📚 ESAT Chemistry: Organic Chemistry Fundamentals | ESAT 化学：有机化学基础专题
Organic chemistry is a major component of the ESAT Chemistry syllabus, testing your ability to recall functional group reactions, mechanisms, and structural principles. This guide condenses the essential knowledge into a focused, bilingual revision resource tailored to the exam’s style.
有机化学是 ESAT 化学考纲的重要组成部分，考查你对官能团反应、机理和结构原理的记忆与理解。本指南将核心知识浓缩为一份贴合考试风格的中英双语复习资料。
1. Carbon Bonding and Hybridisation | 碳的成键与杂化
Carbon has four valence electrons and always forms four covalent bonds in organic compounds. The type of bond depends on the hybridisation of the carbon orbitals.
碳有四个价电子，在有机化合物中总是形成四个共价键。键的类型取决于碳原子轨道的杂化方式。
In sp³ hybridisation, three p-orbitals and one s-orbital mix to form four equivalent orbitals. The shape is tetrahedral with bond angles of 109.5°. Examples include alkanes such as methane (CH₄) and ethane (C₂H₆).
在 sp³ 杂化中，一个 s 轨道与三个 p 轨道混合形成四个等价轨道。空间构型为四面体，键角为 109.5°。例如甲烷（CH₄）和乙烷（C₂H₆）等烷烃。
In sp² hybridisation, one s-orbital and two p-orbitals mix, leaving one unhybridised p-orbital. The shape is trigonal planar (120°) and the unhybridised p-orbital overlaps sideways to form a π-bond. Alkenes, such as ethene (C₂H₄), exhibit this.
在 sp² 杂化中，一个 s 轨道与两个 p 轨道混合，留下一个未杂化 p 轨道。空间构型为平面三角形（120°），未杂化 p 轨道侧面重叠形成 π 键。烯烃如乙烯（C₂H₄）属于此类。
In sp hybridisation, one s-orbital and one p-orbital mix; the molecule is linear with a 180° angle, and two unhybridised p-orbitals form two π-bonds. Alkynes like ethyne (C₂H₂) are examples.
在 sp 杂化中，一个 s 轨道与一个 p 轨道混合；分子呈直线形，键角为 180°，两个未杂化 p 轨道形成两个 π 键。乙炔（C₂H₂）等炔烃即属此类。
Single bonds are σ-bonds, allowing free rotation, whereas double and triple bonds restrict rotation and create rigid planar or linear structures.
单键是 σ 键，可自由旋转；而双键和三键限制旋转，形成刚性平面或直线结构。
2. Functional Groups and Nomenclature | 官能团与命名
Organic compounds are classified by their functional groups, which determine their chemical reactivity. For ESAT, you must know the names, formulas, and IUPAC suffixes (or prefixes) of key families.
有机化合物依据其官能团分类，官能团决定了化学反应活性。ESAT 考试要求你掌握重要类别的名称、化学式以及相应的 IUPAC 后缀（或前缀）。
| Class | 类别 | Functional Group | 官能团 | Suffix/Prefix | 后缀/前缀 |
|---|---|---|
| Alkane 烷烃 | C–C single bond | -ane |
| Alkene 烯烃 | C=C | -ene |
| Haloalkane 卤代烃 | –X (F, Cl, Br, I) | fluoro-, chloro-, etc. |
| Alcohol 醇 | –OH | -ol |
| Aldehyde 醛 | –CHO | -al |
| Ketone 酮 | R–CO–R’ | -one |
| Carboxylic Acid 羧酸 | –COOH | -oic acid |
| Ester 酯 | –COOR | -oate |
| Amine 胺 | –NH₂ | -amine |
| Amide 酰胺 | –CONH₂ | -amide |
When naming, select the longest carbon chain containing the principal functional group. Number the chain to give the functional group the lowest possible locant, then list substituents in alphabetical order.
