📚 Spectroscopic Identification of Organic Compounds | 有机化合物的光谱鉴定方法
Spectroscopic techniques are among the most powerful tools available to chemists for determining the structure of organic molecules. In the IB Chemistry syllabus, students are expected to interpret mass spectra, infrared (IR) spectra, and proton nuclear magnetic resonance (¹H NMR) spectra to deduce structural information about unknown compounds. Mastering these techniques requires both conceptual understanding and systematic problem-solving practice.
光谱技术是化学家用来确定有机分子结构的最强大工具之一。在IB化学课程中，学生需要能够解读质谱、红外光谱和质子核磁共振谱，从而推断未知化合物的结构信息。掌握这些技术既需要概念性的理解，也需要系统性的解题练习。
1. Why Spectroscopy Matters | 为什么光谱鉴定如此重要
Spectroscopic methods allow chemists to ‘see’ the invisible architecture of molecules. Unlike chemical tests that often destroy samples and provide only limited information, spectroscopy is typically non-destructive and reveals detailed structural features. For IB Chemistry, three main techniques are emphasized: mass spectrometry (MS), infrared spectroscopy (IR), and proton nuclear magnetic resonance (¹H NMR) spectroscopy.
光谱方法使化学家能够”看见”分子不可见的架构。与通常破坏样品且仅提供有限信息的化学测试不同，光谱分析通常是非破坏性的，并能揭示详细的结构特征。在IB化学中，重点强调三种主要技术：质谱法（MS）、红外光谱法（IR）和质子核磁共振谱法（¹H NMR）。
The key principle is that different functional groups and structural environments absorb or emit electromagnetic radiation at characteristic frequencies, or fragment in predictable ways. By systematically analyzing these patterns, chemists can piece together the structure of even complex organic molecules. This is why spectroscopy forms a significant component of the IB Chemistry examination, particularly in Paper 3 and the extended essay.
关键原理在于，不同的官能团和结构环境会以特征频率吸收或发射电磁辐射，或者以可预测的方式碎裂。通过系统地分析这些模式，化学家甚至可以拼凑出复杂有机分子的结构。这就是为什么光谱鉴定在IB化学考试中占据重要地位，特别是在试卷三和拓展论文中。
2. Mass Spectrometry: The Molecular Fingerprint | 质谱法：分子指纹
Mass spectrometry begins with the vaporization and ionization of a sample. Molecules are bombarded with high-energy electrons, which knock off electrons and form molecular ions (M⁺). The mass spectrometer then separates these ions based on their mass-to-charge ratio (m/z). The resulting spectrum plots relative abundance against m/z, providing two crucial pieces of information.
质谱分析首先将样品汽化并电离。分子被高能电子轰击，失去电子形成分子离子（M⁺）。质谱仪随后根据质荷比（m/z）分离这些离子。所得谱图以相对丰度对m/z作图，提供两条关键信息。
The first is the molecular ion peak (M⁺), which corresponds to the intact molecule and gives the relative molecular mass (Mr). The second is the fragmentation pattern, where the molecular ion breaks into smaller fragments. These fragments, along with characteristic losses, provide structural clues. For example, a loss of 15 mass units (CH₃⁺) suggests the presence of a methyl group, while a loss of 17 units (OH⁺) indicates a hydroxyl group.
第一是分子离子峰（M⁺），它对应于完整的分子并给出相对分子质量（Mr）。第二是碎片化模式，即分子离子分裂成更小的碎片。这些碎片连同特征性丢失提供了结构线索。例如，丢失15个质量单位（CH₃⁺）表明存在甲基，而丢失17个单位（OH⁺）则表示存在羟基。
Consider the mass spectrum of pentan-3-one (C₅H₁₀O). The molecular ion peak appears at m/z = 86. A prominent fragment at m/z = 57 corresponds to the loss of an ethyl group (86 – 29 = 57). Another significant fragment at m/z = 29 (C₂H₅⁺) confirms the presence of ethyl groups. These observations are consistent with the symmetric structure CH₃CH₂COCH₂CH₃.
