📚 AS Chemistry: Reaction Mechanisms from Unit 3 (June 19) | AS 化学：从2019年6月Unit 3看反应机理
Reaction mechanisms form the beating heart of organic chemistry at AS level. They reveal how electrons reorganise during a chemical reaction and explain why certain products predominate. The June 2019 Unit 3 question paper placed heavy emphasis on mechanism drawing, curly arrow conventions, and the ability to predict products from a given set of conditions. This article revisits the key mechanisms examined in that paper and builds a solid revision framework for any AS candidate.
反应机理是AS有机化学的灵魂。它们揭示了电子在化学反应中如何重新排布，并解释了为什么某些产物占优势。2019年6月的Unit 3试卷对机理绘制、弯箭头的使用规则以及根据给定条件预测产物提出了很高的要求。本文重温了该试卷中考查的关键机理，并为所有AS考生搭建扎实的复习框架。
1. What is a Reaction Mechanism? | 什么是反应机理？
A reaction mechanism describes the step-by-step movement of electrons that transforms reactants into products. It uses curly arrows to show the flow of electron pairs, and often involves intermediates such as carbocations or free radicals. Understanding the mechanism allows chemists to control reactions and predict the products formed.
反应机理描述了电子一步一步移动、将反应物转化为产物的过程。它用弯箭头表示电子对的流动，并常常涉及碳正离子或自由基这样的中间体。理解机理能让化学家控制反应并预测所形成的产物。
2. Bond Breaking and Bond Making | 键的断裂与形成
Mechanisms start with either homolytic or heterolytic bond fission. Homolytic fission produces two radicals; each atom takes one electron from the bond. Heterolytic fission produces ions; the more electronegative atom takes both electrons. Curly arrows always start from an electron-rich site (a lone pair or a bond) and move toward an electron-deficient site.
反应机理以均裂或异裂开始。均裂产生两个自由基，每个原子从键中取走一个电子。异裂产生离子，电负性更大的原子取走两个电子。弯箭头总是从富电子位点（孤对电子或化学键）出发，移向缺电子位点。
3. Free Radical Substitution – Alkane Halogenation | 自由基取代——烷烃卤代
The June 19 paper often tested the radical chain mechanism for the chlorination of methane. Initiation requires UV light to split Cl₂ into two chlorine radicals. Propagation steps keep the chain going: a chlorine radical abstracts a hydrogen from CH₄, forming HCl and a methyl radical; the methyl radical then attacks another Cl₂ molecule, forming chloromethane and regenerating a chlorine radical. Termination removes radicals.
2019年6月的试卷经常考查甲烷氯化反应中的自由基链式机理。链引发需要紫外光将Cl₂分裂成两个氯自由基。链增长步骤维持链条：氯自由基从CH₄中夺走氢，生成HCl和甲基自由基；甲基自由基再进攻另一个Cl₂分子，生成氯甲烷并再生氯自由基。链终止步骤消除自由基。
The overall equation for monochlorination is CH₄ + Cl₂ → CH₃Cl + HCl. In mechanism diagrams, fish-hook curly arrows (half arrows) are used to show single-electron movement.
生成一氯甲烷的总反应方程式为 CH₄ + Cl₂ → CH₃Cl + HCl。在机理图中，使用弯鱼钩箭头（半箭头）来表示单电子移动。
4. Electrophilic Addition – Alkenes with HBr and Br₂ | 亲电加成——烯烃与HBr和Br₂
Electrophilic addition was a dominant theme in the June 19 Unit 3 paper. Ethene reacts with HBr: the π bond acts as a nucleophile, attacking the partially positive hydrogen. A carbocation forms, and the bromide ion then attacks the carbocation to complete the addition. The final product is bromoethane.
亲电加成是2019年6月Unit 3试卷的核心主题。乙烯与HBr反应：π键充当亲核试剂，进攻带部分正电荷的氢。生成一个碳正离子，然后溴离子进攻碳正离子完成加成。最终产物是溴乙烷。
With bromine, the mechanism is similar but the initial electrophile is the induced dipole on Br₂. The π electrons attack Br₂, forming a cyclic bromonium ion (if drawn carefully) and a bromide ion. The bromide ion then attacks from the opposite side, giving trans addition. Many mark schemes for June 19 expected a clear diagram of the bromonium ion intermediate.
与溴反应时，机理相似，但最初的亲电试剂是Br₂上诱导出的偶极。π电子进攻Br₂，形成环状溴鎓离子（需仔细绘制）和溴离子。溴离子随后从背面进攻，实现反式加成。2019年6月的评分标准大多要求清晰地画出溴鎓离子中间体。
5. Carbocation Stability and Markovnikov’s Rule | 碳正离子稳定性与马氏规则
When an unsymmetrical alkene such as propene adds HBr, two carbocations can form: a secondary carbocation (CH₃–C⁺H–CH₃) and a primary carbocation (CH₃–CH₂–C⁺H₂). The secondary carbocation is more stable because alkyl groups donate electron density and stabilise the positive charge. Therefore, the bromide ion attacks the more stable carbocation, leading to 2-bromopropane as the major product – this is Markovnikov’s rule.
