📚 A-Level OCR Chemistry: Alkenes Key Points | A-Level OCR 化学：烯烃 考点精讲
Alkenes are a fundamental homologous series in organic chemistry, characterised by the presence of at least one carbon–carbon double bond. Understanding their structure, reactivity, and reaction mechanisms is essential for success in the OCR A-Level Chemistry specification. This guide breaks down the most important concepts, from nomenclature and stereoisomerism to electrophilic addition and polymerisation, providing clarity and depth for revision.
烯烃是有机化学中一个基本同系物，其特征是分子中至少含有一个碳碳双键。理解其结构、反应性和反应机理，是在OCR A-Level化学考试中取得成功的关键。本文将对烯烃最重要的知识点进行精讲，涵盖命名、立体异构、亲电加成和聚合反应，帮助同学们深入复习。
1. Introduction to Alkenes | 烯烃简介
Alkenes are unsaturated hydrocarbons with the general formula CnH2n. They contain a carbon–carbon double bond (C=C) which consists of one σ (sigma) bond and one π (pi) bond. The presence of the π bond makes alkenes much more reactive than alkanes, as the π electrons are more exposed and can be attacked by electrophiles.
烯烃是不饱和烃，通式为CnH2n。它们含有一个碳碳双键（C=C），该双键由一个σ键和一个π键组成。π键的存在使烯烃比烷烃活泼得多，因为π电子更加暴露，容易受到亲电试剂的进攻。
The double bond is a region of high electron density, which governs the characteristic reactions of alkenes: electrophilic addition. In the OCR specification, you need to be able to explain the structure and bonding in ethene as a model alkene.
双键是一个高电子密度区域，这决定了烯烃的特征反应：亲电加成。在OCR考纲中，你需要能够以乙烯为例解释烯烃的结构和成键。
2. Nomenclature of Alkenes | 烯烃的命名
Alkenes are named using the suffix ‘-ene’. The longest carbon chain containing the double bond is selected, and the chain is numbered from the end that gives the double bond the lowest possible locant. The position of the double bond is indicated by a number before the ‘-ene’ suffix. Substituents are listed as prefixes in alphabetical order.
烯烃的命名以后缀“-ene”结尾。选择包含双键的最长碳链作为主链，并从离双键最近的一端开始给碳链编号，使双键的位置编号最小。双键的位置用后缀“-ene”前的数字表示。取代基按字母顺序作为前缀列出。
For example, CH2=CHCH2CH3 is but-1-ene, while CH3CH=CHCH3 is but-2-ene. In cyclic alkenes, the double bond is assumed to be between carbon 1 and 2, and numbering proceeds to give substituents the lowest numbers.
例如，CH2=CHCH2CH3 是丁-1-烯，而 CH3CH=CHCH3 是丁-2-烯。在环烯烃中，双键通常位于1号和2号碳之间，编号从双键开始以使取代基编号最小。
3. E/Z Isomerism in Alkenes | 烯烃的E/Z异构现象
Stereoisomerism occurs in alkenes due to the restricted rotation around the C=C double bond. If each carbon of the double bond is attached to two different groups, the alkene can exist as two distinct geometric isomers: E (entgegen, opposite) and Z (zusammen, together). This is known as E/Z isomerism.
由于碳碳双键不能自由旋转，烯烃存在立体异构现象。如果双键上的每个碳原子都连接着两个不同的基团，烯烃就可以以两种不同的几何异构体存在：E（对位）和 Z（同侧），即E/Z异构。
The CIP (Cahn–Ingold–Prelog) priority rules are used to assign priorities: higher atomic number attached directly to the double-bonded carbon gets higher priority. If the two highest priority groups are on opposite sides of the double bond, the isomer is E; if they are on the same side, it is Z. Do not confuse this with the outdated cis/trans system, which only works when there are identical groups.
通过CIP（Cahn–Ingold–Prelog）次序规则确定优先基团：直接连接到双键碳上的原子序数越大，优先权越高。如果两个优先基团在双键异侧，则为E构型；如果在同侧，则为Z构型。注意不要将它与过时的顺反异构混淆，后者只适用于存在相同基团的情况。
4. Physical Properties of Alkenes | 烯烃的物理性质
Alkenes are non-polar molecules with only weak van der Waals’ forces between molecules. Their boiling points increase with increasing chain length due to greater surface contact and stronger induced dipole–induced dipole interactions. Branched alkenes have lower boiling points than their straight-chain isomers because of reduced surface area.
烯烃是非极性分子，分子间仅存在弱的范德华力。随着碳链增长，分子间接触面积增大，诱导偶极–诱导偶极作用增强，沸点升高。支链烯烃的沸点低于直链异构体，因为其表面积较小。
Alkenes are insoluble in water but dissolve in non-polar organic solvents. The E and Z isomers of an alkene may have slightly different boiling points due to differences in molecular packing, but these are often very close together.
