OCR有机合成 卤代烃 醇 羰基 羧酸酯 路线
有机合成是 OCR A Level Chemistry A（H432）里少数能把整本有机化学串成一条线的题型。试卷 H432/02 的名称就叫 Synthesis and analytical techniques，一道题可以从某个醇出发，要求你写出四步反应的试剂、条件、中间体和最终产物。只背反应清单在这里帮不上太多忙，因为题目真正考的是从目标分子往回推的能力。下面按官能团转化的顺序，把 specification 里 4.2.3 与 6.2.5 要求的合成路线拆开讲，并给出每一步在答题纸上必须写到的关键词。
Organic synthesis is one of the few topics in OCR A Level Chemistry A (H432) that pulls the whole organic course into a single line of reasoning. Paper H432/02 is called Synthesis and analytical techniques, and a single question can start from an alcohol and ask for four steps, marking the reagent, the conditions, the intermediate and the final product separately. Memorising a list of reactions is not enough here. The paper tests whether you can work backwards from the target molecule, then justify every step with the correct reagent and condition. This article follows the functional group sequence in the specification, particularly learning outcomes 4.2.3 and 6.2.5, and collects the exact wording that earns marks on the answer line.
合成路线的起点是倒推，不是正推
很多学生以为合成题必须从原料一路想到产物，结果走到第二步就卡住。更省力的做法是先看目标分子上有哪些官能团，再判断最后一步最可能是什么反应。目标里有酯基，最后一步往往是酯化，或者酰氯与醇的反应；目标里有胺基，最后一步往往从腈或卤代烃得到；目标比原料多一个碳，那一步只能是腈的水解，或者羰基与 HCN 的加成。H432 的 6.2.4 专门讲碳碳键形成，考官在这里用来用去就是两种素材：卤代烃与 KCN 生成腈，羰基化合物与 HCN 生成羟腈。倒推到原料之后再补齐每一步的条件，比从头硬推要快得多，也更不容易在中间某一步凭空造出一个不存在的反应。
The efficient entry point into a synthesis question is the functional groups on the target molecule, not the starting material. Work out the most likely final step first, then walk backwards. A target containing an ester group usually comes from esterification, or from an acyl chloride reacting with an alcohol. A target containing an amine group usually comes from a nitrile or a haloalkane. A target with one carbon more than the starting material must involve either hydrolysis of a nitrile or nucleophilic addition of HCN to a carbonyl. OCR keeps carbon-carbon bond formation (6.2.4) deliberately narrow, so those two tools cover almost every route you will meet. Once the sequence is fixed, adding the conditions is straightforward, and you avoid inventing a reaction that does not exist in the specification.
官能团转化图：从烯到酯的主干道
把 H432 要求掌握的转化画成一条主干道，大致是：烯通过 HBr 或 HCl 加成得到卤代烃；卤代烃用 NaOH 水溶液加热水解得到醇；醇在酸性重铬酸钾(VI) 下氧化，一级醇先变醛、继续氧化变羧酸；羧酸与醇在浓硫酸催化下回流得到酯；腈水解得到羧酸，腈还原得到一级胺。这条线上每个箭头都要能写出反应名称、试剂和条件，缺一项就少一分。真正需要记的不是箭头本身，而是箭头旁括号里的那几个字：水溶液还是乙醇溶剂、加热还是回流、蒸馏还是分液。把这些字写对了，路线题的一半分数就到手了。
Draw the OCR functional group conversions as one main road and the route becomes easier to hold in your head. An alkene with HBr or HCl gives a haloalkane. That haloalkane hydrolyses to an alcohol in aqueous sodium hydroxide with heating. The alcohol is oxidised by acidified potassium dichromate(VI): a primary alcohol gives an aldehyde first, and further oxidation gives a carboxylic acid. Carboxylic acids and alcohols are heated under reflux with concentrated sulfuric acid to make an ester. Nitriles hydrolyse to carboxylic acids, and reduction of a nitrile gives a primary amine. Every arrow on this road needs a reaction name, a reagent and a condition. The marks usually sit in the small print beside the arrow rather than in the arrow itself: aqueous or ethanolic solvent, warming or reflux, distillation or use of a separating funnel.
