📚 Combined Analytical Techniques 110: Solving Organic Structures with MS, IR and NMR | 综合分析技术110：利用质谱、红外与核磁共振解析有机结构
In Edexcel A‑Level Chemistry, you are often asked to deduce the structure of an unknown organic compound using a combination of spectral data. Mass spectrometry (MS) provides the molecular mass and fragmentation pattern, infrared (IR) spectroscopy reveals functional groups, and nuclear magnetic resonance (NMR) spectroscopy maps the carbon–hydrogen framework. This article walks you through each technique and, crucially, shows how to combine them systematically to solve structure‑elucidation problems.
在Edexcel A‑Level化学中，你经常需要利用多种谱图数据推断未知有机化合物的结构。质谱（MS）给出分子量和碎片信息，红外光谱（IR）揭示官能团，核磁共振波谱（NMR）则描绘碳‑氢骨架。本文将逐一介绍这些技术，更重要的是，教你如何系统地把它们结合起来解决结构解析题。
1. The Power of Combined Techniques | 综合分析技术的威力
No single spectroscopic method can unambiguously determine an organic structure on its own. Mass spectrometry tells you the molar mass but not the connectivity; IR confirms the presence of certain bonds but not the full skeleton; NMR provides the atomic connectivity but can be ambiguous if you do not know the functional groups. Only by integrating all available data can you confidently propose a unique structure.
没有哪一种谱学方法能单独明确地确定有机结构。质谱告诉你摩尔质量，却无法说明连接方式；红外确认某些键的存在，但无法描绘整体骨架；核磁共振给出原子连接性，但若不知道官能团也可能模棱两可。唯有综合所有可用数据，你才能自信地提出唯一的分子结构。
2. Mass Spectrometry: Molecular Ion and Fragmentation | 质谱：分子离子与碎片离子
In electron‑impact mass spectrometry, the molecule is bombarded with high‑energy electrons, ejecting an electron to form a radical cation M⁺•. The peak at the highest m/z (ignoring tiny isotope peaks) corresponds to the molecular ion, giving the relative molecular mass Mᵣ. Fragmentation occurs when bonds break, producing smaller cations that appear as a pattern of peaks. For example, an alcohol often loses a methyl (•CH₃) or an ethyl (•C₂H₅) radical, yielding characteristic M–15 or M–29 peaks.
在电子轰击质谱中，分子受到高能电子轰击，失去一个电子形成自由基阳离子 M⁺•。最高m/z处的峰（忽略微小的同位素峰）对应分子离子峰，给出相对分子质量Mᵣ。化学键断裂产生碎片，形成一系列阳离子碎片峰。例如，醇常常失去甲基自由基（•CH₃）或乙基自由基（•C₂H₅），产生特征的M–15或M–29峰。
3. Interpreting a Mass Spectrum: Key Points | 质谱解析：要点
Identify the molecular ion peak – check that the m/z value is consistent with the expected formula. Look for the base peak (the most abundant fragment, set to 100 %) which indicates the most stable cation. Recognise common loss patterns: loss of 15 (CH₃), 17 (OH), 18 (H₂O), 28 (CO or C₂H₄), 29 (C₂H₅ or CHO), 31 (OCH₃) and so on. The nitrogen rule is also useful: if the molecular ion has an odd m/z, the molecule contains an odd number of nitrogen atoms.
找出分子离子峰——检查m/z值是否与预期分子式相符。观察基峰（丰度最高，设为100%），它对应于最稳定的阳离子。识别常见脱落模式：失去15（CH₃）、17（OH）、18（H₂O）、28（CO或C₂H₄）、29（C₂H₅或CHO）、31（OCH₃）等。氮规则也很实用：若分子离子峰m/z为奇数，则分子含有奇数个氮原子。
4. Infrared Spectroscopy: Detecting Functional Groups | 红外光谱：官能团识别
ly bonds with a changing dipole moment absorb IR. Key regions: the fingerprint region (below ~1500 cm⁻¹) is complex and unique to each molecule, while the functional‑ absorptions for specific bonds. For Edexcel, you must know the prominent absorptions of O–H, N–H, C=O, C=C, C≡N, C–O, and C–X bonds.
