Electrophilic Addition | 亲电加成 考点精讲

📚 Electrophilic Addition | 亲电加成 考点精讲

Electrophilic addition is the characteristic reaction of alkenes and other unsaturated compounds. For IGCSE Edexcel Chemistry, you need to understand the mechanism, the role of the electrophile, why carbocations form, and how to predict major and minor products using Markovnikov’s rule. This article breaks down every essential point, pairing definitions with step-by-step explanations, and includes common exam applications like bromine water testing and the addition of hydrogen halides and water. Mastering this topic will secure those marks on reaction pathways, curly‑arrow diagrams, and product prediction.

亲电加成是烯烃及其他不饱和化合物的特征反应。在IGCSE Edexcel化学中，你需要理解反应机理、亲电试剂的作用、碳正离子形成的原因，以及如何运用马氏规则预测主产物和次产物。本文拆解每一个核心考点，定义与分步解释配对呈现，并涵盖溴水检验、卤化氢加成和水合反应等常考应用。掌握这一主题，就能稳稳拿下反应路径、弯箭头图示和产物预测相关的分数。

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

Electrophilic addition is a reaction in which an electrophile attacks a carbon–carbon double bond (C=C), leading to the formation of a saturated product. The π bond in the alkene is broken, and two new σ bonds are formed with the electrophile and a nucleophile respectively.

亲电加成是指亲电试剂进攻碳碳双键（C=C），从而生成饱和产物的反应。烯烃中的π键断裂，亲电试剂和亲核试剂分别与两个碳原子形成两个新的σ键。

The reaction is ‘addition’ because two atoms or groups are added across the double bond. It is ‘electrophilic’ because the first attacking species is electron‑deficient and seeks electrons from the electron‑rich π bond.

称为“加成”是因为两个原子或基团加在双键的两端；称为“亲电”是因为首先进攻的物种缺电子，倾向于从富含电子的π键中获取电子。

The general equation for electrophilic addition to ethene can be written as: C₂H₄ + X–Y → CH₂X–CH₂Y (for symmetrical alkenes).

乙烯发生亲电加成的一般方程式可写作：C₂H₄ + X–Y → CH₂X–CH₂Y（对称烯烃的情况）。

2. The Electrophile: Electron‑Deficient Species | 亲电试剂：缺电子物种

An electrophile is a species that is attracted to regions of high electron density. It is electron‑deficient and can accept a pair of electrons. Common electrophiles in IGCSE Edexcel include H⁺ (from H–X), Br⁺ (or the polarised Brᵟ⁺ in Br₂), and Hᵟ⁺ in polarised water during acid‑catalysed hydration.

亲电试剂是被高电子密度区域吸引的物种。它缺电子，能够接受一对电子。IGCSE Edexcel中常见的亲电试剂包括H⁺（来自H–X）、Br⁺（或Br₂中极化的Brᵟ⁺），以及在酸催化水合反应中极化水分子里的Hᵟ⁺。

In the case of Br₂, the Br–Br bond becomes polarised when it approaches the electron‑rich π bond, making one bromine atom partially positive. This δ⁺ bromine acts as the electrophile. Hydrogen halides, such as H–Cl, are already polar molecules, with H being δ⁺ due to the difference in electronegativity.

对于Br₂而言，当其接近富电子的π键时，Br–Br键发生极化，其中一个溴原子带部分正电荷，这个δ⁺溴就充当亲电试剂。卤化氢，如H–Cl，已经是极性分子，由于电负性差异，H带有δ⁺。

3. Mechanism Overview: Two Steps | 机理概述：两步反应

The mechanism for electrophilic addition always occurs in two main steps. First, the electrophile is attracted to the C=C double bond and accepts a pair of electrons from the π bond, forming a carbocation intermediate. Second, a nucleophile (often the negatively charged counter‑ion or a water molecule) attacks the carbocation, forming the final addition product.

亲电加成机理始终分两步进行。第一步，亲电试剂被C=C双键吸引，从π键接受一对电子，形成碳正离子中间体。第二步，亲核试剂（通常是带负电的平衡离子或水分子）进攻碳正离子，生成最终加成产物。

Curly arrows must be drawn in the exam to show the movement of electron pairs. In step one, the arrow starts from the middle of the C=C bond and points towards the electrophile. Simultaneously, a curly arrow shows the bond between X and Y breaking heterolytically, with the electron pair going to Y. In step two, a curly arrow is drawn from the nucleophile’s lone pair towards the carbocation’s positive carbon.

考试中必须画出弯箭头来表示电子对的移动。第一步，箭头从C=C键的中间出发，指向亲电试剂；与此同时，另一弯箭头表示X–Y键发生异裂，电子对归Y。第二步，弯箭头从亲核试剂的孤对电子出发，指向碳正离子上带正电的碳。

4. Step 1: Formation of Carbocation Intermediate | 第一步：碳正离子中间体的生成

When the electrophile (E⁺) approaches the alkene, the high electron density of the π bond attracts it. Two electrons from the π bond are used to form a new σ bond between one carbon of the double bond and the electrophile. The other carbon atom becomes electron‑deficient and carries a positive charge, forming a carbocation (carbonium ion).

