Organic Analysis | 有机分析

📚 Organic Analysis | 有机分析

Organic analysis is the branch of analytical chemistry that identifies the functional groups and molecular structure of organic compounds. In the AQA A-Level specification, this topic focuses on three core techniques: test-tube reactions for functional group identification, mass spectrometry for relative molecular mass and structure, and infrared spectroscopy for bond detection.

有机分析是分析化学中用于鉴别有机化合物官能团和分子结构的重要分支。在 AQA A-Level 考纲中，本主题聚焦三大核心技术：试管反应鉴定官能团、质谱法测定相对分子质量与结构、红外光谱法检测化学键。

1. Chemical Tests for Alcohols | 醇的化学检验

Primary and secondary alcohols can be oxidised by acidified potassium dichromate(VI) (K₂Cr₂O₇/H⁺). The orange dichromate ion (Cr₂O₇²⁻) is reduced to the green chromium(III) ion (Cr³⁺), providing a clear colour change. Primary alcohols are oxidised first to aldehydes and then to carboxylic acids under reflux; secondary alcohols are oxidised to ketones. Tertiary alcohols do not undergo oxidation because there is no hydrogen atom attached to the carbon bearing the −OH group.

伯醇和仲醇均可被酸化重铬酸钾（K₂Cr₂O₇/H⁺）氧化。橙色的重铬酸根离子（Cr₂O₇²⁻）被还原为绿色的三价铬离子（Cr³⁺），产生明显的颜色变化。伯醇先被氧化为醛，再在回流条件下继续氧化为羧酸；仲醇被氧化为酮。叔醇由于连接 −OH 的碳原子上没有氢原子，故不能发生氧化反应。

CH₃CH₂OH + [O] → CH₃CHO + H₂O
CH₃CHO + [O] → CH₃COOH

A positive test result is shown by an orange-to-green colour change, which indicates the presence of a primary or secondary alcohol. To distinguish between the two, the product can be tested further with Tollens’ reagent (see Section 2).

阳性结果表明颜色由橙色变为绿色，说明存在伯醇或仲醇。若要进一步区分二者，可使用 Tollen’s 试剂检验氧化产物（见第 2 节）。

Key exam point: the reagent must be acidified K₂Cr₂O₇; without acid, the oxidation is not effective. Also, distillation is used to collect the aldehyde before further oxidation occurs.

考试要点：试剂必须是酸化 K₂Cr₂O₇；无酸条件下氧化无法有效进行。此外，需采用蒸馏法在醛被进一步氧化之前将其收集出来。

2. Distinguishing Aldehydes from Ketones | 区分醛与酮

Aldehydes and ketones both contain the carbonyl group (C=O), but aldehydes have the carbonyl carbon bonded to at least one hydrogen atom. This structural difference means aldehydes can be oxidised while ketones cannot. Two classic reagents exploit this difference: Tollens’ reagent and Fehling’s solution.

醛和酮均含有羰基（C=O），但醛的羰基碳至少连接一个氢原子。这一结构差异导致醛可被氧化而酮不能。两种经典试剂正是利用了这一差异：Tollen’s 试剂和 Fehling 溶液。

Tollens’ reagent is an alkaline solution of silver nitrate in ammonia, containing the diamminesilver(I) ion [Ag(NH₃)₂]⁺. When warmed with an aldehyde, the aldehyde is oxidised to a carboxylate salt while Ag⁺ is reduced to metallic silver, depositing as a silver mirror on the test tube wall. Ketones do not react.

Tollen’s 试剂是硝酸银的氨性碱溶液，含有二氨合银(I)配离子 [Ag(NH₃)₂]⁺。与醛共热时，醛被氧化为羧酸盐，同时 Ag⁺ 被还原为金属银，在试管壁上沉积形成银镜。酮不发生反应。

RCHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → RCOO⁻ + 2Ag↓ + 2NH₄⁺ + 2NH₃ + H₂O

Fehling’s solution contains Cu²⁺ ions complexed in alkaline tartrate solution. On warming with an aldehyde, a brick-red precipitate of copper(I) oxide (Cu₂O) forms. Ketones give no reaction, so the blue solution remains unchanged. Note that aromatic aldehydes (e.g. benzaldehyde) give a positive Tollens’ test but do not react with Fehling’s solution.

