📚 A-Level Chemistry Insert 4 (Jun22): Reaction Mechanisms | A-Level 化学插页4（2022年6月）：反应机理
Reaction mechanisms are the core of organic chemistry, revealing how electrons move during chemical transformations. The AQA Insert 4 from June 2022, a typical data sheet, outlines fundamental mechanisms such as electrophilic addition, nucleophilic substitution, and free radical substitution. This article unpacks these mechanisms with detailed explanations, curly arrow notation, and exam-focused insights to help you master A-Level reaction mechanisms.
反应机理是有机化学的核心，揭示了化学变化中电子的移动方式。2022年6月AQA插页4是一份典型的数据表，概述了亲电加成、亲核取代和自由基取代等基本机理。本文将详细解释这些机理，配合弯曲箭头表示法和考试重点讲解，助你掌握A-Level阶段的反应机理。
1. What is a Reaction Mechanism? | 什么是反应机理？
A reaction mechanism describes the step-by-step sequence of bond-breaking and bond-making events, including the movement of electron pairs or single electrons. In A-Level Chemistry, you are expected to illustrate mechanisms using curly arrows that show the flow of electrons from a nucleophile or a base to an electrophile or a proton. These diagrams reveal why reactions happen, predict products, and explain the role of conditions like polar solvents or UV light.
反应机理描述了化学键断裂和形成的逐步过程，包括电子对或单个电子的移动。在A-Level化学中，你需要用弯曲箭头画出电子从亲核试剂或碱流向亲电试剂或质子的过程，以此说明机理。这些示意图揭示了反应发生的原因，可预测产物，并解释极性溶剂或紫外光等条件的作用。
2. Heterolytic and Homolytic Bond Fission | 异裂与均裂
Bond breaking is the first step in any mechanism. Heterolytic fission occurs when a covalent bond breaks unevenly, giving both electrons to one atom to form a cation and an anion. For example, when a hydrogen halide approaches an alkene, the H–X bond breaks heterolytically to generate H⁺ and X⁻. This is typical of polar reactions.
化学键的断裂是所有机理的第一步。异裂是指共价键不均匀断裂，两个电子全部归一个原子，形成阳离子和阴离子。例如，卤化氢接近烯烃时，H–X键异裂生成H⁺和X⁻。这是极性反应的典型方式。
Homolytic fission, on the other hand, splits the bonding electrons equally, producing two radicals. This requires energy input, often from UV light, as seen in the chlorination of methane: Cl–Cl → 2 Cl•. Homolytic fission is the key initiation step in free radical substitution mechanisms.
均裂则使成键电子均等分配，生成两个自由基。这需要外界能量，常由紫外光提供，例如甲烷氯化中的Cl–Cl → 2 Cl•。均裂是自由基取代机理的关键引发步骤。
3. Curly Arrow Rules | 弯曲箭头表示法则
Curly arrows show the movement of an electron pair. The arrow tail starts at the electron source – a lone pair on a nucleophile or a bond pair in a π‑bond – and the arrow head points to the electron-deficient atom (electrophile) or to the region where a new bond will form. A full arrow indicates two electrons; a half‑headed ‘fishhook’ arrow is used for the movement of a single electron in radical mechanisms. Always draw arrows from negative to positive character.
弯曲箭头表示电子对的移动。箭头尾端始于电子源（亲核试剂的孤对电子或π键的成键电子），箭头指向缺电子的原子（亲电试剂）或将形成新键的位置。完整箭头代表两个电子，半箭头“鱼钩”用于自由基机理中单个电子的移动。箭头始终从富电子区域指向缺电子区域。
4. Electrophilic Addition to Alkenes | 烯烃的亲电加成
Alkenes react with electrophiles because the π‑bond is an electron-rich region. In the addition of HBr to ethene, the mechanism proceeds in two steps. First, the π‑electrons attack H⁺, forming a carbocation intermediate (CH₃CH₂⁺) and Br⁻. The curly arrow goes from the C=C bond to the H atom, and then the H–Br bond breaks heterolytically.