命名时，选择包含主要官能团的最长碳链，并给官能团尽可能低的编号；取代基则按字母顺序排列。
3. Isomerism | 同分异构现象
Isomers are compounds with the same molecular formula but different structures. For ESAT, you must recognise three main types: chain, position, and functional group isomerism.
同分异构体是指分子式相同但结构不同的化合物。ESAT 要求你识别三种主要类型：碳链异构、位置异构和官能团异构。
Chain isomerism arises from different branching of the carbon skeleton. For example, butane (C₄H₁₀) can be n-butane or 2-methylpropane.
碳链异构源于碳骨架的不同分支。例如丁烷（C₄H₁₀）可以是正丁烷或 2-甲基丙烷。
Position isomerism occurs when a functional group or substituent occupies a different position on the same chain, such as propan-1-ol vs propan-2-ol.
位置异构是指相同碳链上官能团或取代基位于不同位置，例如丙-1-醇与丙-2-醇。
Functional group isomerism occurs when compounds with the same formula belong to different families, e.g. propanal (C₃H₆O) and propanone (C₃H₆O).
官能团异构是指分子式相同的化合物属于不同类别，例如丙醛（C₃H₆O）和丙酮（C₃H₆O）。
Stereoisomerism includes E/Z (geometric) and optical isomers. E/Z arises from restricted rotation around the C=C bond: if the higher-priority groups are on opposite sides, it is E; if on the same side, Z.
立体异构包括 E/Z（几何）和旋光异构体。E/Z 源于 C=C 双键的旋转受限：若高优先基团位于相反侧则为 E，同侧则为 Z。
Optical isomers occur when a carbon is bonded to four different groups – a chiral centre. They exist as non-superimposable mirror images (enantiomers), which rotate plane-polarised light in opposite directions.
旋光异构发生在碳原子连接四个不同基团时——即手性中心。它们以不能重叠的镜像对（对映异构体）存在，使平面偏振光向相反方向旋转。
4. Free Radical Substitution in Alkanes | 烷烃的自由基取代
Alkanes are relatively unreactive but undergo free radical substitution with halogens in the presence of ultraviolet light. The reaction proceeds in three stages.
烷烃相对不活泼，但在紫外光存在下可与卤素发生自由基取代反应。反应经历三个阶段。
Initiation: UV light breaks the Cl–Cl bond homolytically, producing two chlorine radicals.
链引发：紫外光使 Cl–Cl 键均裂，产生两个氯自由基。
Cl–Cl → 2 Cl•
Propagation: the chlorine radical abstracts a hydrogen atom from methane, forming HCl and a methyl radical. The methyl radical then reacts with another chlorine molecule to form chloromethane and regenerate the chlorine radical.
链增长：氯自由基从甲烷中夺取氢原子，生成 HCl 和甲基自由基；甲基自由基再与另一个氯分子反应，生成氯甲烷并再生氯自由基。
CH₄ + Cl• → •CH₃ + HCl
•CH₃ + Cl₂ → CH₃Cl + Cl•
Termination: two radicals combine to form a stable molecule, e.g. •CH₃ + Cl• → CH₃Cl.
链终止：两个自由基结合形成稳定分子，例如 •CH₃ + Cl• → CH₃Cl。
Remember that multiple substitution can occur, producing dichloromethane (CH₂Cl₂), trichloromethane (CHCl₃), and tetrachloromethane (CCl₄). The product is always a mixture.
注意可能发生多元取代，生成二氯甲烷（CH₂Cl₂）、三氯甲烷（CHCl₃）和四氯甲烷（CCl₄）。产物总是混合物。
5. Electrophilic Addition to Alkenes | 烯烃的亲电加成
The C=C double bond is electron-rich, making alkenes nucleophiles that react with electrophiles. A typical example is the addition of bromine water, which decolourises from orange to colourless – a test for unsaturation.
C=C 双键富含电子，使烯烃作为亲核体与亲电体反应。典型例子是与溴水加成，溴水从橙色褪为无色——这是不饱和键的检验方法。
C₂H₄ + Br₂ → CH₂BrCH₂Br
With hydrogen halides, the electrophile is H⁺ from H–X. For unsymmetrical alkenes, Markovnikov’s rule applies: the hydrogen adds to the carbon with more hydrogens, forming the more stable carbocation intermediate.