红外辐射使化学键在其特征频率处振动。谱图以波数（cm⁻¹）为横轴、百分透过率为纵轴。只有偶极矩发生变化的键才吸收红外。关键区域：指纹区（约低于1500 cm⁻¹）复杂且对每种分子独特，而官能团区（高于1500 cm⁻¹）给出特定键的明确吸收。对Edexcel，你必须掌握O–H、N–H、C=O、C=C、C≡N、C–O和C–X键的显著吸收。
5. Characteristic IR Absorption Table | 特征红外吸收表
| Bond / Functional Group | Wavenumber Range (cm⁻¹) | Intensity / Shape |
|---|---|---|
| O–H (alcohols, H‑bonded) | 3200 – 3600 | Strong, broad |
| O–H (carboxylic acids) | 2500 – 3300 | Very broad, overlapping C–H |
| N–H (amines, amides) | 3300 – 3500 | Medium, sharp (1° amines: two peaks) |
| C≡N (nitriles) | 2210 – 2260 | Medium, sharp |
| C=O (carbonyl) | 1680 – 1750 | Strong, sharp |
| C=C (alkene/aromatic) | 1620 – 1690 | Variable, usually medium |
| C–O (alcohol, ester, ether) | 1000 – 1300 | Strong |
Use this table to quickly spot the main functional groups. Remember that carbonyls in aldehydes and ketones absorb near 1720 cm⁻¹, while esters and carboxylic acids are slightly higher (~1730–1750 cm⁻¹) and amides lower (~1680 cm⁻¹). Look for C–O stretches to confirm esters or alcohols.
使用此表快速识别主要官能团。记住醛酮的羰基吸收在约1720 cm⁻¹，酯和羧酸略高（~1730–1750 cm⁻¹），酰胺更低（~1680 cm⁻¹）。寻找C–O伸缩振动可确认酯或醇。
6. NMR Spectroscopy: The Chemical Environment | 核磁共振波谱：化学环境
Proton NMR (¹H NMR) and carbon‑13 NMR (¹³C NMR) are powerful tools for mapping the hydrogen and carbon framework. In a given molecule, chemically equivalent protons or carbons give the same signal. The chemical shift δ (measured in ppm) tells you about the electronic environment: protons near electronegative atoms (O, N, halogen) are deshielded and appear at higher δ values. Integration traces in ¹H NMR give the relative number of protons responsible for each signal.
质子核磁共振（¹H NMR）和碳‑13核磁共振（¹³C NMR）是描绘氢、碳骨架的强有力工具。在给定分子中，化学等价的质子或碳给出相同的信号。化学位移δ（以ppm为单位）反映了电子环境：靠近电负性原子（O、N、卤素）的质子被去屏蔽，出现在较高δ值。¹H NMR的积分曲线给出了各信号对应的相对质子数。
7. Proton NMR: Spin–Spin Splitting | 质子核磁共振：自旋‑自旋裂分
Each proton signal is split into multiplets by neighbouring non‑equivalent protons (n + 1 rule). A proton with n equivalent neighbours on adjacent carbons splits into n + 1 peaks. Common patterns: singlet (s, 0 neighbours), doublet (d, 1 neighbour), triplet (t, 2 neighbours), quartet (q, 3 neighbours), and multiplet (m, more than 4 neighbours or overlapping). Coupling constants J are measured in Hz and identical for interacting partners – a useful check of connectivity.
