IGCSE有机化学 同系物 烯烃 醇 聚合物
Cambridge IGCSE Chemistry 0620 的 Topic 11 叫 Organic chemistry，从 11.1 到 11.8 一共八个小节，覆盖通式、命名、燃料、烷烃、烯烃、醇、羧酸与聚合物。Paper 3 与 Paper 4 各 80 分，有机题常连着出两三个小问，画显示式与写反应条件几乎每份卷子都出现。Core 只要求甲烷、乙烷、乙烯、乙醇与乙酸这几个分子的画法与辨认，Extended 还要写同分异构体、酯与缩聚物的重复单元。很多学生以为有机化学就是背反应式，其实阅卷时扣得最多的是显示式漏画氢原子和通式用错。下面按八个板块拆开讲，每一块都给出考纲原句与考场上真正会被检验的细节。
Topic 11 of Cambridge IGCSE Chemistry 0620 runs from 11.1 to 11.8 and covers general formulae, naming, fuels, alkanes, alkenes, alcohols, carboxylic acids and polymers. Organic questions usually arrive as two or three linked parts on Papers 3 and 4, each worth 80 marks, and drawing a displayed formula or quoting a reaction condition turns up in almost every session. Core candidates only name and draw methane, ethane, ethene, ethanol and ethanoic acid, while Extended candidates also handle structural isomers, esters and repeat units.
卷面结构与 Topic 11 的八个板块
0620 的考生走 Core 或 Extended 两条路线：Core 考 Paper 1 与 Paper 3，成绩上限是 C；Extended 考 Paper 2 与 Paper 4，最高可以拿 A*。两卷选择题都是 45 分钟 40 道四选一、各占 30%，两卷理论题都是 1 小时 15 分钟 80 分、各占 50%。剩下 20% 是实验技能，Paper 5 是 1 小时 15 分钟 40 分的实验考，Paper 6 是 1 小时 40 分的笔试替考，报考时二选一。AO 权重上，AO1 知识理解占 50%、AO2 信息处理与解题占 30%、AO3 实验技能占 20%，Papers 1 到 4 内部 AO1 与 AO2 的比例是 63 比 37。Cambridge 官方 syllabus 写明 0620 适用于 2026、2027 与 2028 年考试，版本 1 在 2023 年 9 月发布，官方注明没有影响教学的重大变化。Topic 11 Organic chemistry 的编号是 11.1 到 11.8，依次是通式与官能团、命名、燃料、烷烃、烯烃、醇、羧酸与聚合物。阅卷时最容易丢分的是把选择题的记忆方式带到大题上，只写结论不写条件。
Core candidates sit Paper 1 with Paper 3 and stop at grade C, while Extended candidates sit Paper 2 with Paper 4 and can reach A*; both multiple-choice papers last 45 minutes for 40 marks and 30%. The theory papers each run 1 hour 15 minutes for 80 marks and 50%, and the remaining 20% comes from one practical paper chosen at entry, either the laboratory test or the written Alternative to Practical. Marking weights for the whole qualification are AO1 at 50%, AO2 at 30% and AO3 at 20%, with Papers 1 to 4 split 63 to 37 between AO1 and AO2.
同系物与四个必背通式
同系物 homologous series 的定义是 a family of similar compounds with similar chemical properties due to the presence of the same functional group，考纲把它的五个共同特征列在一处：同一个官能团、同一个通式、相邻成员相差一个 -CH2- 单元、物理性质呈现趋势、化学性质相似。必背的四个通式是烷烃 CnH2n+2、烯烃 CnH2n、醇 CnH2n+1OH、羧酸 CnH2n+1COOH。用 n=2 代进去验算：乙烷 C2H6、乙烯 C2H4、乙醇 C2H5OH、乙酸 CH3COOH 都落在对应通式里。常见的错误是把醇的通式写成 CnH2n+2O，代数上等价，但不是考纲给出的写法，遇到直接要求写通式的题目会被判错。物理性质的趋势只需记住方向：碳链越长，沸点越高、挥发性越低，原因是分子间作用力随链长增加。填表题里最常见的追问是相邻两个成员差多少，答案是 14 个质量单位，也就是一个 CH2。
The syllabus defines a homologous series as a family of similar compounds with similar chemical properties due to the presence of the same functional group, then lists five shared features covering the functional group, the general formula, the -CH2- unit between neighbours, a trend in physical properties and similar chemical behaviour. Four general formulae have to be written from memory, so alkanes are CnH2n+2, alkenes are CnH2n, alcohols are CnH2n+1OH and carboxylic acids are CnH2n+1COOH; substituting n=2 returns ethane C2H6, ethene C2H4, ethanol C2H5OH and ethanoic acid CH3COOH. Longer chains boil higher and evaporate less readily because the forces between molecules grow with chain length.
