Electrophilic Addition in IGCSE CIE Chemistry | IGCSE CIE 化学：亲电加成考点精讲

📚 Electrophilic Addition in IGCSE CIE Chemistry | IGCSE CIE 化学：亲电加成考点精讲

Electrophilic addition is one of the most characteristic reaction types of alkenes. In the IGCSE CIE Chemistry syllabus, understanding this mechanism not only helps you explain why the bromine water test works, but also allows you to predict the products when unsymmetrical alkenes react with hydrogen halides. This article breaks down every key point you need to master for the exam, from the nature of the carbon–carbon double bond to Markovnikov’s rule.

亲电加成是烯烃最具特征性的反应类型之一。在 IGCSE CIE 化学大纲中，理解这一机理不仅能帮助你解释溴水试验的原理，还能让你预测不对称烯烃与卤化氢反应时的产物。本文将从碳碳双键的本质到马氏规则，逐一剖析考试中必须掌握的所有要点。

1. What Is Electrophilic Addition? | 什么是亲电加成？

Electrophilic addition is a reaction in which an electrophile attacks an electron‑rich part of a molecule, leading to the addition of atoms or groups across a multiple bond. In the context of IGCSE, it almost exclusively refers to the addition reactions of alkenes, where the π bond of the C=C double bond is broken and two new σ bonds are formed.

亲电加成是一种反应类型：亲电试剂进攻分子中电子云密度较高的部位，导致原子或基团加成到多重键两端。在 IGCSE 范围内，它几乎专指烯烃的加成反应，即 C=C 双键中的 π 键断裂，形成两个新的 σ 键。

2. Alkenes and the C=C Double Bond | 烯烃与 C=C 双键

Alkenes contain a carbon–carbon double bond consisting of one σ bond and one π bond. The π bond is formed by the sideways overlap of p orbitals and is relatively weak and exposed. This region of high electron density makes the double bond susceptible to attack by electrophiles – species that are attracted to electrons.

烯烃含有碳碳双键，由一个 σ 键和一个 π 键组成。π 键由 p 轨道的侧向重叠形成，相对较弱且暴露在外。这一高电子云密度的区域使得双键容易受到亲电试剂（即被电子吸引的物种）的进攻。

• The π electrons are less tightly held than σ electrons, so they act as a source of electrons for electrophiles.

  π 电子的束缚比 σ 电子更弱，因此它们充当了给亲电试剂提供电子的来源。

• Alkenes are unsaturated hydrocarbons, meaning they have the capacity to make more bonds by opening the double bond.

  烯烃是不饱和烃，意味着它们可以通过打开双键来形成更多的化学键。

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3. Electrophiles – the Attacking Species | 亲电试剂——进攻物种

An electrophile is a species that is electron‑deficient and can accept a pair of electrons to form a new covalent bond. Common electrophiles in IGCSE chemistry include the following:

亲电试剂是指缺电子并能接受一对电子形成新共价键的物种。IGCSE 化学中常见的亲电试剂包括：

• Br₂ – although non‑polar in its elemental state, it becomes polarised when approaching the electron‑rich double bond, inducing a dipole moment that makes one bromine atom electrophilic.

  Br₂（溴单质）——在单质状态下是非极性的，但当它接近电子云密度高的双键时会被极化，产生一个偶极矩，使得其中一个溴原子表现出亲电性。

• HBr – hydrogen bromide contains a polar H–Br bond; the hydrogen end is electron‑deficient and acts as the electrophile.

  HBr（溴化氢）——含有极性的 H–Br 键；氢端是缺电子的，作为亲电试剂。

• H₂SO₄ – in some contexts, the concentrated sulfuric acid can protonate alkenes, generating a carbocation, which then reacts further.

  H₂SO₄（浓硫酸）——在某些情况下，浓硫酸可以使烯烃质子化，生成碳正离子，随后进一步反应。

• H⁺ from acids – in hydration reactions, H⁺ ions (from phosphoric acid catalyst) act as electrophiles.

  来自酸的 H⁺——在水合反应中，H⁺ 离子（来自磷酸催化剂）作为亲电试剂。

Remember: A nucleophile is the opposite – an electron‑rich species that donates electrons. Electrophilic addition mechanisms involve electrophiles, not nucleophiles, in the first step.