烯烃与亲电试剂反应，因为π键是富电子区域。在HBr与乙烯的加成中，反应分两步进行。首先，π电子进攻H⁺，生成碳正离子中间体（CH₃CH₂⁺）和Br⁻。弯曲箭头从C=C键指向H原子，H–Br键随之异裂。
In the second step, the bromide ion acts as a nucleophile, donating a lone pair to the positively charged carbon, forming bromoethane. Both steps are shown with curly arrows. If the alkene is unsymmetrical, Markovnikov’s rule applies: the more stable carbocation (tertiary > secondary > primary) determines the major product.
第二步中，溴离子作为亲核试剂，将孤对电子提供给带正电的碳，形成溴乙烷。这两步都需要用弯曲箭头表示。若烯烃不对称，则遵循马氏规则：较稳定的碳正离子（叔 > 仲 > 伯）决定主产物。
When bromine (Br₂) is used, the mechanism involves a cyclic bromonium ion intermediate to avoid carbocation rearrangements. The π‑electrons attack one Br atom, forming a three‑membered ring with a positive charge on bromine, and a Br⁻ is released. The bromide ion then attacks from the opposite side, giving anti addition.
使用溴（Br₂）时，机理经过环状溴鎓离子中间体，以避免碳正离子重排。π电子进攻一个溴原子，形成带正电的三元环溴鎓离子，并释放出Br⁻。接着溴离子从背面进攻，导致反式加成。
5. Nucleophilic Substitution: SN1 and SN2 | 亲核取代：SN1 和 SN2
Haloalkanes undergo nucleophilic substitution when attacked by nucleophiles such as OH⁻, CN⁻ or NH₃. The mechanism can be SN2 (bimolecular) or SN1 (unimolecular), depending on the structure of the haloalkane.
卤代烷受到OH⁻、CN⁻或NH₃等亲核试剂进攻时，发生亲核取代。依据卤代烷的结构，机理可以是SN2（双分子）或SN1（单分子）。
| Feature | SN2 | SN1 |
|---|---|---|
| Molecularity | Bimolecular – rate depends on [haloalkane] and [nucleophile] | Unimolecular – rate depends only on [haloalkane] |
| Preferred substrate | Primary haloalkanes (little steric hindrance) | Tertiary haloalkanes (stable carbocation) |
| Stereochemistry | Inversion of configuration (Walden inversion) | Racemisation (planar carbocation, attack from either side) |
| Intermediate | Transition state with five bonds around carbon | Carbocation intermediate |
For SN2, the nucleophile attacks the carbon bearing the halogen from the back side, pushing the halogen off in a single step. The curly arrow starts from the nucleophile’s lone pair and goes to the carbon, while another arrow shows the C–X bond breaking. For a primary haloalkane like CH₃CH₂Br with NaOH(aq), the reaction is CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻, proceeding via a pentacoordinate transition state.
在SN2中，亲核试剂从卤素背后的方向进攻碳原子，一步将卤素推开。弯曲箭头始于亲核试剂的孤对电子，指向碳；另一箭头表示C–X键断裂。对于伯卤代烷如CH₃CH₂Br与NaOH水溶液反应，方程式为CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻，经过五配位过渡态。
In SN1, the C–halogen bond breaks first (rate-determining step) to form a planar carbocation. The nucleophile then quickly attacks the carbocation from either face, leading to a mixture of enantiomers if the carbon is chiral. Tertiary haloalkanes, such as (CH₃)₃CBr, react via SN1 with aqueous NaOH to give (CH₃)₃COH.
SN1中，C–卤键首先断裂（决速步），生成平面型碳正离子。亲核试剂随后从任一面快速进攻碳正离子，若碳为手性则得到外消旋混合物。叔卤代烷（如(CH₃)₃CBr）与NaOH水溶液通过SN1机理生成(CH₃)₃COH。
6. Elimination to Form Alkenes | 消除反应生成烯烃
When a haloalkane is heated with ethanolic OH⁻, elimination competes with substitution. The hydroxide ion acts as a base, abstracting a β‑hydrogen, while the C–halogen bond breaks, forming a C=C double bond. The mechanism is often E2 (bimolecular elimination) for primary and secondary haloalkanes.
