Electrophilic Addition in Alkenes | 烯烃的亲电加成反应考点精讲

📚 Electrophilic Addition in Alkenes | 烯烃的亲电加成反应考点精讲

Electrophilic addition is the most characteristic reaction of alkenes, underpinning many industrial processes and laboratory tests in organic chemistry. For OCR IGCSE Chemistry, understanding the mechanism, conditions, and key reagents is essential for both multiple-choice and structured questions. This article covers everything you need to know, from the nature of the C=C double bond to Markovnikov’s rule and exam-ready tips.

亲电加成是烯烃最具特征的反应，也是有机化学中许多工业流程与实验室检测的基础。对 OCR IGCSE 化学而言，理解反应机理、条件与关键试剂，对于选择题和简答题都至关重要。本文涵盖从 C=C 双键的本质到马氏规则，再到备考技巧的全部内容。

1. Introduction to Electrophilic Addition | 亲电加成简介

Electrophilic addition is a reaction where an electron-poor species (electrophile) attacks the electron-rich carbon-carbon double bond of an alkene, breaking the π‑bond and forming two new σ‑bonds. The result is a saturated molecule containing the atoms of the added reagent.

亲电加成是一种反应，其中缺电子的物种（亲电试剂）进攻烯烃中富电子的碳碳双键，使 π 键断裂并形成两个新的 σ 键。产物是一个含有加成试剂原子的饱和分子。

This reaction type contrasts with substitution reactions typical of alkanes. While alkanes require UV light and a free-radical mechanism, alkenes undergo rapid addition even in the dark, simply because the double bond is electron‑rich and highly susceptible to electrophiles.

这一反应类型与烷烃典型的取代反应不同。烷烃需要紫外光与自由基机理，而烯烃即使在黑暗中也能快速发生加成，只因双键富电子且极易受到亲电试剂的进攻。

2. Structure of Alkenes – the C=C Double Bond | 烯烃结构——碳碳双键

The double bond in an alkene consists of one σ‑bond and one π‑bond. The σ‑bond is formed by head‑on overlap of sp² hybrid orbitals, while the π‑bond results from sideways overlap of p‑orbitals. The electron density of the π‑bond lies above and below the plane of the molecule, creating a region of high electron density that is exposed to electrophilic attack.

烯烃中的双键由一个 σ 键和一个 π 键组成。σ 键由 sp² 杂化轨道头对头重叠形成，而 π 键由 p 轨道侧向重叠产生。π 键的电子密度分布在分子平面上下方，形成了一片暴露在外、易受亲电进攻的高电子密度区域。

Because the π‑bond is weaker than the σ‑bond, it breaks more easily during a reaction. This makes alkenes much more reactive than alkanes, despite the overall double bond being stronger than a single bond. The two sp² carbon atoms adopt a trigonal planar geometry with bond angles of about 120°, keeping the π‑electrons accessible.

由于 π 键比 σ 键弱，反应中更易断裂。这使得烯烃比烷烃活泼得多，尽管双键的总强度大于单键。两个 sp² 碳原子采用平面三角形构型，键角约 120°，使 π 电子保持可接近的状态。

3. What is an Electrophile? | 什么是亲电试剂？

An electrophile is a species that is attracted to regions of high electron density and can accept a pair of electrons to form a new covalent bond. Common electrophiles in alkene addition reactions include H⁺ (from acids), the partially positive end of polar molecules like HBr or H₂O, and polarised molecules such as Br₂ when influenced by the double bond.

亲电试剂是一类被高电子密度区域吸引、并能接受一对电子形成新共价键的物种。烯烃加成中常见的亲电试剂包括 H⁺（来自酸）、极性分子（如 HBr 或 H₂O）中带部分正电荷的一端，以及受双键影响而产生极化的 Br₂ 等。

In the case of bromine, although the Br–Br molecule is non‑polar, the approach of the electron‑rich double bond induces a temporary dipole, making one bromine atom electrophilic. This induced polarity is enough to trigger the addition. Positive ions such as NO₂⁺ or SO₃ can also act as electrophiles, but these are beyond the IGCSE scope.

对于溴而言，虽然 Br–Br 分子是非极性的，但富电子的双键靠近时会诱导出瞬时偶极，使其中一个溴原子具有亲电性。这种诱导极性足以引发加成。NO₂⁺ 或 SO₃ 等正离子也可作为亲电试剂，但已超出 IGCSE 范围。

4. Mechanism Overview – The Two Steps | 机理概述——两步过程

Electrophilic addition to alkenes generally proceeds via a two‑step ionic mechanism, often illustrated using curly arrows in exams. The first step is the attack of the electrophile on the π‑bond, forming a carbocation intermediate and a negatively charged counter‑ion. The second step is the rapid attack of the nucleophile (often the counter‑ion) on the carbocation to give the final product.

烯烃的亲电加成通常按两步离子机理进行，考试中常用弯箭头表示。第一步是亲电试剂进攻 π 键，生成碳正离子中间体和一个带负电的平衡离子。第二步是亲核试剂（常为平衡离子）快速进攻碳正离子，得到最终产物。

For example, with HBr: Step 1 – The H⁺ electrophile accepts a pair of π‑electrons, forming a C–H bond and leaving a carbocation on the other carbon, while Br⁻ is released. Step 2 – The Br⁻ ion donates a pair of electrons to the positive carbon, forming a C–Br bond. The overall rate is determined by the first, slower step.

以 HBr 为例：第一步——H⁺ 亲电试剂接受一对 π 电子，形成 C–H 键，另一个碳上留下碳正离子，同时释放出 Br⁻；第二步——Br⁻ 离子向带正电的碳提供一对电子，形成 C–Br 键。总反应速率由较慢的第一步决定。

CH₂=CH₂ + H⁺ → CH₃–C⁺H₂ then CH₃–C⁺H₂ + Br⁻ → CH₃CH₂Br

Understanding this carbocation intermediate is key to explaining regioselectivity and Markovnikov’s rule.

理解这一碳正离子中间体是解释区域选择性与马氏规则的关键。

5. Addition of Hydrogen Halides (HBr) | 卤化氢加成（HBr）

Alkenes react with hydrogen halides such as HBr or HCl at room temperature to produce haloalkanes. The reaction is exothermic and proceeds with the electrophilic addition mechanism described above. For symmetrical alkenes like ethene, there is only one possible product: bromoethane.

烯烃在室温下与卤化氢如 HBr 或 HCl 反应，生成卤代烷。该反应放热，按上述亲电加成机理进行。对于对称烯烃如乙烯，只能得到一种产物：溴乙烷。

CH₂=CH₂ + HBr → CH₃CH₂Br

For unsymmetrical alkenes such as propene, two products are possible – 1‑bromopropane and 2‑bromopropane. The major product is determined by the stability of the carbocation intermediate, leading to Markovnikov’s rule.

对于不对称烯烃如丙烯，可能产生两种产物——1-溴丙烷和2-溴丙烷。主要产物由碳正离子中间体的稳定性决定，这就引出了马氏规则。

HCl is less reactive than HBr because the H–Cl bond is stronger, but the addition still occurs. HI reacts even more vigorously, though it is less commonly used at IGCSE level. All these reactions produce a single haloalkane when the alkene is symmetrical.

HCl 的反应性弱于 HBr，因为 H–Cl 键更强，但加成仍能发生。HI 反应更为剧烈，不过在 IGCSE 阶段较少见。当烯烃对称时，这些反应均生成单一卤代烷。

6. Markovnikov’s Rule | 马氏规则

Markovnikov’s rule states that when an unsymmetrical reagent adds to an unsymmetrical alkene, the hydrogen (or electropositive part) of the reagent attaches to the carbon of the double bond that already carries the greater number of hydrogen atoms. In other words, “the rich get richer” – the carbon with more hydrogens gets another hydrogen.

马氏规则指出，当不对称试剂与不对称烯烃加成时，试剂中的氢（或正电部分）会连接到双键上原本含氢较多的碳原子上。也就是说，“富者愈富”——原来氢多的碳获得另一个氢。

This outcome is explained by carbocation stability. A secondary carbocation (R₂CH⁺) is more stable than a primary carbocation (RCH₂⁺) due to positive inductive effects from alkyl groups. During the addition of HBr to propene, the intermediate CH₃–C⁺H–CH₃ (secondary) is favoured over C⁺H₂–CH₂–CH₃ (primary), so the major product is 2‑bromopropane.

这一结果可用碳正离子的稳定性解释。由于烷基的正诱导效应，二级碳正离子（R₂CH⁺）比一级碳正离子（RCH₂⁺）更稳定。HBr 与丙烯加成时，中间体 CH₃–C⁺H–CH₃（二级）优于 C⁺H₂–CH₂–CH₃（一级），因此主要产物为 2-溴丙烷。

CH₃–CH=CH₂ + HBr → CH₃–CHBr–CH₃ (major) + CH₃–CH₂–CH₂Br (minor)

Markovnikov’s rule is observed in the addition of HX, H₂O (acid‑catalysed), and other unsymmetrical electrophiles. It is a reliable guide for predicting organic products in OCR IGCSE exam questions.

马氏规则适用于 HX、酸催化下 H₂O 以及其他不对称亲电试剂的加成，是预测 OCR IGCSE 考试题中有机产物的可靠指南。

7. Addition of Halogens (Br₂) | 卤素加成（溴）

Alkenes readily decolourise bromine at room temperature in the dark, without the need for UV light. The reaction with pure bromine yields a vicinal dibromoalkane. Unlike the addition of HBr, this reaction proceeds through a cyclic bromonium ion intermediate (not required in detail for IGCSE), but the overall stoichiometry is straightforward.

烯烃在室温、黑暗条件下即可使溴褪色，无需紫外光。与纯溴反应得到邻二溴代烷。与 HBr 加成不同，该反应经过环状溴鎓离子中间体（IGCSE 不要求细节），但总化学计量式很简单。

CH₂=CH₂ + Br₂ → CH₂Br–CH₂Br

The reaction also occurs with chlorine, though chlorine is more hazardous and less often demonstrated. Iodine addition is less favourable due to thermodynamics, but bromination remains the classic test for unsaturation. The colour change from orange/brown to colourless is immediate and striking.

该反应同样适用于氯气，但氯气更危险，较少演示。碘的加成在热力学上不利，但溴化仍是不饱和度的经典检验。溶液从橙/棕色变为无色的过程迅速且明显。

8. Test for Unsaturation – Bromine Water | 不饱和检验——溴水试验

Shaking an alkene with orange/yellow bromine water leads to a rapid loss of colour, producing a colourless dibromoalcohol and HBr if water participates, or simply the dibromide if bromine water is used in excess. Alkanes and other saturated compounds do not react under these conditions, so the decolourisation confirms the presence of a C=C double bond.

将烯烃与橙/黄色的溴水一起振荡会快速褪色，若水参与反应则生成无色溴代醇和 HBr，或当溴水过量时仅生成二溴化物。烷烃等饱和化合物在此条件下不反应，因此褪色现象可确认 C=C 双键的存在。

In the lab, this test is performed by adding a few drops of bromine water to a test tube containing the unknown hydrocarbon and shaking. A positive result is the disappearance of the orange colour within seconds. This is a favourite OCR IGCSE practical chemistry question and is often linked to identifying alkenes from alkanes.

实验室中，可在盛有未知烃的试管中加入几滴溴水并振荡，若橙色在数秒内消失即为阳性结果。这是 OCR IGCSE 实验化学中常见的考题，常与区别烯烃和烷烃相关联。

Note that UV light can cause alkanes to slowly decolourise bromine via free‑radical substitution, so the test must be done quickly and in the absence of strong light to avoid a false positive.

注意，紫外光可使烷烃通过自由基取代缓慢褪色，因此测试必须快速并在无强光下进行，以免出现假阳性。

9. Addition of Water (Steam) – Hydration | 水加成（水蒸气）——水合反应

Alkenes can be converted directly into alcohols by the addition of steam in the presence of a phosphoric(V) acid catalyst. This reaction is called catalytic hydration and follows Markovnikov’s rule when the alkene is unsymmetrical. For ethene, it yields ethanol; for propene, it gives propan‑2‑ol as the major product.

烯烃可在磷酸(V)催化剂存在下与蒸汽加成，直接转化为醇。该反应称为催化水合，当烯烃不对称时遵循马氏规则。乙烯得到乙醇；丙烯主要得到丙-2-醇。

CH₂=CH₂ + H₂O(g) ⇌ CH₃CH₂OH

Typical industrial conditions are a temperature of 300 °C, a pressure of 60 atm, and a solid phosphoric acid catalyst supported on silica. The hydration of ethene is a major industrial route to ethanol, competing with fermentation. This equilibrium reaction is favoured by high pressure and relatively low temperature, but the temperature must be high enough to achieve a reasonable rate.

典型的工业条件为温度 300 °C、压力 60 atm，以及负载在二氧化硅上的固体磷酸催化剂。乙烯水合是工业制乙醇的主要途径，与发酵法形成竞争。这一平衡反应受高压和相对低温有利，但温度必须足够高以获得合理的速率。

OCR IGCSE expects you to be able to write the equation, name the catalyst, and recall the conditions, as well as explain why this is an addition reaction and how it links to sustainability (using crude oil fractions vs renewable resources).

OCR IGCSE 要求能书写方程式、命名催化剂并回忆条件，同时解释为何这是加成反应，以及它与可持续性（使用原油馏分与可再生资源）的关系。

10. Addition of Hydrogen – Hydrogenation | 氢气加成——加氢反应

The addition of hydrogen across a C=C double bond is known as hydrogenation. This reaction requires a finely divided metal catalyst such as nickel, platinum, or palladium, and is usually carried out at around 150 °C. It converts unsaturated alkenes into saturated alkanes.

氢气跨 C=C 双键的加成称为加氢反应。该反应需要细微分散的金属催化剂如镍、铂或钯，通常在约 150 °C 下进行，将不饱和烯烃转化为饱和烷烃。

CₙH₂ₙ + H₂ → CₙH₂ₙ₊₂

For example, ethene is hydrogenated to ethane, and propene to propane. The process is used industrially to harden unsaturated vegetable oils into margarine – partially hydrogenating C=C bonds raises the melting point. In this context, the catalyst is often nickel at 60 °C.

例如，乙烯加氢生成乙烷，丙烯生成丙烷。该过程在工业上用于将不饱和植物油硬化为人造黄油——部分加氢 C=C 键可提高熔点。此时常用镍催化剂，温度约 60 °C。

Hydrogenation is an important addition reaction that demonstrates the conversion of an unsaturated to a saturated compound. Be prepared to explain the role of the catalyst in adsorbing H₂ and weakening the H–H bond, although detailed catalytic theory is beyond IGCSE.

加氢反应是体现不饱和向饱和化合物转化的重要加成反应。准备好解释催化剂在吸附 H₂ 并削弱 H–H 键中的作用，尽管详细的催化理论已超出 IGCSE 范围。

11. Summary of Key Reactions | 关键反应总结

The table below summarises the four main electrophilic addition reactions of alkenes required for OCR IGCSE, along with reagents, conditions, and products.

下表总结了 OCR IGCSE 要求的烯烃四种主要亲电加成反应，包括试剂、条件与产物。

Reagent	Conditions	Product Type	Example
Hydrogen, H₂	Ni catalyst, 150 °C	Alkane	Ethene → ethane
Halogen, X₂ (e.g. Br₂)	Room temperature, dark	Dihaloalkane	Ethene → 1,2‑dibromoethane
Hydrogen halide, HX	Room temperature	Haloalkane	Propene → 2‑bromopropane
Steam, H₂O	H₃PO₄ catalyst, 300 °C, 60 atm	Alcohol	Ethene → ethanol

Make sure you can write balanced equations for each of these reactions, including structural or displayed formulas where required. The ability to link the conditions to industrial processes, such as ethanol production and margarine hardening, is often examined.

确保能够为每一反应书写配平的化学方程式，必要时包括结构式或展示式。将条件与工业过程（如乙醇生产与人造黄油硬化）联系起来的能力经常会被考查。

12. Exam Tips for OCR IGCSE | OCR IGCSE考试技巧

When answering exam questions on electrophilic addition, always identify the electrophile and explain why the double bond is susceptible. Use correct terminology: ‘electrophile’, ‘π‑bond’, ‘carbocation’, and ‘Markovnikov’s rule’. Avoid vague phrases like ‘the molecule breaks’ without specifying which bond.

在回答亲电加成的考题时，务必指出亲电试剂并解释双键为何易受攻击。使用正确术语：“亲电试剂”、“π 键”、“碳正离子”和“马氏规则”。避免使用“分子断裂”等模糊说法而不指明具体是哪个键。

For mechanistic questions, draw curly arrows starting from the π‑bond or a lone pair towards the electrophilic atom. Show the formation of the carbocation and the final attack by the nucleophile. Practice using displayed formulas to avoid ambiguity – a common mistake is to lose a hydrogen atom or misplace the positive charge.

对于机理题，弯箭头应从 π 键或孤对电子出发指向亲电原子。展示碳正离子的形成以及最后亲核试剂的进攻。多练习使用展示式以避免歧义——常见错误是丢失氢原子或标错正电荷位置。

In questions on bromine water, mention both the colour change and the fact that no UV light is required. Contrast this with the behaviour of alkanes. When discussing hydration, explicitly mention the phosphoric acid catalyst and reversible reaction arrow; if asked about conditions, state the temperature and pressure values as given in the syllabus.

在涉及溴水的题目中，同时提及颜色变化以及不需要紫外光这一事实。对此与烷烃的行为进行对比。讨论水合反应时，明确提及磷酸催化剂和可逆反应箭头；若问到条件，则给出教学大纲中的温度与压力数值。

Finally, always read the question carefully: if it asks for the ‘major product’ of an unsymmetrical alkene, apply Markovnikov’s rule. If a displayed formula is requested, ensure every bond and atom is shown clearly. Time spent mastering addition reactions will pay off in both Paper 2 (theory) and Paper 6 (alternative to practical).

最后，仔细审题：如果题目问不对称烯烃的“主要产物”，要应用马氏规则。如果要求画出展示式，确保所有键和原子清晰呈现。花时间掌握加成反应，将有助于在 Paper 2（理论）与 Paper 6（实验替代）中取得好成绩。

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Electrophilic Addition in Alkenes | 烯烃的亲电加成反应 考点精讲