Edexcel IGCSE化学 烷烃 烯烃 醇 羧酸 聚合物
有机化学在 Edexcel IGCSE 化学里排在第 4 个主题，编号 4.1 到 4.50，跨八个子题。它的难点不在概念多，而在两份试卷的分工：带 C 编号的内容只在 Paper 2 出现，Paper 1 只考非加粗、无 C 编号的段落。下面的节标题按考纲顺序排，公式、温度与分值都对照 Pearson 官网现行 specification（Issue 3，2024 年 9 月）核对过。
The organic chemistry section of the Pearson Edexcel International GCSE in Chemistry runs from 4.1 to 4.50 across eight sub-topics, and the difficulty lies less in the ideas than in how the two papers divide the work. Statements carrying a C reference are examined only in Paper 2, while Paper 1 sticks to core content that is neither bold nor referenced, so the order below follows the specification itself. Figures, temperatures and mark allocations were checked against Issue 3 of September 2024.
C 标记决定两卷的复习顺序
Paper 1 的代码是 4CH1/1C，考试时间两小时，满分 110 分，占总成绩 61.1%。Paper 2 的代码是 4CH1/2C，时间一小时十五分钟，满分 70 分，占 38.9%。两卷合计 180 分，按总分给 9 到 1 的等级，不分层，也不折算。Paper 1 只考不加粗、没有 C 编号的内容，Paper 2 覆盖全部内容，加粗与 C 编号的部分会考得更深。有机化学里，(a) 到 (d) 与 (h) 前半段的编号不带 C，两卷都可能出现。(e) 醇、(f) 羧酸、(g) 酯，以及 (h) 里的缩聚部分全部带 C，只在 Paper 2 出现。11 月与 6 月都有考试安排，两卷都允许使用计算器。很多学生以为两卷只是题量差别，按 110 比 70 分配时间，结果把醇与酯压到最后一轮才看。
Paper 1C runs for two hours and carries 110 marks at 61.1% of the qualification, while Paper 2C lasts one hour and fifteen minutes for 70 marks and 38.9%. Both are sat in the same series and feed a single 180-mark total with one 9 to 1 grade, and calculators are allowed in each; the crucial point for revision planning is that alcohols, carboxylic acids, esters and condensation polymers carry a C reference and therefore appear only in Paper 2. Treating the two papers as a simple matter of length is where many candidates misjudge the workload.
五种式子与同分异构的写法
4.2 要求能写经验式、分子式、通式、结构式与显示式五种表示法。4.5 反过来要求由分子式画出所有可能的结构式与显示式。4.3 把同系物、官能团与同分异构三个术语并列，4.4 要求按 IUPAC 规则命名，碳原子数上限是六个。4.6 只要求把有机反应归类为取代、加成与燃烧，明确写了不需要反应机理。C4H10 有两个同分异构体，正丁烷与 2 甲基丙烷，画显示式时每一根碳氢键都要画出来。分子式相同的两种物质性质不同，这一问常被当成送分题，实际失分不少。常见的错误是把分子式当结构式，或者显示式里只画碳骨架、把氢省掉。评分时显示式少画一根键，那一空就是零分。
Statements 4.2 and 4.5 ask for five ways of representing a molecule and for drawing every structural and displayed formula that fits a given molecular formula, with IUPAC naming capped at six carbon atoms. Isomerism sits alongside homologous series and functional groups in 4.3, and 4.6 only requires reactions to be classified as substitution, addition or combustion, since mechanisms are explicitly out of scope. C4H10 gives two isomers, butane and 2-methylpropane, and a displayed formula that omits a single bond scores nothing, so draw hydrogen out in full instead of relying on a skeletal sketch.