每个质子信号会被相邻的非等价质子裂分成多重峰（n + 1规则）。若一个质子有n个等价邻位质子，它就会裂分成n + 1重峰。常见模式：单峰（s，0个邻位）、二重峰（d，1个邻位）、三重峰（t，2个邻位）、四重峰（q，3个邻位）和多重峰（m，4个以上邻位或重叠）。耦合常数J以Hz为单位，对相互作用的一对质子相等——这是判断连接性的有用检验。
8. Carbon‑13 NMR Spectroscopy | 碳‑13核磁共振波谱
Unlike ¹H NMR, ¹³C NMR is not usually integrated and shows no splitting (broad‑band decoupled). Instead it simply displays one peak for each chemically distinct carbon environment. The number of signals tells you how many distinct carbon types exist in the molecule, giving a direct view of molecular symmetry. Chemical shifts range from 0 to 220 ppm; carbonyl carbons appear above 160 ppm, alkene/aromatic carbons 100–150 ppm, oxygen‑bearing carbons 50–90 ppm, and saturated carbons 0–50 ppm.
与¹H NMR不同，¹³C NMR通常不积分，也不出现裂分（宽带去耦谱）。它只是为每个化学上不同的碳环境显示一个峰。信号的数量告诉你分子中存在几种不同类型的碳，直接反映分子对称性。化学位移范围0–220 ppm；羰基碳出现在160 ppm以上，烯碳/芳碳100–150 ppm，连氧碳50–90 ppm，饱和碳0–50 ppm。
9. Step‑by‑Step Structure Determination | 分步结构鉴定
Follow a logical order: (1) Use the mass spectrum to obtain the molecular mass and deduce the molecular formula – you may be given percentage composition or combustion data. (2) Examine the IR spectrum to identify functional groups. (3) Count the number of signals in ¹³C NMR to understand symmetry and carbon types. (4) Analyse the ¹H NMR spectrum: note the chemical shifts, integration ratios, and splitting patterns to construct fragments. (5) Assemble the fragments into a structure that satisfies all data. (6) Check your proposed structure against the IR, mass spectrum and the nitrogen rule.
遵循逻辑顺序：（1）利用质谱得到分子质量，并推导分子式——你可能需要结合元素百分比或燃烧数据。（2）检查红外光谱以识别官能团。（3）计数¹³C NMR信号数，了解对称性和碳类型。（4）分析¹H NMR谱：关注化学位移、积分比和裂分模式，构造氢碎片。（5）将碎片拼成满足所有数据的结构。（6）用IR、质谱和氮规则检验你提出的结构。
10. Worked Example: Combining All Data | 例题演示：综合所有数据
shows three signals at δ 15, 36, and 68 ppm. The ¹H NMR spectrum gives: δ 0.9 (t, 3H), 1.4 (m, 2H), 2.3 (s, 1H, exchanges with D₂O), and 3.6 (t, 2H). Step‑by‑step, we deduce: the broad IR at 3350 cm⁻¹ indicates O–H (alcohol); the C–O at 1120 cm⁻¹ supports an alcohol. Mᵣ = 88, suggesting C₄H₈O₂? No, that gives 88 but not matching saturation. Combine: OH gave 17, 88–17=71, suggest C₄H₉O? Actually C₄H₉OH = C₄H₁₀O (Mᵣ 74). Missing 14 – another CH₂ – so C₅H₁₂O? Mᵣ 88 indeed (C₅H₁₂O). ¹³C shows 3 signals, so symmetry: possible 1‑pentanol? 1‑pentanol has 5 carbons but only 3 signals? No, 1‑entanol has 5 distinct carbons, 5 signals. Thus the carbon chain must be symmetric, like 3‑methyl‑1‑butanol? arbons but only 4 signals. Wait C₅H₁₂O with 3 carbon signals means high symmetry, e.g. 2,2‑dimethyl‑1‑ 7; that would be 3 signals. ¹H NMR: δ 0.9 (t, 3H) indicates terminal methyl coupled to CH₂? But a triplet for 3H at 0.9 suggests CH₃–CH₂– 3 (1H, exchangeable, OH). A triplet at 3.6 (2H) for CH₂ attached to O. This pattern fits 1‑pentanol: CH₃CH₂CH₂CH₂CH₂OH. But 1‑ nals, not 3. So the triplet at 0.9 and triplet at 3.6 with the integration suggests CH₃CH₂– and –CH₂CH₂OH, but there is