Fehling 溶液含有碱性酒石酸络合铜离子 Cu²⁺。与醛共热时，生成砖红色的氧化亚铜（Cu₂O）沉淀。酮无反应，蓝色溶液保持不变。须注意：芳香醛（如苯甲醛）可发生银镜反应，但不与 Fehling 溶液反应。

3. Testing for Carboxylic Acids | 羧酸的检验

Carboxylic acids are weak acids that react with carbonates and hydrogencarbonates to release carbon dioxide gas. Adding solid sodium carbonate (Na₂CO₃) or sodium hydrogencarbonate (NaHCO₃) to an unknown compound produces effervescence (bubbles of CO₂) if a carboxylic acid is present.

羧酸是弱酸，能与碳酸盐和碳酸氢盐反应放出二氧化碳气体。向未知化合物中加入固体碳酸钠（Na₂CO₃）或碳酸氢钠（NaHCO₃），若产生气泡（CO₂）则表明存在羧酸。

RCOOH + NaHCO₃ → RCOONa + H₂O + CO₂↑

The gas can be confirmed by passing it through limewater, which turns milky. Phenols also contain an −OH group but are too weakly acidic to react with NaHCO₃, so this test cleanly distinguishes carboxylic acids from phenols and alcohols.

可将生成的气体通入石灰水中进行确认，石灰水变浑浊。苯酚虽含 −OH 基团，但酸性太弱，不与 NaHCO₃ 反应，因此该试验可清晰地鉴别羧酸与酚类和醇类物质。

Exam note: carboxylic acids should also be tested with pH indicator paper (pH 4–5 for typical aliphatic acids) and by their neutralisation reaction with NaOH, which is simply a qualitative confirmation. The NaHCO₃ effervescence test is the required AQA test-tube reaction.

考试提示：羧酸还可用 pH 试纸测试（常见脂肪族酸 pH 为 4–5），以及与 NaOH 的中和反应作为定性确认。NaHCO₃ 起泡实验是 AQA 要求掌握的试管反应。

4. Testing for Haloalkanes | 卤代烃的检验

Haloalkanes are not directly reactive with silver nitrate in the cold because the carbon–halogen bond is covalent. To test for the presence of a halogen atom, the haloalkane must first be hydrolysed by warming with aqueous sodium hydroxide. The halide ion is then released into solution and can be detected by adding nitric acid followed by silver nitrate solution.

卤代烃在常温下不与硝酸银直接反应，因为碳—卤键是共价键。为检验卤原子的存在，必须先加入氢氧化钠水溶液并加热使卤代烃水解，将卤离子释放到溶液中，再加入硝酸酸化，然后滴加硝酸银溶液进行检测。

R–X + OH⁻ → R–OH + X⁻

The precipitate formed identifies the halogen:

生成的沉淀颜色可鉴别卤素的种类：

Chloride (Cl⁻): white precipitate of AgCl, soluble in dilute ammonia
氯化物（Cl⁻）：AgCl 白色沉淀，可溶于稀氨水
Bromide (Br⁻): cream precipitate of AgBr, soluble in concentrated ammonia
溴化物（Br⁻）：AgBr 淡奶油色沉淀，可溶于浓氨水
Iodide (I⁻): yellow precipitate of AgI, insoluble in ammonia
碘化物（I⁻）：AgI 黄色沉淀，不溶于氨水

Nitric acid must be added first to remove excess OH⁻ ions, which would otherwise react with Ag⁺ to give a brown precipitate of Ag₂O and interfere with the result.

必须先加入硝酸以除去过量的 OH⁻ 离子，否则 OH⁻ 会与 Ag⁺ 反应生成棕色 Ag₂O 沉淀，干扰实验结果。

5. Alkenes: Bromine Water Test | 烯烃：溴水检验

Alkenes contain a C=C double bond, which undergoes an addition reaction with bromine. When bromine water (an orange-brown solution) is added to an alkene, the colour rapidly decolorises as 1,2-dibromo compounds form. Saturated compounds such as alkanes do not decolorise bromine water in the dark, although they do react with bromine vapour in UV light via substitution.

烯烃含有 C=C 双键，能与溴发生加成反应。将橙棕色的溴水加入烯烃中，随着 1,2-二溴代物的生成，溴水迅速褪色。饱和化合物如烷烃在暗处不能使溴水褪色（烷烃仅在紫外光下与溴蒸气发生取代反应）。

C₂H₄ + Br₂ → CH₂BrCH₂Br

This test is a quick, reliable way to distinguish an alkene from an alkane or arene. The disappearance of the orange colour is the positive result; do not confuse it with the orange-to-green oxidation test used for alcohols.

