WJEC化学 有机化学 烷烃 烯烃 卤代烃 机理
WJEC A-Level 化学的有机部分横跨 AS 与 A2 两个阶段。AS 的 Unit 2 建立官能团、同系物与三大反应机理的骨架，A2 的 Unit 4 把同一套机理逻辑推进到苯环、羰基化合物与胺类。很多同学把有机化学当成"要背的反应清单"，这是最大的误区。WJEC 评分方案真正考察的是三件事：能否从结构推出性质，能否用卷曲箭头把电子转移写清楚，能否在合成路线里选择合适的条件。
Organic chemistry in WJEC A-Level Chemistry spans both stages of the course. AS Unit 2 builds the framework of functional groups, homologous series and the three core mechanisms. A2 Unit 4 extends the same mechanistic logic to aromatic rings, carbonyl compounds and amines. Students who treat organic chemistry as a list of reactions to memorise tend to plateau, because WJEC mark schemes reward three specific skills: deducing properties from structure, drawing curly arrows that show electron movement correctly, and selecting the right conditions in a multi-step synthesis.
一、官能团与同系物：整个有机模块的地基
官能团决定了分子的化学行为。WJEC 要求你熟记烷烃、烯烃、卤代烃、醇、醛、酮、羧酸、酯、胺、酰胺这几类官能团，并能从分子式直接判断它属于哪一族。同系物的定义要写完整：通式相同、结构相似、每相邻两个成员相差一个 CH2 单元、化学性质相似的化合物系列。物理性质随碳链增长呈现的趋势也要能解释：碳链越长，范德华力越强，沸点越高；但支链越多，分子间接触面积越小，沸点反而下降。
The functional group determines how a molecule behaves. WJEC expects secure knowledge of alkanes, alkenes, halogenoalkanes, alcohols, aldehydes, ketones, carboxylic acids, esters, amines and amides, plus the ability to identify the family from a molecular formula. The definition of a homologous series must be complete: same general formula, similar structure, each successive member differs by one CH2 unit, and members show similar chemical properties. Trends in physical properties need explanation too: a longer carbon chain means stronger van der Waals forces and a higher boiling point, while increased branching reduces the contact area between molecules and lowers the boiling point.
命名是送分题，也是失分题。WJEC 采用 IUPAC 规则：先找出最长的含官能团碳链，从离官能团最近的一端编号，取代基按字母顺序排列。常见陷阱有三个：把支链位置编号搞反、把醇写成烷基、忘记标注双键位置。举例来说，CH3CH(OH)CH2CH3 的正确名称是 butan-2-ol，不是 2-butanol，也不是 2-hydroxybutane。
Nomenclature is easy marks that are frequently lost. WJEC follows IUPAC rules: find the longest chain containing the functional group, number from the end nearest that group, and list substituents alphabetically. Three common mistakes are numbering from the wrong end, naming an alcohol as an alkane, and omitting the position of a double bond. For example, CH3CH(OH)CH2CH3 is butan-2-ol, not 2-hydroxybutane.
二、异构现象：从结构异构到光学异构
AS 阶段需要掌握结构异构的三种形式：碳链异构（如 C4H10 有正丁烷与 2-甲基丙烷两种）、位置异构（如丙醇与丙-2-醇）、官能团异构（如乙醇与二甲醚）。烯烃还要额外掌握 E-Z 异构，判断依据是两个双键碳上较高优先级的基团是否在同侧；优先级的比较规则是原子序数大者优先。
At AS level you need the three forms of structural isomerism: chain isomerism, such as the two isomers of C4H10, position isomerism, such as propan-1-ol and propan-2-ol, and functional group isomerism, such as ethanol and methoxymethane. Alkenes add E-Z isomerism, decided by whether the higher priority groups on each carbon of the double bond lie on the same side. Priority follows atomic number, with the larger atom taking precedence.