当不对称烯烃如丙烯与HBr加成时，可以生成两种碳正离子：二级碳正离子 (CH₃–C⁺H–CH₃) 和一级碳正离子 (CH₃–CH₂–C⁺H₂)。二级碳正离子更稳定，因为烷基可以推电子，稳定正电荷。因此溴离子进攻更稳定的碳正离子，主产物为2-溴丙烷——这就是马氏规则。
Many June 19 candidates lost marks by failing to justify why the major product formed. Always mention carbocation stability in your answer.
许多参加2019年6月考试的学生因未能解释为什么生成主产物而失分。答案中务必提到碳正离子稳定性。
6. Nucleophilic Substitution – Haloalkanes | 亲核取代——卤代烷
Haloalkanes undergo nucleophilic substitution with reagents such as NaOH, KCN, and NH₃. The C–Br bond is polar, with a partial positive charge on carbon, making it susceptible to attack by nucleophiles. The leaving group is the halide ion. In the June 19 Unit 3, a question required students to draw the mechanism for 1-bromopropane reacting with hydroxide ions.
卤代烷与NaOH、KCN、NH₃等试剂发生亲核取代。C–Br键是极性的，碳带部分正电荷，易受亲核试剂进攻。离去基团是卤离子。在2019年6月Unit 3中，有一道题要求学生画出1-溴丙烷与氢氧根离子反应的机理。
For primary haloalkanes, the SN2 mechanism operates: the nucleophile attacks the carbon from the opposite side of the bromine, and the C–Br bond breaks simultaneously. The transition state involves both the nucleophile and the leaving group. A single curly arrow from the nucleophile to carbon and another from the C–Br bond to bromine must be drawn cleanly.
对于伯卤代烷，发生SN2机理：亲核试剂从溴的背面进攻碳，C–Br键同时断裂。过渡态涉及亲核试剂和离去基团。必须清晰地画出从亲核试剂指向碳的一根弯箭头，以及从C–Br键指向溴的另一根弯箭头。
7. Comparing SN1 and SN2 Mechanisms | SN1与SN2机理的比较
At AS level, the distinction between SN1 and SN2 is introduced through tertiary and primary haloalkanes. Tertiary haloalkanes undergo SN1. The C–Br bond breaks heterolytically first, forming a stable tertiary carbocation, which is then attacked by the nucleophile. This two-step mechanism gives racemic mixtures if the carbon is chiral. Primary haloalkanes favour SN2, a single-step backside attack.
在AS层面，通过叔卤代烷和伯卤代烷来引入SN1与SN2的区别。叔卤代烷经历SN1。C–Br键首先异裂，形成稳定的三级碳正离子，然后亲核试剂进攻该碳正离子。若碳为手性中心，两步机理将得到外消旋混合物。伯卤代烷倾向于SN2，是一步完成的背面进攻。
| Feature | 特征 | SN1 | SN2 |
|---|---|---|
| Steps | 步骤 | Two | 两步 | One | 一步 |
| Preferred substrate | 适宜底物 | Tertiary | 叔卤代烷 | Primary | 伯卤代烷 |
| Intermediate | 中间体 | Carbocation | 碳正离子 | Transition state | 过渡态 |
| Stereochemistry | 立体化学 | Racemisation | 外消旋化 | Inversion | 构型翻转 |
In the June 19 context, a question about 2-bromo-2-methylpropane reacting with water highlighted the SN1 route and tested the concept of carbocation stability.
在2019年6月的试卷中，一道关于2-溴-2-甲基丙烷与水反应的题目突出显示了SN1途径，并考查了碳正离子稳定性的概念。
8. Nucleophilic Substitution with Cyanide and Ammonia | 与氰化物和氨的亲核取代
KCN in ethanol provides the nucleophile CN⁻, allowing the synthesis of nitriles from haloalkanes. The mechanism is identical to that with hydroxide: a lone pair on CN⁻ attacks the δ+ carbon, displacing bromide. For example, CH₃CH₂Br + CN⁻ → CH₃CH₂CN + Br⁻. Nitriles can be hydrolysed to carboxylic acids, so this is a useful chain-lengthening step.
KCN的乙醇溶液提供亲核试剂CN⁻，可从卤代烷合成腈。其机理与氢氧根完全相同：CN⁻上的孤对电子进攻带δ+的碳，置换出溴离子。例如，CH₃CH₂Br + CN⁻ → CH₃CH₂CN + Br⁻。腈可以水解成羧酸，因此这是延长碳链的有用步骤。
With ammonia, the initial nucleophilic attack of NH₃ on a haloalkane yields an alkylammonium salt, which loses a proton to another NH₃ molecule, finally giving a primary amine. However, the amine can further react, leading to a mixture of products unless a large excess of ammonia is used.
