Infrared Spectroscopy for IB Chemistry: Key Points Explained | IB 化学：红外光谱考点精讲

📚 Infrared Spectroscopy for IB Chemistry: Key Points Explained | IB 化学：红外光谱考点精讲

Infrared spectroscopy is a powerful analytical technique used in IB Chemistry to identify functional groups in organic molecules by measuring the absorption of infrared radiation. This article provides a concise yet comprehensive review of key concepts, including molecular vibrations, characteristic absorption bands, and spectral interpretation, tailored for IB exam success.

红外光谱是IB化学中用于鉴定有机分子官能团的一种强大分析技术，通过测量红外辐射的吸收来实现。本文简明扼要地全面复习了核心概念，包括分子振动、特征吸收带和谱图解析，旨在帮助学生在IB考试中取得成功。

1. Introduction to Infrared Spectroscopy | 红外光谱简介

Infrared (IR) spectroscopy exploits the interaction between infrared radiation and matter. Molecules absorb IR radiation at specific frequencies, causing bonds to vibrate. The resulting spectrum provides a ‘molecular fingerprint’ that can be used to determine the functional groups present in an unknown substance.

红外光谱利用红外辐射与物质的相互作用。分子在特定频率下吸收红外辐射，引起化学键振动。产生的光谱提供了一种“分子指纹”，可用于确定未知物质中存在的官能团。

The IR region of interest for organic analysis lies approximately between 4000 and 400 cm⁻¹. The spectrum is usually presented as percent transmittance versus wavenumber, where each downward peak corresponds to the absorption of energy by a particular vibrational mode.

用于有机分析的红外区域大致在4000至400 cm⁻¹之间。光谱图通常以透过率百分比对波数作图，图中每个向下的峰对应于特定振动模式对能量的吸收。

2. Molecular Vibrations and Absorption of IR Radiation | 分子振动与红外辐射吸收

For a molecule to absorb IR radiation, the vibration must result in a change in the dipole moment of the bond. Symmetric stretches of non-polar bonds (e.g., O₂, N₂) are IR-inactive because no dipole change occurs. Conversely, the asymmetric stretching and bending vibrations of CO₂ are IR-active since they do alter the overall dipole moment.

分子要吸收红外辐射，振动必须引起键偶极矩的改变。非极性键的对称伸缩振动（如O₂、N₂）由于没有偶极矩变化，是红外非活性的。相反，CO₂的不对称伸缩和弯曲振动会改变总偶极矩，因此是红外活性的。

The frequency of a bond’s vibration can be approximated using Hooke’s law, which relates it to the bond strength (force constant k) and the reduced mass μ of the two atoms:

键的振动频率可以用胡克定律近似描述，该定律将频率与键强度（力常数k）以及两个原子的约化质量μ联系起来：

ν̃ = 1/(2πc) · √(k/μ)

where c is the speed of light. Stronger bonds (larger k) and lighter atoms (smaller μ) lead to higher vibrational frequencies and therefore higher wavenumber absorptions.

其中c为光速。更强的键（k更大）和更轻的原子（μ更小）导致更高的振动频率，从而在更高波数处产生吸收。

3. Types of Vibrations: Stretching and Bending | 振动类型：伸缩与弯曲

Molecular vibrations are broadly classified into stretching (change in bond length) and bending (change in bond angle). Stretching can be symmetric or asymmetric, while bending includes scissoring, rocking, wagging, and twisting. Generally, stretching vibrations absorb at higher wavenumbers (above ~1500 cm⁻¹) than bending vibrations.

分子振动大致分为伸缩振动（键长变化）和弯曲振动（键角变化）。伸缩可以是称的或不对称的，而弯曲振动包括剪式、摇摆、面外摇摆和扭曲。通常，伸缩振动比弯曲振动在更高波数（约1500 cm⁻¹以上）处吸收。

A single molecule like water shows an asymmetric stretch at ~3756 cm⁻¹, a symmetric stretch at ~3657 cm⁻¹, and a bending vibration at ~1595 cm⁻¹. Recognizing these modes helps in interpreting complex spectra.

以水分子为例，其不对称伸缩振动在~3756 cm⁻¹，对称伸缩在~3657 cm⁻¹，弯曲振动在~1595 cm⁻¹。识别这些振动模式有助于解析复杂光谱。

4. The Infrared Spectrum: Transmittance and Wavenumber | 红外光谱图：透过率与波数

An IR spectrum plots % transmittance (or absorbance) against wavenumber in cm⁻¹, typically ranging from 4000 to 400 cm⁻¹. Peaks pointing downward represent absorption bands. The horizontal axis is directly proportional to frequency and energy, and inversely proportional to wavelength.

红外光谱图以透过率百分比（或吸光度）对波数（cm⁻¹）作图，通常范围为4000至400 cm⁻¹。向下的峰代表吸收带。横轴与频率和能量成正比，与波长成反比。

The region above 1500 cm⁻¹ is used to identify specific functional groups through characteristic stretching frequencies. The region below 1500 cm⁻¹ is the ‘fingerprint region’. Always pay attention to the shape (broad or sharp) and relative intensity of the peaks.

1500 cm⁻¹以上的区域用于通过特征伸缩频率识别特定官能团。1500 cm⁻¹以下的区域是“指纹区”。务必注意峰的峰形（宽或尖）和相对强度。

5. Characteristic Absorption Bands for Functional Groups | 官能团的特征吸收带

The table below summarises the most important IR absorption bands you need to know for the IB examination. Recognising these will allow you to deduce the functional groups present in an unknown compound.

下表总结了IB考试中需要掌握的最重要的红外吸收带。识别这些吸收带将有助于推断未知化合物中存在的官能团。

Functional Group	Bond	Wavenumber (cm⁻¹)	Intensity / Remarks
O–H (alcohol)	O–H stretch	3200–3600	Strong, broad
O–H (carboxylic acid)	O–H stretch	2500–3300	Very broad, overlaps C–H
N–H (amine, amide)	N–H stretch	3300–3500	Medium; primary amines show 2 peaks
C–H (alkane)	C–H stretch	2850–2950	Medium to strong
=C–H (alkene/arene)	C–H stretch	3000–3100	Weak to medium
C=O (carbonyl)	C=O stretch	1700–1750	Strong, sharp
C=C (alkene)	C=C stretch	1620–1680	Variable, weaker for symmetrical
C≡C (alkyne)	C≡C stretch	2100–2260	Sharp; absent if symmetrical
C≡N (nitrile)	C≡N stretch	2220–2260	Medium, sharp
C–O (alcohol, ether, ester)	C–O stretch	1000–1300	Strong

Note that carboxylic acids exhibit both a very broad O–H band and a slightly lower C=O stretch (around 1700 cm⁻¹). Esters show C=O around 1735–1750 cm⁻¹ and strong C–O absorption. Aldehydes often give two weak C–H stretches around 2720 and 2820 cm⁻¹ for the aldehyde hydrogen.

请注意，羧酸同时表现出非常宽的O–H谱带和稍低的C=O伸缩峰（约1700 cm⁻¹）。酯的C=O在1735–1750 cm⁻¹附近，并伴有强C–O吸收。醛的醛基氢通常在2720和2820 cm⁻¹附近给出两个弱C–H伸缩峰。

6. Key Absorption Regions: The Fingerprint Region | 关键吸收区域：指纹区

The region below 1500 cm⁻¹ is called the fingerprint region. It contains a complex pattern of absorptions that is unique to each individual molecule, much like a human fingerprint. While it is difficult to assign every peak here, the fingerprint region is extremely useful for confirming the identity of a compound by direct comparison with a reference spectrum.

1500 cm⁻¹以下的区域称为指纹区。该区域包含每个分子特有的复杂吸收图样，类似于人的指纹。虽然很难指认此区域的每个峰，但指纹区通过与标准谱图直接比对，对于确认化合物的身份极为有用。

In an IB exam setting, you are not expected to interpret the fingerprint region in detail. Instead, focus on the functional group region above 1500 cm⁻¹. However, be aware that two isomers will show differences in their fingerprint patterns even if their functional group absorptions are similar.