A2 阶段新增光学异构：分子含有手性碳（连有四个不同基团）时就可能出现一对对映体，它们互为镜像但不能重合。手性碳的识别方法是逐个检查每个碳原子的四个取代基是否互不相同。药物化学里这一点极其重要，因为两种对映体可能有完全不同的生物活性。
A2 introduces optical isomerism. A molecule containing a chiral carbon, meaning a carbon attached to four different groups, may exist as a pair of enantiomers that are mirror images and cannot be superimposed. To find the chiral centre, check each carbon in turn and confirm that all four attached groups differ. This matters greatly in pharmaceutical chemistry, where the two enantiomers of a drug can have completely different biological activity.
三、烷烃：自由基取代反应机理
烷烃的主要反应是与卤素在紫外光下发生自由基取代。机理分三步，每一步的专有名词都要写对：链引发（initiation）、链增长（propagation）、链终止（termination）。以甲烷与氯气为例，链引发是 Cl2 在 UV 照射下均裂成两个氯自由基；链增长是氯自由基夺取氢原子生成 HCl 与甲基自由基，甲基自由基再与 Cl2 反应生成一氯甲烷与新的氯自由基；链终止是两个自由基结合成稳定分子。
Alkanes react mainly by free radical substitution with halogens under ultraviolet light. The mechanism has three named stages that must be labelled correctly: initiation, propagation and termination. Taking methane and chlorine as the example, initiation is the homolytic fission of Cl2 under UV light to give two chlorine radicals. Propagation is a chlorine radical abstracting a hydrogen atom to form HCl and a methyl radical, which then reacts with Cl2 to give chloromethane and a fresh chlorine radical. Termination is the combination of two radicals to form a stable molecule.
考试里最容易丢分的是卷曲箭头画法。自由基机理中的单电子转移要用"鱼钩箭头"（半箭头）表示，而不是普通双电子箭头。另一个高频考点是为什么产物是混合物：因为一氯甲烷会继续被取代，生成二氯、三氯、四氯甲烷，同时链终止也可能生成乙烷。实验条件限制了选择性，这正是自由基取代在合成上价值有限的原因。
The most common lost mark is arrow drawing. Single electron movement in radical mechanisms uses a half headed arrow, often called a fish hook, rather than the ordinary double headed arrow. Another frequent question asks why the product is a mixture: chloromethane continues to be substituted to give dichloro, trichloro and tetrachloromethane, while termination can also produce ethane. Poor selectivity is exactly why free radical substitution has limited value in synthesis.
四、烯烃：亲电加成与检验反应
烯烃的核心反应是亲电加成（electrophilic addition）。以乙烯与溴反应为例：双键的电子云使溴分子极化，形成诱导偶极；接近的溴端带部分正电荷，被双键的电子对进攻；结果是形成一个环状溴鎓离子与溴离子，随后溴离子从背面进攻打开三元环，得到 1,2-二溴乙烷。WJEC 要求你画出完整的卷曲箭头，并且箭头必须从电子对出发、指向形成新键的位置。
The central reaction of alkenes is electrophilic addition. With ethene and bromine, the electron rich double bond polarises the bromine molecule and creates an induced dipole. The partially positive bromine is attacked by the pi electrons, forming a cyclic bromonium ion and a bromide ion, which then attacks from the opposite side to open the three membered ring and give 1,2-dibromoethane. WJEC expects fully drawn curly arrows that start from an electron pair and point to the position of the new bond.
不对称烯烃的加成方向遵循马氏规则：氢原子加到含氢较多的双键碳上。更本质的解释是碳正离子稳定性，因为中间体碳正离子的稳定性顺序为叔碳正离子大于仲碳正离子大于伯碳正离子，烷基的诱导效应与超共轭效应分散了正电荷。如果中间体可能重排成更稳定的碳正离子，反应就会走向重排产物，这一点在 A2 的合成题中经常出现。
The direction of addition to an unsymmetrical alkene follows Markovnikov's rule: hydrogen adds to the carbon of the double bond that already carries more hydrogens. The deeper explanation is carbocation stability, because the intermediate carbocation is more stable when it is tertiary or secondary rather than primary, since alkyl groups spread the positive charge through inductive and hyperconjugative effects. If a rearrangement can produce a more stable carbocation, the reaction may follow the rearranged pathway, and A2 synthesis questions test exactly this.
