📚 IR Spectroscopy Exam Focus for CIE A-Level Chemistry | A-Level CIE 化学：红外光谱 考点精讲
Infrared (IR) spectroscopy is a powerful analytical tool that probes the vibrations of chemical bonds. When a molecule absorbs infrared radiation, its bonds stretch and bend at characteristic frequencies. The resulting spectrum acts as a molecular fingerprint, enabling chemists to identify functional groups and deduce the structure of unknown compounds. For CIE A-Level Chemistry, you must be able to interpret IR spectra, correlate absorption bands with specific bond vibrations, and explain how modern breathalysers and greenhouse gas monitoring rely on this technique.
红外光谱是一种强大的分析工具，它探测化学键的振动。当分子吸收红外辐射时，其化学键会以特征频率发生伸缩和弯曲振动。所得光谱就像分子的指纹，使化学家能够识别官能团并推断未知化合物的结构。对于 CIE A-Level 化学，你必须能够解读红外光谱、将吸收峰与特定的键振动关联起来，并解释现代酒精呼气测试仪和温室气体监测如何依赖这一技术。
1. The Principle of IR Spectroscopy | 红外光谱的基本原理
Covalent bonds are not rigid rods; they behave like tiny springs that can vibrate. These vibrations include stretching (symmetrical and asymmetrical) and bending (scissoring, rocking, wagging, twisting). When a molecule is irradiated with infrared light, it will only absorb energy if the frequency of the IR radiation exactly matches the natural vibrational frequency of a particular bond, and if the vibration causes a change in the bond’s dipole moment.
共价键并非刚性杆，它们像微小的弹簧一样可以振动。这些振动包括伸缩振动（对称和不对称）和弯曲振动（剪式、摇摆、面外摇摆、扭曲）。当分子受到红外光照射时，只有当红外辐射的频率恰好与特定键的固有振动频率相匹配，并且该振动能引起键偶极矩的变化时，分子才会吸收能量。
For a vibration to be ‘IR active’, the dipole moment of the bond must change during the vibration. Symmetrical diatomic molecules like O₂ and N₂ have no dipole moment change when they stretch, so they do not absorb IR radiation. Polar bonds such as C=O, O–H, and C–Cl exhibit strong IR absorptions.
要使振动具有“红外活性”，键的偶极矩必须在振动过程中发生变化。对称的双原子分子，如 O₂ 和 N₂，在伸缩时偶极矩不发生变化，因此它们不吸收红外辐射。而像 C=O、O–H 和 C–Cl 这样的极性键则表现出强烈的红外吸收。
A typical IR spectrum plots percentage transmittance (or absorbance) against wavenumber (cm⁻¹). The wavenumber is directly proportional to frequency and inversely proportional to wavelength. The functional group region (above 1500 cm⁻¹) contains characteristic peaks for specific bonds, while the fingerprint region (below 1500 cm⁻¹) is unique to each individual compound.
典型的红外光谱图以百分透过率（或吸光度）为纵轴，波数（cm⁻¹）为横轴。波数与频率成正比，与波长成反比。官能团区（1500 cm⁻¹ 以上）包含特定键的特征峰，而指纹区（1500 cm⁻¹ 以下）对每个化合物都是独一无二的。
Wavenumber (ν̃) = 1 / λ
2. Bond Vibrations and Hooke’s Law | 键振动与胡克定律
The approximate wavenumber of an absorption can be predicted using Hooke’s Law, which treats a diatomic bond as two masses connected by a spring. The fundamental equation shows that wavenumber is proportional to the square root of the bond force constant (k) and inversely proportional to the square root of the reduced mass (μ).
吸收峰的大致波数可以用胡克定律预测，该定律将双原子键视为由弹簧连接的两个质量。基本方程表明，波数与键力常数 (k) 的平方根成正比，与折合质量 (μ) 的平方根成反比。
ν̃ ∝ √(k / μ)
Stronger bonds (higher bond order) have larger force constants and therefore absorb at higher wavenumbers. For example, C≡C absorbs around 2200 cm⁻¹, C=C around 1650 cm⁻¹, and C–C around 1000 cm⁻¹. Lighter atoms also lead to higher wavenumbers due to the smaller reduced mass; C–H stretches appear near 3000 cm⁻¹ while C–Cl stretches are found much lower, around 700 cm⁻¹.
