📚 GCSE CIE Chemistry: Infrared Spectroscopy | GCSE CIE 化学：红外光谱 考点精讲
the most useful analytical tools in organic chemistry. By measuring how strongly a sample absorbs infrared radiation at different wavenumbers, chemists can work out which covalent bonds are present in a molecule, such as O– hemistry students, learning to read a simple IR spectrum is an essential skill for tackling questions on organic analysis questions. This article covers all the key concepts, a step-by-step interpretation method, worked examples, the full data table you need for exams, and the common pitfalls that cost students marks.
红外光谱是有机化学中最常用的分析工具之一。通过测量样品在不同波数处对红外辐射的吸收强度，化学家可以判断分子中存在哪些共价键，例如 O–H、C=O 和 C=C。对于 CIE IGCSE 与 GCSE 化学考生来说，学会解读简单的红外光谱是有机分析题的必备技能。本文涵盖全部核心概念、逐步解析方法、典型例题、考试必备数据表，以及最容易失分的常见陷阱。
1. What is Infrared Spectroscopy? | 什么是红外光谱？
Infrared (IR) spectroscopy is an analytical technique that exploits the fact that covalent bonds in molecules absorb infrared energy at characteristic frequencies. When IR radiation passes through a sample, certain bonds vibrate more vigorously if the frequency of the radiation matches the natural vibration frequency of that bond. The resulting absorption pattern is plotted as a spectrum, and that spectrum reveals which functional groups the molecule contains.
红外（IR）光谱是一种分析技术，它利用分子中的共价键在特征频率处吸收红外能量的性质。当红外辐射穿过样品时，如果辐射频率与某个键的固有振动频率匹配，该键就会更剧烈地振动。所得的吸收图形被绘制成光谱，而这张光谱可以揭示分子中含有哪些官能团。
In CIE IGCSE and GCSE Chemistry papers you will be given an IR spectrum or a table of absorption data, about absorptions and asked to identify the presence of alcohols, carboxylic acids, alkenes, carbonyl compounds or other functional groups. You are not expected to explain the quantum mechanics of the absorption - you are expected to read the data, match the absorption value to a bond, and name the functional group correctly.
在 CIE IGCSE 与 GCSE 化学试卷中，题目会给你一张红外光谱图或一组吸收数据，要求你判断分子中是否存在醇、羧酸、烯烃、羰基化合物或其他官能团。你不需要解释吸收背后的量子力学原理——你需要做的是读取数据、把吸收值与化学键对应起来，并准确说出官能团的名称。
9701). At IGCSE level, focus on identifying a small number of very distinctive absorptions, especially O–H and C=O.
考纲提示：红外光谱属于有机分析的一部分，在剑桥国际 A Level（9701）中会考得更深入。在 IGCSE 阶段，重点掌握少数几个极其典型的吸收峰，尤其是 O–H 和 C=O。
2. The Physics Behind the Absorption: Bond Vibrations | 吸收背后的原理：化学键的振动
A covalent bond behaves rather like a spring connecting two atoms. The bond can stretch (the atoms move further apart and closer together along the bond axis) and it can bend (the angle between bonds changes). Each of these vibrations has its own natural frequency, which depends on the masses of the atoms and the strength of the bond.
共价键的行为很像连接两个原子的弹簧。化学键可以伸缩（原子沿键轴方向相互靠近和远离），也可以弯曲（键与键之间的夹角发生变化）。每一种振动都有其固有频率，该频率取决于原子的质量和键的强度。
When infrared radiation of exactly the right energy hits the bond, the bond absorbs that energy and vibrates with a larger amplitude. If the energy does not match, the radiation passes straight through and is not absorbed. This is why the spectrum shows discrete absorption peaks rather than a continuous blur.
当恰好具有合适能量的红外辐射照射到该键上时，键会吸收这部分能量并以更大的振幅振动。如果能量不匹配，辐射就会直接穿过样品而不被吸收。这正是光谱上出现的是一个个离散吸收峰、而不是一片连续模糊的原因。
change in the dipole moment of the bond. In practice this means that polar bonds such as O–H, C=O and C– at all. This explains why the C=O peak is always strong while the C=C peak is often weak.