考虑3-戊酮（C₅H₁₀O）的质谱。分子离子峰出现在m/z = 86处。一个显著的碎片峰在m/z = 57处，对应于丢失一个乙基（86 – 29 = 57）。另一个重要的碎片在m/z = 29处（C₂H₅⁺），证实了乙基的存在。这些观察结果与对称结构CH₃CH₂COCH₂CH₃一致。
M⁺ = Mr of molecule → molecular formula
Fragmentation → structural information
3. Mass Spectrometry: Isotopes and High-Resolution Data | 质谱法：同位素与高分辨数据
In addition to determining molecular mass, mass spectrometry can also reveal the presence of certain elements through isotope patterns. Chlorine and bromine, for instance, have distinctive isotopic signatures. Chlorine exists as ³⁵Cl (75%) and ³⁷Cl (25%), so a molecule containing one chlorine atom will show an M+2 peak approximately one-third the height of the M⁺ peak. Bromine (⁷⁹Br and ⁸¹Br, roughly 1:1) produces an M+2 peak almost equal in height to the M⁺ peak.
除确定分子质量外，质谱法还可以通过同位素模式揭示某些元素的存在。例如，氯和溴具有独特的同位素特征。氯以³⁵Cl（75%）和³⁷Cl（25%）的形式存在，因此含有一个氯原子的分子将显示一个约为M⁺峰高度三分之一的M+2峰。溴（⁷⁹Br和⁸¹Br，约为1:1）产生的M+2峰的高度几乎与M⁺峰相等。
High-resolution mass spectrometry provides even greater precision. By measuring m/z values to four or more decimal places, chemists can determine the exact molecular formula. For example, a molecular ion at m/z 44.0262 could correspond to C₂H₄O (calculated exact mass = 44.0262) but not to CO₂ (43.9898) or C₃H₈ (44.0626). This distinction is impossible with nominal mass alone.
高分辨质谱法提供了更高的精度。通过将m/z值测量到小数点后四位或更多，化学家可以确定精确的分子式。例如，m/z 44.0262的分子离子可能对应于C₂H₄O（计算精确质量=44.0262），但不对应CO₂（43.9898）或C₃H₈（44.0626）。这种区分仅靠标称质量是无法实现的。
In IB examinations, students are usually expected to identify the Mr from the M⁺ peak and suggest structures for fragment ions. It is essential to remember that the base peak (the tallest peak) is the most stable fragment, not necessarily the molecular ion. The M⁺ peak for some compounds may be very weak or even absent if the molecular ion is highly unstable.
在IB考试中，学生通常需要从M⁺峰确定Mr，并为碎片离子提出结构。必须记住，基峰（最高峰）是最稳定的碎片，而不一定是分子离子。对于某些化合物，如果分子离子极不稳定，M⁺峰可能非常弱甚至不存在。
4. Infrared Spectroscopy: Detecting Functional Groups | 红外光谱法：检测官能团
Infrared spectroscopy works by passing IR radiation through a sample and measuring which frequencies are absorbed. Covalent bonds absorb IR energy at characteristic frequencies, causing them to vibrate in specific ways – stretching, bending, or wagging. The absorption frequency depends on the bond strength and the masses of the atoms involved.
红外光谱法的工作原理是将红外辐射穿过样品，并测量哪些频率被吸收。共价键在特征频率处吸收红外能量，导致它们以特定方式振动——伸缩、弯曲或摇摆。吸收频率取决于键的强度和相关原子的质量。
For IB Chemistry, students are required to identify key functional groups from IR spectra using a data booklet. The most important absorptions to remember are: O-H (alcohols) at 3200-3600 cm⁻¹ as a broad peak, O-H (carboxylic acids) at 2500-3300 cm⁻¹ also broad, C=O at 1680-1750 cm⁻¹, C-O at 1000-1300 cm⁻¹, and C≡N at 2200-2260 cm⁻¹. The carbonyl group is particularly distinctive because its absorption is strong and appears in a relatively clear region of the spectrum.
