to be determined directly from the accurate mass. For example, CO and N₂ both have nominal Mr 28, but their accurate masses differ, so HRMS can distinguish them. In normal low-resolution spectra, only integer m/z values are shown.

高分辨率质谱 (HRMS) 可以将 m/z 值测量到小数点后几位，从而根据精确质量直接确定分子式。例如，CO 和 N₂ 的名义 Mr 都是 28，但它们的精确质量不同，因此 HRMS 可以区分它们。在普通的低分辨率谱图中，只能看到整数 m/z 值。

3. Fragmentation Patterns in Mass Spectra | 质谱中的碎片模式

When the molecular ion is formed, it often has excess internal energy and can break apart into smaller fragments. The most abundant fragment gives the base peak, assigned 100 % intensity. Common fragmentation peaks include m/z 15 (CH₃⁺), m/z 29 (C₂H₅⁺ or CHO⁺), and m/z 43 (C₃H₇⁺ or CH₃CO⁺). Recognising these peaks helps identify alkyl chains or carbonyl groups.

分子离子形成时，往往带有过剩的内能，会分解成较小的碎片。丰度最高的碎片给出基峰，强度定为 100 %。常见的碎片峰包括 m/z 15 (CH₃⁺)、m/z 29 (C₂H₅⁺ 或 CHO⁺) 和 m/z 43 (C₃H₇⁺ 或 CH₃CO⁺)。识别这些峰有助于确定烷基链或羰基。

• m/z 15: CH₃⁺ | m/z 15：CH₃⁺
• m/z 29: C₂H₅⁺ or CHO⁺ | m/z 29：C₂H₅⁺ 或 CHO⁺
• m/z 43: C₃H₇⁺ or CH₃CO⁺ | m/z 43：C₃H₇⁺ 或 CH₃CO⁺
• m/z 77: C₆H₅⁺, a phenyl cation from aromatic compounds | m/z 77：C₆H₅⁺，芳香化合物的苯基阳离子

A fragmentation pattern can also help distinguish isomers. For example, a peak at m/z 71 suggests a loss of CH₃ from a C₄H₉ chain, which is common in branched alkanes. You do not need to memorise every fragment, but you should be able to interpret simple cleavages at functional group positions.

碎片模式也可以帮助区分同分异构体。例如，m/z 71 处的峰表明 C₄H₉ 链失去 CH₃，这在支链烷烃中很常见。你不需要记住所有碎片，但应该能够解释官能团位置附近的简单断裂。

4. Infrared Spectroscopy: How Bonds Absorb IR Radiation | 红外光谱：化学键如何吸收红外辐射

Infrared spectroscopy exploits the fact that covalent bonds vibrate at characteristic frequencies. When IR radiation of the same frequency is absorbed, the bond stretches or bends more energetically. The spectrum plots percentage transmittance against wavenumber (cm⁻¹). The absorptions appear as downward peaks. The region most useful for organic analysis is roughly 4000–1500 cm⁻¹, where functional groups show distinctive bands.

红外光谱利用共价键以特征频率振动这一事实。当相同频率的红外辐射被吸收时，化学键会以更高的能量伸缩或弯曲。光谱图以百分透过率对波数 (cm⁻¹) 作图。吸收表现为向下的峰。对有机分析最有用的区域大约在 4000–1500 cm⁻¹，官能团在该区域显示出特征谱带。

The fingerprint region below about 1500 cm⁻¹ is complex and unique to each molecule. It can be used to identify a compound by comparison with reference spectra, but in A-Level exams you are normally expected to focus on the functional group region above 1500 cm⁻¹.

大约 1500 cm⁻¹ 以下的指纹区非常复杂，对每个分子都是唯一的。通过与参考谱图对比，它可以用于鉴定化合物。但在 A-Level 考试中，通常只要求关注 1500 cm⁻¹ 以上的官能团区域。

5. Key IR Absorptions for Common Functional Groups | 常见官能团的主要红外吸收

You must be able to match absorption ranges to functional groups. A broad O-H stretch in alcohols or carboxylic acids appears at about 2500–3550 cm⁻¹. The C=O stretch in aldehydes, ketones, carboxylic acids and esters appears at about 1680–1750 cm⁻¹. C-H stretches in alkanes occur just below 3000 cm⁻¹, while C=C stretches in alkenes appear around 1620–1680 cm⁻¹.

你必须能够将吸收范围与官能团匹配。醇或羧酸中的 O-H 伸缩振动为宽峰，大约在 2500–3550 cm⁻¹。醛、酮、羧酸和酯中的 C=O 伸缩振动大约在 1680–1750 cm⁻¹。烷烃中的 C-H 伸缩振动出现在 3000 cm⁻¹ 以下，而烯烃中的 C=C 伸缩振动约在 1620–1680 cm⁻¹。

Aldehydes and ketones can sometimes be distinguished by C-H stretches: aldehydes often show two weak C-H stretches near 2720 and 2820

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