与氨反应时，NH₃首先对卤代烷进行亲核进攻，生成烷基铵盐，该盐再向另一分子NH₃失去质子，最终得到伯胺。但生成的胺可以继续反应，导致产物为混合物，除非使用大过量的氨。
9. Elimination Reactions – Producing Alkenes | 消除反应——生成烯烃
When a haloalkane is heated with ethanolic NaOH under reflux, elimination competes with nucleophilic substitution. The hydroxide ion acts as a base, removing a β-hydrogen, while the C–Br bond breaks, forming a double bond. For example, 2-bromopropane gives propene. Ethene can be produced from bromoethane by this method.
卤代烷与NaOH的乙醇溶液回流加热时，消除反应会与亲核取代竞争。此时氢氧根离子充当碱，夺取β-氢，C–Br键同时断裂，形成双键。例如，2-溴丙烷生成丙烯。溴乙烷可通过此方法制备乙烯。
In the June 19 Unit 3, a question asked students to write the mechanism for the elimination of HBr from 2-bromobutane, showing the formation of both but-1-ene and but-2-ene. The curly arrow from the C–H bond to the C–C bond and from the C–Br bond to bromine had to be drawn precisely. Major product prediction relied on Zaitsev’s rule.
2019年6月的Unit 3中，有一道题要求学生写出2-溴丁烷消除HBr的机理，展示丁-1-烯和丁-2-烯的生成。必须精确画出从C–H键指向C–C键的弯箭头，以及从C–Br键指向溴的弯箭头。主产物的预测依赖札伊采夫规则。
10. Curly Arrow Rules and Common Mistakes in the June 19 Paper | 弯箭头规则与6月试卷中的常见错误
Examiners repeatedly penalised misdirected arrows. Arrows must start from a bond or a lone pair, never from a positive charge. For electrophilic addition, the arrow must start from the middle of the double bond, not from a carbon atom. In radical mechanisms, fish-hook arrows with a single barb are required. Many candidates lost marks for using full curly arrows where half arrows were needed.
考官反复对画错的箭头扣分。箭头必须从化学键或孤对电子出发，绝不能从正电荷出发。在亲电加成中，箭头必须从双键中间出发，而不是从某个碳原子。在自由基机理中，需要使用单钩的弯鱼钩箭头。许多考生在该用半箭头的地方用了全弯箭头，因而失分。
Another typical error was omitting the partial charges on the electrophile before the attack. For example, H–Br should be drawn as Hᵟ⁺–Brᵟ⁻ to justify the initial attack. Label lone pairs clearly when they act as nucleophiles.
另一个常见错误是在亲电试剂进攻前没有标出部分电荷。例如，H–Br应当画成 Hᵟ⁺–Brᵟ⁻，这样才能合理化初始进攻。同时，当孤对电子充当亲核试剂时，务必清晰画出。
11. Constructing a Mechanism from a June 19 Exam Question | 从6月考题中构建机理
Let us reconstruct a typical mechanism-based problem: ‘Propene reacts with concentrated sulfuric acid followed by water to form propan-2-ol. Draw the mechanism.’ The first step is electrophilic addition of H⁺ from H₂SO₄ across the double bond, giving the more stable secondary carbocation. The second step is nucleophilic attack by water (or HSO₄⁻) on the carbocation, followed by loss of a proton to regenerate the acid catalyst.
让我们重现一道典型的机理考题：“丙烯与浓硫酸反应，再与水作用，生成丙-2-醇。画出反应机理。”第一步是H₂SO₄中的H⁺对双键进行亲电加成，生成更稳定的二级碳正离子。第二步是水（或HSO₄⁻）对碳正离子亲核进攻，再脱去质子，再生出酸催化剂。
This composite mechanism integrates electrophilic addition and nucleophilic substitution, a style widely examined in the June 19 Unit 3. Candidates must show the regeneration of H⁺ to score the full marks. The overall hydration of propene is an indirect but important industrial route.
这个组合机理融合了亲电加成和亲核取代，正是2019年6月Unit 3广泛考查的类型。考生必须展示H⁺的再生才能拿到满分。丙烯的间接水合法是一条重要工业路线。
12. Summary and Exam Tips | 总结与考试建议
Mastering reaction mechanisms for AS Unit 3 means practising curly arrow drawing until it becomes second nature. Always identify the electron-rich species (nucleophile or π bond) and the electron-poor site (δ+ atom or carbocation). Check that your arrows originate precisely, and that any intermediate drawn is chemically reasonable. When unsure, refer back to the classic mechanisms: free radical substitution, electrophilic addition, SN1, SN2, and elimination.
掌握AS Unit 3的反应机理，意味着反复练习弯箭头绘制，直到成为本能。始终先识别富电子物种（亲核试剂或π键）和缺电子位点（δ+原子或碳正离子）。检查箭头起点是否精确，所画的中间体在化学上是否合理。当拿不准时，回归到经典机理：自由基取代、亲电加成、SN1、SN2和消除。
The June 19 paper rewarded deep understanding over memorisation. Spend time explaining why a certain pathway is favoured, using stability arguments. With these strategies, you can convert mechanism questions into guaranteed marks.
2019年6月的试卷奖励深入理解而非死记硬背。花时间用稳定性论证来解释为什么某一反应途径占优势。掌握这些策略，你就能将机理题转化为必得分的题目。
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