只有当振动引起键的偶极矩发生变化时，才会发生吸收。实际表现为：O–H、C=O、C–O 等极性键给出强吸收峰，而像乙烯中完全对称的非极性 C=C 键则吸收很弱甚至不吸收。这就解释了为什么 C=O 峰总是很强，而 C=C 峰往往很弱。
E = hν = hc ÷ λ
Each absorption corresponds to a specific energy gap ΔE between vibrational energy levels, and that energy gap is fixed for a given bond in a given environment.
每一个吸收峰都对应振动能级之间一个特定的能量差 ΔE，而对于处在特定化学环境中的某个键，这个能量差是固定的。
3. Wavenumber: The Unit You Must Know | 波数：必须掌握的单位
IR spectra are almost always plotted against wavenumber rather than wavelength or frequency. The wavenumber is simply the reciprocal of the wavelength, expressed in cm⁻¹. It is used because the numbers are convenient and because wavenumber is directly proportional to energy.
红外光谱几乎总是以波数而不是波长或频率作为横坐标。波数就是波长的倒数，单位为 cm⁻¹。使用波数是因为数值方便，而且波数与能量成正比。
wavenumber = 1 ÷ λ (unit: cm⁻¹)
A larger wavenumber means a higher frequency, a shorter wavelength and a higher energy. So a bond that absorbs at 3300 cm⁻¹ is absorbing much more energetic radiation than one absorbing at 1000 cm⁻¹.
波数越大，表示频率越高、波长越短、能量越大。因此，在 3300 cm⁻¹ 处吸收的键，所吸收的辐射能量远高于在 1000 cm⁻¹ 处吸收的键。
In school-level IR spectroscopy the useful range runs from about 4000 cm⁻¹ down to 400 cm⁻¹. Almost every functional group you need to identify absorbs somewhere between 4000 cm⁻¹ and 1000 cm⁻¹, so this is the region worth memorising.
在学校层面的红外光谱中，有用的范围大约是从 4000 cm⁻¹ 到 400 cm⁻¹。你需要识别的几乎每一个官能团都在 4000 cm⁻¹ 到 1000 cm⁻¹ 之间产生吸收，因此这一段范围最值得记忆。
4. Reading the Axes of an IR Spectrum | 看懂红外光谱的坐标轴
The horizontal axis is wavenumber in cm⁻¹, and it is drawn with the values decreasing from left to right - 4000 on the left, 400 on the right. Many students lose marks simply because they misread the direction of the axis. The vertical axis is percentage transmittance, so a strong absorption appears as a deep downward peak.
横坐标是波数，单位为 cm⁻¹，其数值从左到右递减——左边是 4000，右边是 400。很多学生仅仅因为看反了坐标轴方向而失分。纵坐标是透过率百分比，因此强吸收表现为一个向下的深峰。
The top of the chart represents 100 percent transmittance, which means no absorption at all - the sample is transparent to that wavenumber. A peak that drops all the way towards the baseline means nearly all of that radiation has been absorbed.
图表顶端代表 100% 透过率，意味着完全没有吸收——样品对该波数是透明的。如果一个峰一直下探到接近基线，说明该波数处的辐射几乎被完全吸收。
When describing a peak in an exam answer, always give three pieces of information: the wavenumber range, the intensity (strong, medium or weak) and the shape (sharp or broad). For example: "a strong, broad absorption between 3200 and 3600 cm⁻¹, showing the presence of an O–H group in an alcohol."
在考试作答时描述一个峰，一定要给出三条信息：波数范围、强度（强、中、弱）和形状（尖锐或宽）。例如："在 3200 至 3600 cm⁻¹ 之间有一个强而宽的吸收峰，表明存在醇中的 O–H 基团。"
5. The Characteristic Absorption Data Table | 特征吸收数据表
The table below lists the absorptions you are most likely to be asked about. Learn the bold entries first: they are the ones examiners use most often.