卤代烷与氢氧化钠的乙醇溶液共热时，消除反应与取代反应竞争。氢氧根离子作为碱，夺取β-氢，同时C–卤键断裂，形成C=C双键。对伯、仲卤代烷，通常按E2（双分子消除）机理进行。
The curly arrows show the base attacking the β‑hydrogen, the electrons from the C–H bond moving to form the π‑bond, and the halogen leaving. For 2‑bromopropane, CH₃CHBrCH₃ + OH⁻ → CH₃CH=CH₂ + H₂O + Br⁻. Saytzeff’s rule predicts the more substituted alkene as the major product when the haloalkane is unsymmetrical.
弯曲箭头显示：碱进攻β-氢，C–H键的电子移向形成π键，卤素离去。对于2-溴丙烷：CH₃CHBrCH₃ + OH⁻ → CH₃CH=CH₂ + H₂O + Br⁻。当卤代烷不对称时，依据扎伊采夫规则，取代更多的烯烃为主产物。
7. Free Radical Substitution in Alkanes | 烷烃的自由基取代
Alkanes react with halogens in the presence of UV light via a free radical chain mechanism. The overall reaction for methane with chlorine is CH₄ + Cl₂ → CH₃Cl + HCl. The mechanism has three stages: initiation, propagation, and termination.
烷烃在紫外光照下与卤素按自由基链式机理反应。甲烷与氯的总反应为CH₄ + Cl₂ → CH₃Cl + HCl。机理包含三个阶段：引发、增长、终止。
Initiation: Cl–Cl bond undergoes homolytic fission to give two chlorine radicals. Cl₂ → 2 Cl• (UV light provides energy).
引发：Cl–Cl键均裂产生两个氯自由基。Cl₂ → 2 Cl•（紫外光提供能量）。
Propagation steps: A chlorine radical abstracts a hydrogen from methane, forming HCl and a methyl radical (CH₃•). Then the methyl radical reacts with a Cl₂ molecule, producing chloromethane and regenerating a chlorine radical. These two steps repeat, sustaining the chain.
增长步：氯自由基夺取甲烷的一个氢，生成HCl和甲基自由基（CH₃•）。甲基自由基与Cl₂分子反应，生成氯甲烷并再生氯自由基。这两步反复进行，维持链式反应。
Termination occurs when any two radicals combine: Cl• + Cl• → Cl₂, CH₃• + Cl• → CH₃Cl, or CH₃• + CH₃• → C₂H₆. Termination reduces the concentration of radicals and stops the chain.
终止步发生在任意两个自由基结合时：Cl• + Cl• → Cl₂，CH₃• + Cl• → CH₃Cl，或CH₃• + CH₃• → C₂H₆。终止使自由基浓度下降，链反应结束。
8. Nucleophilic Addition to Carbonyl Compounds | 羰基化合物的亲核加成
Carbonyl compounds (aldehydes and ketones) have a polar C=O bond. The carbon is electron-deficient and susceptible to nucleophilic attack. A classic A-Level example is the addition of hydrogen cyanide, HCN, to form cyanohydrins. The nucleophile is the cyanide ion, CN⁻, generated from KCN and dilute acid.
羰基化合物（醛和酮）具有极性的C=O键。碳原子缺电子，易受亲核试剂进攻。A-Level的典型例子是氰化氢HCN的加成，生成氰醇。亲核试剂氰根离子CN⁻由KCN与稀酸产生。
The mechanism: In the first step, the cyanide ion uses its lone pair to attack the carbonyl carbon. A curly arrow shows the π‑electrons of the C=O bond moving onto the oxygen, forming an alkoxide intermediate (O⁻). In the second step, this negatively charged oxygen abstracts a proton from HCN (or H⁺ from the solvent) to give the final alcohol.