在IB化学中，学生需要使用数据手册从红外谱图中识别关键官能团。需要记住的最重要吸收包括：O-H（醇）在3200-3600 cm⁻¹处呈宽峰，O-H（羧酸）在2500-3300 cm⁻¹处同样为宽峰，C=O在1680-1750 cm⁻¹处，C-O在1000-1300 cm⁻¹处，以及C≡N在2200-2260 cm⁻¹处。羰基特别具有辨识度，因为其吸收强且出现在谱图中相对清晰的区域。
Key IR absorptions: C=O ≈ 1700 cm⁻¹ | O-H (alcohol) ≈ 3300 cm⁻¹ broad | O-H (acid) ≈ 3000 cm⁻¹ very broad | C≡N ≈ 2250 cm⁻¹
To distinguish between functional groups, students must consider both the position and shape of absorption bands. For example, an alcohol and a carboxylic acid both contain O-H bonds, but the acid’s O-H absorption is broader and extends to lower wavenumbers. The presence of a strong C=O absorption at approximately 1700 cm⁻¹ differentiates the carboxylic acid from the alcohol.
为了区分官能团，学生必须同时考虑吸收带的位置和形状。例如，醇和羧酸都含有O-H键，但酸的O-H吸收更宽并且延伸到更低的波数。在约1700 cm⁻¹处出现强C=O吸收，可将羧酸与醇区分开来。
5. IR Spectroscopy: Practical Interpretation Strategy | 红外光谱法：实用解读策略
When approaching an IR spectrum in an examination, adopt a systematic strategy. First, check for the absence or presence of a broad O-H stretch. A very broad band centered around 3300 cm⁻¹ indicates an alcohol or carboxylic acid (the latter being even broader). Second, look for the sharp, strong C=O band around 1700 cm⁻¹. Its presence rules out alcohols, alkanes, and amines, and points toward aldehydes, ketones, carboxylic acids, esters, or amides.
在考试中处理红外谱图时，应采用系统性的策略。首先，检查是否存在宽大的O-H伸缩峰。以3300 cm⁻¹为中心的非常宽的峰表示醇或羧酸（羧酸的峰更宽）。其次，寻找在1700 cm⁻¹附近的尖锐强C=O带。其存在排除了醇、烷烃和胺，指向醛、酮、羧酸、酯或酰胺。
Third, verify the identity with secondary absorptions. An aldehyde shows a distinctive C-H stretch around 2700-2800 cm⁻¹ (two small peaks). An ester has a strong C-O stretch at 1200-1300 cm⁻¹. A carboxylic acid shows a very broad O-H that overlaps with C-H absorptions. The absence of C=O combined with a broad O-H suggests an alcohol.
第三，用次要吸收来验证判断。醛在2700-2800 cm⁻¹附近显示出特征性的C-H伸缩（两个小峰）。酯在1200-1300 cm⁻¹处有强C-O伸缩。羧酸显示出与C-H吸收重叠的非常宽的O-H。没有C=O且同时存在宽的O-H则表明是醇。
The following table summarizes the key IR absorptions that IB students must know. Note that the data booklet provides exact ranges; memorizing approximate values is useful for quick elimination during exams.
下表总结了IB学生必须掌握的关键红外吸收。请注意，数据手册提供了精确范围；记住近似值有助于在考试中快速排除。
| Functional Group | 官能团 | Wavenumber (cm⁻¹) | 波数 | Appearance | 外观 |
| O-H (alcohol) | 3200-3600 | broad, strong |
| O-H (carboxylic acid) | 2500-3300 | very broad, strong |
| C=O | 1680-1750 | sharp, strong |
| C-O | 1000-1300 | strong |
| C≡N | 2200-2260 | sharp, medium |
| C=C (alkene) | 1620-1680 | medium |
| C-H (alkane) | 2850-2960 | strong |
6. ¹H NMR: The Language of Hydrogen Environments | ¹H核磁共振：氢环境的语言
Proton NMR spectroscopy provides the most detailed structural information of all three techniques. It examines how hydrogen nuclei (protons) behave in a magnetic field. When protons are in different chemical environments – meaning they have different neighboring atoms or groups – they experience slightly different magnetic fields and absorb radiation at slightly different frequencies. This phenomenon is called chemical shift (δ), measured in parts per million (ppm).
质子核磁共振波谱法提供了三种技术中最详细的结构信息。它研究氢核（质子）在磁场中的行为。当质子处于不同的化学环境中——即它们具有不同的相邻原子或基团——它们会感受到略微不同的磁场，并以略微不同的频率吸收辐射。这种现象称为化学位移（δ），以百万分之一（ppm）为单位测量。
In IB Chemistry, students must interpret four key features of an ¹H NMR spectrum: the number of signals (distinct hydrogen environments), the chemical shift of each signal (identifying the type of environment), the integration (relative number of protons, shown by peak area), and the splitting pattern (number of adjacent non-equivalent hydrogens). The number of signals tells us how many different types of hydrogen atoms exist in the molecule.