检验反应的细节也要牢记。溴水褪色是烯烃等不饱和化合物的经典检验，现象是橙色褪去；酸化高锰酸钾溶液褪色也能证明不饱和键存在。注意区分：烷烃不会使溴水褪色，这是鉴别烷烃与烯烃最直接的方法。
Practical tests need precise detail. The decolourisation of bromine water is the classic test for a double bond, with the orange colour disappearing. Acidified potassium manganate(VII) also loses its colour in the presence of an unsaturated bond. Note the contrast: alkanes do not decolourise bromine water, which is the simplest way to distinguish an alkane from an alkene.
五、卤代烃：亲核取代与消除反应
卤代烃中碳卤键是极性键，碳带部分正电荷，因此容易受到亲核试剂进攻。这里存在一个条件决定产物的分岔路：与 NaOH 水溶液共热，发生亲核取代，生成醇；与 NaOH 的乙醇溶液共热，发生消除反应，脱去卤化氢生成烯烃。题目里出现 aqueous 还是 ethanolic 这一个词，就决定了整道题的答案。
The carbon halogen bond in a halogenoalkane is polar, leaving the carbon partially positive and open to nucleophilic attack. This creates a fork in the road decided entirely by conditions: heating with aqueous sodium hydroxide gives nucleophilic substitution and an alcohol, while heating with sodium hydroxide in ethanol gives elimination and an alkene. A single word, aqueous or ethanolic, decides the whole answer in an exam question.
反应活性方面，C-I 键键能最小，也最容易断裂，因此碘代烷反应最快，氟代烷最慢，顺序为 C-I 大于 C-Br 大于 C-Cl 大于 C-F。这是因为碳卤键的键能随卤素原子半径增大而减小。伯卤代烷倾向 SN2 单步机理，反应速率与卤代烷浓度和亲核试剂浓度都成正比；叔卤代烷倾向 SN1 两步机理，先离解出碳正离子，速率只取决于卤代烷浓度。
Reactivity follows bond strength. The carbon iodine bond is the weakest and breaks most easily, so iodoalkanes react fastest and fluoroalkanes slowest, in the order C-I greater than C-Br greater than C-Cl greater than C-F, because bond enthalpy falls as the halogen atom gets larger. Primary halogenoalkanes favour the single step SN2 pathway, whose rate depends on the concentration of both the halogenoalkane and the nucleophile. Tertiary halogenoalkanes favour the two step SN1 pathway, which begins with ionisation to a carbocation and whose rate depends only on the halogenoalkane concentration.
六、醇：氧化、脱水与合成价值
醇的化学性质由羟基所在碳上的取代情况决定。伯醇可以被酸化重铬酸钾氧化成醛，继续氧化成羧酸；仲醇氧化成酮，且酮不易被进一步氧化；叔醇在常规条件下不被氧化。区分蒸馏与回流是考试高频点：要得到醛必须用蒸馏装置，边生成边蒸出，防止醛被继续氧化；要得到羧酸则用回流装置，让氧化充分进行。现象是橙色重铬酸钾变成绿色铬离子。
The chemistry of an alcohol depends on how many carbon groups are attached to the carbon bearing the hydroxyl group. A primary alcohol is oxidised by acidified dichromate(VI) first to an aldehyde and then to a carboxylic acid. A secondary alcohol gives a ketone, which resists further oxidation. A tertiary alcohol is not oxidised under normal conditions. Distinguishing distillation from reflux is a high frequency exam point: distillation is used to obtain the aldehyde and remove it as it forms, preventing further oxidation, while reflux is used to obtain the carboxylic acid because it keeps the oxidising agent in contact with the alcohol. The visible change is orange dichromate turning green.
醇的脱水是制备烯烃的重要路线：在浓硫酸或氧化铝催化下加热，发生消除反应生成烯烃。此外醇还能与羧酸在浓硫酸催化下发生酯化反应，生成有果香味的酯，这个反应是可逆的，需要用浓硫酸吸水并加热来推动平衡。合成题里常见的一条链路是：卤代烷经取代变醇，醇氧化成酸，酸与醇酯化，每一步的条件都要写清楚。
Dehydration of an alcohol is an important route to an alkene: heating with concentrated sulfuric acid or an aluminium oxide catalyst brings about elimination. Alcohols also esterify with carboxylic acids under concentrated sulfuric acid catalysis to give sweet smelling esters. The reaction is reversible, so the acid catalyst also removes water and heating drives the equilibrium forward. A common synthesis chain is halogenoalkane to alcohol by substitution, alcohol to acid by oxidation, then acid plus alcohol to ester, with conditions stated precisely at every step.