该试验是鉴别烯烃与烷烃、芳烃的快速可靠方法。橙色消失即为阳性结果；注意勿将其与醇类的橙色变绿色氧化试验混淆。

6. Mass Spectrometry: Molecular Ion | 质谱法：分子离子峰

In a mass spectrometer, the molecule is bombarded by high-energy electrons, causing it to lose an electron and form a molecular ion M⁺. The m/z value (mass-to-charge ratio) of the molecular ion gives the relative molecular mass (Mᵣ). The molecular ion peak is usually the peak with the highest m/z value, excluding the M+1 and M+2 isotope peaks.

在质谱仪中，分子被高能电子轰击，失去一个电子形成分子离子 M⁺。分子离子的 m/z 值（质荷比）即为相对分子质量（Mᵣ）。分子离子峰通常是 m/z 值最大的峰，不含 M+1 和 M+2 同位素峰。

The M+1 peak arises from molecules containing one ¹³C atom instead of ¹²C. Since ¹³C has a natural abundance of about 1.1%, the M+1 peak height relative to M⁺ can indicate the number of carbon atoms: approximately 1.1% per carbon atom. For a compound containing n carbon atoms, the M+1 peak is roughly 1.1n% of the M⁺ peak.

M+1 峰来自含有 ¹³C 而非 ¹²C 的分子。由于 ¹³C 的天然丰度约为 1.1%，M+1 峰相对于 M⁺ 峰的高度可用于推断碳原子数：每含一个碳原子约贡献 1.1%。含 n 个碳原子的化合物的 M+1 峰约为 M⁺ 峰的 1.1n%。

High-resolution mass spectrometry can measure m/z values to four decimal places, allowing the precise atomic masses of elements (e.g. ¹²C = 12.0000, ¹⁶O = 15.9949, ¹⁴N = 14.0031) to be used to determine the molecular formula uniquely.

高分辨质谱可将 m/z 值精确至小数点后四位，利用元素精确原子质量（如 ¹²C = 12.0000、¹⁶O = 15.9949、¹⁴N = 14.0031）唯一确定分子式。

7. Fragmentation Patterns | 碎片离子峰型

When the molecular ion is formed, it often possesses excess energy and fragments into smaller pieces. Only positively charged fragments are detected, giving rise to a fragmentation pattern characteristic of the compound. These fragments appear as peaks at lower m/z values and are crucial for structural elucidation.

分子离子形成后通常带有过剩能量，会碎裂成更小的碎片。只有带正电荷的碎片才能被检测到，从而形成化合物特有的碎片峰型。这些碎片以较低的 m/z 值出现在谱图中，对结构解析至关重要。

Common fragmentation patterns include:

常见的碎片模式包括：

Loss of a methyl radical (•CH₃): M − 15 peak
失去甲基自由基（•CH₃）：产生 M − 15 峰
Loss of water from alcohols: M − 18 peak
醇失去水分子：产生 M − 18 峰
Loss of carbon monoxide from aldehydes: M − 28 peak
醛失去一氧化碳：产生 M − 28 峰
Formation of acylium ion RCO⁺ from ketones/aldehydes
酮/醛生成酰基阳离子 RCO⁺
Loss of HCl from chloroalkanes: M − 36
氯代烷失去 HCl：产生 M − 36 峰

Worked example: Propanal (Mᵣ = 58) shows peaks at m/z 57 (loss of H), m/z 43 (CH₃CO⁺), m/z 29 (CHO⁺), and m/z 15 (CH₃⁺). The peak at m/z 43 is particularly abundant because the acylium cation is resonance-stabilised.

实例：丙醛（Mᵣ = 58）的质谱显示 m/z 57（失去 H）、m/z 43（CH₃CO⁺）、m/z 29（CHO⁺）和 m/z 15（CH₃⁺）等峰。其中 m/z 43 峰特别强，因为酰基阳离子具有共振稳定化作用。

8. Infrared Spectroscopy: Fundamentals | 红外光谱：基本原理

Infrared (IR) spectroscopy measures the absorption of infrared radiation by covalent bonds. When IR radiation matches the vibrational frequency of a bond, the bond absorbs energy and vibrates more vigorously. The absorption frequency (reported as wavenumber, ṽ, in cm⁻¹) depends on the bond type and the atoms involved.