烯烃不溶于水，但可溶于非极性有机溶剂。烯烃的E和Z异构体可能由于分子堆积方式的差异而有略微不同的沸点，但通常非常接近。
5. Electrophilic Addition Mechanism | 亲电加成反应机理
The most important reaction of alkenes is electrophilic addition. In this mechanism, the electron-rich double bond attracts an electrophile (an electron-deficient species). The π bond breaks, and the electrophile attaches to one carbon, generating a carbocation intermediate. A nucleophile then quickly bonds to the positively charged carbon, completing the addition.
烯烃最重要的反应是亲电加成。该机理中，富电子的双键吸引亲电试剂（缺电子物种）。π键断裂，亲电试剂连到其中一个碳上，生成碳正离子中间体。随后，亲核试剂迅速与带正电荷的碳成键，完成加成。
For OCR, you must be able to draw the curly arrow mechanism: a curly arrow from the C=C bond to the electrophile, showing heterolytic fission of the E–Nu bond if the electrophile is generated from a polar molecule (e.g., H–Br). A second curly arrow from the nucleophile (e.g., Br⁻) to the carbocation completes the reaction.
对OCR考试，你必须能够画出弯箭头机理：从C=C键指向亲电试剂的弯箭头，若亲电试剂来自极性分子（如H–Br），需展示E–Nu键的异裂。第二个弯箭头从亲核试剂（如Br⁻）指向碳正离子，完成反应。
6. Reaction with Hydrogen Halides & Markovnikov’s Rule | 与卤化氢的反应及马氏规则
Alkenes react with hydrogen halides (HCl, HBr, HI) to form halogenoalkanes. When the alkene is unsymmetrical, two possible products can form. Markovnikov’s rule predicts that the hydrogen atom attaches to the carbon with the greater number of hydrogen atoms already attached, and the halide attaches to the more substituted carbon. This is explained by carbocation stability: tertiary > secondary > primary > methyl. Alkyl groups are electron-donating and stabilise the positive charge through inductive effects.
烯烃与卤化氢（HCl、HBr、HI）反应生成卤代烷。当烯烃不对称时，可能形成两种产物。马氏规则预测，氢原子会加在原来连有较多氢原子的碳上，卤素加在取代较多的碳上。这可以用碳正离子稳定性解释：叔碳正离子 > 仲碳正离子 > 伯碳正离子 > 甲基碳正离子。烷基具有给电子效应，通过诱导效应稳定正电荷。
The intermediate carbocation may also undergo rearrangement to a more stable carbocation if possible, which explains some unexpected products. It is important to understand the underlying reason, not just the rule itself.
碳正离子中间体如果可能的话，还可能重排为更稳定的碳正离子，这解释了一些非预期产物的生成。重要的是理解其背后的原因，而不仅仅是规则本身。
7. Reaction with Halogens | 与卤素的反应
Alkenes react rapidly with bromine or chlorine at room temperature via an electrophilic addition mechanism. The reaction with bromine water (orange → colourless) is the classic test for unsaturation. The mechanism involves the polarisation of the Br–Br bond as it approaches the double bond, generating an induced dipole. One bromine atom acts as the electrophile, forming a cyclic bromonium ion intermediate (for OCR, the simple carbocation mechanism is often accepted, but the bromonium ion better explains anti-addition).
烯烃在室温下与溴或氯迅速发生亲电加成反应。与溴水的反应（橙色→无色）是经典的不饱和性检验。机理包括Br–Br键在接近双键时发生极化，产生诱导偶极。一个溴原子作为亲电试剂，生成环状溴鎓离子中间体（OCR常接受简单的碳正离子机理，但溴鎓离子能更好地解释反式加成）。
The product is a vicinal dihalogenoalkane. Care should be taken with curly arrow diagrams: the arrow from the double bond goes to the slightly positive bromine, and the Br⁻ released then attacks the back side of the intermediate.
产物是邻二卤代烷。画弯箭头示意图时要注意：从双键出发的箭头指向略带正电的溴，释放的Br⁻进而从中间体的背面进攻。
8. Reaction with Sulfuric Acid & Hydration | 与硫酸的反应及水合反应
Alkenes react with cold concentrated sulfuric acid to form alkyl hydrogensulfates. This is an electrophilic addition where the electrophile is H⁺ (generated from H2SO4), and the nucleophile is HSO4⁻. The alkyl hydrogensulfate can then be hydrolysed by warming with water to produce an alcohol. This is an industrial method for the hydration of ethene to ethanol. The overall reaction is the addition of H–OH across the double bond.
烯烃与冷的浓硫酸反应生成烷基硫酸氢酯。这是一个亲电加成反应，亲电试剂为H⁺（由H2SO4提供），亲核试剂为HSO4⁻。生成的烷基硫酸氢酯再与水共热水解，得到醇。这是工业上乙烯水合制乙醇的方法。总反应相当于H–OH加成到双键上。
Direct hydration of ethene with steam and a phosphoric acid catalyst (H3PO4 on a solid support) is also an essential industrial process. Conditions: 300 °C, 60 atm. This produces ethanol without the need for the acid hydrolysis step.
乙烯的直接水合使用水蒸气和磷酸催化剂（负载于固体载体上），也是一个重要的工业过程。条件为：300°C、60 atm。这种方法无需经过酸水解步骤即可制得乙醇。
9. Oxidation of Alkenes | 烯烃的氧化反应
Alkenes can be oxidised by cold, dilute, acidified potassium manganate(VII) to form diols. The purple colour of MnO4⁻ disappears, and a brown precipitate of MnO2 forms. This reaction can also be used as a test for unsaturation, alongside the bromine water test.
烯烃可被冷的、稀的酸性高锰酸钾氧化生成二醇。MnO4⁻的紫色褪去，并生成棕色MnO2沉淀。与溴水检验一样，该反应也可用于不饱和性检验。
With hot, concentrated acidified KMnO4, the double bond is completely cleaved, leading to the formation of carbonyl compounds or carbon dioxide, depending on the substitution pattern. This oxidative cleavage is used to deduce the position of the double bond in an unknown alkene.
在热的、浓的酸性KMnO4作用下，双键被完全切断，生成羰基化合物或二氧化碳，具体产物取决于取代情况。这种氧化断裂可用于推断未知烯烃中双键的位置。
10. Addition Polymerisation | 加成聚合
Alkenes can undergo addition polymerisation to form long-chain polymers. In this process, the π bond of many alkene monomers opens up and the monomers link together through σ bonds. Poly(ethene) and poly(propene) are common examples. The reaction is initiated by free radicals or by a catalyst, depending on the industrial method.
烯烃可发生加成聚合反应，形成长链高分子。在此过程中，许多烯烃单体的π键打开，单体通过σ键连接在一起。聚乙烯和聚丙烯就是常见的例子。根据工业方法的不同，反应可由自由基或催化剂引发。
You must be able to represent the polymerisation using structural formulae, and identify the repeat unit from a given polymer chain. The polymer’s properties can be modified by using different alkenes, chain lengths, and branching. Disposal issues and the environmental impact of non-biodegradable polymers also form part of the OCR syllabus.
你必须能够用结构式表示聚合反应，并从给定的聚合物链识别重复单元。通过使用不同的烯烃、链长和支化度可以改变聚合物的性质。不可降解聚合物的处理问题及环境影响也属于OCR考纲内容。
11. Test for Unsaturation | 不饱和性检验
The standard test for an alkene is to add bromine water (orange-brown) and shake. If an alkene is present, the solution rapidly decolourises. This is an addition reaction forming a colourless dibromo compound. This test distinguishes alkenes from alkanes and other saturated compounds.
检验烯烃的标准方法是加入溴水（橙棕色）并振荡。如果存在烯烃，溶液将迅速褪色。这是一个生成无色二溴化合物的加成反应。该检验可将烯烃与烷烃及其他饱和化合物区分开。
Acidified potassium manganate(VII) can also be used: the purple colour disappears and a brown precipitate forms. Remember that many other reducing agents (e.g., aldehydes) also decolourise KMnO4, so the bromine water test is more specific for C=C bonds.
也可以使用酸性高锰酸钾：紫色消失并生成棕色沉淀。但要记住，许多其他还原剂（如醛类）也能使KMnO4褪色，因此溴水检验对C=C键更具特异性。
12. Summary of Key Reactions | 关键反应总结
The following table summarises the principal reactions of alkenes that you are expected to know for the OCR A-Level examination. Being able to recall these reagents, conditions, and organic products is vital.
下表总结了OCR A-Level考试中你需要掌握的烯烃主要反应。记住这些试剂、条件和有机产物至关重要。
| Reaction | Reagents / Conditions | Product |
|---|---|---|
| Electrophilic addition of HX | HCl or HBr, room temp. | Halogenoalkane |
| Addition of Br2 or Cl2 | Bromine or chlorine, room temp. | Dihalogenoalkane |
| Hydration (indirect) | Conc. H2SO4, then H2O, warm | Alcohol |
| Hydration (direct) | H2O(g), H3PO4 catalyst, 300°C, 60 atm | Alcohol |
| Oxidation (mild) | Cold, dilute KMnO4, acidified | Diol |
| Oxidation (vigorous) | Hot, concentrated KMnO4, acidified | Carbonyls / CO2 |
| Polymerisation | Catalyst / initiator, high pressure | Poly(alkene) |
Practise drawing mechanisms for each reaction and be prepared to explain stereochemical outcomes where relevant. A solid understanding of alkene chemistry will not only help in the organic chemistry sections but also in synthesis and analytical problem-solving questions.
练习画出每个反应的机理，并准备好解释相关的立体化学结果。扎实掌握烯烃化学，不仅有助于有机化学部分，还有助于合成和分析解题。
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