International A-Level Chemistry: Reaction Mechanisms – CH03 Unit 3 Example Responses | 国际A-Level化学：反应机理 – 第三单元示例答案

📚 International A-Level Chemistry: Reaction Mechanisms – CH03 Unit 3 Example Responses | 国际A-Level化学：反应机理 – 第三单元示例答案

Reaction mechanisms are at the heart of organic chemistry, showing exactly how bonds break and form to convert reactants into products. In Unit 3 of the International A-Level Chemistry specification, you are expected to describe and illustrate mechanisms using curly arrows, identify key types of bond fission and predict products. This article provides clear example responses to typical exam questions, guiding you through free radical substitution, electrophilic addition and nucleophilic substitution mechanisms – the three core families tested in CH03. Each section includes a model answer with detailed explanations to help you master the logic behind every curly arrow.

反应机理是有机化学的核心，它精确展示了化学键如何断裂和生成，将反应物转化为产物。在国际A-Level化学第三单元中，你需要用卷曲箭头描述并图示机理，识别键断裂的关键类型并预测产物。本文围绕自由基取代、亲电加成和亲核取代这三类核心机理，提供典型考题的范例答案。每部分都配以模型答案和详细解析，帮助你掌握每个卷曲箭头背后的逻辑。

1. Defining Reaction Mechanisms | 定义反应机理

A reaction mechanism is a step‑by‑step sequence of elementary steps by which an overall chemical change occurs. It shows the movement of electron pairs (or single electrons) during bond‑breaking and bond‑making. The key to any mechanism is the curly arrow, which represents the movement of two electrons from an electron‑rich site (nucleophile or double bond) to an electron‑poor site (electrophile or leaving group).

反应机理是化学变化过程中一系列基元步骤的逐步序列。它展示了键断裂和形成过程中电子对（或单电子）的运动。任何机理的核心都是卷曲箭头，代表两个电子从富电子位点（亲核试剂或双键）移向缺电子位点（亲电试剂或离去基团）。

2. Homolytic vs Heterolytic Fission | 均裂与异裂

Bond breaking can occur in two distinct ways. In heterolytic fission, both bonding electrons move to one atom, forming a cation and an anion. This is common in polar reactions and is shown using a double‑headed curly arrow. In homolytic fission, each atom receives one electron from the bond, producing two free radicals. This is shown with a single‑headed (fish‑hook) arrow. You must be able to distinguish these in exam questions.

键的断裂有两种截然不同的方式。异裂中，成键的两个电子都移向一个原子，生成一个阳离子和一个阴离子。这常见于极性反应，用双头卷曲箭头表示。均裂中，每个原子从键中获得一个电子，生成两个自由基。这用单头（鱼钩）箭头表示。你必须能在考题中区分这两种情况。

3. Free Radical Substitution – General Pattern | 自由基取代 – 通用模式

Alkanes react with halogens in the presence of UV light via a free radical chain mechanism. The overall process involves three stages: initiation, propagation and termination. Initiation: UV light causes homolytic fission of the halogen molecule, e.g. Cl₂ → 2 Cl•. Propagation: Cl• abstracts a hydrogen atom from the alkane to form HCl and an alkyl radical; the alkyl radical then reacts with a Cl₂ molecule to regenerate Cl•. Termination: two radicals combine.

烷烃在紫外光下与卤素通过自由基链式机理反应。整个历程包含三个阶段：引发、增长和终止。引发：紫外光使卤素分子均裂，如 Cl₂ → 2 Cl•。增长：Cl• 从烷烃中夺取一个氢原子，生成 HCl 和一个烷基自由基；烷基自由基再与 Cl₂ 分子反应，再生 Cl•。终止：两个自由基结合。

4. Example Response: Chlorination of Methane | 范例答案：甲烷的氯化

Question: Show the mechanism for the reaction of methane with chlorine in UV light. Identify the initiation, propagation and termination steps.

问题： 图示甲烷与氯气在紫外光下的反应机理，标出引发、增长和终止步骤。

Model Answer:
Initiation: Cl₂ → 2 Cl• (homolytic fission, UV light).
Propagation: CH₄ + Cl• → CH₃• + HCl; CH₃• + Cl₂ → CH₃Cl + Cl•.
Termination: Cl• + Cl• → Cl₂; CH₃• + Cl• → CH₃Cl; CH₃• + CH₃• → C₂H₆.
Overall equation: CH₄ + Cl₂ → CH₃Cl + HCl. Note that further substitution can occur, so a mixture of products is obtained.

模型答案：
引发：Cl₂ → 2 Cl•（均裂，紫外光）。
增长：CH₄ + Cl• → CH₃• + HCl；CH₃• + Cl₂ → CH₃Cl + Cl•。
终止：Cl• + Cl• → Cl₂；CH₃• + Cl• → CH₃Cl；CH₃• + CH₃• → C₂H₆。
总方程式：CH₄ + Cl₂ → CH₃Cl + HCl。注意可能发生进一步取代，因此得到产物混合物。

5. Electrophilic Addition – Alkenes with Hydrogen Halides | 亲电加成 – 烯烃与卤化氢

Alkenes undergo addition reactions because the π‑bond is electron‑rich and can be attacked by electrophiles. With hydrogen halides (HBr, HCl), the mechanism proceeds via two steps. First, the electrophilic H⁺ adds to the less substituted carbon of the double bond to form the more stable carbocation intermediate (Markovnikov’s rule). Then the halide ion attacks the carbocation to complete the addition. You must use curly arrows to show electron movement from the C=C bond to H⁺, and from Br⁻ to the carbocation.

烯烃由于π键富电子，容易被亲电试剂进攻，发生加成反应。与卤化氢（HBr、HCl）反应时，机理分两步进行。首先，亲电的H⁺加在双键中取代较少的碳上，形成较稳定的碳正离子中间体（马氏规则）。然后卤负离子进攻碳正离子完成加成。你必须用卷曲箭头显示从C=C双键到H⁺，以及从Br⁻到碳正离子的电子转移。

6. Example Response: Addition of HBr to Propene | 范例答案：丙烯与HBr加成

Question: Illustrate the electrophilic addition of HBr to propene. State the major product and justify your answer.

问题： 图示丙烯与HBr的亲电加成反应。说出主要产物并解释。

Model Answer: The C═C bond attacks HBr; the H⁺ adds to CH₂ (less substituted) forming the secondary carbocation CH₃–CH⁺–CH₃. Then Br⁻ attacks this carbocation to give 2‑bromopropane. The alternative primary carbocation CH₃–CH₂–CH₂⁺ is less stable, so the secondary pathway dominates. Major product: 2‑bromopropane. Curly arrows: one from the C═C bond to H⁺, one from Br⁻ to the carbocation carbon.

模型答案： C═C双键进攻HBr；H⁺加在CH₂上（取代较少）生成二级碳正离子 CH₃–CH⁺–CH₃。然后 Br⁻ 进攻该碳正离子得到2‑溴丙烷。另一途径的一级碳正离子 CH₃–CH₂–CH₂⁺ 稳定性较低，因此二级途径占主导。主要产物：2‑溴丙烷。卷曲箭头：一个从C═C到H⁺，一个从Br⁻到碳正离子碳。

7. Electrophilic Addition – Alkenes with Halogens | 亲电加成 – 烯烃与卤素

When alkenes react with halogens (Br₂, Cl₂), the mechanism also proceeds via electrophilic addition but with a key intermediate: a cyclic halonium ion. As the π‑electrons approach the Br₂ molecule, polarisation occurs (induced dipole), the Br–Br bond breaks heterolytically, and a three‑membered bromonium ion forms. Then the halide ion attacks from the opposite face, giving overall anti‑addition. This stereochemistry is a classic exam point.

烯烃与卤素（Br₂、Cl₂）反应时，机理也是亲电加成，但其中有一个关键中间体：环状卤鎓离子。当π电子靠近Br₂分子时，发生极化（诱导偶极），Br–Br键异裂，形成一个三元环溴鎓离子。然后卤负离子从背面进攻，给出反式加成产物。这种立体化学是经典考点。

8. Example Response: Bromination of Ethene | 范例答案：乙烯的溴化

Question: Describe the electrophilic addition of Br₂ to ethene. Explain why the product is trans‑1,2‑dibromoethane if a cyclic structure is considered, though for ethene the stereochemistry is not observed.

问题： 描述Br₂与乙烯的亲电加成。如果考虑环状结构，解释为什么产物是反‑1,2‑二溴乙烷，尽管乙烯中观察不到立体化学。

Model Answer: The π‑electrons of ethene attack one Br atom; the Br–Br bond breaks, forming a bromonium ion (C₂H₄Br⁺) and a Br⁻ ion. The Br⁻ then attacks the bromonium ion from the backside, opening the ring to give BrCH₂–CH₂Br. For substituted alkenes, the anti‑addition leads to trans stereochemistry. Curly arrows: two arrows for formation of the bromonium ion (one from C=C to Br, one from Br–Br to Br⁻), and one from Br⁻ to a carbon of the ring.

模型答案： 乙烯的π电子进攻一个Br原子；Br–Br键断开，形成溴鎓离子（C₂H₄Br⁺）和Br⁻离子。然后Br⁻从背面进攻溴鎓离子，开环得到BrCH₂–CH₂Br。对于取代烯烃，反式加成导致反式立体化学。卷曲箭头：生成溴鎓离子用两个箭头（C=C到Br一个，Br–Br到Br⁻一个），Br⁻到环碳一个箭头。

9. Nucleophilic Substitution – Haloalkanes with Hydroxide | 亲核取代 – 卤代烷与氢氧根

Haloalkanes contain a polar C–X bond, making the carbon electron‑deficient and susceptible to attack by nucleophiles. With aqueous hydroxide ions, the reaction is a nucleophilic substitution producing an alcohol. The mechanism can follow either an S_N2 (bimolecular) or S_N1 (unimolecular) pathway depending on the structure of the haloalkane. Primary haloalkanes favour S_N2, tertiary favour S_N1, and secondary can follow both under different conditions.

卤代烷含有极性的C–X键，使碳缺电子，易受亲核试剂进攻。与氢氧根水溶液反应时，发生亲核取代生成醇。机理可根据卤代烷结构采取S_N2（双分子）或S_N1（单分子）途径。伯卤代烷倾向于S_N2，叔卤代烷倾向于S_N1，仲卤代烷在不同条件下可走两种途径。

10. Example Response: S_N2 vs S_N1 for 2‑Bromobutane | 范例答案：2‑溴丁烷的S_N2和S_N1

Question: Using 2‑bromobutane, illustrate both the S_N2 and S_N1 mechanisms with OH⁻. Explain the difference in stereochemical outcomes.

问题： 以2‑溴丁烷为例，分别图示与OH⁻的S_N2和S_N1机理。解释立体化学结果的差异。

Model Answer:
S_N2: OH⁻ attacks the carbon bearing Br from the opposite side of the C–Br bond (backside attack). Transition state involves a pentacoordinate carbon; Br leaves as the OH bonds. The product is butan‑2‑ol with inversion of configuration.
S_N1: First, the C–Br bond breaks to form a planar carbocation (CH₃CH₂CH⁺CH₃) and Br⁻. Then OH⁻ attacks the planar cation from either face, giving a racemic mixture (50:50 R and S).
Curly arrows: S_N2 – one arrow from OH⁻ to C, one from C–Br bond to Br. S_N1 – two steps, first arrow from C–Br to Br, second arrow from OH⁻ to carbocation carbon.

模型答案：
S_N2：OH⁻从C–Br键的对面进攻带溴的碳（背面进攻）。过渡态涉及五配位碳；Br在OH键合时离去。产物为构型翻转的丁‑2‑醇。
S_N1：首先C–Br键断裂，生成平面碳正离子（CH₃CH₂CH⁺CH₃）和Br⁻。然后OH⁻从平面两侧均可进攻，得到外消旋混合物（R和S各50%）。
卷曲箭头：S_N2 – 一个从OH⁻到C，一个从C–Br键到Br。S_N1 – 两步，第一步从C–Br到Br，第二步从OH⁻到碳正离子碳。

11. Nucleophilic Substitution with Other Nucleophiles | 与其他亲核试剂的取代

Common nucleophiles tested include cyanide ions (CN⁻) and ammonia (NH₃). With CN⁻, haloalkanes yield nitriles, extending the carbon chain by one atom. With NH₃, a primary amine is formed, but further alkylation can occur yielding secondary and tertiary amines. Both reactions follow S_N2 for primary substrates. In your mechanism, always show the nucleophile attacking the α‑carbon and the halide leaving.

常考的亲核试剂包括氰根离子（CN⁻）和氨（NH₃）。与CN⁻反应，卤代烷生成腈，碳链延长一个原子。与NH₃反应生成伯胺，但可能进一步烷基化得到仲胺和叔胺。这两种反应对伯卤代烷均遵循S_N2。在机理图示中，总要显示亲核试剂进攻α‑碳，卤离子离去。

12. Curly Arrow Rules and Common Mistakes | 卷曲箭头规则与常见错误

Examiners are strict about curly arrow usage. Arrows must start from a lone pair, a bond or a negative charge, and point to an atom or space between atoms. Never start an arrow from a positive charge. Always draw the arrows for each step separately and label the rate‑determining step if required. Common errors include forgetting to show the leaving group departing with an arrow, using single‑headed arrows for polar reactions, and omitting necessary charges on intermediates.

考官对卷曲箭头的使用非常严格。箭头必须从孤对电子、化学键或负电荷出发，指向一个原子或原子之间。绝不能从正电荷出发画箭头。每一步的箭头要分开绘制，必要时标出决速步骤。常见错误包括忘记用箭头表示离去基团离去、在极性反应中使用单头箭头，以及遗漏中间体上必要的电荷。

Published by TutorHao | Chemistry Revision Series | aleveler.com

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