在IB考试中，你不需要详细解析指纹区。应重点关注1500 cm⁻¹以上的官能团区域。但要明白，即使两个异构体的官能团吸收相似，它们的指纹区图样也会不同。

7. Factors Affecting Absorption Frequencies | 影响吸收频率的因素

Several factors shift the position of an absorption band:

有几个因素会使吸收带的位置发生移动：

Bond strength: Triple bonds (C≡C) are stronger than double bonds (C=C), which are stronger than single bonds (C–C). Hence, C≡C absorbs at ~2100–2260 cm⁻¹, C=C at ~1620–1680 cm⁻¹, and C–C at lower frequencies that fall in the fingerprint region.

键强度：三键（C≡C）比双键（C=C）强，双键又比单键（C–C）强。因此C≡C在~2100–2260 cm⁻¹吸收，C=C在~1620–1680 cm⁻¹吸收，而C–C则在较低频率（落入指纹区）吸收。

Reduced mass: Lighter atoms vibrate faster. Compare C–H (~3000 cm⁻¹) with C–D (~2200 cm⁻¹) or O–H (~3400 cm⁻¹) with O–D (~2500 cm⁻¹). This isotopic shift is consistent with Hooke’s law.

约化质量：原子越轻，振动越快。比较C–H（~3000 cm⁻¹）与C–D（~2200 cm⁻¹），以及O–H（~3400 cm⁻¹）与O–D（~2500 cm⁻¹）。这种同位素位移符合胡克定律。

Hydrogen bonding: Intermolecular hydrogen bonds weaken the O–H or N–H bond, leading to broadening and shifting to lower wavenumbers. For example, a free O–H stretch appears sharp at ~3600 cm⁻¹, whereas hydrogen-bonded O–H in alcohols gives a broad band around 3300–3400 cm⁻¹. In carboxylic acids, extremely strong hydrogen bonding pushes the O–H absorption as low as 2500 cm⁻¹ and makes it very wide.

氢键：分子间氢键会弱化O–H或N–H键，导致峰变宽并向低波数移动。例如，游离的O–H伸缩在~3600 cm⁻¹处呈尖峰，而醇中形成氢键的O–H则在3300–3400 cm⁻¹附近呈现宽峰。羧酸中极强的氢键作用将O–H吸收推低至2500 cm⁻¹，并使其非常宽。

Conjugation and resonance: Conjugation of a C=O group with a C=C bond or an aromatic ring lowers the double-bond character, shifting the C=O stretch to a lower wavenumber (e.g., around 1680 cm⁻¹ for conjugated ketones vs. 1715 cm⁻¹ for unconjugated ones). Electron-withdrawing groups can increase the frequency.

共轭与共振：C=O基团与C=C键或芳环共轭会降低双键性质，使C=O伸缩向低波数移动（例如共轭酮在~1680 cm⁻¹，而非共轭酮在~1715 cm⁻¹）。吸电子基团可使频率升高。

8. Interpreting an IR Spectrum: Step-by-Step Guide | 解读红外光谱：分步指南

Step 1: Look for a broad, strong O–H absorption in the 3200–3600 cm⁻¹ region. A very broad peak that extends to 2500 cm⁻¹ suggests a carboxylic acid. A moderate broad peak around 3300 cm⁻¹ suggests an alcohol. If you also see a strong C=O peak, think carboxylic acid. If you see C–O but no C=O, it is likely an alcohol or ether.

步骤1：在3200–3600 cm⁻¹区域寻找宽而强的O–H吸收。若宽峰延伸至2500 cm⁻¹，提示为羧酸。若在3300 cm⁻¹附近有中等宽峰，提示为醇。如果同时看到强C=O峰，则考虑羧酸。若有C–O而无C=O，则可能为醇或醚。

Step 2: Check for N–H stretches around 3300–3500 cm⁻¹. Primary amines show two peaks (symmetric and asymmetric stretching), while secondary amines show one. Amides also exhibit a C=O stretch, so the combination helps distinguish them.

步骤2：检查3300–3500 cm⁻¹附近的N–H伸缩。伯胺显示两个峰

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Infrared Spectroscopy for IB Chemistry: Key Points Explained | IB 化学：红外光谱 考点精讲