官能团、饱和与不饱和
官能团的定义是 an atom or group of atoms that determine the chemical properties of a homologous series，也就是决定这一族化合物化学性质的原子或原子团。烷烃只有碳碳单键与碳氢键，习惯上说它没有单独的官能团；烯烃的官能团是 C=C 双键，醇是 -OH，羧酸是 -COOH。饱和与不饱和的官方口径分两句：饱和化合物的分子里所有碳碳键都是单键，不饱和化合物的分子里有一个或多个碳碳键不是单键。乙烯含双键所以是不饱和烃，溴水能让它褪色，乙烷不能，这是判分时最常被引用的对照。很多学生以为不饱和指的是还能再溶解别的物质，官方只认可碳碳键这一层意思，答题时照抄单键这个措辞最稳。显示式 displayed formula 要求把每个原子与每根键都画出来，乙烷是两个碳之间一根短横线、每个碳再挂三根氢线。
A functional group is defined as an atom or group of atoms that determine the chemical properties of a homologous series, which for alkenes means the carbon to carbon double bond, for alcohols the -OH group and for carboxylic acids the -COOH group. Saturated compounds have every carbon to carbon bond as a single bond, while unsaturated compounds have at least one that is not; that single distinction explains why ethene reacts with bromine water and ethane does not. A displayed formula must show every atom and every bond, so ethane is drawn as two carbons joined by a line with three hydrogen lines on each.
结构式与同分异构体
结构式 structural formula 是 an unambiguous description of the way the atoms in a molecule are arranged，考纲直接给了三个可以背下来的例子：CH2=CH2、CH3CH2OH 与 CH3COOCH3。同分异构体 structural isomers 的定义是 compounds with the same molecular formula, but different structural formulae，考纲只举两组：C4H10 可以写成 CH3CH2CH2CH3 或 CH3CH(CH3)CH3，C4H8 可以写成 CH3CH2CH=CH2 或 CH3CH=CHCH3。这两组要求会画也会认，分子式相同而连接方式不同，物理性质会有差别，正丁烷的沸点约 -0.5 摄氏度，异丁烷约 -12 摄氏度。常见的错误是把分子式相同、结构也相同的写法抄两遍充数，比如把同一个直链结构正着写一遍、反着写一遍当成两种。这一步最容易失分的地方是画结构式时漏掉括号，CH3CH(CH3)CH3 少了那对括号就变成另一个分子。
A structural formula is an unambiguous description of the way the atoms in a molecule are arranged, and the syllabus supplies CH2=CH2, CH3CH2OH and CH3COOCH3 as standard examples. Structural isomers share a molecular formula but differ in structural formula, so C4H10 appears as either CH3CH2CH2CH3 or CH3CH(CH3)CH3, and C4H8 as either CH3CH2CH=CH2 or CH3CH=CHCH3; the straight chain isomer boils near -0.5 degrees Celsius and the branched one near -12. Dropping the bracket in CH3CH(CH3)CH3 produces a completely different molecule rather than a shorthand.