在IB化学中，学生必须解读¹H NMR谱图的四个关键特征：信号数量（不同的氢环境数量）、每个信号的化学位移（识别环境类型）、积分（质子相对数量，由峰面积显示），以及裂分模式（相邻不等价氢的数量）。信号数量告诉我们分子中存在多少种不同类型的氢原子。
Consider ethanol (CH₃CH₂OH). Its ¹H NMR spectrum shows three signals in a 3:2:1 ratio, corresponding to the CH₃ protons, CH₂ protons, and OH proton respectively. The OH proton is often exchangeable and may appear as a broad singlet. The CH₃ protons are split into a triplet by the two adjacent CH₂ protons, while the CH₂ protons are split into a quartet by the three adjacent CH₃ protons. The following rule applies:
考虑乙醇（CH₃CH₂OH）。其¹H NMR谱图显示三个信号，比例为3:2:1，分别对应于CH₃质子、CH₂质子和OH质子。OH质子通常是可交换的，可能表现为宽的单峰。CH₃质子被相邻的两个CH₂质子裂分为三重峰，而CH₂质子被相邻的三个CH₃质子裂分为四重峰。适用以下规则：
n + 1 rule: a proton with n equivalent neighboring protons is split into (n + 1) peaks
7. Chemical Shifts and Characteristic Values | 化学位移与特征值
Chemical shift values are affected by electronegativity, hybridization, and magnetic anisotropy. Protons attached to carbon atoms that are bonded to electronegative atoms (like O, N, or halogens) are deshielded and appear at higher δ values (downfield). Conversely, protons in electron-rich environments are shielded and appear at lower δ values (upfield).
化学位移值受电负性、杂化方式和磁各向异性的影响。与连接电负性原子（如O、N或卤素）的碳原子相连的质子被去屏蔽，出现在更高的δ值处（低场）。相反，处于富电子环境中的质子被屏蔽，出现在较低的δ值处（高场）。
For IB-level problems, the following approximate chemical shift ranges are essential. Methyl protons on an alkane (R-CH₃) appear at δ 0.9-1.0 ppm. Protons on a carbon adjacent to a carbonyl or aromatic ring (R-CH₂-CO or R-CH₂-Ar) appear at δ 2.1-2.5 ppm. Protons on a carbon bonded to oxygen (R-CH₂-O) appear at δ 3.3-3.7 ppm. Aldehydic protons (R-CHO) appear at a distinctive δ 9-10 ppm, and carboxylic acid protons at δ 10-12 ppm.
对于IB级别的问题，以下近似化学位移范围至关重要。烷烃上的甲基质子（R-CH₃）出现在δ 0.9-1.0 ppm处。与羰基或芳环相邻的碳上的质子（R-CH₂-CO或R-CH₂-Ar）出现在δ 2.1-2.5 ppm处。与氧键合的碳上的质子（R-CH₂-O）出现在δ 3.3-3.7 ppm处。醛基质子（R-CHO）出现在特征性的δ 9-10 ppm处，羧酸质子出现在δ 10-12 ppm处。
| Type of Proton | 质子类型 | Chemical Shift δ (ppm) | Approximate | 近似值 |
| R-CH₃ | 0.9-1.0 | 0.9 |
| R-CH₂-R | 1.2-1.5 | 1.3 |
| R-CH₂-CO | 2.1-2.5 | 2.2 |
| R-CH₂-O | 3.3-3.7 | 3.5 |
| R-CHO | 9.0-10.0 | 9.7 |
| R-COOH | 10.0-12.0 | 11.0 |
8. Integration and Splitting Patterns | 积分与裂分模式
The area under each NMR signal is proportional to the number of protons responsible for that signal. Modern NMR spectra display these areas as integral curves or numerical values. In IB exams, you will often be given the integration ratio or asked to deduce it from the spectrum. This ratio is crucial for determining molecular structure, as it tells you how many hydrogen atoms exist in each environment.
每个NMR信号下的面积与该信号对应的质子数成正比。现代NMR谱图将面积显示为积分曲线或数值。在IB考试中，通常会给出积分比，或要求从谱图中推断出来。这个比值对于确定分子结构至关重要，因为它告诉你在每个环境中存在多少个氢原子。
Splitting patterns follow the n+1 rule. A signal appears as a singlet (1 peak) if the proton has no neighboring non-equivalent hydrogens. It appears as a doublet (2 peaks) with one neighboring proton, a triplet (3 peaks) with two, and a quartet (4 peaks) with three. The relative peak heights within a multiplet follow Pascal’s triangle: 1:1 for a doublet, 1:2:1 for a triplet, 1:3:3:1 for a quartet, and 1:4:6:4:1 for a quintet.
裂分模式遵循n+1规则。如果质子没有相邻的不等价氢，信号表现为单峰（1个峰）。有一个相邻质子时为二重峰（2个峰），有两个相邻质子时为三重峰（3个峰），有三个相邻质子时为四重峰（4个峰）。多重峰内的相对峰高遵循帕斯卡三角形：二重峰为1:1，三重峰为1:2:1，四重峰为1:3:3:1，五重峰为1:4:6:4:1。
For example, in propanone (CH₃COCH₃), all six hydrogens are equivalent, so the spectrum shows only a single signal at δ 2.2 ppm. In contrast, propanal (CH₃CH₂CHO) shows three signals: a triplet at δ 1.1 (CH₃), a multiplet at δ 2.5 (CH₂), and a triplet at δ 9.8 (CHO). The aldehyde proton appears as a triplet due to coupling with the two CH₂ protons.
例如，在丙酮（CH₃COCH₃）中，所有六个氢都是等价的，因此谱图仅在δ 2.2 ppm处显示一个信号。相比之下，丙醛（CH₃CH₂CHO）显示三个信号：δ 1.1处的三重峰（CH₃），δ 2.5处的多重峰（CH₂），以及δ 9.8处的三重峰（CHO）。醛基质子因与两个CH₂质子耦合而表现为三重峰。
9. Combined Spectroscopic Analysis: A Worked Example | 组合光谱分析：实例解析
Examinations often provide data from multiple techniques and ask students to deduce the full structure. Let us work through a typical IB-style problem. An unknown compound X has the molecular formula C₃H₆O₂. Its IR spectrum shows a very broad absorption from 2500-3300 cm⁻¹ and a strong peak at 1710 cm⁻¹. The ¹H NMR spectrum shows three signals in a 3:2:1 ratio at δ 1.2 (triplet), δ 4.1 (quartet), and δ 11.0 (singlet).
考试通常会提供多种技术的数据，并让学生推断完整结构。让我们解析一个典型的IB风格问题。未知化合物X的分子式为C₃H₆O₂。其红外谱图显示从2500-3300 cm⁻¹的非常宽的吸收和在1710 cm⁻¹处的强峰。¹H NMR谱图显示三个信号，比例为3:2:1，分别在δ 1.2（三重峰）、δ 4.1（四重峰）和δ 11.0（单峰）处。
Begin with the IR spectrum. The very broad O-H absorption (2500-3300 cm⁻¹) combined with a C=O peak at 1710 cm⁻¹ confirms a carboxylic acid. The molecular formula C₃H₆O₂ matches a saturated carboxylic acid: propanoic acid (CH₃CH₂COOH) or methyl ethanoate (CH₃COOCH₃). The IR spectrum rules out the ester because the ester has no O-H absorption.
从红外谱图开始。非常宽的O-H吸收（2500-3300 cm⁻¹）结合在1710 cm⁻¹处的C=O峰，证实了羧酸的存在。分子式C₃H₆O₂匹配饱和羧酸：丙酸（CH₃CH₂COOH）或乙酸甲酯（CH₃COOCH₃）。红外谱图排除了酯，因为酯没有O-H吸收。
Now examine the NMR data. The signal at δ 11.0 is the carboxylic acid proton (COOH), appearing as a singlet. The triplet at δ 1.2 (3H) is the CH₃ group, split by two neighboring CH₂ protons. The quartet at δ 4.1 (2H) might seem unexpected for propanoic acid, since CH₂ protons adjacent to COOH typically appear around δ 2.3. However, this quartet appears at δ 4.1 because the CH₂ is bonded to an oxygen atom. Therefore, X is not propanoic acid but ethyl methanoate?
