A-Level Chemistry: Electrophilic Addition Reactions of Alkenes | A-Level 化学：烯烃的加成反应详解

📚 A-Level Chemistry: Electrophilic Addition Reactions of Alkenes | A-Level 化学：烯烃的加成反应详解

Alkenes are hydrocarbons that contain a carbon-carbon double bond (C=C), which consists of one sigma (σ) bond and one pi (π) bond. The presence of this π bond makes alkenes significantly more reactive than alkanes, and their most characteristic chemical behaviour is the addition reaction. In this article, we will explore the mechanism, regiochemistry, and key examples of electrophilic addition reactions of alkenes, tailored specifically for the CIE A-Level Chemistry syllabus.

烯烃是含有碳碳双键（C=C）的烃类化合物，该双键由一个 sigma（σ）键和一个 pi（π）键组成。π 键的存在使烯烃的化学反应活性远高于烷烃，而其最具特征性的化学行为便是加成反应。在本文中，我们将针对 CIE A-Level 化学考纲，深入探讨烯烃亲电加成反应的机理、区域选择性及重要实例。

1. Structure of the C=C Double Bond | 碳碳双键的结构

The carbon-carbon double bond in alkenes is formed by the overlap of atomic orbitals in two distinct ways. The sigma bond arises from the end-on overlap of two sp² hybrid orbitals, while the pi bond is formed by the side-on overlap of two unhybridised p orbitals. The electron density in the π bond is located above and below the plane of the molecule, making it exposed and readily available for interaction with electrophiles.

烯烃中的碳碳双键通过原子轨道的两种不同重叠方式形成。sigma 键由两个 sp² 杂化轨道的头碰头重叠形成，而 pi 键则由两个未杂化 p 轨道的肩并肩重叠形成。π 键的电子云分布在分子平面的上方和下方，使其暴露在外，易于与亲电试剂发生相互作用。

The π bond is weaker than the σ bond, with a bond energy of approximately 264 kJ mol⁻¹, compared to about 348 kJ mol⁻¹ for a C–C σ bond. However, the total bond energy of the C=C double bond (about 612 kJ mol⁻¹) is greater than that of a single bond, which explains why addition reactions are favoured over direct cleavage of the double bond.

π 键比 σ 键弱，键能约为 264 kJ mol⁻¹，而 C–C σ 键的键能约为 348 kJ mol⁻¹。然而，C=C 双键的总键能（约 612 kJ mol⁻¹）大于单键，这解释了为什么加成反应优先于双键的直接断裂。

C=C: one σ bond + one π bond

2. The General Mechanism: Electrophilic Addition | 通用机理：亲电加成

The general mechanism of electrophilic addition to alkenes proceeds in two main steps. In the first step, the π electrons of the double bond attack an electrophile (E⁺), forming a new σ bond between the electrophile and one of the alkene carbon atoms. The other carbon atom acquires a positive charge, generating a carbocation intermediate.

烯烃亲电加成反应的通用机理主要分为两步。第一步，双键中的 π 电子进攻亲电试剂（E⁺），在亲电试剂与烯烃的一个碳原子之间形成新的 σ 键。另一个碳原子则带上正电荷，生成碳正离子中间体。

In the second step, a nucleophile (Nu⁻) attacks the positively charged carbocation, forming a second σ bond. This completes the addition process, and both carbon atoms of the former double bond now have four single bonds, consistent with sp³ hybridisation.

第二步，亲核试剂（Nu⁻）进攻带正电的碳正离子，形成第二个 σ 键。至此加成过程完成，原来双键的两个碳原子均形成四个单键，符合 sp³ 杂化。

C=C + E⁺ → C⁺–C–E → Nu–C–C–E

For the CIE syllabus, typical electrophiles include H⁺ (from hydrogen halides or acids), Br⁺/Cl⁺ (from halogens), and H⁺ from water in acid-catalysed hydration. It is essential to draw the full curly arrow mechanism in examinations, showing the movement of electron pairs clearly.

