International A-Level Chemistry Example Responses: CH04 Unit 4 Reaction Mechanisms | 国际A-Level化学例题解答：CH04第四单元反应机理

📚 International A-Level Chemistry Example Responses: CH04 Unit 4 Reaction Mechanisms | 国际A-Level化学例题解答：CH04第四单元反应机理

In International A-Level Chemistry, Unit 4 (CH04) focuses on the fundamental principles of organic reaction mechanisms. Understanding how to write and interpret mechanisms for free radical substitution, electrophilic addition, nucleophilic substitution, and electrophilic substitution is essential for exam success. This article provides clear explanations and example responses to typical exam questions, helping you to master the application of curly arrows, identify intermediates, and justify major products.

在国际A-Level化学中，第四单元（CH04）聚焦于有机反应机理的基本原理。掌握如何书写和解释自由基取代、亲电加成、亲核取代和亲电取代的机理对于考试成功至关重要。本文提供了清晰的解释和针对典型考题的示例回答，帮助你熟练掌握卷曲箭头的使用、识别中间体，并合理解释主要产物的生成。

1. Introduction to Reaction Mechanisms | 反应机理简介

Reaction mechanisms detail the step-by-step sequence of elementary reactions through which an overall chemical change occurs. In Unit 4, you are required to propose plausible mechanisms, use curly arrows to depict electron movement, identify intermediates such as free radicals and carbocations, and link mechanisms to energy profiles or rate equations. A mechanism must account for all bond-breaking and bond-forming events, and curly arrows always start from a source of electrons (a bond, a lone pair, or a negative charge) and point towards an electron-deficient atom.

反应机理详细描述了整体化学变化所经历的基元反应序列。在第四单元中，你需要提出合理的机理，用卷曲箭头表示电子移动，识别自由基和碳正离子等中间体，并将机理与能量分布图或速率方程联系起来。一个机理必须解释所有的断键和成键过程，而卷曲箭头始终从电子源（化学键、孤对电子或负电荷）出发，指向缺电子的原子。

2. Types of Mechanisms in Unit 4 | 第四单元中的机理类型

The four main mechanistic classes covered in Unit 4 are free radical substitution (e.g., halogenation of alkanes), electrophilic addition (e.g., addition of HBr to alkenes), nucleophilic substitution (SN1 and SN2 of haloalkanes), and electrophilic substitution (e.g., nitration of benzene). Each follows a characteristic pathway with distinct intermediates, and you should be able to compare them in terms of rate laws, stereochemical outcomes, and conditions.

第四单元涵盖的四种主要机理类型分别是：自由基取代（如烷烃的卤化）、亲电加成（如卤化氢与烯烃的加成）、亲核取代（卤代烷的SN1与SN2反应）以及亲电取代（如苯的硝化）。每类反应遵循特有的路径并生成不同的中间体，你应当能够从速率方程、立体化学结果和反应条件等角度对它们进行比较。

3. Free Radical Substitution: Chlorination of Methane | 自由基取代：甲烷的氯化

This photochemical reaction proceeds via a chain mechanism initiated by UV light. Chlorine molecules undergo homolytic fission to generate chlorine radicals. These radicals then abstract hydrogen from methane, producing a methyl radical which reacts further with Cl₂. The overall reaction replaces a C–H bond with a C–Cl bond.

该光化学反应在紫外光引发下通过链式机理进行。氯分子发生均裂生成氯自由基，这些自由基从甲烷中夺取氢原子，形成甲基自由基，后者再与氯分子反应。总反应将C–H键转化为C–Cl键。

Initiation: Cl–Cl → 2Cl•

引发：Cl–Cl → 2Cl•

Propagation: CH₄ + Cl• → •CH₃ + HCl

增长：CH₄ + Cl• → •CH₃ + HCl

•CH₃ + Cl₂ → CH₃Cl + Cl•

Termination: 2Cl• → Cl₂ / 2•CH₃ → C₂H₆ / Cl• + •CH₃ → CH₃Cl

终止：2Cl• → Cl₂ / 2•CH₃ → C₂H₆ / Cl• + •CH₃ → CH₃Cl

The propagation steps are the heart of the chain reaction, regenerating the chlorine radical so that many cycles occur per initiation event. The termination steps are rare because radical concentrations are low, but they must be included to complete the mechanism.

增长步骤是链反应的核心，它们再生氯自由基，使每个引发事件能引发多次循环。终止步骤因自由基浓度低而很少发生，但必须列出以使机理完整。

4. Example Response: Free Radical Substitution Question | 示例解答：自由基取代问题

Question: Methane reacts with chlorine in the presence of UV light to form a mixture of products. Write the mechanism for the formation of chloromethane, and suggest why further substitution can occur.

题目：甲烷在紫外光下与氯气反应，生成一系列产物。写出生成氯甲烷的机理，并说明为何会发生进一步取代。

Model answer: The initiation step is the homolytic fission of chlorine:

标准答案：引发步骤是氯分子的均裂：

Cl–Cl → 2Cl•

Propagation: a chlorine radical abstracts a hydrogen atom from methane, forming HCl and a methyl radical. Then the methyl radical reacts with a Cl₂ molecule to produce chloromethane and regenerate a Cl•.