更强的键（键级更高）具有更大的力常数，因此在更高波数处吸收。例如，C≡C 在约 2200 cm⁻¹ 处吸收，C=C 约在 1650 cm⁻¹，C–C 约在 1000 cm⁻¹。较轻的原子由于折合质量较小，也会导致更高的波数；C–H 伸缩峰出现在约 3000 cm⁻¹ 附近，而 C–Cl 伸缩峰则出现在低得多的波数，约 700 cm⁻¹。
3. Regions of the IR Spectrum | 红外光谱的区域划分
The IR spectrum is conventionally divided into two main regions. The functional group region spans from 4000 cm⁻¹ down to about 1500 cm⁻¹. Here, absorptions are largely due to stretching vibrations of specific functional groups. The fingerprint region lies below 1500 cm⁻¹ and consists of complex vibration patterns including bending, skeletal vibrations, and combination bands that are highly specific to the molecule as a whole.
红外光谱通常分为两个主要区域。官能团区从 4000 cm⁻¹ 向下延伸至约 1500 cm⁻¹。在此区域内，吸收主要是由特定官能团的伸缩振动引起的。指纹区位于 1500 cm⁻¹ 以下，由复杂的振动模式组成，包括弯曲振动、骨架振动和组合频带，这些对分子整体具有高度特异性。
When identifying a compound, you must look at the functional group region for characteristic peaks. However, to confirm whether two samples are the same compound, you would compare the entire fingerprint region. In CIE exams, you may be asked to match an unknown spectrum to reference spectra using the fingerprint region, or to identify functional groups present in a given spectrum.
在鉴定化合物时，必须查看官能团区中的特征峰。然而，要确认两个样品是否为同一种化合物，则需比较整个指纹区。在 CIE 考试中，你可能会被要求利用指纹区将未知光谱与参考光谱进行匹配，或者识别给定光谱中存在的官能团。
4. Characteristic Absorptions of Alkanes, Alkenes and Arenes | 烷烃、烯烃和芳烃的特征吸收
For saturated hydrocarbons (alkanes), the dominant absorptions are C–H stretches in the range 2850–2960 cm⁻¹. Bending vibrations for C–H in methyl and methylene groups appear around 1375–1465 cm⁻¹. There is no absorption in the multiple-bond region, making the spectrum relatively simple above 1500 cm⁻¹.
对于饱和烃（烷烃），主要的吸收是位于 2850–2960 cm⁻¹ 范围的 C–H 伸缩振动。甲基和亚甲基中 C–H 的弯曲振动出现在约 1375–1465 cm⁻¹。多重键区域没有吸收，因此 1500 cm⁻¹ 以上的光谱相对简单。
Alkenes show C–H stretches slightly above 3000 cm⁻¹ (sp² C–H). The very important diagnostic peak for C=C stretching appears in the range 1620–1680 cm⁻¹. Conjugation can shift this to slightly lower wavenumbers. Out-of-plane bending of alkene C–H bonds in the fingerprint region can also indicate the substitution pattern, but this detail is generally beyond the CIE syllabus.
烯烃在略高于 3000 cm⁻¹ (sp² C–H) 处显示 C–H 伸缩振动。非常重要的 C=C 伸缩诊断峰出现在 1620–1680 cm⁻¹ 范围内。共轭作用会使其略微向低波数移动。指纹区中烯烃 C–H 的面外弯曲振动也能指示取代模式，但这一细节通常超出了 CIE 大纲范围。
Aromatic compounds show C–H stretches just above 3000 cm⁻¹, often as a weak to medium band. Characteristic C=C ring stretching vibrations (often called ‘skeletal vibrations’) appear as a pair or trio of sharp peaks around 1450–1600 cm⁻¹. An overtone pattern in the 1660–2000 cm⁻¹ region may indicate the substitution pattern but is not routinely assessed.
芳香族化合物在刚好高于 3000 cm⁻¹ 处显示 C–H 伸缩振动，通常为弱至中等强度的谱带。特征的 C=C 环伸缩振动（常称为“骨架振动”）在约 1450–1600 cm⁻¹ 处呈现一对或三个尖锐峰。1660–2000 cm⁻¹ 区域的泛频模式可以指示取代模式，但通常不作考核要求。
5. Carbonyl Compounds: Aldehydes, Ketones, Carboxylic Acids and Esters | 羰基化合物：醛、酮、羧酸和酯
The carbonyl group C=O gives one of the strongest and most easily recognised absorptions in IR spectroscopy. The stretching vibration typically appears in the range 1700–1750 cm⁻¹. The exact position varies with the type of carbonyl compound, and CIE candidates are expected to know qualitative trends.
