AS Chemistry Paper 2 Exam Report: Reaction Mechanisms | AS化学试卷二考试报告：反应机理

📚 AS Chemistry Paper 2 Exam Report: Reaction Mechanisms | AS化学试卷二考试报告：反应机理

The AS Chemistry Paper 2 examiners’ report highlights that reaction mechanisms remain a challenging area for many candidates. Common pitfalls include drawing curly arrows from the wrong source, omitting lone pairs, and failing to show the correct intermediate or transition state. This article analyses these errors and provides clear guidance to help you secure full marks on mechanism questions.

AS化学试卷二的考官报告指出，反应机理对许多考生来说仍是一个棘手领域。常见错误包括弯箭头绘制起点错误、遗漏孤对电子以及未能正确显示中间体或过渡态。本文分析这些错误并提供清晰的指导，帮助你在机理题上获得满分。

1. Understanding the Electrophilic Addition Mechanism | 理解亲电加成机理

Electrophilic addition is the characteristic reaction of alkenes. The π-bond of the alkene acts as a nucleophile and attacks an electrophile such as H⁺ from HBr or Br⁺ from Br₂. Examiners expect you to show a curly arrow from the middle of the C=C double bond towards the electrophile, forming a carbocation intermediate. A second curly arrow must then be drawn from a halide ion (Br⁻) or other nucleophile to the positively charged carbon.

亲电加成是烯烃的特征反应。烯烃的π键作为亲核试剂进攻亲电试剂，例如来自HBr的H⁺或来自Br₂的Br⁺。考官要求你用弯箭头从C=C双键中间指向亲电试剂，形成碳正离子中间体。然后必须再画一个弯箭头从卤离子(Br⁻)或其他亲核试剂指向带正电荷的碳。

Many candidates lose marks by showing the arrow starting from a single carbon atom rather than the bond itself. Always start the curly arrow from the bond or lone pair, not from an atom. For the addition of HBr to propene, you must consider the stability of the carbocation: the secondary carbocation is more stable than the primary, leading to 2-bromopropane as the major product according to Markovnikov’s rule.

许多考生因为将箭头起点标在单个碳原子上而不是键上而失分。请务必从键或孤对电子开始绘制弯箭头，而不是从原子。对于丙烯与HBr的加成，必须考虑碳正离子的稳定性：仲碳正离子比伯碳正离子更稳定，因此根据Markovnikov规则，主要产物为2-溴丙烷。

2. Free-Radical Substitution Step by Step | 逐步解析自由基取代

Free-radical substitution is tested frequently in Paper 2. The mechanism requires clear identification of the initiation step: homolytic fission of Cl₂ by UV light to give two Cl· radicals. Propagation steps must show a chlorine radical abstracting a hydrogen atom from an alkane, producing HCl and an alkyl radical; that alkyl radical then reacts with another Cl₂ molecule, yielding the halogenoalkane and a new Cl· radical.

自由基取代在试卷二中频繁出现。其机理要求清晰标明引发步骤：Cl₂在紫外光下发生均裂，生成两个Cl·自由基。增长步骤必须展示氯

Published by TutorHao | AS Chemistry Revision Series | aleveler.com

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