增长：一个氯自由基从甲烷中夺取一个氢原子，生成HCl和甲基自由基。接着甲基自由基与Cl₂分子反应，生成氯甲烷并再生一个Cl•。

CH₄ + Cl• → •CH₃ + HCl

•CH₃ + Cl₂ → CH₃Cl + Cl•

Termination involves the combination of any two radicals, e.g., 2Cl• → Cl₂. Further substitution occurs because the chloromethane product still possesses C–H bonds that can be attacked by chlorine radicals, leading to polychlorinated products.

终止步骤包括任意两个自由基的结合，例如2Cl• → Cl₂。进一步取代之所以会发生，是因为产物氯甲烷仍具有可被氯自由基进攻的C–H键，从而生成多氯代产物。

5. Electrophilic Addition: Bromination of Ethene | 亲电加成：乙烯的溴化

Electrophilic addition to alkenes begins with the attack of the π bond on an electrophile. When ethene reacts with bromine, the Br–Br bond becomes polarised as it approaches the electron-rich double bond. A curly arrow from the C=C bond to the δ⁺ bromine atom breaks the Br–Br bond heterolytically, generating a cyclic bromonium ion (or a carbocation, depending on the mechanism variant taught) and a bromide ion. The bromide ion then attacks the positively charged intermediate to give 1,2-dibromoethane.

烯烃的亲电加成始于π键对亲电试剂的进攻。当乙烯与溴反应时，Br–Br键在靠近富电子的双键时发生极化。来自C=C键的卷曲箭头指向δ⁺ Br原子，使Br–Br键发生异裂，生成一个环状溴鎓离子（或碳正离子，取决于所教授的机理版本）和一个溴离子。随后溴离子进攻带正电荷的中间体，得到1,2-二溴乙烷。

Overall: H₂C=CH₂ + Br₂ → BrCH₂–CH₂Br

总反应：H₂C=CH₂ + Br₂ → BrCH₂–CH₂Br

The key curly arrows: one from the π bond to Brδ⁺, and another from the Br–Br bond to the departing Br⁻. In the second step, a curly arrow goes from the lone pair on Br⁻ to the positive carbon centre.

关键的卷曲箭头：一个是从π键指向Brδ⁺，另一个是从Br–Br键指向离去的Br⁻。在第二步中，一个卷曲箭头从Br⁻的孤对电子指向带正电荷的碳中心。

6. Example Response: Electrophilic Addition Question | 示例解答：亲电加成问题

Question: Propene reacts with hydrogen bromide. Draw the mechanism and explain why 2-bromopropane is the major product.

题目：丙烯与溴化氢反应。画出反应机理，并解释为何主要产物是2-溴丙烷。

Model answer: The HBr molecule is polarised: Hδ⁺–Brδ⁻. The π electrons of the C=C bond attack the Hδ⁺, breaking the H–Br bond heterolytically and forming a carbocation and a bromide ion. Two carbocations are possible: a primary carbocation and a secondary carbocation. Because a secondary carbocation is more stable (due to the electron-donating effect of the adjacent alkyl groups), it forms preferentially. In the second step, Br⁻ attacks the secondary carbocation, yielding 2-bromopropane as the major product.

标准答案：HBr分子发生极化：Hδ⁺–Brδ⁻。C=C键的π电子进攻Hδ⁺，使H–Br键异裂，生成一个碳正离子和一个溴离子。可能生成两种碳正离子：伯碳正离子和仲碳正离子。由于仲碳正离子更稳定（因相邻烷基的给电子效应），优先生成该中间体。第二步中，Br⁻进攻仲碳正离子，得到主产物2-溴丙烷。

Intermediate stability: 3° > 2° > 1° carbocation

中间体稳定性：3° > 2° > 1° 碳正离子

This is an example of Markovnikov’s rule, where the hydrogen adds to the less substituted carbon of the double bond. The mechanism must show the curly arrow from C=C to H, and then from Br⁻ to the carbocation.

这是马氏规则的实例，氢加在双键中取代较少的碳原子上。机理必须展示出从C=C指向H的卷曲箭头，以及从Br⁻指向碳正离子的箭头。

7. Nucleophilic Substitution: SN1 and SN2 | 亲核取代：SN1与SN2

Haloalkanes undergo nucleophilic substitution with reagents such as aqueous NaOH. The mechanism can be either SN2 (bimolecular) or SN1 (unimolecular), depending on the structure of the haloalkane. SN2 involves a concerted backside attack by the nucleophile with simultaneous departure of the leaving group, proceeding through a single transition state and inverting the configuration at carbon. SN1 proceeds via a carbocation intermediate formed in a slow, rate-determining step, followed by rapid attack of the nucleophile, which can lead to racemisation.