当亲电试剂（E⁺）接近烯烃时，π键的高电子密度吸引它。π键中的两个电子被用来在双键的一个碳与亲电试剂之间形成一个新的σ键。另一个碳原子变得缺电子，带正电荷，形成碳正离子。

This step is the rate‑determining step because it involves breaking the π bond and forming the high‑energy carbocation intermediate. The stability of the resulting carbocation greatly influences the reaction rate and the product distribution.

这一步是决速步骤，因为它涉及断裂π键并生成高能量的碳正离子中间体。所生成的碳正离子的稳定性极大地影响着反应速率和产物分布。

5. Step 2: Attack by the Nucleophile | 第二步：亲核试剂进攻

In the second step, a nucleophile (Nu⁻), rich in electrons, attacks the positively charged carbon of the carbocation. A new σ bond forms between the nucleophile and that carbon. This step is fast and completes the addition process.

第二步中，富含电子的亲核试剂（Nu⁻）进攻碳正离子上带正电的碳。亲核试剂与该碳之间形成一个新的σ键。这一步速度快，完成加成过程。

The identity of the nucleophile depends on the reagents used. For the addition of H–X, the nucleophile is the halide ion (e.g., Cl⁻, Br⁻). For the addition of Br₂, the nucleophile is Br⁻ (produced in the first step). In hydration, water acts as the nucleophile, and a subsequent deprotonation step yields the alcohol.

亲核试剂的种类取决于所用的试剂。对于H–X加成，亲核试剂是卤离子（如Cl⁻、Br⁻）；对于Br₂加成，亲核试剂是第一步产生的Br⁻；在水合反应中，水分子充当亲核试剂，随后经过脱质子步骤得到醇。

6. Major and Minor Products: Markovnikov’s Rule | 主产物与次产物：马氏规则

When an unsymmetrical alkene (such as propene) reacts with an unsymmetrical reagent (such as H–Br), two carbocation intermediates are possible in theory, depending on which carbon atom the H⁺ attaches to. The major product is formed via the more stable carbocation. Markovnikov’s rule states: ‘In the addition of a hydrogen halide to an unsymmetrical alkene, the hydrogen atom attaches to the carbon that already has the greater number of hydrogen atoms.’ This leads to the formation of the more stable carbocation intermediate.

当不对称烯烃（如丙烯）与不对称试剂（如H–Br）反应时，理论上可能有两种碳正离子中间体，取决于H⁺加在哪个碳上。主产物经由更稳定的碳正离子生成。马氏规则指出：“在卤化氢与不对称烯烃的加成中，氢原子加在原本氢原子数目较多的碳上。” 这导致生成更稳定的碳正离子中间体。

For example, addition of HBr to propene gives mainly 2‑bromopropane (via the secondary carbocation) rather than 1‑bromopropane (via the primary carbocation). The exam expects you to identify the major and minor products and justify using carbocation stability.

例如，HBr与丙烯加成，主要得到2‑溴丙烷（经由二级碳正离子），而非1‑溴丙烷（经由一级碳正离子）。考试要求你识别主产物和次产物，并用碳正离子的稳定性加以解释。

7. Carbocation Stability and Rearrangements | 碳正离子稳定性与重排

Carbocation stability increases in the order: primary (1°) < secondary (2°) < tertiary (3°). Alkyl groups are electron‑donating, and they help to stabilise the positive charge on the carbon atom. More alkyl substituents on the charged carbon result in greater stability.

碳正离子稳定性顺序为：一级（1°）<二级（2°）<三级（3°）。烷基是给电子基团，有助于稳定碳原子上的正电荷。带电荷的碳上的烷基取代基越多，稳定性越高。

At the IGCSE level, you are not required to draw carbocation rearrangements in detail, but you should know that if a more stable carbocation can be formed by a 1,2‑shift of a hydride (H⁻) or an alkyl group, it will occur, leading to unexpected products. Mentioning the stability order is sufficient for Edexcel IGCSE answers.

在IGCSE阶段，不要求详细画碳正离子重排，但你应该知道，如果能通过氢负离子（H⁻）或烷基的1,2‑迁移形成更稳定的碳正离子，重排就会发生，导致生成意料之外的产物。Edexcel IGCSE答案中提及稳定性顺序就足够了。

The key takeaway: always identify the type of carbocation that would be formed when H⁺ adds to each carbon of the double bond, and choose the pathway that yields the more stable carbocation.

关键要点：当H⁺加到双键两个碳上时，始终识别可能形成的碳正离子类型，并选择生成较稳定碳正离子的途径。

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

Alkenes react with hydrogen halides such as HCl, HBr, and HI at room temperature to produce haloalkanes. The reaction rate depends on the hydrogen halide: HI reacts fastest, followed by HBr, then HCl, due to the decreasing bond strength of H–X.

烯烃与卤化氢（如HCl、HBr、HI）在室温下反应生成卤代烷。反应速率取决于卤化氢的种类：HI反应最快，其次是HBr，然后才是HCl，这是因为H–X键的强度依次增强。

The mechanism follows the two‑step pathway: H⁺ adds to the C=C, forming the more stable carbocation, and the halide ion then attacks. For symmetrical alkenes, only one product is formed; for unsymmetrical alkenes, Markovnikov’s rule predicts the major product.