七、A2 延伸：苯、羰基化合物与胺
Unit 4 把 AS 的机理逻辑推广到新体系。苯的稳定性来自离域的大派键，因此它发生亲电取代而不是加成，典型反应包括硝化、卤代与傅-克反应。羰基化合物中碳氧双键极化，碳带部分正电荷，因此醛酮易受亲核加成：与 HCN 加成生成羟基腈，与 2,4-二硝基苯肼反应生成橙色沉淀用于鉴定。醛还能被托伦试剂（银镜）与费林试剂氧化，酮则不能，这是区分醛酮的经典方法。
Unit 4 extends the mechanistic logic of AS into new systems. Benzene is stabilised by a delocalised pi system, so it undergoes electrophilic substitution rather than addition, with nitration, halogenation and Friedel-Crafts reactions as the classic examples. In carbonyl compounds the polarised carbon oxygen double bond leaves the carbon partially positive, so aldehydes and ketones undergo nucleophilic addition, for example with HCN to give hydroxynitriles, or with 2,4-dinitrophenylhydrazine to give an orange precipitate used for identification. Aldehydes are also oxidised by Tollens reagent, giving a silver mirror, and by Fehling solution, while ketones are not, which is the standard way to tell them apart.
胺类与氨基酸部分要抓住碱性来源与肽键形成。胺的氮原子有一对孤对电子，可以接受质子，因此显碱性；芳香胺的孤对电子参与苯环共轭，碱性明显减弱。氨基酸同时含有氨基与羧基，因此具有两性，在水溶液中以两性离子形式存在，等电点是其净电荷为零时的 pH。两个氨基酸脱水缩合形成肽键，多个氨基酸连成多肽与蛋白质。
The amine and amino acid section hinges on the origin of basicity and on peptide bond formation. The nitrogen in an amine carries a lone pair that accepts a proton, so amines are basic. In aromatic amines that lone pair is delocalised into the ring, which weakens the basicity noticeably. Amino acids contain both an amino group and a carboxyl group, so they are amphoteric, exist as zwitterions in aqueous solution, and have an isoelectric point at which the net charge is zero. Two amino acids condense with the loss of water to form a peptide bond, and many repeat units build up polypeptides and proteins.
八、机理书写与考试技巧
WJEC 的有机大题分值集中在机理与推理，而不是纯记忆。给分点通常包括：正确标注部分电荷、卷曲箭头起点与终点正确、中间体结构正确、命名或条件描述准确。写卷曲箭头时务必记住两条铁律：箭头必须从电子对（孤对电子或键中电子）出发，必须指向新键形成的位置；正电荷与负电荷要在中间体上标出来。
The organic questions in WJEC concentrate marks on mechanisms and reasoning rather than pure recall. Mark points usually include correctly labelled partial charges, curly arrows drawn from the right source to the right destination, a correctly drawn intermediate, and accurate naming or conditions. Two rules govern every curly arrow: it must start from an electron pair, whether a lone pair or a bonding pair, and it must point to where the new bond forms. Charges on the intermediate should always be shown.
复习策略上建议按官能团建立反应地图，而不是按章节背诵。每学一个新官能团，就问自己三个问题：它可以由什么转化而来，它可以转化为什么，转化需要什么条件。把这三问串起来，合成路线题就变成了路径搜索。同时把每个机理画在卡片上，正面写反应物与条件，背面写机理名称与中间体，每周默画一遍，两个月就能形成稳定的肌肉记忆。
A better revision strategy is to build a reaction map organised by functional group rather than memorising chapter by chapter. For every new functional group, ask three questions: what can it be made from, what can it be turned into, and what conditions are needed. Chaining these questions turns synthesis questions into a path finding exercise. Draw each mechanism on a revision card with the reactants and conditions on the front and the mechanism name plus intermediate on the back, redraw them weekly, and the recall becomes automatic within about two months.
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