红外光谱（IR）测量共价键对红外辐射的吸收。当红外辐射频率与化学键的振动频率匹配时，化学键吸收能量并增强振动。吸收频率（以波数 ṽ 表示，单位 cm⁻¹）取决于键的类型及所涉及的原子。

Key absorption ranges to memorise for AQA:

AQA 需记忆的关键吸收范围：

Bond	Wavenumber range / cm⁻¹	Structural context
O–H (alcohol)	3230–3550 (broad)	Hydrogen-bonded hydroxyl
O–H (carboxylic acid)	2500–3300 (very broad)	Strong hydrogen bonding
C=O	1630–1820 (sharp, strong)	Aldehydes, ketones, acids
C–O	1000–1300	Esters, ethers, alcohols
C–H	2850–3100	Alkane / alkene / arene
N–H (amine/amide)	3200–3500	Primary amine: two peaks
C≡N (nitrile)	2220–2260	Sharp, medium intensity

A carbonyl group is a powerful diagnostic feature: its sharp, strong absorption around 1700 cm⁻¹ is usually the most prominent peak in the spectrum. The exact position of the C=O stretch varies: aldehydes absorb near 1725 cm⁻¹, ketones near 1715 cm⁻¹, and carboxylic acids near 1710 cm⁻¹ due to conjugation and hydrogen bonding effects.

羰基是极强的诊断特征：其 1700 cm⁻¹ 附近的吸收尖锐且强，通常是谱图中最显著的峰。C=O 伸缩振动的准确位置随结构而变：醛约在 1725 cm⁻¹，酮约在 1715 cm⁻¹，羧酸约在 1710 cm⁻¹，这是由于共轭和氢键效应所致。

9. Interpreting IR Spectra | 红外谱图解析

To interpret an IR spectrum systematically, follow these steps. First, check the region above 3000 cm⁻¹ for O–H or N–H stretches. A broad peak at 2500–3300 cm⁻¹ indicates a carboxylic acid — this is a highly distinctive signature. A narrower broad peak at 3230–3550 cm⁻¹ suggests an alcohol. Then, look for a strong C=O peak around 1700 cm⁻¹, which confirms a carbonyl-containing functional group.

系统解析红外谱图可遵循以下步骤。首先检查 3000 cm⁻¹ 以上的区域，寻找 O–H 或 N–H 伸缩振动。2500–3300 cm⁻¹ 处的宽峰说明是羧酸——这是极具特征性的指纹信号。3230–3550 cm⁻¹ 处的较窄宽峰提示醇的存在。随后寻找 1700 cm⁻¹ 附近的强 C=O 峰，以确认是否存在含羰基官能团。

Worked example: A compound with Mᵣ = 60 gives an IR spectrum with a very broad peak at 2500–3300 cm⁻¹ and a strong peak at 1710 cm⁻¹, but no peak above 3300 cm⁻¹. The molecular formula could be C₂H₄O₂ or C₃H₈O. The IR spectrum excludes the alcohol (C₃H₈O would show an O–H at 3230–3550 but no C=O), so the compound is ethanoic acid, CH₃COOH.

实例解析：某化合物 Mᵣ = 60，红外谱图显示 2500–3300 cm⁻¹ 处有极宽峰、1710 cm⁻¹ 处有强峰，3300 cm⁻¹ 以上无吸收。可能的分子式为 C₂H₄O₂ 或 C₃H₈O。红外谱图排除了醇类（C₃H₈O 应在 3230–3550 cm⁻¹ 出现 O–H，但无 C=O），因此该化合物为乙酸 CH₃COOH。

Exam tip: in AQA questions, you will often be given the molecular formula and two candidate structures; use IR to choose between them. The presence or absence of C=O and the exact O–H region are the deciding factors.

考试技巧：AQA 考题常提供分子式与两个候选结构，要求用红外光谱选出正确答案。C=O 的有无以及 O–H 吸收峰的具体位置是判断的关键。

10. Combining Analytical Techniques | 综合分析技术联用

Modern structure determination relies on combining multiple techniques. The complete analytical strategy for an unknown organic compound typically proceeds as follows:

现代结构测定依赖多种技术联用。对未知有机化合物的完整分析策略通常按以下步骤进行：

Elemental analysis / combustion analysis: provides the empirical formula.
元素分析 / 燃烧分析：提供实验式。
Mass spectrometry: provides Mᵣ and hence the molecular formula from the empirical formula; fragmentation helps identify alkyl chains and functional groups.
质谱法：提供 Mᵣ，从而结合实验式确定分子式；碎片峰有助于识别烷基链和官能团。
Infrared spectroscopy: confirms functional groups such as O–H, C=O, C≡N, N–H.
红外光谱：确认 O–H、C=O、C≡N、N–H 等官能团。
Chemical tests: confirm or eliminate specific functional groups (e.g. Tollens’ test for aldehydes).
化学检验：确认或排除特定官能团（如银镜反应检验醛）。

Integrated worked example: An unknown compound X has Mᵣ = 74. Elemental analysis gives C: 64.9%, H: 13.5%, O: 21.6%. IR shows a broad O–H at 3350 cm⁻¹ but no C=O. Mass spectrum shows peaks at m/z 74 (M⁺), 59 (M − 15), 45 (M − 29), and 31 (CH₂OH⁺). Determine the structure.

