📚 Core Principles in AS Chemistry Insert 2, January 2021 | AS化学插入页2（2021年1月）核心原理
The AQA AS Chemistry Paper 2 Insert for January 2021 provides a concise collection of spectroscopic and analytical data tables essential for solving structural problems. Mastering the core principles behind infrared (IR) spectroscopy, mass spectrometry (MS), and nuclear magnetic resonance (NMR) spectroscopy – both 1H and 13C – is vital for interpreting the spectra and deducing molecular structures accurately under exam conditions. This article dissects the fundamental concepts embedded in that insert, linking each data table to the underlying physical and chemical principles.
AQA AS化学卷二（2021年1月）的插入页提供了一组简洁的光谱与分析数据表，是解答结构推断题的重要工具。掌握红外光谱（IR）、质谱（MS）以及核磁共振波谱（1H和13C NMR）背后的核心原理，对于在考试条件下准确解析谱图、推导分子结构至关重要。本文将深入剖析该插入页中蕴含的基本概念，将每张数据表与其背后的物理和化学原理联系起来。
1. Overview of the Insert 2 Data | 插入页2数据概览
The insert typically contains four key reference sections: characteristic infrared absorption frequencies, proton NMR chemical shift data, carbon-13 NMR chemical shift data, and a table of common mass spectrometry fragments. Together, these allow you to piece together the identity of an unknown organic compound. Each table is grounded in well-defined principles: bond vibrations, nuclear spin behaviour, and electron-impact fragmentation.
该插入页通常包含四个关键参考部分：特征红外吸收频率、质子核磁共振化学位移数据、碳-13核磁共振化学位移数据以及常见质谱碎片表。这些数据共同帮助你拼凑出未知有机化合物的身份。每一张表都建立在明确的原理之上：键的振动、核自旋行为以及电子轰击碎裂。
Effective use requires not just memorising numbers but understanding why a carbonyl group absorbs near 1700 cm⁻¹ or why a proton on a benzene ring appears at δ 7.2. The principles that follow equip you with that understanding.
有效利用这些数据不仅需要记住数字，更需要理解为什么羰基在约1700 cm⁻¹处吸收，或者为什么苯环上的质子在δ 7.2处出峰。接下来的原理将为你提供这种理解。
2. Infrared Spectroscopy – The Principle of Bond Vibrations | 红外光谱 – 键振动原理
Infrared spectroscopy probes the vibrational energy levels of covalent bonds. When a molecule absorbs IR radiation of a frequency that matches the natural vibrational frequency of a bond, the bond is excited to a higher vibrational state. The frequency of absorption depends on the bond strength (force constant) and the reduced mass of the atoms involved – stronger bonds and lighter atoms vibrate at higher frequencies.
红外光谱探测的是共价键的振动能级。当分子吸收的红外辐射频率与某个化学键的固有振动频率匹配时，该键便跃迁到更高的振动能级。吸收频率取决于键的强度（力常数）和所涉及原子的约化质量——键越强、原子越轻，振动频率越高。
This is why a C≡N triple bond (strong, with a light N atom) absorbs at a high wavenumber around 2220–2260 cm⁻¹, while a C–O single bond stretches at a much lower 1000–1300 cm⁻¹. The insert’s table presents these diagnostic frequencies as ranges, reflecting the slight variations caused by adjacent groups.
这就是为什么C≡N三键（键很强且N原子较轻）在约2220–2260 cm⁻¹的高波数处吸收，而C–O单键的伸缩振动出现在低得多的1000–1300 cm⁻¹。插入页的表格以范围的形式给出这些诊断性频率，反映了相邻基团引起的细微变化。
ν ∝ √(k / μ)
where k is the bond force constant and μ is the reduced mass. Only vibrations that cause a change in the dipole moment of the molecule are IR active.
其中k为键力常数，μ为约化质量。只有引起分子偶极矩变化的振动才具有红外活性。
For a bond to absorb IR radiation, the vibration must alter the dipole moment. Symmetrical stretches in molecules like CO₂ (symmetric stretch) are IR inactive and do not appear in the spectrum, though the insert’s data focuses on functional groups that commonly show strong absorptions.
一个化学键要吸收红外辐射，其振动必须改变偶极矩。像CO₂的对称伸缩振动这类运动是红外非活性的，不会在光谱中出现，不过插入页的数据重点在于那些通常表现出强吸收的官能团。
3. Characteristic IR Absorption Frequencies – Interpreting the Table | 特征红外吸收频率 – 解读数据表
The insert lists absorption ranges for key bonds: O–H (alcohols, carboxylic acids), C=O, C–O, C=C, C≡N, and C–H (alkanes, alkenes). Each range is a fingerprint of a functional group. For example, a broad, strong absorption at 3230–3550 cm⁻¹ indicates an O–H bond, typically in alcohols or phenols, while a sharp peak at 1680–1750 cm⁻¹ signals a C=O group in aldehydes, ketones, acids, or esters.
插入页列出了关键键的吸收范围：O–H（醇、羧酸）、C=O、C–O、C=C、C≡N和C–H（烷烃、烯烃）。每个范围都是一个官能团的“指纹”。例如,在3230–3550 cm⁻¹处的宽而强的吸收表明存在O–H键，通常属于醇或酚；而在1680–1750 cm⁻¹处的尖峰则意味着存在C=O基团，可能来自醛、酮、酸或酯。
The exact position gives further clues: an acid chloride C=O stretch appears at around 1775–1810 cm⁻¹, slightly higher than most carbonyls due to electron-withdrawing inductive effects that increase the C=O bond order. An ester C=O is typically near 1735 cm⁻¹, whereas a ketone sits around 1715 cm⁻¹. Carboxylic acids show a broad O–H envelope overlapping the C=O peak, often extending from 2500–3300 cm⁻¹.
精确的位置能提供更多线索：酰氯的C=O伸缩振动出现在约1775–1810 cm⁻¹，略高于大多数羰基化合物，这是因为吸电子诱导效应增加了C=O的键级。酯的C=O通常在1735 cm⁻¹附近，而酮则位于约1715 cm⁻¹。羧酸会呈现一个宽大的O–H包络峰，与C=O峰重叠，通常从2500延伸到3300 cm⁻¹。
Alkene C=C stretches are observed at 1620–1680 cm⁻¹, often weaker than carbonyl absorptions. Aromatic C=C bonds appear in a similar region but typically show multiple bands. The C–H stretching region (2850–3100 cm⁻¹) can distinguish alkanes (below 3000 cm⁻¹) from alkenes/arenes (above 3000 cm⁻¹), a nuance reflected in some editions of the insert.
烯烃的C=C伸缩振动出现在1620–1680 cm⁻¹，通常比羰基吸收峰弱。芳香族C=C键落在相似的区域，但往往呈现多重谱带。C–H伸缩振动区（2850–3100 cm⁻¹）可以区分烷烃（低于3000 cm⁻¹）和烯烃/芳烃（高于3000 cm⁻¹），这一细微差别在某些版本的插入页中有所反映。
4. The Fingerprint Region and Compound Identification | 指纹区与化合物鉴定
Below about 1500 cm⁻¹ lies the fingerprint region, where complex bending and stretching modes produce a pattern unique to each molecule. While the insert does not tabulate these bands, the principle is important: two compounds with identical functional groups can still be distinguished by their fingerprint patterns. In AS level exam contexts, you compare the fingerprint region of an unknown with that of a reference.
约1500 cm⁻¹以下的区域为指纹区，此处的复杂弯曲振动和伸缩振动模式为每个分子提供了独一无二的图样。虽然插入页并没有将这一区域的谱带列表，但其原理至关重要：两个具有相同官能团的化合物，依然可以通过指纹区图谱加以区分。在AS考试情境中，你需要将未知物的指纹区与参比图谱进行比较。
The fingerprint region arises from coupled vibrations, such as C–C skeletal modes and C–H rocking, which involve many atoms. Since no two molecules have identical arrays of bonds and angles, the fingerprint region acts as a molecular ID. The insert itself focuses on the functional group region (above 1500 cm⁻¹), but recognising that below 1500 cm⁻¹ you must look for a perfect match to confirm identity is a key interpretative skill.
指纹区源于各种耦合振动，如C–C骨架振动和C–H面内摇摆，这些振动牵涉众多原子。由于没有两个分子具有完全相同的键和键角排列，指纹区就起到了分子身份证的作用。插入页本身重点关注官能团区（1500 cm⁻¹以上），但认识到必须检查1500 cm⁻¹以下区域的完全匹配以确认化合物身份，是一项关键的解谱技能。
5. Mass Spectrometry – Electron Ionisation and Ion Separation | 质谱 – 电子电离与离子分离
Mass spectrometry, as referenced in the AS insert, relies on electron impact ionisation: a high-energy electron beam knocks out an electron from sample molecules, creating positively charged molecular ions (M⁺). The ions are then accelerated through an electric field and deflected by a magnetic field; the radius of deflection depends on the mass-to-charge ratio (m/z).
AS插入页中所引用的质谱分析依赖于电子轰击电离：高能电子束从样品分子中打出一个电子，生成带正电的分子离子（M⁺）。这些离子随后在电场中加速，并在磁场中偏转；偏转半径取决于质荷比（m/z）。
The molecular ion peak provides the relative molecular mass of the compound, assuming the charge is 1. Occasionally, a small M+1 peak is observed due to the presence of the 13C isotope. The insert’s data focus on common fragment ions – species formed when the molecular ion breaks apart – which give structural information.
分子离子峰给出了化合物的相对分子质量，前提是电荷数为1。有时，由于13C同位素的存在，会观察到一个小的M+1峰。插入页的数据侧重于常见碎片离子——即分子离子碎裂时形成的物种——它们提供了结构信息。
The mass spectrometer detects ions and plots relative abundance against m/z. The base peak is the most intense peak and is assigned a relative abundance of 100%. All other peaks are scaled relative to it. Understanding the fragmentation patterns behind the insert’s table is essential for deducing the skeleton of an organic molecule.
质谱仪检测离子，并绘制相对丰度对m/z的图谱。基峰是强度最高的峰，其相对丰度被指定为100%。所有其他峰的强度以此为基准进行标度。理解插入页表格背后的碎裂规律对于推断有机分子的骨架至关重要。
6. Fragmentation Patterns and the M⁺ Peak – Principles | 碎片模式与M⁺峰 – 原理
The molecular ion M⁺ is formed directly from the sample molecule. Its m/z value equals the relative molecular mass, Mᵣ. Fragmentation occurs because the molecular ion possesses excess internal energy; it can break covalent bonds, often at weak points, to yield a positively charged fragment and a neutral radical. The observed fragment ions correspond to the charged pieces.
分子离子M⁺直接由样品分子形成，其m/z值等于相对分子质量Mᵣ。碎裂的发生是由于分子离子具有过多的内能；它可以在某些位点（常为薄弱点）打断共价键，生成一个带正电的碎片和一个中性自由基。观测到的碎片离子对应于带电的碎片。
Common fragments listed in the insert, such as CH₃⁺ (m/z = 15), C₂H₅⁺ (29), and C₆H₅⁺ (77), arise from the cleavage of carbon–carbon bonds. The stability of the resulting carbocation governs the abundance of each fragment: secondary and tertiary carbocations are more stable and thus produce more intense peaks than primary ones. This principle helps rationalise why, for example, 2-methylpropane gives a prominent fragment at m/z = 43 ((CH₃)₂CH⁺) but a weaker peak at m/z = 57.
插入页中列出的常见碎片如CH₃⁺ (m/z = 15)、C₂H₅⁺ (29) 和 C₆H₅⁺ (77)，是由碳碳键断裂产生的。所生成碳正离子的稳定性决定了各个碎片的丰度：二级和三级碳正离子比一级的更稳定，因此产生的峰强度更高。这一原理有助于解释为什么，例如，2-甲基丙烷会在m/z = 43 ((CH₃)₂CH⁺) 处产生一个突出的碎片峰，而m/z = 57的峰则较弱。
A typical AS question might ask you to identify a compound from its mass spectrum. You note the M⁺ peak for Mᵣ, then look for mass differences between peaks. The loss of 15 units (CH₃), 29 (C₂H₅), or 17 (OH) corresponds to recognisable fragments. The insert’s table of common fragments gives quick reference to these building blocks.
典型的AS考题可能会要求你根据质谱图推断化合物。你需要先找出M⁺峰以确定Mᵣ，然后观察峰与峰之间的质量差。失去15个单位（CH₃）、29（C₂H₅）或17（OH）对应着可识别的碎片。插入页的常见碎片表为这些结构单元提供了快捷参考。
7. Proton NMR – The Principle of Chemical Shift | 质子核磁共振 – 化学位移原理
Proton nuclear magnetic resonance spectroscopy exploits the magnetic properties of 1H nuclei. In an external magnetic field, the spins of protons align either with or against the field; radiofrequency radiation can flip the spin state. The precise frequency required depends on the electronic environment around each proton: electrons shield the nucleus, reducing the effective magnetic field it experiences.
质子核磁共振波谱利用了1H核的磁性。在外磁场中，质子的自旋要么顺着磁场方向，要么逆着磁场方向排列；射频辐射可以使自旋态发生翻转。所需射频的精确频率取决于每个质子周围的电子环境：电子对核产生屏蔽作用，降低了其感受到的有效磁场强度。
Electronegative atoms or electron-withdrawing groups deshield nearby protons, requiring a higher frequency for resonance. This effect is expressed as a chemical shift, δ, measured in parts per million (ppm) relative to tetramethylsilane (TMS). The insert’s 1H NMR table organises shifts by proton environment: alkyl protons (0.7–1.6 ppm), protons adjacent to a carbonyl (2.0–3.0 ppm), alkene protons (4.5–6.0 ppm), aromatic protons (6.0–8.5 ppm), and aldehyde protons (9.0–10.0 ppm).
电负性原子或吸电子基团会对邻近质子产生去屏蔽作用，需要更高的频率才能引发共振。这种效应用化学位移δ表示，以四甲基硅烷（TMS）为参照，单位为百万分之一（ppm）。插入页的1H NMR表格按照质子环境列出了化学位移：烷基氢（0.7–1.6 ppm）、邻羰基氢（2.0–3.0 ppm）、烯烃氢（4.5–6.0 ppm）、芳香氢（6.0–8.5 ppm）和醛基氢（9.0–10.0 ppm）。
The O–H or N–H protons (0.5–5.5 ppm, exchangeable) are more variable due to hydrogen bonding and concentration effects. Carboxylic acid protons appear far downfield, often 10.0–12.0 ppm. By matching experimental shifts to the table, you can map protons onto a proposed structure.
O–H或N–H质子（0.5–5.5 ppm，可交换）由于氢键和浓度效应，化学位移变化更大。羧酸质子出现在很低的磁场区域，通常在10.0–12.0 ppm。通过将实验化学位移与表格匹配，你可以将质子对应到所提出的结构上。
8. Spin–Spin Coupling and Integration – Interpreting NMR Patterns | 自旋–自旋耦合与积分 – 核磁共振谱图的解析
Proton NMR spectra show splitting due to neighbouring non-equivalent protons, following the n+1 rule: a proton with n equivalent neighbours splits into (n+1) peaks. This coupling, measured as the J constant in Hz, offers valuable connectivity information. The insert does not list coupling constants, but the concept of splitting is essential for using the shift table effectively.
质子核磁共振波谱中，由于邻近不等价质子的存在而产生裂分，遵循n+1规则：一个质子若有n个等价邻近质子，其信号将裂分成(n+1)重峰。这种耦合以耦合常数J（Hz）来度量，提供了宝贵的连接信息。插入页虽然不列出耦合常数，但裂分的概念对于有效使用化学位移表至关重要。
Integration traces (or numerical ratios) show the relative number of protons contributing to each signal. Together, splitting and integration allow you to deduce fragments like CH₃–CH₂–, CH₃–CH–, or isolated –CH₂–. For example, a triplet integrating to 3H at ~1.0 ppm and a quartet integrating to 2H at ~3.5 ppm strongly suggest an ethyl group attached to an electronegative atom (like –O–).
积分曲线（或数字比例）显示出每个信号所对应的质子相对数目。结合裂分和积分，你可以推断出诸如CH₃–CH₂–、CH₃–CH–或孤立的–CH₂–等片段。例如，在~1.0 ppm处有一个积分为3H的三重峰，且在~3.5 ppm处有一个积分为2H的四重峰，这强烈暗示存在一个连接在电负性原子（如–O–）上的乙基。
The insert’s chemical shift ranges help you confirm these assignments. A quartet at 4.1 ppm is typical of a CH₂ attached to an ester oxygen, whereas a quartet at 2.5 ppm suggests a CH₂ next to a carbonyl. Layering splitting knowledge on top of the shift data is the core of structural elucidation.
插入页的化学位移范围可以帮助你确认这些归属。位于4.1 ppm的四重峰是连接在酯氧上的CH₂的典型特征，而位于2.5 ppm的四重峰则表明CH₂邻接一个羰基。在化学位移数据的基础上叠加裂分知识，是结构解析的核心。
9. Carbon-13 NMR Spectroscopy Principles | 碳-13核磁共振光谱原理
Carbon-13 NMR follows the same fundamental principle as proton NMR but observes the less abundant 13C isotope (1.1% natural abundance). It provides direct information about the carbon skeleton. Each chemically distinct carbon atom gives one peak in the spectrum; the number of peaks equals the number of non-equivalent carbon environments.
碳-13核磁共振遵循与质子核磁共振相同的基本原理，但观测的是丰度较低的13C同位素（天然丰度1.1%）。它提供了关于碳骨架的直接信息。每个化学环境不同的碳原子在谱图中给出一个峰；峰的数量等于不等价碳环境的数目。
The insert’s 13C table categorises shifts: alkane carbons (0–50 ppm), carbons adjacent to oxygen or halogens (50–90 ppm), alkene/aromatic carbons (110–160 ppm), and carbonyl carbons (160–220 ppm). Note that carbonyl carbons in aldehydes and ketones appear around 190–210 ppm, whereas ester and acid carbonyls are slightly more shielded at 160–185 ppm.
插入页的13C表格对化学位移进行了分类：烷烃碳（0–50 ppm）、邻氧或卤素的碳（50–90 ppm）、烯烃/芳烃碳（110–160 ppm）和羰基碳（160–220 ppm）。需要注意的是，醛和酮中的羰基碳出现在约190–210 ppm，而酯和酸的羰基碳由于屏蔽稍强，出现在160–185 ppm。
There is no splitting in 13C NMR (due to proton decoupling in exam spectra), so you simply count peaks and assign environments. A molecule with the formula C₄H₈O giving three 13C peaks in the carbonyl region strongly suggests an aldehyde or ketone with symmetry. The insert is indispensable for confirming these assignments.
在考试涉及的13C核磁共振谱中（质子去耦），不会出现裂分，因此你只需数峰并归属化学环境。一个分子式为C₄H₈O的化合物如果在羰基区给出三个13C峰，则强烈暗示它是一个具有对称性的醛或酮。插入页对于确证这些归属不可或缺。
10. Combined Spectral Analysis – Integrating Insert Data | 综合光谱分析 – 整合插入页数据
The true power of the Insert 2 comes from combining all three techniques. A typical problem presents IR, MS, and NMR data; you must use the insert’s tables to deduce the structure. The logical sequence is: (1) use MS to find Mᵣ and key fragments; (2) use IR to identify functional groups; (3) use 1H NMR for hydrogen environments and connectivity; and (4) use 13C NMR to confirm the carbon framework.
插入页2的真正威力在于将三种技术结合起来。一个典型的问题会给出红外、质谱和核磁共振数据；你必须利用插入页的表格去推断结构。合理的逻辑顺序是：（1）利用质谱确定Mᵣ和关键碎片；（2）利用红外识别官能团；（3）利用1H NMR判定氢环境及连接方式；（4）利用13C NMR确认碳骨架。
For example, an infrared absorption at 1740 cm⁻¹ and a mass spectrum with M⁺ at 88, coupled with proton NMR signals at δ 1.2 (t, 3H), δ 2.3 (q, 2H), and δ 3.7 (s, 3H), points to ethyl acetate: CH₃COOCH₂CH₃. The insert’s ester C=O range (1735–1750 cm⁻¹) and the chemical shifts for –COOCH₂– protons confirm this. The integration ratios and the absence of an –OH proton support the ester assignment.
例如，红外吸收在1740 cm⁻¹，质谱显示M⁺为88，同时质子核磁共振信号出现在δ 1.2（三重峰，3H）、δ 2.3（四重峰，2H）和 δ 3.7（单峰，3H），指向乙酸乙酯：CH₃COOCH₂CH₃。插入页中酯的C=O范围（1735–1750 cm⁻¹）以及–COOCH₂–质子的化学位移证实了这一推断。积分比例和缺少–OH质子信号也支持酯的判断。
Always check the number of 13C peaks against the proposed structure – ethyl acetate should show four peaks. The insert’s carbon data can quickly flag symmetrical or equivalent carbons. Consistent cross-referencing of all insert tables eliminates alternative isomers and solidifies your answer.
始终要将13C峰的数量与所提结构进行核对——乙酸乙酯应显示四个峰。插入页的碳谱数据能快速指示出对称或等价的碳原子。对所有插入页表格进行一致的交叉参照，可以排除异构体的其他可能，并巩固你的答案。
11. Using the Insert for Common Functional Group Identification | 利用插入页识别常见官能团
Certain functional groups produce highly characteristic patterns across multiple techniques. Aldehydes, for instance, show a C=O stretch at 1720–1740 cm⁻¹, a proton NMR signal for –CHO at 9.0–10.0 ppm (often a singlet or a triplet with small coupling), and a 13C peak near 190–200 ppm. Primary alcohols exhibit a broad O–H stretch around 3350 cm⁻¹, a proton signal for –OH (variable, often broad), and deshielded –CH₂–O– protons at 3.3–4.0 ppm.
某些官能团在多个技术中产生高度特征性的模式。例如，醛会在1720–1740 cm⁻¹处显示C=O伸缩振动，在9.0–10.0 ppm出现–CHO的质子NMR信号（通常是单峰或具有小耦合的三重峰），并在190–200 ppm附近呈现13C峰。伯醇则在约3350 cm⁻¹处展现宽的O–H伸缩峰，出现–OH的质子信号（位置可变，通常较宽），以及位于3.3–4.0 ppm的去屏蔽–CH₂–O–质子信号。
The insert’s data are deliberately organised to aid this pattern recognition. By learning the principles behind why a carboxylic acid O–H appears downfield or why a C=C bond absorbs near 1650 cm⁻¹, you move beyond rote
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