羰基 C=O 是红外光谱中最强、最易识别的吸收之一。其伸缩振动通常出现在 1700–1750 cm⁻¹ 范围内。确切位置因羰基化合物类型而异，CIE 考生应了解其定性趋势。
| Compound | C=O stretch (cm⁻¹) | Other key absorptions |
|---|---|---|
| Aldehyde | 1720–1740 | C–H aldehyde peak at ~2720–2820 cm⁻¹ (often doublet) |
| Ketone | 1705–1725 | No aldehyde C–H; simple spectrum |
| Carboxylic acid | 1700–1725 | Very broad O–H stretch ~2500–3300 cm⁻¹ |
| Ester | 1735–1750 | C–O stretch ~1000–1300 cm⁻¹ (often two bands) |
The aldehyde C–H stretch is a critical diagnostic feature that distinguishes aldehydes from ketones. It appears as a distinct absorption around 2720–2820 cm⁻¹, often as a doublet due to Fermi resonance. This, together with the carbonyl peak, positively identifies an aldehyde.
醛基的 C–H 伸缩振动是区分醛和酮的关键诊断特征。它在约 2720–2820 cm⁻¹ 处呈现一个独特的吸收，由于费米共振常表现为双峰。这与羰基峰一起可明确鉴定为醛。
Carboxylic acids show a notoriously broad O–H stretch that can extend from 2500 cm⁻¹ all the way to 3300 cm⁻¹, completely obscuring the C–H stretch region. This broadness is due to extensive hydrogen bonding in liquid and solid samples. The C=O stretch is at the lower end of the typical carbonyl range.
羧酸显示出极其宽泛的 O–H 伸缩峰，可从 2500 cm⁻¹ 一直延伸到 3300 cm⁻¹，完全掩盖了 C–H 伸缩区域。这种宽峰是由于液态和固态样品中存在广泛氢键。C=O 伸缩振动位于典型羰基范围的较低端。
6. Alcohols, Phenols and Ethers | 醇、酚和醚
Alcohols and phenols are characterised by the O–H stretch. In the gas phase (or very dilute solution), free O–H absorbs as a sharp peak around 3600–3650 cm⁻¹. However, in condensed phases, hydrogen bonding broadens this dramatically, producing a broad, rounded peak centred around 3200–3400 cm⁻¹. This broadness is a key identification clue.
醇和酚以 O–H 伸缩振动为特征。在气相（或极稀溶液）中，游离的 O–H 在约 3600–3650 cm⁻¹ 处呈现一个尖锐峰。然而，在凝聚相中，氢键使其显著变宽，产生一个以 3200–3400 cm⁻¹ 为中心的宽而圆的峰。这种宽峰是一个关键的识别线索。
The C–O stretching absorption is found in the fingerprint region, typically between 1000 and 1300 cm⁻¹. Primary alcohols show a distinct band near 1050 cm⁻¹, while secondary and tertiary alcohols shift to around 1100 cm⁻¹ and 1150 cm⁻¹ respectively. Phenols exhibit a C–O stretch at a slightly higher wavenumber, around 1200–1250 cm⁻¹, due to conjugation with the aromatic ring.
C–O 伸缩吸收位于指纹区，通常在 1000 至 1300 cm⁻¹ 之间。伯醇在约 1050 cm⁻¹ 处有一个显著的谱带，而仲醇和叔醇分别移至约 1100 cm⁻¹ 和 1150 cm⁻¹。酚的 C–O 伸缩振动因与芳环共轭而出现在略高的波数，约 1200–1250 cm⁻¹。
Ethers lack the O–H stretch and are identified primarily by the strong C–O–C asymmetric stretch, which appears as a broad, intense band around 1060–1150 cm⁻¹. Symmetrical ethers may show a weaker symmetric stretch. The absence of a carbonyl peak helps rule out esters.
醚没有 O–H 伸缩峰，主要通过位于约 1060–1150 cm⁻¹ 的强而宽的 C–O–C 不对称伸缩谱带来识别。对称醚可能显示出较弱的不对称伸缩。没有羰基峰有助于排除酯类。
7. Nitrogen-Containing Compounds: Amines, Amides and Nitriles | 含氮化合物：胺、酰胺和腈
Primary amines display two N–H stretching absorptions in the region 3300–3500 cm⁻¹, corresponding to symmetric and asymmetric stretching. Secondary amines show only a single N–H stretch in this region. The peaks are typically sharper than O–H stretches. Tertiary amines lack N–H bonds and cannot be identified by characteristic IR peaks above the fingerprint region.
伯胺在 3300–3500 cm⁻¹ 区域显示两个 N–H 伸缩吸收，分别对应于对称和不对称伸缩。仲胺在该区域仅显示一个 N–H 伸缩峰。这些峰通常比 O–H 伸缩峰更锐利。叔胺缺乏 N–H 键，因此无法通过高于指纹区的特征红外峰进行识别。
Amides combine the C=O stretch and N–H stretches. Primary amides show two N–H stretches around 3350 and 3180 cm⁻¹, and the carbonyl (Amide I) band appears at a lower wavenumber than in ketones, around 1630–1690 cm⁻¹, due to resonance. The Amide II band (mixed N–H bending and C–N stretching) is seen near 1500–1600 cm⁻¹.
酰胺结合了 C=O 伸缩和 N–H 伸缩。伯酰胺在约 3350 和 3180 cm⁻¹ 处显示两个 N–H 伸缩峰，且由于共振作用，其羰基（酰胺 I 带）出现在比酮更低的波数，约 1630–1690 cm⁻¹。酰胺 II 带（N–H 弯曲与 C–N 伸缩混合振动）出现在约 1500–1600 cm⁻¹ 附近。
Nitriles (C≡N) have a very sharp and distinct stretch in the triple bond region at 2200–2260 cm⁻¹. This is an unmistakable peak, as few other functional groups absorb in this region (alkynes, C≡C, absorb around 2100–2260 cm⁻¹ but usually weaker). The presence of a sharp peak near 2250 cm⁻¹ is strong evidence for a nitrile group.
腈 (C≡N) 在三键区域 2200–2260 cm⁻¹ 处有一个非常尖锐且独特的伸缩峰。这是一个无可争议的峰，因为在此区域吸收的其他官能团很少（炔烃 C≡C 在 2100–2260 cm⁻¹ 附近吸收，但通常较弱）。约 2250 cm⁻¹ 附近出现一个尖锐峰是腈基的有力证据。
8. Interpreting Spectra: A Stepwise Approach | 光谱解读：逐步分析方法
When faced with an IR spectrum in an exam, follow a systematic approach. First, check for the presence or absence of a broad O–H peak around 3200–3600 cm⁻¹. If a very broad peak that extends into the C–H region is present, it is likely a carboxylic acid. If a rounded but not as extremely broad peak is seen, an alcohol or phenol is indicated.
在考试中面对红外光谱时，要遵循系统的方法。首先，检查 3200–3600 cm⁻¹ 附近是否存在宽的 O–H 峰。如果存在一个延伸到 C–H 区域的极宽峰，则很可能是羧酸。如果看到的是一个圆润但不如前述极宽的峰，则表示醇或酚。
Second, examine the carbonyl region around 1700–1750 cm⁻¹. A strong, sharp peak here indicates the presence of C=O. Then check for accompanying peaks: an aldehyde C–H near 2720–2820 cm⁻¹, or a broad O–H for acid, or N–H for amide. If C=O is absent, consider alcohols, ethers, or simple hydrocarbons.
其次，检查约 1700–1750 cm⁻¹ 的羰基区域。此处一个强而尖锐的峰表明存在 C=O。然后检查伴生峰：约 2720–2820 cm⁻¹ 处是否有醛基 C–H，或是否有酸性 O–H 或酰胺 N–H 的宽峰。如果没有 C=O，则考虑醇、醚或简单的烃类。
Third, look at the triple bond region (2000–2500 cm⁻¹). A sharp peak here could be C≡N or C≡C. Fourth, note the C–H stretch region just below 3000 cm⁻¹ for alkanes, or just above 3000 cm⁻¹ for alkenes/aromatics. Finally, use the fingerprint region to confirm identity by matching with a reference spectrum if provided.