卤代烃：取代还是消去，取决于溶剂
常见的错误是把 NaOH 和 KOH 的醇溶液混为一谈。卤代烃与 NaOH 水溶液一起加热，走的是取代，OH 进攻带部分正电荷的碳，生成醇。同样的卤代烃与热的 KOH 乙醇溶液反应，走的是消去，脱掉一分子 HX 生成烯。溶剂不是可有可无的细节，水能稳定生成的离子，让亲核试剂充分溶剂化；乙醇让碱性更强而亲核性较弱的 KOH 更容易拔掉 β 位氢。考试里常把 NaOH 水溶液写成 NaOH 溶液当作干扰项，看清括号再动笔。卤代烃与 KCN 的乙醇溶液回流得到腈，碳链延长一个碳；与过量氨的乙醇溶液在密封管中加热得到胺，因为生成的胺会继续与卤代烃反应，氨必须过量才能把反应压在伯胺这一步。
A common error is to treat aqueous NaOH and ethanolic KOH as interchangeable. Heating a haloalkane with aqueous sodium hydroxide gives substitution: the hydroxide ion attacks the partially positive carbon and an alcohol is formed. The same haloalkane with hot ethanolic potassium hydroxide eliminates HX and gives an alkene instead. The solvent is not a detail. Water stabilises the ions formed and solvates the nucleophile, while ethanol favours the stronger but less nucleophilic hydroxide, which pulls a hydrogen off the carbon next to the halogen. Examination papers frequently write sodium hydroxide solution without brackets as a distractor, so read the state symbols before you write anything. Refluxing a haloalkane with KCN in ethanol extends the chain by one carbon and gives a nitrile, while heating it with excess ammonia in ethanol in a sealed tube gives a primary amine, because the amine produced can react again with the haloalkane.
醇的氧化：蒸馏还是回流，决定拿到醛还是酸
一级醇被酸性重铬酸钾(VI) 氧化时先生成醛，而醛很容易继续被氧化成羧酸。想让反应停在醛，必须把生成的醛及时蒸出，装置用蒸馏，收集沸点较低的馏出液。想拿到羧酸，就把混合物加热回流一段时间，装置是带垂直冷凝管的全回流，让醛留在体系里继续氧化。溶液颜色从橙色变成绿色，是 Cr(VI) 被还原为 Cr(III) 的证据，实验题常要求写出这个现象。二级醇氧化得到酮，酮在这个条件下不再被氧化，所以二级醇不必考虑过氧化的问题；三级醇没有可以被夺走的 α 氢，不能与酸性重铬酸钾(VI) 反应，这一点常用来区分三个同级醇的样品。
Oxidising a primary alcohol with acidified potassium dichromate(VI) gives an aldehyde first, and that aldehyde is easily oxidised further to a carboxylic acid. To stop at the aldehyde you must distil the product out as it forms, collecting the lower boiling fraction as it leaves the flask. To reach the carboxylic acid, heat the mixture under reflux for a period so the aldehyde stays in the flask and is oxidised again. The colour change from orange to green is evidence that chromium(VI) has been reduced to chromium(III), and practical questions often ask for exactly that observation. A secondary alcohol gives a ketone, which is not oxidised further under these conditions, so over-oxidation is not a concern. A tertiary alcohol has no hydrogen on the carbon bearing the OH group and does not react with acidified potassium dichromate(VI), which is a handy way to tell three isomeric alcohols apart.
羰基的鉴别与还原：从 2,4-DNP 到银镜
2,4-二硝基苯肼（2,4-DNP，也叫 Brady 试剂）与醛、酮都生成橙黄色沉淀，所以它只能证明样品里有羰基，不能区分醛和酮。区分要靠氧化性试剂：Tollens 试剂与醛在温水浴中生成银镜，Fehling 溶液与醛生成砖红色沉淀，酮对两者都不反应。把沉淀滤出后测熔点，与文献值对照，就能确定是哪一个羰基化合物，这一步在 H432/02 的结构推断题里经常出现。还原方面，NaBH4 在甲醇或水中就能把醛还原成一级醇、把酮还原成二级醇，条件比 LiAlH4 温和，是合成题里最常出现的还原剂。带 CH3CO 片段的甲基酮和乙醛与碱性碘溶液加热会生成黄色碘仿沉淀，这个现象可以用来确认结构片段。
2,4-Dinitrophenylhydrazine, usually shortened to 2,4-DNP or Brady reagent, gives an orange-yellow precipitate with both aldehydes and ketones. That confirms a carbonyl group but cannot tell the two apart. Oxidising agents do the separating: Tollens reagent gives a silver mirror with an aldehyde in a warm water bath, Fehling solution gives a brick-red precipitate with an aldehyde, and a ketone reacts with neither. Filtering the precipitate and comparing its melting point with a data book value identifies which carbonyl compound it is, and this step appears regularly in structure determination questions on H432/02. For reduction, sodium borohydride in methanol or water converts an aldehyde to a primary alcohol and a ketone to a secondary alcohol under much milder conditions than lithium aluminium hydride, which is why it is the reducing agent most often quoted in route questions. Methyl ketones and ethanal, both containing a CH3CO unit, give a yellow precipitate of triiodomethane with alkaline iodine solution, a test that confirms that particular structural fragment.