分馏塔里六个馏分的沸点与黏度
4.8 要求描述工业分馏怎样把原油分成不同馏分。4.9 给出六个名称：炼厂气、汽油、煤油、柴油、燃料油与沥青。4.10 要求掌握颜色、沸点与黏度三个趋势。碳链越长，沸点越高，黏度越大，颜色越深，从塔顶的炼厂气到塔底的沥青依次排列。分馏塔内温度由下往上递减，沸点高的馏分在塔底冷凝，沸点低的从塔顶逸出。学生最常把趋势方向记反，以为黏度大意味着更易流动。把六个馏分按碳链长度写在一条竖线上，颜色与黏度跟着一起记，比单背名词省时间。考题常让判断用途，柴油对应卡车，沥青对应路面。
Fractional distillation separates crude oil into six named fractions, from refinery gases at the top of the column to bitumen at the bottom, and 4.10 asks for the trend in colour, boiling point and viscosity rather than a list of names. Longer chains boil higher, flow less readily and look darker, while the temperature inside the column falls towards the top so that the heaviest fraction condenses lowest. Writing the six fractions on a vertical line, with uses such as diesel for lorries and bitumen for roads beside them, stops the direction of the trend from being reversed under exam pressure.
完全燃烧与不完全燃烧的产物
4.11 把燃料定义为燃烧时释放热能的物质。4.12 要求写出完全燃烧与不完全燃烧的产物：氧气充足时生成二氧化碳与水，氧气不足时生成一氧化碳、碳粒与水。4.13 要求解释一氧化碳为什么有毒，说法落在它会降低血液输送氧气的能力，specification 写明不需要提到血红蛋白。4.14 指出汽车发动机内温度足够高，空气中的氮与氧会反应生成氮氧化物。4.15 与 4.16 把硫杂质生成的二氧化硫与氮氧化物一起连到酸雨。阅卷时最常见的失分是把不完全燃烧的产物写成二氧化碳，或者把碳粒整个漏掉。一氧化碳无色无味，这一点常在实验安全题里单独考。酸雨的化学名词要写 English 原文时统一用 acid rain。
4.12 asks for the products of complete and incomplete combustion, which are carbon dioxide and water when oxygen is plentiful and carbon monoxide, soot and water when it is not. Carbon monoxide is poisonous because it reduces the blood's capacity to carry oxygen, and the specification states that haemoglobin need not be mentioned, while nitrogen oxides form inside hot engines when nitrogen and oxygen from the air combine. Sulphur impurities give sulphur dioxide, and both gases end up in acid rain, so naming carbon dioxide as the product of incomplete combustion is a regular way to drop a mark.
催化裂化的温度 催化剂与供需
4.17 要求描述长链烷烃如何在催化裂化中变成烯烃与更短的烷烃。催化剂用二氧化硅或氧化铝，温度在 600 到 700 摄氏度之间。一个常见写法是 C10H22 -> C8H18 + C2H4，两边碳氢数都对得上。4.18 要求解释为什么需要裂化，答案落在供需：市场对重馏分需求小，对汽油与烯烃需求大。裂化把低需求的长链变成高需求的短链与烯烃，这一步是有机合成的原料入口。很多学生以为催化剂会被消耗，把它写进反应物一侧。催化剂只改变反应速率，不进入方程式。分数常丢在温度区间被写成大约数，600 到 700 这个范围要写全。
Catalytic cracking converts long-chain alkanes into shorter alkanes and alkenes over a silica or alumina catalyst at 600 to 700 degrees Celsius, and 4.18 frames the process as a response to the balance between supply and demand. Heavy fractions are in low demand while petrol and alkenes are wanted in quantity, so cracking turns an unwanted product into feedstocks for fuels and polymers. A line such as C10H22 -> C8H18 + C2H4 balances on both sides, and a catalyst is never written as a reactant because it is not consumed.