反应遵循两步机理：H⁺加到C=C上，形成较稳定的碳正离子，然后卤离子进攻。对于对称烯烃，只生成一种产物；对于不对称烯烃，马氏规则预测主产物。

Example: Propene + HBr → 2‑bromopropane (major) + 1‑bromopropane (minor). Use curly arrows to show the heterolytic fission of H–Br and the attack of Br⁻ on the carbocation.

示例：丙烯 + HBr → 2‑溴丙烷（主）+ 1‑溴丙烷（次）。用弯箭头表示H–Br的异裂以及Br⁻对碳正离子的进攻。

9. Addition of Halogens (Br₂, Cl₂) | 卤素（Br₂、Cl₂）的加成

Alkenes react with bromine or chlorine at room temperature to form di‑halogenated alkanes. This reaction is commonly used to test for unsaturation, as the orange/brown colour of bromine is decolourised instantly.

烯烃与溴或氯在室温下反应，生成二卤代烷。该反应常用于检验不饱和性，因为溴的橙／棕色会立即褪去。

The mechanism for bromination: the Br–Br bond is polarised by the approaching π electrons. The δ⁺ bromine acts as the electrophile and forms a cyclic bromonium ion intermediate (not required for IGCSE, but the concept of a three‑membered ring may be mentioned for extension). For IGCSE, the simplified carbocation mechanism is acceptable: Br–Br breaks heterolytically, giving Br⁺ (electrophile) and Br⁻. Br⁺ adds to one carbon, leaving a carbocation, which is then attacked by Br⁻. This yields 1,2‑dibromoalkane.

溴化反应机理：接近的π电子使Br–Br键极化。δ⁺溴充当亲电试剂，形成环状溴鎓离子中间体（IGCSE不要求，但可提及三元环的概念供拓展）。对于IGCSE，简化的碳正离子机理亦可接受：Br–Br异裂，生成Br⁺（亲电试剂）和Br⁻。Br⁺加到一个碳上，留下碳正离子，后者再被Br⁻进攻，得到1,2‑二溴代烷。

Disappearance of bromine colour without UV light distinguishes alkenes from alkanes. Alkanes only react with bromine in the presence of UV light via free‑radical substitution.

无需紫外光溴水就褪色，这是区分烯烃和烷烃的关键。烷烃只有在紫外光存在下才通过自由基取代与溴反应。

10. Addition of Water (Hydration) | 水合反应

Alkenes can be converted to alcohols by the addition of water across the double bond. This is called hydration and requires an acid catalyst, typically concentrated phosphoric acid (H₃PO₄) or sulfuric acid, at high temperature (e.g., 300 °C) and pressure (60–70 atm).

烯烃可以通过在双键上加成水转化为醇，这称为水合反应，需要酸催化剂（通常是浓磷酸或硫酸），并在高温（如300 °C）和高压（60–70 atm）下进行。

The mechanism proceeds via protonation of the alkene by H⁺ from the acid, forming a carbocation (Markovnikov pathway), followed by attack of water acting as a nucleophile. A final deprotonation regenerates the acid catalyst and gives the alcohol.

该机理历程为：酸提供的H⁺使烯烃质子化，形成碳正离子（马氏规则途径），然后水作为亲核试剂进攻，最后经过脱质子步骤再生酸催化剂并得到醇。

For unsymmetrical alkenes, the major alcohol product follows Markovnikov’s rule: e.g., hydration of propene gives propan‑2‑ol as the major product, not propan‑1‑ol. This industrial method is used to manufacture ethanol from ethene.

对于不对称烯烃，主产物醇遵循马氏规则：例如，丙烯水合生成2‑丙醇为主产物，而非1‑丙醇。这一工业方法用于从乙烯制备乙醇。

11. Testing for Unsaturation: Bromine Water Test | 不饱和性检验：溴水实验

The bromine water test is a core practical in IGCSE Edexcel. When an alkene (or any compound containing a C=C bond) is shaken with orange‑brown bromine water, the solution turns colourless immediately. This indicates the presence of a double bond undergoing electrophilic addition.

溴水实验是IGCSE Edexcel的核心实验。将烯烃（或任何含C=C键的化合物）与橙棕色溴水一起振荡，溶液立即变为无色，表明存在发生了亲电加成的双键。

Alkanes and other saturated compounds do not react with bromine water in the dark; the orange colour persists. If a colour change occurs under UV light, it is due to a substitution reaction, not addition.

烷烃及其他饱和化合物在黑暗条件下不与溴水反应，橙色保持不变。如果在紫外光下发生颜色变化，那是取代反应而非加成反应。

Remember: bromine water is aqueous and contains Br₂, but also some Br⁻ and HOBr. The simplified exam answer can state that Br₂ adds across the C=C bond, decolourising the bromine water.

记住：溴水是水溶液，含有Br₂，也含有一些Br⁻和次溴酸。考试简化的说法可以表述为Br₂加成在C=C键上，使溴水褪色。

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