卤代烷与NaOH水溶液等试剂发生亲核取代反应。机理可以是SN2（双分子）或SN1（单分子），取决于卤代烷的结构。SN2涉及亲核试剂的协同背面进攻和离去基团的同时离去，经过一个单一的过渡态，并导致碳手性反转。SN1则通过一个缓慢的速率决定步骤形成碳正离子中间体，然后亲核试剂快速进攻，可能导致外消旋化。

SN2: rate = k[haloalkane][nucleophile]; favoured by primary alkyl halides.

SN2：速率 = k[卤代烷][亲核试剂]；有利于伯卤代烷。
SN1: rate = k[haloalkane]; favoured by tertiary alkyl halides in protic solvents.

SN1：速率 = k[卤代烷]；有利于叔卤代烷在质子性溶剂中。

8. Example Response: Nucleophilic Substitution Question | 示例解答：亲核取代问题

Question: Compare the mechanisms of hydrolysis of 1-bromobutane and 2-bromo-2-methylpropane with aqueous NaOH. Include diagrams of curly arrows and explain how the rate laws differ.

题目：比较1-溴丁烷和2-溴-2-甲基丙烷在NaOH水溶液中的水解机理。用卷曲箭头图示，并解释速率方程如何不同。

Model answer for 1-bromobutane (SN2): The nucleophile OH⁻ attacks the carbon bearing the bromine from the opposite side of the C–Br bond, forming a transition state with partially bonded OH and Br. The C–Br bond breaks simultaneously, yielding butan-1-ol with inversion. Rate = k[1-bromobutane][OH⁻].

1-溴丁烷（SN2）标准答案：亲核试剂OH⁻从C–Br键的背面进攻带有溴的碳原子，形成一个OH和Br部分键合的过渡态。C–Br键同步断裂，得到1-丁醇并伴随构型反转。速率 = k[1-溴丁烷][OH⁻]。

Model answer for 2-bromo-2-methylpropane (SN1): In the slow step, the C–Br bond breaks heterolytically to form a tertiary carbocation. This step is unimolecular and determines the rate. The carbocation is then rapidly attacked by OH⁻ to give 2-methylpropan-2-ol. Rate = k[2-bromo-2-methylpropane].

2-溴-2-甲基丙烷（SN1）标准答案：在慢步骤中，C–Br键异裂形成叔碳正离子。该步骤是单分子的，决定总速率。随后碳正离子被OH⁻快速进攻，生成2-甲基-2-丙醇。速率 = k[2-溴-2-甲基丙烷]。

The difference arises because steric hindrance prevents a direct SN2 attack on the tertiary halide, and the tertiary carbocation is stabilised by alkyl groups.

差异的产生是由于空间位阻阻碍了对叔卤代烷的直接SN2进攻，而叔碳正离子因烷基而得以稳定。

9. Electrophilic Substitution: Nitration of Benzene | 亲电取代：苯的硝化

Benzene reacts with a mixture of concentrated nitric and sulfuric acids to form nitrobenzene. The electrophile is the nitronium ion, NO₂⁺, generated in situ. The delocalised π system of benzene attacks NO₂⁺, forming a positively charged sigma complex (arenium ion). This intermediate then loses a proton to restore the aromatic sextet, yielding nitrobenzene.

苯与浓硝酸和浓硫酸的混合物反应生成硝基苯。亲电试剂是现场生成的硝酰阳离子NO₂⁺。苯的离域π体系进攻NO₂⁺，形成一个带正电荷的σ络合物（芳基正离子）。然后该中间体失去一个质子恢复芳香六隅体，得到硝基苯。

HNO₃ + 2H₂SO₄ → NO₂⁺ + 2HSO₄⁻ + H₃O⁺

The sigma complex is stabilised by resonance, with the positive charge delocalised over several carbon atoms of the ring. The final deprotonation by HSO₄⁻ regenerates the catalyst (H₂SO₄) and yields the substitution product.

σ络合物通过共振稳定，正电荷离域于环上的多个碳原子。最后通过HSO₄⁻去质子化，再生催化剂（H₂SO₄）并得到取代产物。

10. Example Response: Electrophilic Substitution Question | 示例解答：亲电取代问题

Question: Describe the mechanism for the nitration of benzene, clearly showing the formation of the electrophile and all curly arrows.

题目：描述苯的硝化机理，清楚展示亲电试剂的生成及所有卷曲箭头。

Model answer: In a mixture of concentrated HNO₃ and H₂SO₄, the nitronium ion is formed:

标准答案：在浓HNO₃和H₂SO₄混合物中，生成硝酰阳离子：

HNO₃ + 2H₂SO₄ → NO₂⁺ + 2HSO₄⁻ + H₃O⁺

A pair of π electrons from the benzene ring attacks the nitrogen atom of NO₂⁺, forming a C–N bond and giving a positively charged cyclohexadienyl cation intermediate (sigma complex). The curly arrow starts

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International A-Level Chemistry Example Responses: CH04 Unit 4 Reaction Mechanisms | 国际A-Level化学例题解答：CH04第四单元反应机理