烷烃 饱和 与紫外光下的取代
4.19 给出烷烃通式 CnH2n+2。4.20 要求解释饱和的含义：碳原子之间只有单键，没有碳碳双键。4.21 要求画到五个碳原子的结构式与显示式，并能命名无支链异构体。4.22 把烷烃与卤素的反应限定为一取代，条件是紫外光，机理不考。典型方程式是 CH4 + Cl2 -> CH3Cl + HCl，紫外光写在箭头上方。4.6 把这类反应归为取代，因为氢原子被卤素原子换掉。很多学生以为烷烃完全惰性，实际上在紫外光下会持续被取代。名字与结构式对不上是另一个高频错误，甲烷、乙烷、丙烷的碳数要一一对应。
Alkanes follow CnH2n+2 and are described as saturated because every carbon-carbon bond is single, and 4.22 limits their reaction with halogens to mono-substitution in the presence of ultraviolet radiation. CH4 + Cl2 -> CH3Cl + HCl is the standard example, and the reaction counts as substitution because a hydrogen atom is replaced rather than added to. Drawing displayed formulae up to five carbon atoms, and matching each name to its carbon count, accounts for most of the marks this sub-topic carries.
烯烃 双键 加成 与溴水鉴别
4.23 把烯烃的官能团写成 >C=C<。4.24 给出通式 CnH2n，4.25 要求解释不饱和的含义。4.26 要求画到四个碳原子为止，明确不考顺反异构与 E/Z 标记。4.27 的典型反应是烯烃与溴加成生成二溴代烷，CH2=CH2 + Br2 -> CH2BrCH2Br。4.28 是溴水实验：烯烃使橙棕色溴水褪色，烷烃不反应。这个实验是鉴别题的标准答案，两个现象要对比着写。加成与取代的差别在于是否有原子被加进来而不失去任何原子。错得最多的是把褪色写成变浑浊，或者忘记说烷烃那一侧没有变化。
Alkenes contain the >C=C< functional group, follow CnH2n and are unsaturated because the double bond can open up to accept atoms, and 4.26 stops at four carbon atoms while excluding cis-trans and E/Z notation. Bromine adds across the double bond to give a dibromoalkane, and bromine water turns from orange-brown to colourless with an alkene while an alkane leaves it unchanged. That contrast, rather than the word decolourised on its own, is what the distinguishing test question rewards.
醇的三条氧化路线
4.29C 把醇的官能团定为 OH，4.30C 只要求甲醇、乙醇、丙醇与丁醇四种。4.31C 列了三条氧化路线：在空气或氧气中燃烧生成二氧化碳与水，在空气中被微生物氧化成乙酸，与稀硫酸中的重铬酸钾共热生成乙酸。三条路线的产物不同，第一条彻底氧化，后两条停在羧酸。重铬酸钾的颜色由橙色转绿，这是判断反应发生的直观证据。很多学生以为发酵与实验室氧化是同一件事，前者靠酵母中的酶，后者靠强氧化剂。写方程式时乙醇到乙酸要多一个氧，配平常被漏掉。这一节的 C 编号只在 Paper 2 出现。
Alcohols carry the OH group, and 4.31C lists three oxidations of ethanol: burning in air or oxygen, microbial oxidation to ethanoic acid, and heating with potassium dichromate(VI) in dilute sulphuric acid. The first goes all the way to carbon dioxide and water while the other two stop at the carboxylic acid, and the dichromate solution turning from orange to green is the visible sign that the third has happened. Fermentation and laboratory oxidation are not interchangeable, since one relies on yeast enzymes and the other on a strong oxidising agent.
乙醇的两条工业制法
4.32C 给出两条路线。催化水合把乙烯与蒸汽通过磷酸催化，温度约 300 摄氏度，压强约 60 到 70 个大气压。发酵用葡萄糖在无氧条件下进行，最适温度约 30 摄氏度，靠酵母中的酶。方程式是 C6H12O6 -> 2C2H5OH + 2CO2，两边碳氢氧都对得上。4.33C 要求解释为什么必须无氧、为什么有最适温度：有氧时乙醇会被继续氧化，温度过低反应太慢，过高酶会失活。阅卷时常见学生只写高温高压，不给具体数字，分值拿不满。两条路线的原料不同，一条来自石油化工，一条来自农产品。选哪条路线取决于当地原料与能源成本。
Ethanol is made industrially by hydrating ethene with steam over a phosphoric acid catalyst at about 300 degrees Celsius and 60 to 70 atmospheres, or by fermenting glucose with yeast enzymes at an optimum near 30 degrees Celsius in the absence of air. C6H12O6 -> 2C2H5OH + 2CO2 balances completely, and 4.33C asks why air must be excluded and why the temperature is held near the optimum: oxygen oxidises the ethanol further, cool mixtures react too slowly and heat deactivates the enzymes. Answers that say high temperature and pressure without figures rarely earn full marks.