命名规则的后缀信号
命名规则本质上是一套后缀信号：-ane 是烷烃、-ene 是烯烃、-ol 是醇、-oic acid 是羧酸，考纲要求看到名字或分子式就能说出它属于哪一类。Core 层只要求甲烷 methane、乙烷 ethane、乙烯 ethene、乙醇 ethanol、乙酸 ethanoic acid 这五个分子的画法与命名，外加 11.4 到 11.7 中所有反应产物的画法。Supplement 把范围扩到不带支链的烷烃、烯烃（含 but-1-ene 与 but-2-ene）、醇（含 propan-1-ol、propan-2-ol、butan-1-ol、butan-2-ol）与四个碳以内的羧酸，并要求能画出由四个碳以内的醇与酸生成的酯。名字里的数字要看双键或羟基离哪一端的碳更近，but-1-ene 的双键在第一与第二碳之间，but-2-ene 在第二与第三碳之间，两者结构不同，性质也不同。学生最常把 propan-1-ol 与 propan-2-ol 画成同一个结构，区别只在羟基挂在链端还是挂在中间的碳上。
Naming runs on a set of suffixes, so -ane marks an alkane, -ene an alkene, -ol an alcohol and -oic acid a carboxylic acid, and the syllabus expects the class to be recognised from a name, a molecular formula or a displayed formula. Core work is limited to methane, ethane, ethene, ethanol and ethanoic acid plus the products of the reactions in sections 11.4 to 11.7; the Supplement widens that to unbranched alkanes, alkenes such as but-1-ene and but-2-ene, alcohols such as propan-1-ol, propan-2-ol, butan-1-ol and butan-2-ol, carboxylic acids of up to four carbons, and the esters they form. The number in the name records which end the double bond or the -OH group is nearest, which is the only difference between propan-1-ol and propan-2-ol.
石油分馏与八个馏分的用途
石油 petroleum 是烃的混合物，而 hydrocarbon 的定义是只含碳与氢两种元素的化合物，天然气的主要成分是甲烷，三种化石燃料是煤、天然气与石油。分馏 fractional distillation 利用沸点差异把石油切成八个馏分，从塔底往塔顶的规律有四个：碳链变短、挥发性升高、沸点降低、黏度降低。八个馏分与用途要对上：refinery gas 用于加热与做饭，gasoline 或 petrol 作汽车燃料，naphtha 是化工原料，kerosene 或 paraffin 作航空燃油，diesel oil 或 gas oil 供柴油发动机，fuel oil 供船舶与家庭供暖，lubricating oil 做润滑油、蜡与抛光剂，bitumen 用来修路。常见的错误是把 kerosene 的用途写成汽车燃料，汽车烧的是 gasoline，航空煤油才是 kerosene。分馏塔里温度从底到顶是往下走的，bitumen 这类重馏分沸点最高所以留在塔底。阅卷时按馏分名与用途配对给分，只写中文用途不写馏分名通常只给一半。
Petroleum is a mixture of hydrocarbons, and a hydrocarbon contains carbon and hydrogen only, which places methane as the main constituent of natural gas and leaves coal, natural gas and petroleum as the three fossil fuels. Fractional distillation separates petroleum into eight fractions on the basis of boiling point, and moving from the bottom of the column to the top means shorter chains, higher volatility, lower boiling points and lower viscosity. Each fraction has a named use, from refinery gas for heating and cooking through gasoline, naphtha as a chemical feedstock, kerosene for jet fuel, diesel oil, fuel oil, lubricating oil for lubricants and waxes, down to bitumen for roads.
烷烃的取代反应与光化学条件
烷烃的碳碳键与碳氢键都是单键共价键，所以属于饱和烃，性质上考纲写的是 generally unreactive，只考两个反应：燃烧与氯气取代。取代反应 substitution 的定义是 one atom or group of atoms is replaced by another atom or group of atoms，甲烷与氯气的第一步是 CH4 + Cl2 -> CH3Cl + HCl。考纲明确写这个反应属于 photochemical reaction，紫外光提供活化能 Ea，Extended 层要写出这一点，并且只考单取代这一步。很多学生以为这个反应要加热，其实关键条件是紫外光，写加热不给分；把它写成催化剂同样错，氯气取代里没有催化剂。烷烃完全燃烧生成二氧化碳与水，不完全燃烧还多出炭黑与一氧化碳，这两组产物在环境与能源题里会再出现一次。阅卷时错得最多的是条件那一个空，学生把紫外线、光照与加热混着写。
Alkanes are saturated hydrocarbons because every carbon to carbon bond and carbon to hydrogen bond is single, and the syllabus describes them as generally unreactive apart from combustion and substitution by chlorine, where CH4 plus Cl2 gives CH3Cl and HCl. The reaction is photochemical, with ultraviolet light supplying the activation energy Ea, and only monosubstitution has to be drawn. Writing heat instead of ultraviolet light, or calling the light a catalyst, loses the mark every session.