综合实例：未知化合物 X 的 Mᵣ = 74。元素分析给出 C: 64.9%、H: 13.5%、O: 21.6%。红外显示 3350 cm⁻¹ 有宽 O–H 峰，无 C=O。质谱显示 m/z 74（M⁺）、59（M − 15）、45（M − 29）和 31（CH₂OH⁺）峰。请推断其结构。

Step 1: divide by atomic masses to find the mole ratio — C: 64.9/12 = 5.41; H: 13.5/1 = 13.5; O: 21.6/16 = 1.35. Divide by 1.35 gives C₄H₁₀O as the empirical formula, which matches Mᵣ = 74, so molecular formula is C₄H₁₀O. Since IR shows O–H but no C=O, this is a saturated alcohol. The M − 15 peak indicates loss of CH₃; M − 29 suggests loss of CHO or C₂H₅. The fragmentation pattern with CH₂OH⁺ (m/z 31) indicates a primary alcohol. Among the butanol isomers, butan-1-ol (CH₃CH₂CH₂CH₂OH) produces CH₂OH⁺ upon cleavage; the M − 29 loss of C₂H₅ also fits. Therefore X is butan-1-ol.

步骤 1：用原子量除得摩尔比——C: 64.9/12 = 5.41；H: 13.5/1 = 13.5；O: 21.6/16 = 1.35。除以 1.35 得 C₄H₁₀O 为实验式，与 Mᵣ = 74 吻合，故分子式为 C₄H₁₀O。红外有 O–H 无 C=O，说明是饱和醇。M − 15 峰表明失去 CH₃；M − 29 表明失去 CHO 或 C₂H₅。m/z 31 的 CH₂OH⁺ 碎片强烈提示为伯醇。在丁醇异构体中，丁-1-醇（CH₃CH₂CH₂CH₂OH）断裂可产生 CH₂OH⁺；M − 29 失去 C₂H₅ 也吻合。因此 X 为丁-1-醇。

11. Common Errors and Exam Strategy | 常见错误与应试策略

Students frequently lose marks on organic analysis questions due to avoidable mistakes. The most common errors are: stating that tertiary alcohols decolourise acidified K₂Cr₂O₇; confusing the Tollens’ and Fehling’s results; forgetting that the NaHCO₃ test requires bubbles of CO₂, not just any gas; and quoting incorrect IR wavenumber ranges.

学生在有机分析题目中常因可避免的错误而失分。最常见的错误包括：误认为叔醇能使酸化 K₂Cr₂O₇ 褪色；混淆 Tollen’s 与 Fehling 试剂的反应结果；忘记 NaHCO₃ 试验必须产生 CO₂ 气泡而非任意气体；以及红外波数范围记忆错误。

To maximise marks in examinations:

为了在考试中最大化得分：

Always quote the exact observation (colour change, precipitate colour, gas evolved) and the reagent and conditions in test-tube reaction questions.
在试管反应题中，务必准确描述观察结果（颜色变化、沉淀颜色、气体放出），并写明试剂与反应条件。
In mass spectrometry, label the molecular ion peak as M⁺ and explain each fragment peak using correct bond cleavage arguments.
在质谱题中，标注分子离子峰 M⁺，并用正确的断键论据解释每个碎片峰。
In IR questions, connect the peak to the specific bond and functional group, not just “absorption at 1700 cm⁻¹” — state it is the C=O stretch of a carbonyl group.
在红外题中，将吸收峰与具体化学键和官能团关联，不要只写”1700 cm⁻¹ 有吸收”——应指出这是羰基的 C=O 伸缩振动。
Remember that O–H in carboxylic acids is much broader and extends to lower wavenumbers than O–H in alcohols.
牢记羧酸中 O–H 峰比醇中的更宽，且延伸到更低的波数范围。

A structured answering approach — formula → functional groups via IR → confirmation via chemical tests → fragmentation via MS — will earn full marks even for unfamiliar compounds.

采用结构化答题思路——分子式 → 红外确定官能团 → 化学检验确认 → 质谱解析碎片——即使面对未知化合物也能获得满分。

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Organic Analysis | 有机分析