羧酸 酯化 与乙酸乙酯
4.34C 把羧基写成 COOH，4.35C 要求画到四个碳原子的无支链羧酸并命名。4.36C 要求描述羧酸水溶液与金属、金属碳酸盐的反应，与碳酸盐反应会放出二氧化碳。4.37C 说明食醋就是乙酸的水溶液。4.38C 到 4.43C 转到酯：乙醇与乙酸在酸催化下生成乙酸乙酯，反应可逆，产物有特征香味。酯易挥发，用作食品香精与香水。4.43C 是必做实验，制备一份乙酸乙酯。很多学生以为酯化不需要催化剂，实际上缺少酸催化反应速率很慢。命名时先写醇来的烷基再写酸来的酸根，顺序写反就整空不得分。
Carboxylic acids contain COOH, react with metals and metal carbonates, and give carbon dioxide with the carbonates, while vinegar is simply an aqueous solution of ethanoic acid. Esters form when an alcohol and a carboxylic acid are heated with an acid catalyst, and ethyl ethanoate from ethanol and ethanoic acid is the standard example, reversible and recognisable by its smell. Named from the alcohol first and the acid second, esters are volatile and used as flavourings and in perfumes, and 4.43C requires the preparation of a sample in the laboratory.
加聚物 缩聚物 单体与重复单元
4.44 把加聚物定义为许多小分子单体连接成的产物。4.45 要求画聚乙烯、聚丙烯、聚氯乙烯与聚四氟乙烯的重复单元。4.46 是双向题：由重复单元推单体，或由单体写重复单元，重复单元两端要留出延伸的键。4.47 要求解释加聚物难处置的原因：化学惰性、不可生物降解、燃烧时产生有毒气体。4.48C 到 4.50C 转向缩聚：二元羧酸与二元醇生成聚酯，同时脱出一分子水。乙二酸与乙二醇是最常见的例子。学生最常把缩聚的副产物水漏掉，方程式就不平衡。部分聚酯属于可生物降解的生物聚酯，这一点是处置题的对比答案。
Addition polymers are built from monomers joined without losing any atom, and 4.45 requires repeat units for poly(ethene), poly(propene), poly(chloroethene) and PTFE, with two bonds left extending from the unit. 4.46 runs in both directions, from repeat unit to monomer and back, while 4.47 explains the disposal problem through inertness, resistance to biodegradation and toxic gases on burning. Condensation polymerisation joins a dicarboxylic acid to a diol and releases water, and forgetting that water is the commonest reason a polyester equation fails to balance.
塑料处置与六分题的作答落点
4.47 的三个理由要在答案里分开写，惰性、不可降解、燃烧产生有毒气体是三个不同的得分点。可生物降解的生物聚酯提供了一条处理路线，回收与焚烧是另外两条。作答前先看分值，再看命令词，describe 与 explain 的要求不同。有机化学在两卷都可能出六分开放题，六分题通常要求分段结构与具体化学名词。这一步最容易丢分的是只写污染环境，不给机制与化学名词。把每个子题的核心方程式抄在一起，考前只翻这一张纸。180 分的总分里，有机化学的比重随当年题目分布变化，没有固定配额。
Disposal arguments need three separate points, namely inertness, inability to biodegrade and toxic gases on burning, with biodegradable biopolyesters, recycling and incineration as the alternatives. Read the command word before writing, because describe and explain demand different levels of detail, and a six-mark open response usually expects a structured answer built on correct chemical vocabulary. Writing that the environment is polluted, with no mechanism and no named substance, is the most reliable way to lose marks on this topic.
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