在 CIE 考纲中，典型的亲电试剂包括 H⁺（来自卤化氢或酸）、Br⁺/Cl⁺（来自卤素单质），以及酸催化水合反应中的 H⁺。考试中必须画出完整的弯曲箭头机理，清晰展示电子对的流向。

3. Addition of Hydrogen Halides (HX) | 卤化氢的加成

When an alkene reacts with a hydrogen halide (HCl, HBr, or HI), the product is a haloalkane. The reaction can be carried out by bubbling the gaseous hydrogen halide through the liquid alkene, or by mixing them in an inert solvent. The reactivity of hydrogen halides increases in the order HCl < HBr < HI, which correlates with the increasing bond polarity and decreasing bond strength down the group.

当烯烃与卤化氢（HCl、HBr 或 HI）反应时，产物为卤代烷烃。该反应可以将气态卤化氢通入液态烯烃中进行，或在惰性溶剂中混合反应。卤化氢的反应活性按 HCl < HBr < HI 的顺序递增，这与键的极性增强和键能减小相关。

For example, ethene reacts with hydrogen bromide to form bromoethane:

例如，乙烯与溴化氢反应生成溴乙烷：

CH₂=CH₂ + HBr → CH₃–CH₂Br

The reaction proceeds via the attack of H⁺ on the double bond, forming a carbocation, followed by attack of Br⁻ on the carbocation. In the case of symmetrical alkenes like ethene, only one product is possible because the two carbon atoms are equivalent.

该反应通过 H⁺ 进攻双键形成碳正离子，随后 Br⁻ 进攻碳正离子而完成。对于乙烯这样的对称烯烃，两个碳原子是等价的，因此只可能生成一种产物。

4. Markovnikov’s Rule and Regioselectivity | 马尔可夫尼科夫规则与区域选择性

When an unsymmetrical alkene such as propene reacts with a hydrogen halide, two possible products could form. However, in practice, one product dominates. Markovnikov’s rule states that when HX adds to an unsymmetrical alkene, the hydrogen atom attaches to the carbon atom of the double bond that already has the greater number of hydrogen atoms, and the halogen attaches to the carbon with fewer hydrogen atoms.

当丙烯这样的不对称烯烃与卤化氢反应时，理论上可能生成两种产物。然而，实际反应中只有一种产物占主导。马尔可夫尼科夫规则指出：当 HX 与不对称烯烃加成时，氢原子加在双键中原本含氢较多的碳原子上，而卤素加在含氢较少的碳原子上。

For propene, the major product is 2-bromopropane:

对于丙烯，主要产物是 2-溴丙烷：

CH₃–CH=CH₂ + HBr → CH₃–CHBr–CH₃ (major)

The reason for this regioselectivity lies in the stability of the carbocation intermediate. The addition of H⁺ to propene can generate either a primary carbocation (CH₃–CH₂–CH₂⁺) or a secondary carbocation (CH₃–CH⁺–CH₃). Secondary carbocations are more stable than primary carbocations due to the electron-donating inductive effect of alkyl groups, which helps to delocalise and stabilise the positive charge.

这种区域选择性的原因在于碳正离子中间体的稳定性。H⁺ 加成到丙烯上可能生成伯碳正离子（CH₃–CH₂–CH₂⁺）或仲碳正离子（CH₃–CH⁺–CH₃）。仲碳正离子比伯碳正离子更稳定，因为烷基的给电子诱导效应有助于分散和稳定正电荷。

Carbocation stability follows the order: benzylic > allylic > tertiary > secondary > primary > methyl. This ordering is crucial for predicting the major product in electrophilic addition reactions.

碳正离子稳定性顺序为：苄基型 > 烯丙基型 > 叔碳 > 仲碳 > 伯碳 > 甲基。这一顺序对于预测亲电加成反应的主要产物至关重要。

5. Addition of Halogens (Br₂, Cl₂) | 卤素的加成

Alkenes react readily with halogens at room temperature. For example, ethene decolourises bromine water (an orange-brown solution), forming 1,2-dibromoethane, a colourless liquid. This reaction is commonly used as a qualitative test for the presence of a C=C double bond in an unknown compound.

烯烃在室温下即可与卤素迅速反应。例如，乙烯能使溴水（橙棕色溶液）褪色，生成无色液体 1,2-二溴乙烷。该反应常被用作定性检验未知化合物中是否存在 C=C 双键的方法。

CH₂=CH₂ + Br₂ → CH₂Br–CH₂Br

The mechanism involves the polarisation of the Br–Br bond as the π electrons approach the bromine molecule. The bromine molecule becomes polarised, and the electrophilic Br⁺ is attracted to the double bond. A cyclic bromonium ion intermediate is formed, which is more stable than an open carbocation because the positive charge is shared between the two carbon atoms.

该机理涉及 Br–Br 键的极化：当 π 电子靠近溴分子时，溴分子发生极化，亲电性的 Br⁺ 被双键吸引。反应形成环状溴鎓离子中间体，该中间体比开放的碳正离子更稳定，因为正电荷在两个碳原子之间共享。

The Br⁻ ion then attacks the bromonium ion from the opposite side, leading to anti addition. This means the two bromine atoms add to opposite faces of the double bond, an important stereochemical consequence of halogen addition.

随后 Br⁻ 从反面进攻溴鎓离子，导致反式加成。这意味着两个溴原子加在双键的相反两侧，这是卤素加成的重要立体化学后果。

6. Addition of Steam (Hydration) | 与水蒸气加成（水合反应）

Alkenes react with steam in the presence of an acid catalyst, typically concentrated phosphoric acid (H₃PO₄) or sulfuric acid (H₂SO₄), to form alcohols. This is an industrial method for producing ethanol from ethene, and the reaction is reversible and exothermic.

烯烃在酸催化剂（通常为浓磷酸 H₃PO₄ 或浓硫酸 H₂SO₄）存在下与水蒸气反应生成醇。这是工业上由乙烯制备乙醇的方法，该反应是可逆的放热反应。

CH₂=CH₂ + H₂O ⇌ CH₃–CH₂OH

Industrial conditions for this process are typically a temperature of 300 °C, a pressure of 60–70 atm, and phosphoric acid supported on silica as the catalyst. A high yield of ethanol is obtained by continuously removing ethanol from the reaction mixture and recycling unreacted ethene.

该工艺的工业条件通常为：温度 300 °C，压力 60–70 atm，以负载在硅胶上的磷酸为催化剂。通过不断移出反应混合物中的乙醇并循环利用未反应的乙烯，可以获得较高的乙醇产率。

The mechanism is similar to hydrogen halide addition: H⁺ protonates the double bond to form a carbocation, then water acts as a nucleophile to attack the carbocation, and finally deprotonation yields the alcohol. For unsymmetrical alkenes, Markovnikov’s rule applies, giving the more substituted alcohol as the major product.

该机理与卤化氢加成类似：H⁺ 质子化双键形成碳正离子，水作为亲核试剂进攻碳正离子，最后去质子化得到醇。对于不对称烯烃，遵循马尔可夫尼科夫规则，主要产物为取代程度更高的醇。

7. Addition of Sulfuric Acid | 硫酸的加成

Concentrated sulfuric acid adds to alkenes to form hydrogen sulfate esters (alkyl hydrogensulfates). For example, ethene reacts with cold concentrated H₂SO₄ to form ethyl hydrogensulfate (CH₃–CH₂–OSO₃H). This compound can be hydrolysed by warming with water to produce ethanol, providing an alternative route to alcohols.

浓硫酸能与烯烃加成生成硫酸氢酯（烷基硫酸氢酯）。例如，乙烯与冷的浓 H₂SO₄ 反应生成乙基硫酸氢酯（CH₃–CH₂–OSO₃H）。该化合物与水共热发生水解可生成乙醇，为醇的制备提供了另一条途径。

CH₂=CH₂ + H₂SO₄ → CH₃–CH₂–OSO₃H

CH₃–CH₂–OSO₃H + H₂O → CH₃–CH₂OH + H₂SO₄

This indirect hydration method was traditionally used in industry before the development of the direct catalytic hydration process. It also exemplifies the acid-catalysed addition of water, where the acid is regenerated at the end of the overall process. In the CIE examination, students should be able to write the equations for both steps and explain the role of sulfuric acid.

这种间接水合法在直接催化水合工艺发展之前曾被工业界广泛采用。它也是酸催化加水的典型实例，酸在整个过程结束时得以再生。在 CIE 考试中，学生应能写出这两步的化学方程式并解释硫酸的作用。

8. Addition of Hydrogen (Catalytic Hydrogenation) | 氢气的加成（催化加氢）

Alkenes react with hydrogen gas in the presence of a nickel, platinum, or palladium catalyst to form alkanes. This process is called catalytic hydrogenation. The reaction requires gentle heating (around 150 °C) with a finely divided nickel catalyst, although platinum and palladium are more active and can catalyse the reaction at room temperature.

烯烃在镍、铂或钯催化剂存在下与氢气反应生成烷烃，该过程称为催化加氢。反应通常需要在约 150 °C 下使用细粉状镍催化剂进行，而铂和钯的活性更高，可在室温下催化该反应。

CH₂=CH₂ + H₂ → CH₃–CH₃ (Ni catalyst, 150 °C)

Catalytic hydrogenation proceeds via a heterogeneous mechanism on the metal surface. The alkene and hydrogen molecules adsorb onto the catalyst surface, where the H–H bond breaks and hydrogen atoms are transferred to the double bond. Both hydrogen atoms add to the same face of the double bond, resulting in syn addition.

催化加氢通过金属表面上的多相机理进行。烯烃和氢分子吸附在催化剂表面，H–H 键断裂，氢原子被转移到双键上。两个氢原子从同一面加到双键上，即顺式加成。

This reaction has significant industrial importance. For example, vegetable oils (unsaturated fats) are hydrogenated to produce margarine and solid cooking fats. Partial hydrogenation controls the degree of unsaturation and thus the physical properties of the final product.

该反应具有重要的工业价值。例如，植物油（不饱和脂肪）通过加氢生产人造黄油和固体烹调油脂。部分加氢可以控制不饱和程度，从而调节最终产品的物理性质。

9. Addition Polymers | 加成聚合物

Alkenes can undergo addition polymerisation, where thousands of monomer units join together to form a long-chain polymer. This reaction is initiated by free radicals or catalysts and involves the breaking of the π bond without the elimination of any small molecule.

烯烃可以发生加成聚合反应，数千个单体单元通过 π 键断裂连接在一起形成长链聚合物，过程中无小分子脱除。该反应由自由基或催化剂引发。

For example, ethene polymerises to form poly(ethene), commonly known as polyethene or polythene:

例如，乙烯聚合生成聚（乙烯），俗称聚乙烯：

n CH₂=CH₂ → [–CH₂–CH₂–]ₙ

Substituted alkenes such as propene, chloroethene (vinyl chloride), and phenylethene (styrene) polymerise similarly to form poly(propene), poly(chloroethene) or PVC, and poly(phenylethene) or polystyrene, respectively. These polymers have a range of applications, from packaging materials to pipes and insulating foams.

取代烯烃如丙烯、氯乙烯和苯乙烯也以类似方式聚合，分别生成聚丙烯、聚氯乙烯（PVC）和聚苯乙烯。这些聚合物用途广泛，涵盖包装材料、管道和绝缘泡沫等。

The structure of the repeating unit must be identified correctly in examinations. For addition polymers, the repeating unit is the same as the monomer with the double bond opened. Students should be able to draw the repeating unit, deduce the monomer from a given polymer, and recognise that addition polymers are generally non-biodegradable due to their inert C–C and C–H bonds.

考试中必须正确识别重复单元的结构。对于加成聚合物，重复单元就是双键打开后的单体。学生应能绘制重复单元、从给定聚合物推导单体，并认识到加成聚合物通常不可生物降解，因为其 C–C 和 C–H 键为惰性。

10. Summary of Key Reactions | 关键反应总结

The table below summarises the key electrophilic addition reactions of alkenes that you need to know for the CIE A-Level Chemistry examination. Be sure to memorise the reagents, conditions, and products, and practise drawing curly arrow mechanisms for each reaction.

下表总结了 CIE A-Level 化学考试中需要掌握的烯烃关键亲电加成反应。请务必牢记试剂、条件和产物，并练习为每个反应绘制弯曲箭头机理。

Reaction	Reagent / Conditions	Product
Addition of HX	HX (g) or (aq), room temperature	Haloalkane
Addition of halogens	Br₂ in CCl₄ or bromine water	Dihaloalkane
Hydration	Steam, H₃PO₄ catalyst, 300 °C, 60 atm	Alcohol
Addition of H₂SO₄	Cold conc. H₂SO₄	Alkyl hydrogensulfate
Hydrogenation	H₂, Ni/Pt/Pd catalyst, 150 °C	Alkane
Polymerisation	High pressure, catalyst, heat	Polyalkene

Note that the addition of bromine to alkenes is also used as a test for unsaturation. The orange-brown bromine water is decolourised when shaken with an alkene, producing a colourless dibromo compound. Alkanes, in contrast, do not decolourise bromine water under these conditions, which allows the two classes of hydrocarbons to be distinguished.

注意，溴的加成反应也用于检验不饱和性。橙棕色的溴水与烯烃振荡时会褪色，生成无色的二溴化合物。相比之下，烷烃在这些条件下不会使溴水褪色，因此可以用该方法区分两类烃。

11. Common Exam Pitfalls and Tips | 常见考试误区与建议

Students often lose marks in examinations due to careless errors in drawing mechanisms or misapplying Markovnikov’s rule. Here are the most common pitfalls and how to avoid them:

学生在考试中经常因绘制机理粗心或误用马尔可夫尼科夫规则而失分。以下是最常见的误区及避免方法：

Always show full curly arrows from the electron-rich centre to the electron-poor centre. A curly arrow starts at a bond or lone pair and points to the atom that accepts the electrons.

始终从电子富集中心画出完整的弯曲箭头指向电子贫乏中心。弯曲箭头从化学键或孤对电子起始，指向接受电子的原子。
Do not write ‘Br⁺’ as a free ion in the mechanism for bromine addition. Instead, show the polarisation of the Br–Br bond and the formation of the bromonium ion.

在溴加成机理中不要将 Br⁺ 写为自由离子。应展示 Br–Br 键的极化以及溴鎓离子的形成。
Apply Markovnikov’s rule only to the addition of unsymmetrical electrophiles (HX, H₂O) to unsymmetrical alkenes. Halogen addition and hydrogenation are not governed by this rule.

马尔可夫尼科夫规则只适用于不对称亲电试剂（HX、H₂O）对不对称烯烃的加成。卤素加成和加氢反应不受该规则支配。
In hydration, do not forget that the acid catalyst is regenerated. The overall stoichiometric equation shows only alkene + water → alcohol.

在水合反应中，不要忘记酸催化剂会再生。总化学计量方程式仅显示烯烃 + 水 → 醇。
When drawing polymer structures, show the repeating unit in square brackets with an ‘n’ subscript, and ensure that the bonds on either side extend beyond the brackets.

绘制聚合物结构时，用方括号展示重复单元并标注下标 ‘n’，且方括号两侧的键必须向外延伸。

Additionally, always consider the stereochemistry of the addition. Anti addition occurs in halogen addition via the bromonium ion pathway, while syn addition occurs in catalytic hydrogenation. Although the CIE syllabus does not always require detailed stereochemical analysis, understanding this concept can help you answer extension questions confidently.

此外，始终考虑加成的立体化学。卤素加成通过溴鎓离子途径发生反式加成，而催化加氢发生顺式加成。虽然 CIE 考纲并不总是要求详细的立体化学分析，但理解这一概念可以帮助你自信地回答拓展性问题。

12. Conclusion | 结语

Electrophilic addition reactions are one of the most important topics in A-Level organic chemistry. Understanding the structure of the double bond, the stepwise mechanism, and the factors that determine product distribution will allow you to tackle a wide range of examination questions with confidence. Remember to practise drawing mechanisms repeatedly, as muscle memory and accuracy are key to earning full marks in the written papers.

亲电加成反应是 A-Level 有机化学中最重要的课题之一。理解双键结构、分步机理以及决定产物分布的因素，将使你能够自信地应对各类考试问题。记住要反复练习绘制机理，因为熟练度和准确性是书面考试中拿到满分的关键。

The reactions covered in this article — addition of hydrogen halides, halogens, steam, sulfuric acid, hydrogen, and polymerisation — are all recurring themes in CIE examination papers. Master them, understand the underlying principles, and you will be well-prepared for your examinations.

本文涵盖的反应——卤化氢加成、卤素加成、水蒸气加成、硫酸加成、加氢和聚合——都是 CIE 考卷中的常考主题。掌握它们，理解其背后的原理，你将为考试做好充分准备。

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A-Level Chemistry: Electrophilic Addition Reactions of Alkenes | A-Level 化学：烯烃的加成反应详解