记住：亲电试剂的反面是亲核试剂——富电子并能提供电子的物种。亲电加成机理的第一步中投入使用的是亲电试剂，而非亲核试剂。

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4. Electrophilic Addition of Bromine – The Mechanism | 溴的亲电加成——机理

When bromine (Br₂) is added to an alkene at room temperature, the reaction occurs via electrophilic addition. The overall equation for ethene is:

当溴（Br₂）在室温下与烯烃反应时，反应通过亲电加成进行。以乙烯为例的总反应方程式为：

C₂H₄ + Br₂ → C₂H₄Br₂

The mechanism proceeds in two steps:

机理分两步进行：

Step 1: Electrophilic attack and carbocation formation. As a bromine molecule approaches the alkene, the electron‑rich double bond polarises the Br–Br bond. The slightly positive bromine atom accepts a pair of electrons from the π bond, forming a C–Br bond. The other bromine atom departs as a bromide ion (Br⁻). Simultaneously, the carbon that did not bond to bromine becomes electron‑deficient, generating a carbocation intermediate.

第一步：亲电进攻与碳正离子形成。 当溴分子靠近烯烃时，富电子的双键使 Br–Br 键极化。稍带正电的溴原子接受来自 π 键的一对电子，形成 C–Br 键。另一个溴原子以溴负离子（Br⁻）的形式离去。与此同时，未与溴成键的碳原子变得缺电子，生成碳正离子中间体。

Step 2: Nucleophilic attack by bromide ion. The bromide ion, which is an electron‑rich nucleophile, quickly donates an electron pair to the positively charged carbon, forming a second C–Br bond. This gives the dibromoalkane product.

第二步：溴负离子的亲核进攻。 溴负离子是一个富电子的亲核试剂，它迅速向带正电荷的碳提供一对电子，形成第二个 C–Br 键，得到二溴代烷产物。

No catalyst is required for this reaction, and it occurs readily at room temperature. The rapid decolorisation of bromine is the basis of the test for unsaturation.

该反应不需要催化剂，在室温下即可顺利进行。溴水的快速褪色正是检验不饱和烃的依据。

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5. The Bromine Water Test for Unsaturation | 溴水试验检验不饱和键

Shaking an alkene with orange‑brown bromine water results in the immediate decolorisation of the solution. This is a standard test to distinguish alkenes from alkanes: alkanes do not decolorise bromine water under ordinary conditions unless UV light initiates a substitution reaction (free‑radical substitution).

将烯烃与橙黄色的溴水混合振荡，溶液会立即褪色。这是区分烯烃和烷烃的标准试验：烷烃在通常条件下不能使溴水褪色，除非在紫外光引发下发生取代反应（自由基取代）。

The reaction with an alkene is an addition, yielding a colourless dibromo compound, whereas any colour change with alkanes requires the presence of UV light and produces HBr fumes – a different phenomenon.

与烯烃的反应是加成反应，生成无色的二溴化合物；而烷烃即使褪色也需要紫外光并产生 HBr 烟雾——这是完全不同的现象。

In the exam, you may be asked to state the observations and write an equation. Make sure you specify that red‑brown bromine water turns colourless when shaken with an alkene, and that this indicates the presence of a C=C bond.

考试中可能会要求你描述观察结果并书写方程式。务必要明确写出：红棕色的溴水与烯烃振荡后变为无色，这表明存在 C=C 双键。

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6. Addition of Hydrogen Bromide (HBr) | 溴化氢（HBr）的加成

Hydrogen bromide adds across the double bond of an alkene in a similar electrophilic manner. The HBr molecule is already polarised: H carries a partial positive charge, making it the electrophile; Br carries a partial negative charge.

溴化氢以类似的亲电方式与烯烃的双键发生加成。HBr 分子本身是极性的：H 带有部分正电荷，是亲电试剂；Br 带有部分负电荷。

With a symmetrical alkene such as ethene, the product is simply bromoethane:

对于对称烯烃如乙烯，产物只有溴乙烷：

C₂H₄ + HBr → C₂H₅Br

The mechanism follows the same two steps: electrophilic attack by H⁺ to form a carbocation, then rapid nucleophilic attack by Br⁻ to form the haloalkane.

机理遵循同样的两步：H⁺ 作为亲电试剂进攻，生成碳正离子；随后 Br⁻ 快速亲核进攻，生成卤代烷。

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7. Unsymmetrical Alkenes and Markovnikov’s Rule | 不对称烯烃与马氏规则

When an unsymmetrical alkene (e.g., propene, CH₃–CH=CH₂) reacts with an unsymmetrical reagent like HBr, two constitutional isomers are theoretically possible. However, one isomer predominates. Markovnikov’s rule allows us to predict the major product.

当不对称烯烃（如丙烯，CH₃–CH=CH₂）与不对称试剂（如 HBr）反应时，理论上可能得到两种构造异构体。然而，其中一种异构体占主导。马氏规律使我们能预测主要产物。

Markovnikov’s rule states: In the addition of H–X to an unsymmetrical alkene, the hydrogen atom attaches to the carbon that already has more hydrogen atoms (the less substituted carbon), while the halogen attaches to the carbon with fewer hydrogens (the more substituted carbon). This arises because the more stable carbocation intermediate is formed preferentially.

马氏规律指出：在 H–X 与不对称烯烃的加成中，氢原子加在含氢较多的碳原子上（取代度较低的碳），而卤素加在含氢较少的碳原子上（取代度较高的碳）。 这是由于更稳定的碳正离子中间体优先生成。

Alkene	Reagent	Major product	Explanatory note
Propene (CH₃CH=CH₂)	HBr	2‑bromopropane (CH₃CHBrCH₃)	H adds to the end CH₂ group; Br adds to the middle carbon (more substituted carbocation).
But‑1‑ene	HBr	2‑bromobutane	The H attaches to C1; the secondary carbocation at C2 is more stable than the primary carbocation at C1.

For propene, the intermediate carbocations are CH₃–⁺CH–CH₃ (secondary) and CH₃–CH₂–⁺CH₂ (primary). Secondary carbocations are more stable due to the +I (inductive) effect of the alkyl groups, so it forms more quickly, leading to 2‑bromopropane as the major product.

以丙烯为例，中间体碳正离子为 CH₃–⁺CH–CH₃（仲碳正离子）和 CH₃–CH₂–⁺CH₂（伯碳正离子）。仲碳正离子由于烷基的 +I（诱导）效应而更稳定，因此生成更快，导致主要产物为 2‑溴丙烷。

Examiners often ask you to draw the structures of both possible products and label the major one. Always apply the rule by comparing the number of hydrogens on the two carbon atoms of the double bond.

考官常要求你画出两种可能产物的结构并标明主要产物。始终通过比较双键两个碳原子上的氢原子数来应用该规律。

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8. Electrophilic Addition with Sulfuric Acid | 与硫酸的亲电加成

Concentrated sulfuric acid can also add to alkenes in an electrophilic manner, especially at low temperatures. The product is an alkyl hydrogensulfate. For example, ethene reacts with cold concentrated H₂SO₄ to form ethyl hydrogensulfate:

浓硫酸也能以亲电方式与烯烃加成，特别是在低温下。产物是硫酸氢烷酯。例如，乙烯与冷浓硫酸反应生成硫酸氢乙酯：

CH₂=CH₂ + H₂SO₄ → CH₃CH₂OSO₂OH

The mechanism: H₂SO₄ acts as an electrophile, with H⁺ adding to one carbon of the double bond, forming a carbocation; then the HSO₄⁻ ion attacks the carbocation. This reaction is important industrially as it can be followed by hydrolysis to produce ethanol – the two‑step hydration process.

机理：H₂SO₄ 作为亲电试剂，H⁺ 加在双键的一个碳上，形成碳正离子；然后 HSO₄⁻ 离子进攻碳正离子。该反应在工业上很重要，因为它可以随后水解生成乙醇——即两步水合法。

For IGCSE, you do not need to learn the full two‑step industrial process in detail, but you should be able to recognise the addition pattern and the fact that concentrated H₂SO₄ adds across a double bond.

IGCSE 阶段不需要详细学习两步工业生产流程，但应能识别这种加成模式以及浓硫酸能加成到双键两端这一事实。

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9. Addition of Steam (Hydration) and the Role of Electrophilic Addition | 水蒸气加成（水合反应）与亲电加成的角色

The industrial production of ethanol from ethene and steam uses a phosphoric(V) acid catalyst supported on a solid. The reaction is:

由乙烯和水蒸气工业生产乙醇的过程使用负载在固体上的磷酸（V）作催化剂。反应为：

C₂H₄ + H₂O ⇌ C₂H₅OH

The mechanism is electrophilic addition: H⁺ (from the acid catalyst) first adds to ethene, forming the CH₃CH₂⁺ carbocation. Water then acts as a nucleophile and attacks the carbocation, followed by loss of H⁺ to regenerate the catalyst. This explains why an acid catalyst is required – it provides the initial electrophile necessary to start the addition.

该反应的机理是亲电加成：H⁺（来自酸催化剂）首先加成到乙烯上，生成 CH₃CH₂⁺ 碳正离子。随后水作为亲核试剂进攻碳正离子，再脱去 H⁺ 使催化剂再生。这就解释了为何需要酸催化剂——它提供了启动加成所需的初始亲电试剂。

Even though the detailed steps are not always required in IGCSE answers, being able to link the hydration of ethene to electrophilic addition demonstrates a deeper understanding of the reaction type.

虽然 IGCSE 答案不总是要求详细步骤，但能够将乙烯的水合反应与亲电加成联系起来，展示了对反应类型更深层次的理解。

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10. Summary of Reaction Conditions and Reagents | 反应条件与试剂总结

Reaction	Reagent	Conditions	Mechanism type
Bromination	Br₂ (in water or organic solvent)	Room temperature, no catalyst; dark is fine	Electrophilic addition
Hydrobromination	HBr gas	Room temperature	Electrophilic addition (Markovnikov if unsymmetrical)
Hydration	Steam (H₂O)	300 °C, 60–70 atm, H₃PO₄ catalyst	Electrophilic addition
Addition of H₂SO₄	Conc. H₂SO₄	Cold (0–5 °C)	Electrophilic addition

Memorising this table will help you quickly recall the reagents and conditions required for each electrophilic addition, a common exam demand.

记住这张表格能帮助你快速回忆每种亲电加成所需的试剂和条件，这是考试中的常见要求。

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11. Common Misconceptions and Exam Tips | 常见误区与考试技巧

• Misconception: ‘Bromine water decolorisation indicates an alkane.’ Correction: Alkanes do not decolorise bromine water without UV light; decolorisation indicates the presence of unsaturation (C=C).

  误区：“溴水褪色说明存在烷烃。”纠正： 在没有紫外光时，烷烃不会使溴水褪色；褪色表明不饱和键（C=C）的存在。

• Misconception: ‘For unsymmetrical alkenes, both products are formed in equal amounts.’ Correction: Markovnikov’s rule predicts the major product because one carbocation pathway is more stable.

  误区：“不对称烯烃反应时，两种产物等量生成。”纠正： 马氏规律预测主要产物，因为其中一种碳正离子路径更稳定。

• Exam tip: When asked to draw the mechanism for ethene and Br₂, show the curly arrow from the double bond to the slightly positive Br, then the departure of Br⁻, and finally the attack of Br⁻ on the carbocation. Practise drawing the dipoles on Br₂.

  考试技巧： 如果要求画出乙烯与 Br₂ 反应的机理，要画出从双键指向稍微带正电的 Br 的弯箭头，然后是 Br⁻ 的离去，最后是 Br⁻ 进攻碳正离子。练习画出 Br₂ 上的偶极标记。

• Exam tip: For Markovnikov questions, always first identify which carbon of the double bond has more hydrogens. Attach the H of H–X to that carbon. The rest of the reagent goes to the other carbon.

  考试技巧： 对于马氏规律的问题，总是先确定双键哪个碳原子上氢更多。将 H–X 中的 H 加在该碳上。试剂的其余部分加到另一个碳上。

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12. Quick Revision – Electrophilic Addition in a Nutshell | 快速复习——亲电加成核心要点

Electrophilic addition is the characteristic reaction of alkenes. The electron‑rich π bond attracts an electrophile, a carbocation forms, and a nucleophile adds to complete the product. Bromine water serves as a test for C=C bonds. With HBr and unsymmetrical alkenes, Markovnikov’s rule applies: the H goes to the carbon with more hydrogens. Keep these principles in mind, and you will handle any related exam question with confidence.

亲电加成是烯烃的特征反应。富电子的 π 键吸引亲电试剂，生成碳正离子，随后亲核试剂加成得到产物。溴水用作检验 C=C 双键的试剂。对于 HBr 与不对称烯烃，马氏规律适用：H 加到含氢较多的碳上。牢记这些原则，任何相关的考试题目都能从容应对。

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