下表列出了最常考的吸收数据。请优先记住加粗的条目——它们是命题人使用频率最高的。
| Bond | Wavenumber / cm⁻¹ | Appearance |
| O–H (alcohol) | 3200–3600 | strong, broad |
| O–H (carboxylic acid) | 2500–3300 | very broad, often overlapping C–H |
| N–H (amine, amide) | 3300–3500 | medium, sharp-ish |
| C–H (alkane) | 2850–3000 | strong, sharp |
| C–H (alkene, arene) | 3000–3100 | medium, above 3000 |
| C–H (aldehyde) | 2700–2800 | two weak sharp peaks |
| C≡C (alkyne) | 2100–2260 | weak |
| C≡N (nitrile) | 2220–2260 | medium, sharp |
| C=O (aldehyde) | 1720–1740 | strong, sharp |
| C=O (ketone) | 1705–1725 | strong, sharp |
| C=O (carboxylic acid) | 1700–1725 | strong, sharp |
| C=O (ester) | 1735–1750 | strong, sharp |
| C=C (alkene) | 1620–1680 | weak to medium |
| C–O (alcohol, ester, ether) | 1000–1300 | strong |
e carbonyl group is attached to, which is why an ester C=O appears at a slightly higher wavenumber than a ketone C=O.
注意，所有 C=O 的吸收都集中在大约 1700 至 1750 cm⁻¹ 这样一个很窄的区间内。具体位置会根据羰基所连的基团略有偏移，这就是酯的 C=O 波数略高于酮的 C=O 的原因。
6. A Step-by-Step Method for Interpreting Any Spectrum | 解读任何光谱的逐步方法
Follow these five steps every time, and you will rarely miss a functional group.
每次都按以下五个步骤操作，你几乎不会漏掉任何官能团。
- Step 1: Look above 3000 cm⁻¹ first. Is there a broad absorption? Broad O–H means alcohol; an extremely broad absorption spreading from about 2500 to 3300 cm⁻¹ means carboxylic acid.
- Step 2: Look at 1700–1750 cm⁻¹. A strong, sharp peak here means a C=O group is present.
- Step 3: Check 2700–0 cm⁻¹. Two weak peaks here, combined with a C=O at 1720 cm⁻¹ or above, means an aldehyde.
- Step 4: Check 1620–1680 cm⁻¹ for a weak C=C, and 3000–3100 cm⁻¹ for an alkene C–H just above 3000.
- Step 5: Combine the evidence. Never name a functional group from a single peak if the question expects you to compare two spectra.
- 第一步：先看 3000 cm⁻¹ 以上。有没有宽峰？宽而强的 O–H 说明是醇；从 2500 一直到 3300 cm⁻¹ 都极度宽阔的吸收说明是羧酸。
- 第二步：看 1700–1750 cm⁻¹。此处出现强而尖的峰，说明存在 C=O。
- 第三步：检查 2700–2800 cm⁻¹。此处有两个弱峰，同时在 1720 cm⁻¹ 或更高处有 C=O，说明是醛。
- 第四步：检查 1620–1680 cm⁻¹ 是否有弱的 C=C，以及 3000–3100 cm⁻¹ 是否有略高于 3000 的烯烃 C–H。
- 第五步：综合所有证据。如果题目要求比较两张光谱，绝不能只凭一个峰就下结论。
A useful habit is to annotate the spectrum before writing your answer: draw a small arrow to each important peak and label it with the bond it represents. This keeps your reasoning organised and shows the examiner your working.
一个很实用的习惯是：在写答案之前先在光谱图上做标注——给每个重要峰画一个小箭头，并写上它代表的化学键。这样能让你的推理更有条理，也让阅卷老师看到你的思路。
7. Worked Examples | 典型例题精讲
Example 1: Compound A has the molecular formula C₂H₆O and shows a strong, broad absorption at 3200–3600 cm⁻¹ but no absorption in the 1700–1750 cm⁻¹ region. Identify the functional group and suggest the structure.
例题 1：化合物 A 的分子式为 C₂H₆O，在 3200–3600 cm⁻¹ 处有强而宽的吸收峰，但在 1700–1750 cm⁻¹ 区域没有吸收。请判断其官能团并推测结构。
Answer: the broad absorption above 3200 cm⁻¹ is the O–H stretch of an alcohol, and the absence of a peak near 1715 cm⁻¹ rules out a carbonyl group. The compound is therefore ethanol, CH₃CH₂OH. Note that the molecular formula C₂H₆O could also describe dimethyl ether, CH₃OCH₃, which is an ether - it has no O–H bond, so it would show no broad absorption above 3000 cm⁻¹. The IR spectrum therefore distinguishes the two isomers.
答案：3200 cm⁻¹ 以上的宽峰是醇的 O–H 伸缩振动，而 1715 cm⁻¹ 附近没有峰则排除了羰基。因此该化合物是乙醇 CH₃CH₂OH。请注意，分子式 C₂H₆O 也可能对应二甲醚 CH₃OCH₃，它是一种醚——它没有 O–H 键，因此在 3000 cm⁻¹ 以上不会出现宽峰。所以红外光谱可以把这两个同分异构体区分开来。
Example 2: Compound B shows a strong sharp peak at 1715 cm⁻¹, a broad absorption spreading from 2500 to 3300 cm⁻¹, and a strong peak at 1200 cm⁻¹. Identify the compound class.
例题 2：化合物 B 在 1715 cm⁻¹ 处有强而尖的峰，在 2500 至 3300 cm⁻¹ 有很宽的吸收，并在 1200 cm⁻¹ 有强峰。请判断它属于哪一类化合物。
ndicates C=O; the very broad absorption between 2500 and 3300 cm⁻¹ is the characteristic O–H of a carboxylic acid, which is broader than an alcohol O– because of strong hydrogen bonding; the strong peak around 1200 cm⁻¹ corresponds to the C–O single bond. Compound B is a carboxylic acid, for example ethanoic acid, CH₃COOH.
答案：1715 cm⁻¹ 的峰表明有 C=O；2500 至 3300 cm⁻¹ 的极宽吸收是羧酸 O–H 的特征峰，由于存在强氢键，它比醇的 O–H 更宽；1200 cm⁻¹ 附近的强峰对应 C–O 单键。因此化合物 B 是羧酸，例如乙酸 CH₃COOH。
Example 3: A student says "there is a peak at 1650 cm⁻¹, so the compound must be a ketone." Explain why this statement is unsafe.
例题 3：某学生说"在 1650 cm⁻¹ 处有一个峰，所以这个化合物一定是酮"。请解释为什么这个说法不可靠。
Answer: 1650 cm⁻¹ lies below the normal ketone C=O range of 1705–1725 cm⁻¹, but within the C=C alkene range of 1620– rom an amide C=O or from absorbed water. The student must also check whether there is an O–H absorption and where exactly the peak sits, rather than relying on a single number.
答案：1650 cm⁻¹ 低于酮 C=O 的正常范围 1705–1725 cm⁻¹，但落在烯烃 C=C 的 1620–1680 cm⁻¹ 范围之内。这一区域的峰也可能来自酰胺的 C=O 或来自吸收的水分。学生还必须检查是否存在 O–H 吸收，以及峰的确切位置，而不能只凭一个数字下结论。
8. Distinguishing Common Pairs of Compounds | 常见化合物对的区分
Exam questions very often give you two spectra and ask which belongs to which compound. These are the classic pairs.
考试题经常给你两张光谱，问哪一张属于哪个化合物。以下是几组经典搭配。
| Pair | Key difference in the IR spectrum |
| Ethanol vs ethanoic acid | Both have broad O–H, but the acid also has a strong C=O at about 1710 cm⁻¹ and its O–H is much broader
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