📚 Electrophilic Addition: A Comprehensive Guide for IB and OCR Chemistry | 亲电加成：IB 与 OCR 化学考点精讲
Electrophilic addition is a fundamental reaction mechanism in organic chemistry, forming the cornerstone of alkene and alkyne reactivity for IB (both SL and HL) and OCR A-Level specifications. This article systematically breaks down every essential concept – from the nature of the electrophile to the stereochemical outcomes – ensuring you are fully prepared for examination questions that test both knowledge and application.
亲电加成是有机化学中一个基础的反应机理，构成了 IB（SL 和 HL）以及 OCR A-Level 课程中烯烃和炔烃反应活性的核心。本文将系统地分解从亲电试剂本质到立体化学结果的每一个关键概念，确保你为考查知识与应用的试题做好充分准备。
1. What is Electrophilic Addition? | 什么是亲电加成？
Electrophilic addition is a reaction in which an electron-deficient species (an electrophile) attacks an electron-rich centre – typically the carbon-carbon double bond (C=C) of an alkene – resulting in the addition of two groups across the π-bond. The double bond is broken, and two new sigma bonds are formed. It is the characteristic reaction of unsaturated hydrocarbons.
亲电加成是一种反应，其中缺电子物种（亲电试剂）进攻富电子中心——通常是烯烃的碳碳双键（C=C）——导致两个基团加在 π 键两端。双键断裂并形成两个新的 σ 键。这是不饱和烃的特征反应。
2. The Electrophile and the π-Bond | 亲电试剂与 π 键
Alkenes possess a region of high electron density above and below the plane of the C=C bond due to the loosely held π-electrons. This makes the π-bond an attractive target for electrophiles (electron-loving species). Common electrophiles in these reactions include H⁺ (from HX acids), Br⁺ (from polarised Br₂), and the partially positive sulfur atom in concentrated sulfuric acid.
烯烃由于 π 电子云松散地分布在 C=C 键平面的上方和下方，形成了一个高电子密度区域。这使得 π 键成为亲电试剂（亲电子物种）的吸引力靶点。这些反应中常见的亲电试剂包括 H⁺（来自 HX 酸）、Br⁺（来自极化的 Br₂）以及浓硫酸中带部分正电荷的硫原子。
A key skill in both IB and OCR exams is identifying which species acts as the electrophile. For example, when bromine approaches an alkene, the electron cloud of the double bond induces a dipole in the Br₂ molecule, creating Brᵟ⁺–Brᵟ⁻. The Brᵟ⁺ then serves as the electrophile.
IB 和 OCR 考试中的一项关键技能是识别哪个物种充当亲电试剂。例如，当溴靠近烯烃时，双键的电子云在 Br₂ 分子中诱导出偶极，产生 Brᵟ⁺–Brᵟ⁻。随后 Brᵟ⁺ 作为亲电试剂进攻。
3. Mechanism of Electrophilic Addition: General Steps | 亲电加成机理：一般步骤
The reaction proceeds through a two-step ionic mechanism (not a concerted process for most additions). Understanding the sequence is essential for drawing curly arrow diagrams.
该反应通过两步离子机理进行（大多数加成反应不是协同过程）。理解该顺序对于绘制弯箭头图至关重要。
Step 1: Formation of the carbocation intermediate. The electrophile accepts a pair of electrons from the π-bond, forming a new sigma bond to one carbon atom. The other carbon becomes electron-deficient, forming a carbocation (carbonium ion). The curly arrow starts from the middle of the C=C bond and points to the electrophile (e.g., H⁺).
第一步：生成碳正离子中间体。亲电试剂从 π 键接受一对电子，与一个碳原子形成新的 σ 键。另一个碳原子则变得缺电子，形成碳正离子（碳鎓离子）。弯箭头从 C=C 键的中间指向亲电试剂（例如 H⁺）。
Step 2: Nucleophilic attack by the anion. The negatively charged ion (e.g., Br⁻, Cl⁻, HSO₄⁻) then donates a pair of electrons to the positively charged carbon atom of the carbocation, forming the second new sigma bond. The curly arrow starts from a lone pair on the nucleophile and points to the positive carbon.
第二步：阴离子的亲核进攻。带负电荷的离子（如 Br⁻、Cl⁻、HSO₄⁻）向碳正离子上带正电的碳原子提供一对电子，形成第二个新的 σ 键。弯箭头从亲核试剂的孤对电子出发，指向带正电的碳。
OCR examiners are particularly strict about the precise placement of curly arrows. Ensure the arrow in Step 1 originates from the π-bond (between the two carbon atoms) and not from a specific carbon atom, unless you are drawing the arrow from the C=C bond centre.
OCR 考官对弯箭头的精确放置要求尤其严格。确保第一步的箭头从 π 键（两个碳原子之间）出发，而不是从某个特定的碳原子出发，除非你从 C=C 键的中心画起。
4. Addition of Hydrogen Halides (HX) to Alkenes | 卤化氢（HX）与烯烃的加成
When an alkene reacts with a hydrogen halide such as HCl, HBr, or HI, the product is a haloalkane. The reaction is often rapid at room temperature, with HBr reacting faster than HCl due to the weaker H–Br bond and greater polarisability.
