AQA International AS Chemistry Unit 2: Example Responses | AQA国际AS化学第二单元：示例回答

📚 AQA International AS Chemistry Unit 2: Example Responses | AQA国际AS化学第二单元：示例回答

In the AQA International AS Chemistry course, Unit 2 tests your ability to recall, apply and evaluate key concepts from organic and physical chemistry. Examiners often comment that students lose marks not because they lack knowledge, but because they fail to structure their answers or use the correct terminology. This article presents a series of model questions and example responses, each with a breakdown of the essential points needed to gain full marks. By studying these examples, you can learn how to write concise, accurate and well-structured answers for the real examination.

在AQA国际AS化学课程中，第二单元考查你回忆、应用和评估有机化学与物理化学关键概念的能力。考官经常评价说，学生失分并非因为缺乏知识，而是因为没有组织好答案或使用准确的术语。本文呈现一系列典型问题与示例回答，并解析每个答案获得满分所必需的要点。通过研究这些示例，你可以学会如何在真实考试中写出简洁、准确且结构良好的答案。

1. Naming and Isomerism | 命名与异构体

Question: ‘Name the compound CH₃CH(CH₃)CH₂CH₃ and state its molecular formula.’

问题：“命名化合物CH₃CH(CH₃)CH₂CH₃并写出其分子式。”

Example response: The longest continuous carbon chain contains four carbon atoms, so the parent chain is ‘butane’. A methyl group is attached to the second carbon atom. The systematic name is 2-methylbutane. The molecular formula is C₅H₁₂.

示例回答：最长的连续碳链包含四个碳原子，因此主链为“丁烷”。一个甲基连在第二个碳原子上。系统命名为2-甲基丁烷。分子式为C₅H₁₂。

Examiner’s tip: Always number the chain from the end that gives the lowest possible locant to the substituent. In this case, numbering from the left gives the methyl group at position 2; numbering from the right would give position 3, which is incorrect. You must also remember that the name is written as one word with no space after the number.

考官提示：编号时应从使取代基位次最低的一端开始。本例中，从左端编号甲基位于2位；从右端编号则为3位，这是错误的。还要注意名称是一个单词，数字后不加空格。

Common mistake: Some students write “2-methylbutane” but forget that the molecular formula of butane derivatives must be derived from the full structure. For 2-methylbutane, the total carbon count is five and the total hydrogen count is twelve because the saturated alkane formula CₙH₂ₙ₊₂ applies.

常见错误：有些学生写出“2-甲基丁烷”，但忘记分子式必须从完整结构推导。2-甲基丁烷的总碳数为五，氢数为十二，因为它符合饱和烷烃通式CₙH₂ₙ₊₂。

2. Alkenes and Addition Reactions | 烯烃与加成反应

Question: ‘Propene reacts with hydrogen bromide to form two possible products. Explain, with a mechanism, why 2-bromopropane is the major product.’

问题：“丙烯与溴化氢反应可能生成两种产物。请用机理解释为什么2-溴丙烷是主要产物。”

Example response: The double bond in propene is an area of high electron density, which attracts the electrophile H⁺. The H⁺ adds to the less substituted carbon atom at the end of the double bond, so that the positive charge in the intermediate carbocation forms on the central, more substituted carbon. This secondary carbocation, CH₃CH⁺CH₃, is more stable than the primary alternative CH₃CH₂CH₂⁺ because alkyl groups are electron-releasing and stabilise the positive charge. The Br⁻ ion then attacks the carbocation to give 2-bromopropane. This is Markovnikov’s rule.

示例回答：丙烯中的双键是电子云密度高的区域，会吸引亲电试剂H⁺。H⁺加在双键末端取代较少的碳原子上，使中间体碳正离子的正电荷位于中间取代较多的碳原子上。仲碳正离子CH₃CH⁺CH₃比伯碳正离子CH₃CH₂CH₂⁺更稳定，因为烷基是供电子基团，能稳定正电荷。随后Br⁻进攻碳正离子，生成2-溴丙烷。这就是马尔科夫尼科夫规则。

CH₃CH=CH₂ + HBr → CH₃CHBrCH₃

Key point: You must state that the electrophile is H⁺, show the formation of a carbocation intermediate, and explain the stability difference using the terms ‘primary’, ‘secondary’ or ‘tertiary’. Without the stability argument, the answer is incomplete.

关键点：必须指出亲电试剂是H⁺，展示碳正离子中间体的生成，并用“伯”“仲”“叔”来解释稳定性差异。如果没有稳定性论证，答案是不完整的。

Common mistake: A frequent error is to write “the more stable carbocation is formed on the carbon with more hydrogen atoms” – this is the opposite of the true rule. The correct idea is that the more substituted carbon (fewer hydrogen atoms) carries the positive charge because it is more stable.

常见错误：常见的错误是写“更稳定的碳正离子是连有更多氢原子的碳上形成的”——这与真实规则相反。正确理解是：取代越多（即氢越少）的碳带正电荷，因为它更稳定。

3. Alcohols and Oxidation | 醇与氧化

Question: ‘A student oxidises a sample of an alcohol with acidified potassium dichromate(VI). The product turns the orange dichromate to green and gives a colourless liquid that does not react with Tollens’ reagent. Deduce the type of alcohol and explain this observation.’

问题：“学生用酸化的重铬酸钾(VI)氧化某种醇样品。产物使橙色的重铬酸盐变为绿色，所得无色液体不与托伦试剂反应。请推断该醇的类型并解释观察现象。”

Example response: The orange to green colour change shows that the alcohol has been oxidised by the dichromate(VI) ion. Because the product is a ketone, it does not react with Tollens’ reagent, which is a test for aldehydes. A secondary alcohol is oxidised to a ketone, so the original alcohol must be secondary. Primary alcohols would first be oxidised to aldehydes, which would give a silver mirror with Tollens’ reagent, and further oxidation would give a carboxylic acid.