与卤化氢反应时，亲电体是 H–X 产生的 H⁺。对于不对称烯烃，遵循马氏规则：氢加到含氢较多的碳上，形成更稳定的碳正离子中间体。
For example, propene with HBr gives 2-bromopropane, not 1-bromopropane, because the secondary carbocation (CH₃–CH⁺–CH₃) is more stable than the primary one.
例如，丙烯与 HBr 加成生成 2-溴丙烷而非 1-溴丙烷，因为仲碳正离子（CH₃–CH⁺–CH₃）比伯碳正离子更稳定。
Hydration of alkenes (with steam and H₃PO₄ catalyst) produces alcohols, also following Markovnikov’s rule.
烯烃水合（用蒸汽和 H₃PO₄ 作催化剂）生成醇，同样遵循马氏规则。
6. Nucleophilic Substitution in Haloalkanes | 卤代烃的亲核取代
Haloalkanes contain a polar C–X bond, making the carbon atom electron-deficient. They react with nucleophiles, such as OH⁻, CN⁻, and NH₃.
卤代烃含有极性 C–X 键，使碳原子缺电子。它们可与亲核体如 OH⁻、CN⁻ 和 NH₃ 反应。
Hydrolysis with aqueous sodium hydroxide gives an alcohol:
与氢氧化钠水溶液水解得到醇：
C₂H₅Br + NaOH → C₂H₅OH + NaBr
Reaction with alcoholic potassium cyanide yields a nitrile, which is useful for extending the carbon chain:
与氰化钾的醇溶液反应生成腈，可用于延长碳链：
C₂H₅Br + KCN → C₂H₅CN + KBr
Two mechanisms are possible. SN1 (unimolecular) occurs for tertiary haloalkanes: the C–X bond breaks first, forming a carbocation intermediate, which the nucleophile attacks. It is a two-step reaction with a carbocation intermediate.
存在两种机理。SN1（单分子）发生在叔卤代烃中：C–X 键先断裂形成碳正离子中间体，亲核体再进攻。这是两步反应，有碳正离子中间体。
SN2 (bimolecular) occurs for primary haloalkanes: the nucleophile attacks the carbon from behind as the bromide leaves, in one simultaneous step. This causes inversion of configuration at the carbon.
SN2（双分子）发生在伯卤代烃中：亲核体从碳的背面进攻，同时溴离子离去，一步完成。这导致碳原子构型反转。
7. Elimination Reactions | 消除反应
Haloalkanes can also undergo elimination when heated with alcoholic potassium hydroxide. This removes H and X from adjacent carbons, forming an alkene.
卤代烃与氢氧化钾的醇溶液共热时还可发生消除反应；从相邻碳上消去 H 和 X，生成烯烃。
C₂H₅Br + KOH (alc.) → C₂H₄ + KBr + H₂O
For haloalkanes with three or more carbons, multiple alkenes are possible. The major product is the most substituted alkene, according to Saytzeff’s rule – the alkene with the more highly substituted double bond is preferred.
对于含三个或更多碳的卤代烃，可能有多种烯烃产物。根据查依采夫规则，主要产物是取代程度最高的烯烃——即双键上烷基取代更多的烯烃更稳定、更占优。
In ESAT, you must distinguish between substitution and elimination conditions: aqueous NaOH favours substitution, while alcoholic KOH favours elimination.
在 ESAT 中，你须区分取代与消除的条件：NaOH 水溶液有利于取代，而 KOH 醇溶液有利于消除。
8. Alcohols and Their Oxidation | 醇及其氧化
Alcohols are classified as primary, secondary, or tertiary based on the number of carbon atoms attached to the carbon bearing the –OH group. This determines their oxidation behaviour.
醇根据与羟基碳相连的碳原子数分为伯醇、仲醇和叔醇，这决定了它们的氧化行为。
Primary alcohols are oxidised first to aldehydes, then to carboxylic acids. For example, ethanol → ethanal → ethanoic acid.