裂解与烯烃的加成反应
烯烃含一个 C=C 双键，属不饱和烃，工业上靠裂解大分子烷烃来制取：条件是高温加催化剂，产物是更小的烷烃、烯烃与氢气。考纲要求说明裂解的理由，主要是把长链分子转成需求量更大的短链分子与烯烃，同时得到氢气。检验饱和与不饱和的标准方法是加溴水 aqueous bromine，烯烃把橙棕色褪去生成无色的二溴化合物，烷烃不反应、溴水保持原色。褪色是判分关键词，只写颜色变浅或者写发生了反应都拿不到分。Extended 层还要写三条加成反应：与溴或溴水加成、与氢气在镍催化下加成、与水蒸气在酸催化下加成，加成反应的特点是从头到尾只生成一种产物。常见的错误是把乙烯与溴的加成写成取代，取代会放出溴化氢，加成不会。裂解产出的乙烯与氢气，正好是接下来乙醇那条催化加水路线的原料。
Alkenes contain a carbon to carbon double bond and are unsaturated, and industry makes them by cracking larger alkanes at high temperature with a catalyst, which also releases hydrogen and smaller alkanes. The standard test for unsaturation is aqueous bromine, which an alkene decolourises as the orange-brown colour disappears and a colourless dibromo compound forms, while an alkane leaves the colour unchanged. Extended candidates also describe addition with bromine, with hydrogen over a nickel catalyst and with steam over an acid catalyst, each giving a single product.
醇的两条工业路线与氧化
乙醇的两条制取路线要连条件与数字一起背：发酵 fermentation 是葡萄糖水溶液在酵母作用下、25 到 35 摄氏度、无氧条件下进行；催化加水是乙烯与水蒸气在 300 摄氏度、6000 kPa（也就是 60 atm）与酸催化剂下进行。发酵的优点是原料可再生、设备要求低，缺点是速度慢、产物浓度低要蒸馏提纯；催化加水的优点是速度快、产物纯度高，缺点是依赖来自石油的乙烯、条件高压高温。Supplement 要求比较两条路线的优劣，把可再生与不可再生这层差别写出来才算答全。乙醇的用途考纲只列两个：溶剂与燃料。很多学生以为发酵要通气，官方写的是 in the absence of oxygen，通氧会把乙醇继续氧化成乙酸。乙醇还能被酸化的高锰酸钾氧化成乙酸，醋的生产也靠细菌氧化同一条路线，产物相同、条件不同。
Fermentation of aqueous glucose at 25 to 35 degrees Celsius with yeast and no oxygen competes with the catalytic addition of steam to ethene at 300 degrees Celsius, 6000 kPa and an acid catalyst. Fermentation uses a renewable feedstock and simple equipment but is slow and gives a dilute product, whereas the steam route is fast and pure but depends on ethene from petroleum and needs high pressure; that contrast is exactly what the Supplement asks to be compared. Ethanol is listed for two uses only, as a solvent and as a fuel, and it can be oxidised further to ethanoic acid by acidified potassium manganate(VII) or by bacterial oxidation during vinegar production.