also the middle CH₂. 1‑ signals. However, if the ¹³C shows only 3 signals, perhaps the compound is symmetrical: 1‑propanol, but Mᵣ = 60. Not 88. The data given is illustrative – equire solving a real compound for length, but we can present consistent data leading to 3‑methyl‑1‑butanol, which gives four ¹³C signals: two methyls on terminal isopropyl, CH–, CH₂–OH, CH–CH₃? e 4 signals. The worked example can be simplified: Show that the data corresponds to butan‑1‑ol? Mᵣ 74, too low. Let' worked problem that matches typical Edexcel exam style, using a known compound like propan‑1‑ol but with Mᵣ 60. I'll change to Mᵣ = 74 for butan‑1‑ol. Let's craft: Mᵣ = 74; IR: broad 3350, 1120; ¹³C NMR: 4 signals (but we only want a few). I'll use a simple compound like propan‑2‑ol, Mᵣ 60, with ¹³C 2 signals. But then the example won't have integration complexities. To keep within length, I'll present a systematic reasoning without actual compound, but I'll give the solution. Better to write: " m/z = 60; IR: broad band at 3350 cm⁻¹, strong peak at 1150 cm⁻¹; ¹³C NMR: two signals at δ ~25 and δ ~ R: δ 1.2 (d, 6H), δ 3.1 (broad s, 1H, exchange), δ 4.0 (septet, 1H). Deduce the structure." Then the solution is propan‑2‑ol. This is consistent and easy. I'll use that.
化合物X的质谱显示分子离子峰m/z = 60；IR在3350 cm⁻¹处有宽峰，1150 cm⁻¹有强峰；¹³C NMR显示两个信号δ~25和δ~64 ppm；¹H NMR给出：δ 1.2 (d, 6H)，δ 3.1 (宽单峰, 1H, 可被D₂O交换)，δ 4.0 (七重峰, 1H)。推出结构为丙‑2‑醇。数据分析：IR宽峰3350为O–H，1150为C–O，说明是醇。m/z 60对应C₃H₈O。¹³C两个信号表明有两种碳环境，因此分子有对称性。¹H NMR中，δ 1.2处二重峰积分为6H，说明两个等价的CH₃被一个CH质子裂分；δ 4.0七重峰（n+1=7，n=6）表明CH被6个等价的相邻质子裂分，连接两个CH₃；δ 3.1交换峰为OH。因此结构为(CH₃)₂CHOH。
11. Common Mistakes in Combined Problems | 综合分析中的常见错误
er pitfall is misidentifying exchangeable protons (OH, NH): these appear as broad singlets and disappear on D₂O shake. Ignoring the nitrogen rule can cause you to miss nitrogen atoms. Finally, always cross‑check your final structure with all three methods—if the IR shows no carbonyl but your structure contains C=O, something is wrong.
学生常忘了利用积分曲线确定质子比例，导致碎片中氢原子数目错误。另一个陷阱是误判可交换质子（OH、NH）：这些信号呈宽单峰，经D₂O交换后消失。忽视氮规则可能让你遗漏氮原子。最后，永远要用三种方法交叉检验最终结构——如果IR没有羰基吸收但你的结构含C=O，那一定出错了。
12. Summary and Exam Tips | 总结与备考建议
Mastering combined spectral analysis requires practice with a wide range of compounds. For Edexcel exams, always annotate spectra clearly, show your reasoning, and quote wavenumber ranges and chemical shifts. When you propose a structure, draw it out and assign every NMR signal and key IR band. Remember that a correct answer is one that consistently explains all the given data, not just some of it. With systematic thinking, combined structure problems become rewarding puzzles rather than daunting challenges.
掌握综合分析技术需要大量练习。在Edexcel考试中，务必清晰标注谱图，展示推理过程，并引用波数范围和化学位移值。当你提出结构时，把它画出来并为每个NMR信号和关键IR吸收带进行归属。请记住，正确答案必须连贯地解释全部给定数据，而非部分数据。通过系统思考，综合结构题将成为有成就感的智力游戏，而非令人生畏的挑战。
Published by TutorHao | Chemistry Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图