伯醇先被氧化为醛，再氧化为羧酸。例如乙醇 → 乙醛 → 乙酸。
Secondary alcohols are oxidised to ketones. Propan-2-ol gives propanone.
仲醇被氧化为酮。丙-2-醇氧化得到丙酮。
Tertiary alcohols are not oxidised under normal conditions because there is no hydrogen attached to the carbon bearing the –OH.
叔醇在通常条件下不能被氧化，因为羟基碳上没有氢。
The oxidising agent is often acidified potassium dichromate(VI) (K₂Cr₂O₇/H₂SO₄); the orange dichromate turns green, indicating reduction of Cr(VI) to Cr(III).
常用氧化剂是酸化的重铬酸钾（K₂Cr₂O₇/H₂SO₄）；橙色重铬酸根变为绿色，表明 Cr(VI) 被还原为 Cr(III)。
Alcohols also undergo dehydration to alkenes with concentrated H₂SO₄ at high temperature, and substitution with PCl₅ or thionyl chloride to form haloalkanes.
醇还可与浓 H₂SO₄ 在高温下脱水生成烯烃，也可与 PCl₅ 或氯化亚砜发生取代反应生成卤代烃。
9. Carbonyl Compounds: Aldehydes and Ketones | 羰基化合物：醛与酮
The carbonyl group (C=O) is polarised, with the carbon bearing a partial positive charge. Aldehydes (RCHO) have the carbonyl at the chain end, while ketones (RCOR’) have it in the middle.
羰基（C=O）有极性，碳带部分正电荷。醛（RCHO）的羰基在碳链末端，而酮（RCOR’）的羰基在碳链中间。
A key reaction is nucleophilic addition with hydrogen cyanide (HCN) to form hydroxynitriles, which adds one carbon to the chain:
一个关键反应是与氰化氢（HCN）的亲核加成生成羟基腈，从而使碳链增加一个碳：
CH₃CHO + HCN → CH₃CH(OH)CN
Aldehydes are easily oxidised to carboxylic acids by Tollens’ reagent (ammoniacal silver nitrate). A silver mirror forms. Ketones do not give a positive test, so this distinguishes the two.
醛可被托伦试剂（硝酸银的氨溶液）氧化为羧酸，产生银镜。酮没有阳性结果，因此该反应可用于鉴别醛和酮。
Fehling’s solution (blue) also detects aldehydes; it forms a brick-red precipitate of Cu₂O. Ketones do not react.
斐林溶液（蓝色）同样可鉴定醛；它生成砖红色的 Cu₂O 沉淀。酮不反应。
Both aldehydes and ketones can be reduced with NaBH₄ to their corresponding alcohols.
醛和酮都可被 NaBH₄ 还原为相应的醇。
10. Carboxylic Acids and Derivatives | 羧酸及其衍生物
Carboxylic acids contain –COOH, which ionises slightly in water to release H⁺. Their acidity is stronger than alcohols but weaker than mineral acids, and they react with bases, carbonates, and metals.
羧酸含有 –COOH，在水中部分电离释放 H⁺。其酸性强于醇但弱于无机酸，可与碱、碳酸盐和活泼金属反应。
CH₃COOH ⇌ CH₃COO⁻ + H⁺
Esterification occurs when a carboxylic acid reacts with an alcohol in the presence of concentrated H₂SO₄, producing an ester and water.
羧酸与醇在浓 H₂SO₄ 催化下发生酯化反应，生成酯和水。
CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O
Esters have sweet, fruity smells and are used in flavourings and perfumes. They hydrolyse in acid or base; alkaline hydrolysis (saponification) gives a carboxylate salt and alcohol.
酯有甜味和果香，用于香料和调味剂。酯在酸或碱条件下水解；碱性水解（皂化）生成羧酸盐和醇。
Acyl chlorides (RCOCl) are more reactive derivatives and react with water, alcohols, and ammonia to form carboxylic acids, esters, and amides respectively.