羧酸与酯：酰氯为什么比酯化更好用
酯化反应是可逆的，醇和羧酸在浓硫酸催化下回流，产率通常不高，要靠过量的一种反应物或者移走产物来推动平衡移动。酰氯与醇的反应不可逆，常温下就能进行，副产物是氯化氢气体，在水面上形成白雾，所以它更适合放在合成路线的最后一步，用来稳稳地"锁住"酯基。酸酐与醇的反应介于两者之间，PAG6 里合成阿司匹林用的就是乙酐与 2-羟基苯甲酸的反应，产率比直接酯化高。判断该写哪个试剂，看题目给的条件：出现浓硫酸和加热回流就是酯化，出现室温、白雾或者需要干燥装置就是酰氯。酯的碱性水解是不可逆的，皂化反应就是靠这一点把酯彻底拆开。
Esterification is reversible. Heating an alcohol with a carboxylic acid under reflux with concentrated sulfuric acid usually gives a modest yield, and the position of equilibrium has to be shifted by using one reactant in excess or by removing a product. Acyl chlorides react with alcohols irreversibly at room temperature and release hydrogen chloride, which forms a white mist in moist air, so an acyl chloride is the better choice when the ester has to be made cleanly at the end of a route. Acid anhydrides sit between the two, and the aspirin synthesis in PAG6 uses ethanoic anhydride with 2-hydroxybenzoic acid precisely because it gives a better yield than direct esterification. Read the conditions to decide which reagent to write: concentrated sulfuric acid with heating under reflux means esterification, while room temperature, a white mist or a drying tube means an acyl chloride. Alkaline hydrolysis of an ester is irreversible, which is why saponification can break an ester apart completely.
腈：一步延长碳链，两个方向变酸或变胺
卤代烃与 KCN 的取代把碳链延长一个碳，得到的腈是一个分叉口。用稀酸回流水解，氰基变成羧基，得到一个比原料多一个碳的羧酸；用 LiAlH4 还原或者催化加氢，氰基变成伯胺基，得到一个多一个碳的一级胺。羰基化合物与 HCN 的加成走另一条路，醛或酮与 HCN 生成羟腈，水解后得到 2-羟基酸，这也是把醛酮变成酸的一条常用通道。羟腈的生成是亲核加成，CN- 先进攻羰基碳，氧上的负电荷随后被质子化，机理题里这几步箭头要画准，电荷和孤对电子都不能漏。HCN 有毒并且易挥发，实验中常用 KCN 加稀酸现场生成，处理危险试剂类题目时这条信息会派上用场。
Substitution of a haloalkane by KCN in ethanol lengthens the carbon chain by one atom, and the nitrile formed is a fork in the road. Refluxing it with dilute acid hydrolyses the nitrile group to a carboxylic acid with one carbon more than the starting material. Reducing it with lithium aluminium hydride or by catalytic hydrogenation converts the same group into a primary amine, again one carbon longer. Addition of HCN to a carbonyl compound is a different branch: an aldehyde or ketone gives a hydroxynitrile, which hydrolyses to a 2-hydroxy acid, and this is a standard way of turning a carbonyl into an acid. The formation of a hydroxynitrile is nucleophilic addition, so the mechanism must show the cyanide ion attacking the carbonyl carbon first and the resulting negative charge on oxygen being protonated afterwards, with every charge and lone pair accounted for. Hydrogen cyanide is toxic and volatile, so practical work usually generates it in situ from potassium cyanide and dilute acid, which is worth remembering in any question about safety or hazard control.