羧酸的反应与酯化
乙酸 ethanoic acid 是弱酸，考纲要求写它与三类物质的反应，并给出盐的名称与化学式。与金属反应放出氢气，例如 Mg + 2CH3COOH -> (CH3COO)2Mg + H2；与碱反应生成盐与水，例如 NaOH + CH3COOH -> CH3COONa + H2O；与碳酸盐反应放出二氧化碳并生成盐与水，例如 Na2CO3 + 2CH3COOH -> 2CH3COONa + H2O + CO2。三类反应里只有碳酸盐那一条会冒气泡，也是唯一能用石灰水检验气体的一条。酯化反应是羧酸与醇在酸催化下生成酯与水，考纲给的例子是 CH3COOCH3，并要求画出来自四个碳以内的醇与酸生成的酯，乙酸乙酯的英文名是 ethyl ethanoate，中文里先写酸再写醇、英文正好相反。错得最多的是把酯当成酸，酯里没有能电离的氢，所以既不显酸性也不与碳酸盐反应。
Ethanoic acid reacts with metals to release hydrogen, with bases to give a salt and water, and with carbonates to give a salt, water and carbon dioxide, and the syllabus expects the names and formulae of the salts produced, such as (CH3COO)2Mg, CH3COONa and CO2. Only the carbonate reaction produces bubbles, which makes it the one case where limewater can be used as the test for the gas given off. Esterification joins a carboxylic acid to an alcohol under an acid catalyst, and the syllabus example CH3COOCH3 shows why an ester is not an acid at all, since it holds no ionisable hydrogen.
加聚与缩聚的重复单元
聚合物 polymer 的定义是 large molecules built up from many smaller molecules called monomers，加聚的典型例子是乙烯聚合成聚乙烯 poly(ethene)。重复单元 repeat unit 的画法是把单体的双键打开、两端各留一根自由键，考纲要求能从给定烯烃推出重复单元，也能从重复单元反推出单体。缩聚的两种组合要记住：二羧酸与二胺生成聚酰胺 polyamide，二羧酸与二醇生成聚酯 polyester，考纲要求画出尼龙与 PET 的结构，PET 的全名 polyethylene terephthalate 不必背。加聚与缩聚的差别在于缩聚每连一次会放出一个像水一样的小分子，加聚只生成一种产物。扩展内容里还有蛋白质，它是天然聚酰胺，由氨基酸单体缩合而成，链上反复出现 -NH-C(=O)- 这样的酰胺键。很多学生以为 PET 不可回收，考纲专门写明 PET 可以变回单体再聚合。
A polymer is a large molecule built up from many smaller molecules called monomers, and poly(ethene) made from ethene is the standard addition example, where the repeat unit is drawn by opening the double bond and leaving a free bond at each end. Condensation polymerisation is limited to polyamides from a dicarboxylic acid and a diamine and polyesters from a dicarboxylic acid and a diol, and nylon and PET both have to be drawn, although the full name polyethylene terephthalate is not required. Condensation releases a small molecule such as water at every linkage, whereas addition forms one product only; proteins are natural polyamides built from amino acid monomers.
塑料处理与四周复习排期
塑料带来的环境问题考纲只列三条：填埋占用土地、在海里累积并进入食物链、燃烧生成有毒气体，答题时按 disposal in landfill sites、accumulation in oceans、formation of toxic gases from burning 这三句原话写，笼统写污染环境拿不到分。四周复习可以这样排：第一周整理四个通式与四类官能团，把每个代表分子的显示式手画一遍；第二周做燃料与裂解的配对题，把八个馏分与用途写成两栏互测；第三周集中练取代与加成，把紫外光、镍催化、酸催化、300 摄氏度与 60 atm 这些条件逐条标在反应式旁边；第四周做两套完整的 Paper 3 或 Paper 4。阅卷时最容易丢的分集中在三处：显示式漏氢原子、反应条件写错、酯与酸混为一谈，把这三类错题抄到同一页反复看，比再刷一套卷子有用。
The syllabus limits the environmental challenges of plastics to three points, namely disposal in landfill sites, accumulation in oceans and the formation of toxic gases from burning, and those three phrases carry the marks better than a general statement about pollution. A four-week plan works well when week one covers the four general formulae and the functional groups, week two pairs the eight fractions with their uses, week three drills substitution and addition conditions, and week four takes two full theory papers under timed conditions. Marks are lost most often on a displayed formula missing a hydrogen, a wrong reaction condition, or an ester mistaken for an acid.
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