机理：第一步，氰根离子用孤对电子进攻羰基碳，弯曲箭头显示C=O的π电子转移到氧上，形成烷氧负离子中间体（O⁻）。第二步，带负电的氧从HCN（或溶剂中的H⁺）夺取质子，得到最终醇。
For ethanal: CH₃CHO + CN⁻ → CH₃CH(CN)O⁻, then + H⁺ → CH₃CH(OH)CN. The reaction is useful because it extends the carbon chain by one carbon and introduces a functional group that can be hydrolysed to acids or reduced to amines.
以乙醛为例：CH₃CHO + CN⁻ → CH₃CH(CN)O⁻，再与H⁺反应 → CH₃CH(OH)CN。该反应之所以重要，是因为它使碳链延长一个碳原子，并引入可水解为羧酸或还原为胺的官能团。
9. Electrophilic Substitution of Benzene | 苯的亲电取代
Benzene, due to its delocalised π‑electron system, undergoes electrophilic substitution rather than addition. The most common reactions at A-Level are nitration and Friedel‑Crafts alkylation/acylation, as well as halogenation. All proceed via a similar two‑step mechanism: generation of the electrophile, attack by benzene to form a Wheland intermediate, and deprotonation to restore aromaticity.
苯由于其离域π电子体系，发生亲电取代而非加成。A-Level最常见的反应是硝化、弗克烷基化/酰基化以及卤代。这些反应都遵循相似的两步机理：生成亲电试剂，苯进攻形成韦兰德中间体，随后脱除质子恢复芳香性。
For nitration, the electrophile is the nitronium ion NO₂⁺, generated from concentrated HNO₃ and H₂SO₄: HNO₃ + 2H₂SO₄ → NO₂⁺ + 2HSO₄⁻ + H₃O⁺. Benzene’s π‑electrons attack NO₂⁺, forming a carbocation intermediate (the sigma complex). This intermediate then loses a proton (H⁺) to the HSO₄⁻ base, regenerating the benzene ring and giving nitrobenzene (C₆H₅NO₂).
硝化反应中，亲电试剂是硝鎓离子NO₂⁺，由浓HNO₃和浓H₂SO₄生成：HNO₃ + 2H₂SO₄ → NO₂⁺ + 2HSO₄⁻ + H₃O⁺。苯的π电子进攻NO₂⁺，生成碳正离子中间体（σ络合物）。然后该中间体失去一个质子（H⁺）给HSO₄⁻碱，恢复苯环，得到硝基苯（C₆H₅NO₂）。
Halogenation of benzene requires a halogen carrier (e.g., FeBr₃ for bromination) to polarise the Br–Br bond and generate a stronger electrophile, effectively Br⁺. The mechanism is identical: the bromonium‑like electrophile is attacked by benzene, the Wheland intermediate is formed, and loss of H⁺ yields bromobenzene. Chlorination follows the same pattern with AlCl₃.
苯的卤代需要卤素载体（如溴化需FeBr₃）以极化Br–Br键，产生更强的亲电物种，可视为Br⁺。机理完全相同：苯进攻类溴鎓亲电试剂，形成韦兰德中间体，失去H⁺得到溴苯。氯化使用AlCl₃，模式相同。
10. Mechanism Summary and Exam Tips | 机理总结与应考提示
In A-Level exams, you will often be asked to draw mechanisms with curly arrows, state the type of mechanism, and explain the roles of reagents. Keep the following in mind: always draw arrows from electron‑rich to electron‑poor sites; show all intermediates and formal charges; for addition reactions, indicate the major product according to Markovnikov’s rule where applicable; for elimination, apply Saytzeff’s rule; for substitution, justify choice of SN1 vs SN2 based on the haloalkane structure and conditions.
A-Level考试常要求画出带弯曲箭头的机理、判断机理类型并解释试剂的作用。请记住：箭头始终从富电子指向缺电子部位；画出所有中间体和形式电荷；加成反应中当适用时按马氏规则
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