当烯烃与卤化氢（如 HCl、HBr 或 HI）反应时，产物是卤代烷烃。该反应通常在室温下快速进行，HBr 因 H–Br 键较弱且极化性更强，反应速率比 HCl 快。
CH₂=CH₂ + HBr → CH₃CH₂Br
（乙烯与溴化氢反应生成溴乙烷）
With unsymmetrical alkenes, the orientation of addition becomes important (see Markovnikov’s rule). The mechanism always involves the initial attack of H⁺, not the halide ion.
对于不对称烯烃，加成的取向变得重要（参见马氏规则）。该机理始终涉及 H⁺ 的初始进攻，而非卤离子。
5. Markovnikov’s Rule and Carbocation Stability | 马氏规则与碳正离子稳定性
Markovnikov’s rule states that when HX adds to an unsymmetrical alkene, the hydrogen atom attaches to the carbon with the greater number of hydrogen atoms already attached (the ‘ric’ gets richer). In mechanistic terms, the electrophile H⁺ adds to the less substituted carbon, leading to the more stable carbocation intermediate.
马氏规则指出，当 HX 与不对称烯烃加成时，氢原子加在原来连接氢原子数较多的碳原子上（“富者愈富”）。从机理角度，亲电试剂 H⁺ 加在取代度较低的碳上，从而生成更稳定的碳正离子中间体。
The rationale lies in carbocation stability: tertiary (3°) > secondary (2°) > primary (1°) > methyl. Alkyl groups are electron-donating, stabilising the positive charge through positive inductive and hyperconjugation effects. IB examinations frequently ask for an explanation of this stability trend in terms of charge dispersal.
其原因在于碳正离子稳定性：叔（3°）> 仲（2°）> 伯（1°）> 甲基。烷基是给电子基团，通过正诱导效应和超共轭效应稳定正电荷。IB 考试经常要求用电荷分散的原理解释这一稳定性趋势。
For example, in the addition of HBr to propene (CH₃CH=CH₂), the secondary carbocation CH₃CH⁺CH₃ is formed in preference to the primary carbocation CH₃CH₂CH₂⁺ because the former is more stable.
例如，在 HBr 与丙烯（CH₃CH=CH₂）的加成中，优先形成仲碳正离子 CH₃CH⁺CH₃，而非伯碳正离子 CH₃CH₂CH₂⁺，因为前者更稳定。
6. Addition of Halogens (X₂) to Alkenes | 卤素（X₂）与烯烃的加成
Alkenes react with chlorine (Cl₂) or bromine (Br₂) at room temperature to give dihalogenoalkanes. The mechanism is similar to HX addition but the electrophile is the polarised halogen molecule. It is often depicted via the formation of a cyclic halonium ion intermediate (e.g., bromonium ion) in many advanced syllabi, but the IB and OCR core typically present the stepwise carbocation pathway with initial attack by Brᵟ⁺.
烯烃在室温下与氯（Cl₂）或溴（Br₂）反应生成二卤代烷烃。其机理与 HX 加成相似，但亲电试剂是极化的卤素分子。许多高级课程常常将其描绘为通过形成环状卤鎓离子中间体（如溴鎓离子），但 IB 和 OCR 核心课程通常采用逐步碳正离子路径，由 Brᵟ⁺ 初始进攻。
In the case of ethene and bromine:
CH₂=CH₂ + Br₂ → CH₂BrCH₂Br
对于乙烯与溴的情况：
CH₂=CH₂ + Br₂ → CH₂BrCH₂Br
The product is 1,2-dibromoethane. The reaction is regioselective but not governed by Markovnikov’s rule because both carbon centres in a symmetrical alkene are equivalent.
产物是 1,2-二溴乙烷。该反应具有区域选择性，但因对称烯烃中两个碳中心是等价的，不受马氏规则支配。
7. Addition of Water: Hydration of Alkenes | 水的加成：烯烃的水合反应
Alkenes do not react directly with water because water is not a strong enough electrophile. However, in the presence of a strong acid catalyst (commonly concentrated sulfuric acid in OCR, or H₃PO₄ for industrial hydration), the addition of water occurs in two stages.
烯烃不能直接与水反应，因为水不是足够强的亲电试剂。然而，在强酸催化剂（OCR 中常用浓硫酸，工业水合中则用 H₃PO₄）存在下，水的加成分两个阶段进行。
Stage 1 – Cold concentrated H₂SO₄: The alkene reacts with sulfuric acid via electrophilic addition. The H⁺ attacks the π-bond, and the HSO₄⁻ ion then adds to the carbocation, forming an alkyl hydrogen sulfate.
第一阶段 – 冷的浓 H₂SO₄：烯烃通过亲电加成与硫酸反应。H⁺ 进攻 π 键，随后 HSO₄⁻ 离子加到碳正离子上，生成烷基硫酸氢酯。
CH₂=CH₂ + H₂SO₄ → CH₃CH₂OSO₃H
Stage 2 – Warm water addition and hydrolysis: When water is added and the mixture is heated, the alkyl hydrogen sulfate is hydrolysed to yield the alcohol and regenerate the sulfuric acid catalyst.
第二阶段 – 加入温水并水解：当加入水并加热时，烷基硫酸氢酯水解生成醇，同时再生硫酸催化剂。
CH₃CH₂OSO₃H + H₂O → CH₃CH₂OH + H₂SO₄
The overall reaction is the addition of H–OH across the C=C bond, which follows Markovnikov’s rule for unsymmetrical alkenes, giving secondary or tertiary alcohols as major products.
总反应是 H–OH 加在 C=C 键两端，对于不对称烯烃遵循马氏规则，主要产物为仲醇或叔醇。
8. Bromine Water Test for Unsaturation | 溴水检验不饱和键
A classic test for the presence of a carbon-carbon double bond is the bromine water test. Orange-brown bromine water is decolorised rapidly when shaken with an alkene. This observation is a result of electrophilic addition, forming a colourless dibromo compound. This is a key practical skill assessed in OCR PAGs and IB individual investigation.
检验碳碳双键存在的经典方法是溴水试验。将橙棕色的溴水与烯烃振摇，会迅速褪色。该现象是亲电加成生成无色二溴化合物的结果。这是 OCR 实践评估和 IB 个人探究中考查的一项关键实验技能。
It is important to note that alkanes do not decolorise bromine water in the dark unless exposed to UV light, which promotes free-radical substitution. This distinction allows chemists to differentiate between saturated and unsaturated hydrocarbons.
需要注意的是，烷烃在黑暗中不会使溴水褪色，除非在紫外光照射下引发自由基取代反应。这一区别使化学家能区分饱和烃与不饱和烃。
9. Stereochemistry of Addition: Anti and Syn Addition | 加成的立体化学：反式加成与同面加成
For IB HL and OCR students targeting higher grades, understanding the stereochemical outcome is critical. When bromine adds to cyclohexene, for example, the two bromine atoms add from opposite faces of the double bond (anti addition). This occurs because the bromonium ion intermediate shields one face of the molecule, forcing the Br⁻ nucleophile to attack from the opposite side.
对于瞄准高分的 IB HL 和 OCR 学生，理解立体化学结果至关重要。例如，当溴与环己烯加成时，两个溴原子从双键的相对两面加成（反式加成）。这是因为溴鎓离子中间体屏蔽了分子的一面，迫使 Br⁻ 亲核试剂从相反面进攻。
In contrast, catalytic hydrogenation (addition of H₂ with a metal catalyst) is a syn addition: both hydrogen atoms add to the same face of the alkene. Although hydrogenation is not electrophilic addition, comparing these two modes helps clarify important concepts of stereoselectivity.
相反，催化氢化（在金属催化剂下加 H₂）是同面加成：两个氢原子加在烯烃的同一面。尽管氢化不属于亲电加成，但对比这两种模式有助于澄清重要的立体选择性概念。
Electrophilic addition of HX, when the carbocation is planar, generally gives a racemic mixture if it creates a chiral centre, due to equal probability of attack from either face. This is often tested in multi-step synthesis questions.
HX 的亲电加成中，当碳正离子为平面结构时，若生成手性中心，通常得到外消旋混合物，因为从两面进攻的概率相等。这一点常出现在多步合成题中。
10. Summary and Exam Tips | 总结与应试技巧
To excel in IB and OCR questions on electrophilic addition, ensure you can:
要想在 IB 和 OCR 亲电加成相关题目中取得优异成绩，请确保你能做到：
- Draw the full mechanism with curly arrows for addition of HX, X₂, and H₂O (via H₂SO₄).
- 绘制完整的弯箭头机理，包括 HX、X₂ 和 H₂O（经 H₂SO₄）的加成。
- Identify the electrophile in each reaction.
- 识别每种反应中的亲电试剂。
- Apply Markovnikov’s rule and explain it through carbocation stability.
- 应用马氏规则并通过碳正离子稳定性进行解释。
- Predict major and minor products for unsymmetrical alkenes.
- 预测不对称烯烃的主要和次要产物。
- Describe colour change in the bromine water test and write the equation.
- 描述溴水试验的颜色变化并写出化学方程式。
- Explain stereochemical outcomes (anti addition for halogens, racemic mixture for HX when chiral centres form).
- 解释立体化学结果（卤素的反式加成，HX 形成手性中心时的外消旋混合物）。
Consistent practice with past paper questions will solidify these concepts. Remember that careful use of correct chemical notation – including partial charges (ᵟ⁺, ᵟ⁻) and proper curly arrows – can make the difference between a good answer and an excellent one.
持续练习历年真题将巩固这些概念。请记住，谨慎使用正确的化学符号——包括部分电荷（ᵟ⁺、ᵟ⁻）和恰当的弯箭头——可以让你的答案从良好变为优秀。
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