酰氯（RCOCl）是更活泼的衍生物，分别与水、醇和氨反应生成羧酸、酯和酰胺。
11. Spectroscopic Identification | 波谱鉴定
ESAT also tests your ability to identify organic molecules using analytical techniques. Mass spectrometry gives the molecular mass and fragmentation pattern; IR spectroscopy identifies functional groups; and NMR provides carbon‑hydrogen environments.
ESAT 也考查运用分析技术鉴定有机分子的能力。质谱给出相对分子质量和碎片信息；红外光谱识别官能团；核磁共振提供碳‑氢环境信息。
In the mass spectrum, the molecular ion peak (M⁺) gives the relative molecular mass. The presence of a M+2 peak for bromine (ratio ~1:1) or chlorine (~3:1) helps identify halogens.
质谱中分子离子峰（M⁺）给出相对分子质量。溴的 M+2 峰存在（强度比约 1:1）或氯的（约 3:1）有助于识别卤素。
In IR spectroscopy, key absorptions include: O–H (alcohols) at 3200–3600 cm⁻¹ (broad), C=O at 1700–1750 cm⁻¹ (strong), and C–H at 2850–2980 cm⁻¹.
红外光谱中关键吸收：醇 O–H 在 3200–3600 cm⁻¹（宽峰），C=O 在 1700–1750 cm⁻¹（强峰），C–H 在 2850–2980 cm⁻¹。
For ¹H NMR, the number of signals equals the number of chemically distinct hydrogen environments. Integration gives the ratio of hydrogen atoms in each environment, and splitting (n+1 rule) reveals the number of adjacent hydrogens.
在 ¹H NMR 中，信号数目等于化学环境不同的氢原子种类数；积分给出各类氢原子数目之比；裂分（n+1 规则）揭示相邻氢的数目。
Exam tip: always combine molecular formula, mass spec, and NMR data systematically. Start with the molecular ion and degree of unsaturation, then assign fragments.
考试提示：始终结合分子式、质谱和 NMR 数据系统分析。先确定分子离子和不饱和度，再归属碎片。
12. Key Exam Strategies for Organic Chemistry | 有机化学核心应试策略
Success in ESAT organic questions relies on accuracy, reasoning, and speed. Here are targeted strategies for maximum marks.
要在 ESAT 有机题中取得高分，需具备准确性、逻辑推理和速度。以下策略助你拿满分数。
Memorise reaction conditions precisely: catalysts, solvents, temperature, and light are often the only clue to the correct product. For example, aqueous NaOH gives substitution while alcoholic NaOH gives elimination; UV light indicates radical reactions.
准确记忆反应条件：催化剂、溶剂、温度和光照往往是判断产物的唯一线索。例如 NaOH 水溶液对应取代，NaOH 醇溶液对应消除；紫外光暗示自由基反应。
Draw mechanistic arrows carefully. In ESAT, you may be asked to predict intermediates or rate-determining steps. Always show curly arrows from the nucleophile/electrophile to the electron-deficient atom.
仔细画机理箭头。ESAT 可能要求预测中间体或决速步。务必从亲核体/亲电体指向缺电子原子画弯箭头。
Use the degree of unsaturation formula: D.o.U = (2C + 2 + N − H − X)/2. This tells you the number of double bonds or rings, which greatly narrows down structural possibilities.
善用不饱和度公式：D.o.U = (2C + 2 + N − H − X)/2。它告诉你双键或环的数目，迅速缩小结构范围。
Keep a concise reaction map in your revision notes linking functional groups: alkane → haloalkane → alcohol → aldehyde/carboxylic acid → ester. This helps you answer interconversion questions.
在复习笔记中建立简洁的反应转化图：烷烃 → 卤代烃 → 醇 → 醛/羧酸 → 酯。这有助于应对相互转化类题目。
Finally, always check valency and charge balance in equations. A missing hydrogen or incorrect charge is a common source
Published by TutorHao | A-Level Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply