📚 Alkenes: A-Level Edexcel Chemistry Revision | 烯烃 A-Level Edexcel 化学考点精讲
Alkenes are unsaturated hydrocarbons containing at least one carbon-carbon double bond (C=C). They are fundamental to organic chemistry, serving as the feedstock for polymers and a wide variety of synthetic transformations. For Edexcel A-Level Chemistry, mastering the structure, bonding, stereochemistry, and characteristic electrophilic addition reactions of alkenes is essential for both the written examinations and practical assessments.
烯烃是含有至少一个碳碳双键 (C=C) 的不饱和烃。它们是有机化学的基础，是聚合物和多种合成转化的重要原料。对于 Edexcel A-Level 化学而言，掌握烯烃的结构、键合、立体化学以及其典型的亲电加成反应，对笔试和实验评估都至关重要。
1. Structure and Bonding in Alkenes | 烯烃的结构与键合
The carbon atoms of the C=C double bond use sp² hybridisation. Each carbon forms three σ bonds using three sp² hybrid orbitals, which lie in a plane with 120° bond angles. The C=C double bond consists of one σ bond and one π bond. The π bond is formed by the sideways overlap of the remaining unhybridised 2p orbital on each carbon atom, creating an electron density region above and below the plane of the molecule. This π bond restricts rotation, giving rise to stereoisomerism.
碳碳双键中的碳原子采用 sp² 杂化。每个碳原子利用三个 sp² 杂化轨道形成三个 σ 键，这些原子位于同一平面，键角为 120°。碳碳双键由一个 σ 键和一个 π 键组成。π 键由每个碳原子上剩余的未杂化 2p 轨道侧向重叠形成，在分子平面的上方和下方产生电子云区域。该 π 键限制了旋转，从而导致了立体异构现象。
The double bond is an area of high electron density, making alkenes susceptible to attack by electrophiles (electron-pair acceptors). Typical bond lengths and strengths illustrate the difference: a C=C bond (~134 pm) is shorter and stronger than a C–C single bond (~154 pm), but the π bond is weaker than the σ bond and breaks more readily during reactions.
双键区域具有较高的电子云密度，使得烯烃容易受到亲电试剂（电子对受体）的攻击。典型的键长和键能可以说明差异：C=C 键（约 134 pm）比 C–C 单键（约 154 pm）更短、更强，但 π 键比 σ 键弱，在反应中更容易断裂。
2. Sigma and Pi Bonds | σ键与π键
A σ bond forms by the head-on overlap of atomic orbitals along the internuclear axis. All single covalent bonds are σ bonds. In ethene, each carbon forms two σ bonds with hydrogen atoms and one σ bond with the other carbon, giving a trigonal planar arrangement. The π bond is the result of parallel overlap of p orbitals perpendicular to the molecular plane; electron density in a π bond is concentrated above and below the bond axis. This model explains why the molecule is flat and why rotation about the C=C is restricted — rotating would break the π overlap.
σ 键由原子轨道沿核间轴正面重叠形成。所有单共价键都是 σ 键。在乙烯中，每个碳与氢原子形成两个 σ 键，与另一个碳形成一个 σ 键，形成平面三角形结构。π 键是垂直于分子平面的 p 轨道平行重叠的结果；π 键的电子密度集中在键轴的上方和下方。该模型解释了为什么分子是平面的，以及为什么围绕 C=C 的旋转受到限制——旋转会破坏 π 重叠。
The presence of the π bond also explains the reactivity of alkenes. Because the sideways overlap is less effective than head-on overlap, the π electrons are held less tightly and can be donated to an electrophile. In curly arrow mechanisms, the π bond is shown attacking the electrophile, breaking to form a new σ bond.
π 键的存在也解释了烯烃的反应活性。由于侧向重叠不如正面重叠有效，π 电子被束缚得较松，可以提供给亲电试剂。在弯箭头机理中，表现为 π 键进攻亲电试剂，断裂并形成新的 σ 键。
3. Stereoisomerism: E/Z Isomerism | 立体异构：E/Z异构
Stereoisomers have the same structural formula but differ in the spatial arrangement of atoms. Due to the restricted rotation of the C=C double bond, alkenes can exist as E/Z isomers (geometric isomers). This occurs when each carbon of the double bond is attached to two different groups. The traditional cis/trans terminology is used when two identical groups are on the same side (cis) or opposite sides (trans) of the double bond, but E/Z notation is more general and required by IUPAC for all tri- or tetrasubstituted alkenes.
立体异构体具有相同的结构式，但原子在空间中的排列不同。由于 C=C 双键的旋转受限，烯烃可以存在 E/Z 异构体（几何异构体）。当双键上的每个碳都与两个不同的基团相连时，就会发生这种情况。当双键同侧或对侧含有两个相同基团时，可使用传统的顺/反（cis/trans）术语，但 E/Z 标记法更普遍，是 IUPAC 对所有三取代或四取代烯烃的要求。
4. Cahn-Ingold-Prelog (CIP) Rules | Cahn-Ingold-Prelog规则
The CIP rules assign priority to the substituents attached to each carbon of the C=C bond based on atomic number: the higher the atomic number of the atom directly bonded to the double-bonded carbon, the higher its priority. If the atoms are identical, look at the next atoms along the chain until a difference is found. Once priorities are assigned for both carbons, if the two highest-priority groups are on the same side of the double bond, the configuration is Z (from German zusammen, together). If they are on opposite sides, it is E (from German entgegen, opposite).
CIP 规则根据原子序数确定与 C=C 双键上每个碳相连的取代基的优先顺序：直接与双键碳相连的原子原子序数越高，优先级越高。如果原子相同，则沿着链观察后续原子，直至发现差异。为两个碳确定优先级后，如果两个最高优先级的基团处于双键的 同侧，构型为 Z（源自德语 zusammen，意为一起）。如果处于 对侧，则为 E（源自德语 entgegen，意为相反）。
For example, in 1-bromo-1-chloro-2-fluoroethene, bromine (Z=35) has priority over chlorine (Z=17) on one carbon, and fluorine (Z=9) has priority over hydrogen (Z=1) on the other. If Br and F are on the same side, the isomer is Z; if they are opposite, it is E. Edexcel exam questions frequently require students to label or draw E and Z isomers for given alkenes.
例如，在 1-溴-1-氯-2-氟乙烯中，一个碳上溴（Z=35）的优先级高于氯（Z=17），另一个碳上氟（Z=9）的优先级高于氢（Z=1）。如果 Br 和 F 在同侧，异构体为 Z；如果在对侧，则为 E。Edexcel 考试经常要求学生为给定烯烃标注或绘制 E 和 Z 异构体。
5. Electrophilic Addition Mechanism | 亲电加成机理
Alkenes undergo electrophilic addition reactions, in which the π bond attacks an electrophile, leading to addition across the double bond. The mechanism proceeds via the formation of a carbocation intermediate (or an equivalent cyclic intermediate in some cases). The general two-step mechanism for addition of H–X (where X = halide) is:
烯烃发生亲电加成反应，其中 π 键进攻亲电试剂，导致双键加成。其机理通过形成碳正离子中间体（或在某些情况下等价的环状中间体）进行。H–X（X 为卤素）加成的通用两步机理为：
Step 1: The π electrons attract the δ+ hydrogen of H–X, causing heterolytic fission of the H–X bond. The π bond breaks, forming a C–H σ bond and a planar carbocation on the adjacent carbon, while X⁻ is released.
第一步：π 电子吸引 H–X 的 δ+ 氢，引起 H–X 键的异裂。π 键断裂，形成一个 C–H σ 键，并在相邻碳上形成平面结构的碳正离子，同时释放出 X⁻。
Step 2: The halide ion X⁻ donates a lone pair to the carbocation, forming a new C–X σ bond. This completes the addition, producing a saturated haloalkane product. Curly arrows are used to show movement of electron pairs, starting from the π bond or a lone pair and pointing towards the accepting atom.
第二步：卤离子 X⁻ 将孤对电子提供给碳正离子，形成新的 C–X σ 键。至此加成完成，生成饱和卤代烷产物。使用弯箭头表示电子对的移动，从 π 键或孤对电子出发指向接受电子的原子。
The rate-determining step is formation of the carbocation. Therefore, factors that stabilise the carbocation (inductive effects, hyperconjugation) will affect the reaction rate and regioselectivity.
反应的决速步是碳正离子的生成。因此，稳定碳正离子的因素（诱导效应、超共轭）将影响反应速率和区域选择性。
6. Addition of Hydrogen Halides & Markovnikov’s Rule | 卤化氢加成与马氏规则
When HBr, HCl, or HI adds to an unsymmetrical alkene, two isomeric products are possible depending on which carbon the hydrogen attaches to. Markovnikov’s rule states that in the addition of H–X to an unsymmetrical alkene, the hydrogen atom bonds to the carbon of the double bond that already has the greater number of hydrogen atoms (or the fewer alkyl substituents) — meaning the halogen attaches to the more substituted carbon. This is because the more substituted carbocation formed in the mechanism is more stable: tertiary > secondary > primary.
当 HBr、HCl 或 HI 与不对称烯烃加成时，根据氢所连接的碳不同，可能产生两种异构体产物。马氏规则指出，在 H–X 与不对称烯烃的加成中，氢原子连接到双键上本来拥有较多氢原子的碳上（即烷基取代较少的碳）——这意味着卤素连接到取代较多的碳上。这是因为机理中形成的取代较多的碳正离子更稳定：叔 > 仲 > 伯。
For example, propene + HBr → 2-bromopropane (major) via the more stable secondary carbocation, not 1-bromopropane, which would require a primary carbocation. Edexcel students must be able to predict major products using carbocation stability principles.
例如，丙烯 + HBr 主要生成 2-溴丙烷，通过更稳定的仲碳正离子，而不是需要伯碳正离子的 1-溴丙烷。Edexcel 学生必须能够利用碳正离子稳定性原则预测主要产物。
7. Addition of Halogens (Bromination) | 卤素加成（溴化反应）
Alkenes react rapidly with bromine or chlorine at room temperature in an electrophilic addition reaction. The mechanism for bromine addition is slightly different: as the Br₂ molecule approaches the π bond, a dipole is induced (δ+ Br–Br δ−). The π electrons attack the partially positive bromine, leading to the formation of a three-membered cyclic bromonium ion and a bromide ion. This cyclic intermediate then undergoes backside attack by Br⁻, opening the ring to give a vicinal dibromoalkane with anti addition stereochemistry.
烯烃在室温下与溴或氯迅速发生亲电加成反应。溴加成的机理略有不同：当 Br₂ 分子接近 π 键时，会诱导产生偶极（δ+ Br–Br δ−）。π 电子进攻带部分正电荷的溴，形成三元环状溴鎓离子和一个溴离子。然后该环状中间体被 Br⁻ 从背面进攻，开环生成邻二溴代烷，具有反式加成立体化学。
The decolorisation of orange-brown bromine water is a classic test for unsaturation (C=C). Alkanes do not react with bromine water in the dark. This reaction is often used in practical assessments to distinguish an alkene from an alkane.
橙棕色溴水的褪色是检验不饱和度（C=C）的经典方法。烷烃在黑暗条件下不与溴水反应。该反应常用于实验评估中以区分烯烃和烷烃。
8. Addition of Sulfuric Acid and Hydration | 硫酸加成与水合反应
Alkenes react with cold concentrated sulfuric acid to form alkyl hydrogen sulfates. The mechanism is electrophilic addition with H–OSO₂OH acting as the electrophile. The initial product can then be hydrolysed by warming with water to produce an alcohol, with regeneration of sulfuric acid. This is an indirect hydration method. For example, ethene + H₂SO₄ → ethyl hydrogen sulfate, then hydrolysis yields ethanol.
烯烃与冷的浓硫酸反应生成烷基硫酸氢酯。其机理为亲电加成，H–OSO₂OH 作为亲电试剂。然后，初始产物与水一起加热水解可生成醇，同时硫酸再生。这是一种间接水合方法。例如，乙烯 + H₂SO₄ 生成硫酸氢乙酯，然后水解得到乙醇。
Direct industrial hydration of alkenes to alcohols uses steam with a phosphoric(V) acid catalyst (H₃PO₄) at high temperature (300 °C) and high pressure (60 atm). The addition follows Markovnikov’s rule for unsymmetrical alkenes: propan-2-ol is the major product from propene, not propan-1-ol.
烯烃直接工业水合制醇使用水蒸气与磷酸(V)催化剂（H₃PO₄），在高温（300 °C）和高压（60 atm）下进行。对于不对称烯烃，加成遵循马氏规则：丙烯的主要产物为丙-2-醇，而非丙-1-醇。
9. Hydrogenation | 加氢反应
Alkenes react with hydrogen in the presence of a metal catalyst (finely divided nickel, palladium, or platinum) to form alkanes. The hydrogen molecule adsorbs onto the catalyst surface, where the H–H bond breaks and the alkene π bond also weakens. This is an example of heterogeneous catalysis. The reaction is exothermic; the enthalpy change for hydrogenation can be used to compare the stability of isomeric alkenes — the less stable alkene releases more energy upon hydrogenation.
烯烃在金属催化剂（细粉状镍、钯或铂）存在下与氢气反应生成烷烃。氢分子吸附在催化剂表面，在那里 H–H 键断裂，烯烃的 π 键也被削弱。这是多相催化的一个例子。该反应是放热的；加氢反应的焓变可用于比较异构烯烃的稳定性——稳定性较差的烯烃在加氢时会释放更多能量。
Hydrogenation is used industrially to convert unsaturated vegetable oils into solid margarine (partial hydrogenation) and to produce alkanes from cracking products. In equations, it is typically written as: alkene + H₂ → alkane, with Ni catalyst and heat.
加氢反应在工业上用于将不饱和植物油转化为固态人造黄油（部分加氢），以及从裂化产物生产烷烃。在方程式中，通常表示为：烯烃 + H₂ → 烷烃，使用 Ni 催化剂并加热。
10. Oxidation Reactions of Alkenes | 烯烃的氧化反应
Alkenes can be oxidised by potassium manganate(VII) (KMnO₄) under different conditions, resulting in different products. With cold, dilute, alkaline KMnO₄ at room temperature, alkenes are oxidised to diols (glycerol-like products). The purple solution turns green and then a brown precipitate of MnO₂ may form. This reaction is also used as a test for unsaturation.
烯烃可在不同条件下被高锰酸钾 (KMnO₄) 氧化，生成不同产物。在室温下使用冷的、稀的碱性 KMnO₄，烯烃被氧化为二醇（邻二醇）。紫色溶液变绿，然后可能形成 MnO₂ 的棕色沉淀。该反应也可用于不饱和度的检验。
With hot, concentrated, acidified KMnO₄, the double bond undergoes oxidative cleavage. Depending on the substitution pattern of the alkene, products can be ketones, aldehydes (which are further oxidised to carboxylic acids), or carbon dioxide. This reaction is useful in structural determination: for example, oxidation of but-2-ene gives ethanoic acid only, while oxidation of 2-methylpropene gives propanone and carbon dioxide.
在热的、浓的、酸化的 KMnO₄ 条件下，双键发生氧化断裂。根据烯烃的取代形式，产物可以是酮、醛（会进一步氧化为羧酸）或二氧化碳。该反应在结构确定中非常有用：例如，丁-2-烯的氧化仅生成乙酸，而 2-甲基丙烯的氧化生成丙酮和二氧化碳。
11. Addition Polymerisation | 加成聚合
Alkenes and substituted alkenes can undergo addition polymerisation to form long-chain polymers, which are of enormous industrial significance. In the presence of an initiator (e.g., an organic peroxide) and often high pressure, the π bond of many alkene molecules breaks open, and the monomers join together via σ bonds. The polymer has the same empirical formula as the monomer because no other product is formed. Common examples include poly(ethene) (from ethene), poly(propene) (from propene), poly(chloroethene) or PVC (from chloroethene), and poly(tetrafluoroethene) (PTFE).
烯烃及其取代物可以发生加成聚合反应，形成长链聚合物，这在工业上具有重大意义。在引发剂（如有机过氧化物）和通常高压的条件下，许多烯烃分子的 π 键打开，单体通过 σ 键连接起来。由于没有其他产物生成，聚合物的经验式与单体相同。常见的例子包括聚乙烯（由乙烯制得）、聚丙烯（由丙烯制得）、聚氯乙烯（PVC，由氯乙烯制得）和聚四氟乙烯（PTFE）。
Students must be able to represent the repeating unit of a polymer given the monomer, or deduce the monomer from a given polymer segment. The repeating unit is drawn by showing two single bonds extending beyond brackets, and for substituted monomers, the side groups must be correctly positioned.
学生必须能够根据给定单体表示聚合物的重复单元，或者从给定的聚合物片段推断单体。重复单元的绘制方法是在括号外延伸两个单键，对于取代单体，侧基的位置必须正确放置。
12. Testing for Alkenes: Bromine Water | 烯烃检验：溴水
The most straightforward chemical test for the presence of a C=C double bond is shaking the sample with orange-brown bromine water. If an alkene is present, the solution decolorises rapidly from orange-brown to colourless due to the electrophilic addition of bromine across the double bond, forming a dibromoalkane. This reaction does not require UV light, distinguishing alkenes from alkanes, which react with bromine only under UV via free-radical substitution (in which HBr gas is evolved and the colour fades slowly).
检测 C=C 双键最直接的化学方法是将样品与橙棕色溴水一起振荡。如果存在烯烃，由于溴通过亲电加成到双键上形成二溴代烷，溶液会迅速从橙棕色变为无色。此反应无需紫外光，从而将烯烃与烷烃区分开来，后者仅在紫外光下通过自由基取代反应与溴反应（同时释放 HBr 气体，颜色缓慢褪去）。
Similarly, the cold dilute KMnO₄ test (Baeyer’s test) will decolorise from purple with the formation of a brown MnO₂ precipitate, providing a second confirmatory test for unsaturation. Both tests are required knowledge for Edexcel practical work and examinations.
类似地，冷稀的 KMnO₄ 试验（Baeyer 试验）会从紫色褪色，并生成 MnO₂ 棕色沉淀，为不饱和度提供了第二种确认试验。这两项试验均为 Edexcel 实验操作和考试要求掌握的内容。
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