第三，查看三键区域（2000–2500 cm⁻¹）。此处的尖锐峰可能是 C≡N 或 C≡C。第四，注意 C–H 伸缩区域：烷烃在刚好低于 3000 cm⁻¹ 处，烯烃/芳烃在刚好高于 3000 cm⁻¹ 处。最后，如果题目提供了参考光谱，可利用指纹区与参考光谱匹配以确认身份。
9. Common Pitfalls and How to Avoid Them | 常见错误及避免方法
One common mistake is confusing the broad O–H peak of a carboxylic acid with that of an alcohol. Remember: the carboxylic acid O–H is so broad that it often extends past the C–H stretch region and may overlap with it almost completely. The alcohol O–H is still broad but is more centred around 3300 cm⁻¹ and does not typically obscure the C–H stretch at 3000 cm⁻¹ completely.
一个常见错误是将羧酸的宽 O–H 峰与醇的混淆。请记住：羧酸的 O–H 非常宽，通常延伸到 C–H 伸缩区之外，并可能几乎完全与之重叠。醇的 O–H 虽然也宽，但更集中在 3300 cm⁻¹ 附近，通常不会完全掩盖 3000 cm⁻¹ 附近的 C–H 伸缩峰。
Another pitfall is misreading wavenumber values. Always read the scale carefully. A peak that looks like carbonyl may be alkene C=C if it is below 1680 cm⁻¹ and weaker. C=C stretches are generally less intense than C=O stretches. Also, do not assign every tiny wiggle as a functional group; many small peaks in the fingerprint region are combination and overtone bands not directly linked to a single bond type.
另一个误区是误读波数值。请始终仔细阅读刻度。一个看似羰基的峰，如果低于 1680 cm⁻¹ 且较弱，则可能是烯烃的 C=C。C=C 伸缩振动通常不如 C=O 强。另外，不要将每个微小波动都认定为官能团；指纹区中的许多小峰是组合频和泛频带，与单一化学键类型无直接关联。
Students sometimes forget that symmetrical stretching in symmetrical molecules may be IR inactive. For example, the C=C stretch in symmetrically substituted ethylene derivatives can be very weak or absent. Similarly, the triple bond in symmetrical alkynes may show little or no absorption. Always consider molecular symmetry when predicting the presence or absence of peaks.
学生有时会忘记对称分子中的对称伸缩振动可能是红外非活性的。例如，对称取代的乙烯衍生物中 C=C 伸缩可能非常弱或缺失。同样，对称炔烃中的三键可能几乎没有吸收。在预测峰是否存在时，要始终考虑分子对称性。
10. Applications: Breathalysers and Greenhouse Gases | 应用：酒精呼气测试仪与温室气体
CIE expects you to know two practical applications of IR spectroscopy. The first is the modern breathalyser. In a fuel cell breathalyser, the alcohol in breath is oxidised to generate an electric current. However, an IR-based breathalyser passes infrared radiation through the breath sample and measures the absorption at the characteristic C–H stretching frequency of ethanol (around 2950 cm⁻¹). The greater the absorbance, the higher the concentration of ethanol. This method is specific, rapid, and non-destructive.
CIE 要求你了解红外光谱的两个实际应用。第一个是现代酒精呼气测试仪。在燃料电池型呼气测试仪中，呼气中的酒精被氧化以产生电流。然而，基于红外的呼气测试仪是让红外辐射通过呼气样本，并测量乙醇在特征 C–H 伸缩频率（约 2950 cm⁻¹）处的吸收。吸光度越大，乙醇浓度越高。该方法特异性好、快速且无损。
The second important application is the monitoring of greenhouse gases. Molecules such as CO₂, CH₄, and H₂O absorb IR radiation at specific wavelengths. For example, carbon dioxide has a strong C=O asymmetric stretch around 2350 cm⁻¹, and water vapour has broad O–H stretches. Satellites and ground-based sensors measure the absorbance of infrared radiation passing through the atmosphere to determine the concentration of these gases globally.
第二个重要应用是温室气体的监测。CO₂、CH₄ 和 H₂O 等分子在特定波长处吸收红外辐射。例如，二氧化碳在约 2350 cm⁻¹ 处有一个强的 C=O 不对称伸缩峰，水蒸气则有宽的 O–H 伸缩峰。卫星和地面传感器通过测量穿过大气的红外辐射的吸光度来确定这些气体的全球浓度。
The link to global warming lies in the fact that these gases trap outgoing long-wavelength IR radiation from the Earth’s surface. The absorbance intensity in the IR spectrum of the atmosphere correlates with the concentration of greenhouse gases, allowing scientists to track changes over time and validate climate models.
与全球变暖的联系在于，这些气体会捕获来自地球表面的出射长波红外辐射。大气红外光谱中的吸收强度与温室气体浓度相关，使科学家能够追踪浓度随时间的演变并验证气候模型。
11. Exam Tips for CIE Questions | CIE 考试答题技巧
In Paper 2 and Paper 4, you may be given an IR spectrum and asked to identify the compound or functional groups. Always cite specific wavenumber ranges and bond vibrations. For instance, rather than writing ‘OH peak’, state ‘broad absorption at 3200–3400 cm⁻¹ due to O–H stretch’. This precision earns marks.
在试卷 2 和试卷 4 中，你可能会拿到一张红外光谱图并被要求鉴定化合物或官能团。务必引用具体的波数范围和键振动。例如，不要只写“OH 峰”，而应写明“3200–3400 cm⁻¹ 处由 O–H 伸缩振动引起的宽吸收”。这种精确性能为你赢得分数。
You may also be asked to explain why a particular compound cannot be identified by IR alone. The answer is often that isomers with the same functional groups give very similar spectra. For example, butan-1-ol and butan-2-ol both show O–H and C–O stretches, but their fingerprint regions differ. If fingerprint data are not provided, mass spectrometry or NMR would be needed to differentiate them.
你也可能被要求解释为何某种化合物无法仅凭红外光谱进行鉴定。答案通常是，具有相同官能团的异构体给出的光谱非常相似。例如，丁-1-醇和丁-2-醇都显示 O–H 和 C–O 伸缩峰，但它们的指纹区不同。若题目未提供指纹数据，则需要质谱或核磁共振来区分它们。
Sometimes a question will link IR data with other information, such as elemental analysis or chemical test results. Integrate all the evidence: if a compound shows an O–H peak and a carbonyl peak above 1700 cm⁻¹, and it also reacts with sodium carbonate to produce CO₂, it must be a carboxylic acid. Building a logical, evidence-based argument is key.
有时题目会将红外数据与其他信息（如元素分析或化学测试结果）结合。要整合所有证据：如果一个化合物既有 O–H 峰又有 1700 cm⁻¹ 以上的羰基峰，且与碳酸钠反应生成 CO₂，那么它一定是羧酸。构建一个逻辑清晰、基于证据的论证是关键。
12. Summary of Key Absorption Ranges | 关键吸收范围总结
Memorising the following table will allow you to quickly interpret most A-Level IR spectra. Always consider the shape and intensity of the peak in addition to its position.
记住下表将帮助你快速解读大多数 A-Level 红外光谱。除了峰位置外，始终要考虑其形状和强度。
| Bond / Functional group | Wavenumber range (cm⁻¹) | Intensity and shape |
|---|---|---|
| O–H (alcohol, phenol) H-bonded | 3200–3600 | Broad, strong |
| O–H (carboxylic acid) H-bonded | 2500–3300 | Very broad, strong |
| N–H (primary amine/amide) | 3300–3500 | Medium, one or two peaks |
| C–H (alkane) | 2850–2960 | Medium to strong |
| C–H (alkene/arene, sp²) | 3000–3100 | Weak to medium |
| C≡N (nitrile) | 2200–2260 | Sharp, medium |
| C≡C (alkyne, terminal) | 2100–2260 | Weak, sharp (often absent if symmetrical) |
| C=O (carbonyl) | 1700–1750 | Very strong, sharp |
| C=C (alkene) | 1620–1680 | Weak to medium |
| C=C (aromatic ring) | 1450–1600 (multiple) | Medium, sharp pairs |
| C–O (alcohol, ester, ether) | 1000–1300 | Strong |
With a solid grasp of these principles and plenty of practice interpreting spectra, you will be well prepared for the IR spectroscopy questions on the CIE A-Level Chemistry exam. Remember, the best approach is always systematic: O–H/N–H region, then C≡N/C≡C, then C=O, then C=C, then C–H, and finally the fingerprint region.
通过扎实掌握这些原理并进行大量解读光谱的练习，你将为 CIE A-Level 化学考试中的红外光谱题目做好充分准备。请记住，最佳方法始终是系统性的：先看 O–H/N–H 区域，然后 C≡N/C≡C，接着 C=O，然后 C=C，再 C–H，最后是指纹区。
Published by TutorHao | Chemistry Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导