答题模板：试剂、条件、产物、现象
阅卷时最常丢分的地方不是反应写错，而是条件写得不完整。一个合格的合成答句至少有三层信息：加什么试剂、在什么条件下、生成什么。只写"氧化乙醇得到乙醛"拿不到条件分，写成"乙醇与酸性重铬酸钾(VI) 加热蒸馏，收集乙醛"才完整。遇到需要证明产物结构的题，还要补一个现象，例如银镜、砖红色沉淀、使溴水褪色、pH 试纸变色。路线题里如果同一种试剂要做两件事，比如稀硫酸既作腈水解的催化剂又作酯化的催化剂，那就分别写清每一步的浓度和温度，不要让阅卷人替你做判断。写多步路线时把中间体单独写在箭头上方，比揉在一句话里更容易检查。
The marks in a synthesis question are lost on incomplete conditions more often than on wrong chemistry. A complete answer line carries three pieces of information: which reagent is added, under what conditions, and which product is formed. Writing that ethanol is oxidised to ethanal earns nothing for the conditions. Writing that ethanol is heated with acidified potassium dichromate(VI) and the ethanal is distilled off as it forms earns the whole mark. When the question also asks you to confirm a structure, add an observation such as a silver mirror, a brick-red precipitate, decolourisation of bromine water or a change in pH paper colour. If the same reagent does two jobs in one route, for example dilute sulfuric acid acting as the catalyst for both nitrile hydrolysis and esterification, state the concentration and temperature for each step separately rather than leaving the examiner to guess. Writing each intermediate above its arrow keeps a multi-step route readable and much easier to check.
后处理：分液、干燥、重结晶与产率
合成的最后一步常常是提纯。液体产物先倒入分液漏斗，用碳酸钠溶液洗掉残留的酸，静置分层后取有机层，再用无水硫酸钠或无水硫酸镁干燥，最后蒸馏收集沸点正确的馏分。PAG5 里由环己醇制环己烯、以及醇与浓盐酸制卤代烃，走的都是这几步。固体产物用重结晶：用最少量的热溶剂溶解粗产物，趁热过滤掉不溶杂质，缓慢冷却让晶体析出，抽滤并用少量冷溶剂洗涤，杂质留在母液里。纯度用熔点或沸点核对，含有杂质的固体熔点会下降而且熔程变宽。产率用实际产量除以理论产量再乘百分之百计算，算之前先判断哪一种反应物是限量试剂，这一步错了后面的数字全错。
Purification is normally the last stage of a preparation. A liquid product goes into a separating funnel, is washed with sodium carbonate solution to remove residual acid, and the organic layer is taken after the layers settle. It is then dried over anhydrous sodium sulfate or magnesium sulfate and distilled, collecting the fraction with the correct boiling point. The preparation of cyclohexene from cyclohexanol in PAG5, and the preparation of a haloalkane from an alcohol with concentrated hydrochloric acid, both follow this sequence. A solid product is recrystallised instead: dissolve the crude solid in the minimum volume of hot solvent, filter hot to remove insoluble impurities, cool slowly so crystals form, then filter under reduced pressure and wash with a little cold solvent while the impurities stay in the mother liquor. Purity is checked against a melting point or boiling point, and an impure solid melts at a lower temperature over a wider range. Percentage yield is the actual mass divided by the theoretical mass, multiplied by one hundred, and the theoretical value depends on identifying the limiting reagent correctly first.
六周复习安排
第1周先把 H432 的官能团转化图默写一遍，每一步都写上试剂和条件，写完对照 specification 补漏。第2周专攻卤代烃和醇，重点是取代、消去、氧化三条路径在溶剂和温度上的差别。第3周做羰基化合物，把 Tollens、Fehling、2,4-DNP 和碘仿四个鉴别试验的现象整理在同一页上。第4周做羧酸衍生物与腈，练两道多步合成题，每道题先用倒推法写一遍，再用正推核对。第5周回到实验部分，把 PAG5 与 PAG6 的装置图、提纯步骤和产率计算默写出来。第6周做 H432/02 的历年合成与分析题目，限时完成，把写错条件的题目抄进错题本，考前只看这本。
A six-week plan works better than reading the topic repeatedly. In week one, write out the H432 functional group map from memory with a reagent and a condition on every arrow, then check it against the specification and fill the gaps. Week two covers haloalkanes and alcohols, focusing on how solvent and temperature separate substitution, elimination and oxidation. Week three is for carbonyl compounds: put the observations for Tollens reagent, Fehling solution, 2,4-DNP and the triiodomethane test on a single page and learn them together. Week four moves to carboxylic acid derivatives and nitriles, with two multi-step routes done backwards first and checked forwards afterwards. Week five returns to the practical work, writing out the apparatus diagrams, purification steps and yield calculations for PAG5 and PAG6. Week six is timed past-paper practice on H432/02 synthesis and analysis questions, and every condition you got wrong goes into a mistakes notebook that you read the night before the exam.
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