现在检查NMR数据。δ 11.0处的信号是羧酸质子（COOH），表现为单峰。δ 1.2处的三重峰（3H）是CH₃基团，被两个相邻CH₂质子裂分。δ 4.1处的四重峰（2H）对于丙酸来说可能显得出乎意料，因为与COOH相邻的CH₂质子通常出现在δ 2.3左右。然而，这个四重峰出现在δ 4.1处，是因为CH₂与氧原子键合。因此，X不是丙酸而是甲酸乙酯？
Wait – let us reconsider. In ethyl methanoate (HCOOCH₂CH₃), the CH₂ protons appear at δ 4.1 (quartet) and CH₃ at δ 1.2 (triplet), but there is no signal at δ 11. The carboxylic acid signal at δ 11.0 requires a COOH group. The correct answer is that X cannot have molecular formula C₃H₆O₂ and show a COOH proton at δ 11 together with an O-CH₂ quartet. This is a contradiction, which teaches us a valuable lesson: always cross-check consistency between IR and NMR data.
等等——让我们重新考虑。在甲酸乙酯（HCOOCH₂CH₃）中，CH₂质子出现在δ 4.1（四重峰），CH₃在δ 1.2（三重峰），但在δ 11处没有信号。δ 11.0处的羧酸信号需要COOH基团。正确答案是，X不可能具有分子式C₃H₆O₂且同时在δ 11处显示COOH质子并在δ 4.1处显示O-CH₂四重峰。这是一个矛盾，它教导我们宝贵的经验：始终交叉检查IR和NMR数据的一致性。
10. Common Pitfalls and Exam Tips | 常见陷阱与考试技巧
Students frequently make several errors when solving spectroscopy problems. The most common is mistaking the base peak for the molecular ion in mass spectra. Remember that the molecular ion is the peak with the highest m/z that corresponds to the intact molecule (excluding isotope peaks). Always use the data booklet for exact IR ranges, but have approximate values memorized for quick elimination.
学生在解决光谱问题时经常犯几个错误。最常见的是在质谱中把基峰误认为分子离子。请记住，分子离子是对应于完整分子的最高m/z峰（不包括同位素峰）。始终使用数据手册获取精确的红外范围，但应记住近似值以便快速排除。
Another common error involves NMR integration. Students may correctly identify the number of signals but misassign the integration ratio. For example, in ethanol, the integration ratio is 3:2:1, not 1:1:1. Additionally, remember that exchangeable protons (O-H, N-H) may not always show expected splitting. They often appear as broad singlets because they exchange rapidly between molecules.
另一个常见错误涉及NMR积分。学生可能正确识别了信号数量，但错误分配了积分比。例如，在乙醇中，积分比为3:2:1，而不是1:1:1。此外，请记住可交换质子（O-H、N-H）可能并不总是显示预期的裂分。它们通常表现为宽单峰，因为它们在分子间快速交换。
Exam tip: When deducing structure, always combine information from all techniques. Use mass spectrometry to find Mr and molecular formula. Use IR to identify functional groups. Use NMR to determine hydrogen environments, their ratios, and neighboring groups. Finally, verify your proposed structure against all data. The degree of unsaturation (DoU = (2C + 2 – H – X + N)/2) can help confirm the presence of rings or double bonds.
考试技巧：在推断结构时，始终结合所有技术的信息。使用质谱法确定Mr和分子式。使用红外光谱法识别官能团。使用NMR确定氢环境、它们的比例和相邻基团。最后，根据所有数据验证你提出的结构。不饱和度（DoU = (2C + 2 – H – X + N)/2）有助于确认环或双键的存在。
11. Practice Problems | 练习题目
To consolidate your understanding, attempt the following problems. For each, use all the data provided to deduce the structure. Remember to work systematically.
为了巩固理解，请尝试以下问题。对于每个问题，使用提供的所有数据来推断结构。记住要系统地工作。
Problem 1: An organic compound Y has Mr = 74. Its IR spectrum shows a strong broad peak at 3350 cm⁻¹ but no peak around 1700 cm⁻¹. Its ¹H NMR spectrum shows three signals: δ 0.9 (triplet, 3H), δ 1.6 (sextet, 2H), δ 3.6 (triplet, 2H). Deduce the structure.
问题1：有机化合物Y的Mr = 74。其红外谱图在3350 cm⁻¹
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