📚 IB OCR Chemistry: Polymers Revision Guide | IB OCR 化学：聚合物 考点精讲
Polymers are the backbone of modern materials, from simple plastic bags to high‑performance car parts and even the biomolecules that make up life itself. This article condenses the key facts, mechanisms, structures and sustainability issues you need for IB and OCR A‑level Chemistry, pairing clear English explanations with matching Chinese paragraphs to accelerate your revision.
聚合物是现代材料的骨架，从普通的塑料袋到高性能汽车零件，再到构成生命本身的生物分子，无处不在。本文浓缩了IB和OCR A‑level化学中必须掌握的关键事实、反应机理、结构以及可持续发展议题，用清晰的英文解释配上对应的中文段落，帮助大家高效复习。
1. Introduction to Polymers | 聚合物简介
A polymer is a long‑chain molecule made by linking together many small repeating units called monomers. The process that builds them is polymerisation. If the polymer chain is built from just one type of monomer it is a homopolymer; chains containing two or more different monomers are copolymers.
聚合物是由许多称为单体的小重复单元连接而成的长链分子。形成聚合物的过程叫作聚合反应。如果链由单一类型的单体构成，则为均聚物；含有两种或两种以上不同单体的链称为共聚物。
Polymers are classified as either addition polymers or condensation polymers depending on the mechanism that creates them. Addition polymers form without the loss of any atoms, while condensation polymers release a small molecule, usually water or HCl, for each new bond made. You need to be able to spot the difference from the structure of the repeating unit and write the corresponding equation for the polymerisation.
根据形成机理，聚合物被分为加成聚合物和缩合聚合物两大类。加成聚合物形成时不丢失任何原子，而缩合聚合物每形成一个新键就会释放一个小分子，通常是水或氯化氢。你需要能从重复单元的结构中识别出这两类聚合物，并写出相应的聚合反应方程式。
2. Addition Polymerisation | 加成聚合
Addition polymerisation requires monomers that contain a carbon–carbon double bond, C=C. The π‑bond breaks open, and the monomers link together by single bonds to form a saturated carbon backbone. Nothing is eliminated, so the empirical formula of the polymer is exactly the same as that of the monomer.
加成聚合要求单体含有碳碳双键（C=C）。π键断裂打开，单体通过单键连接起来形成饱和的碳骨架。没有分子脱去，因此聚合物的经验式与单体的经验式完全相同。
For example, ethene, CH₂=CH₂, polymerises to form poly(ethene), written as —(CH₂—CH₂)ₙ—. The ‘n’ indicates a large but undefined number of repeats. High‑pressure processes produce low‑density poly(ethene) (LDPE) with branched chains, while Ziegler–Natta catalysts give linear high‑density poly(ethene) (HDPE). The different branching leads to vastly different physical properties: LDPE is flexible and used for films, while HDPE is rigid and used for bottles and pipes.
例如，乙烯 CH₂=CH₂ 聚合生成聚乙烯，写作 —(CH₂—CH₂)ₙ—。下标 n 表示数量很大但不确定的重复单元数。高压工艺制造支链较多的低密度聚乙烯（LDPE），齐格勒-纳塔催化剂则产生线性高密度聚乙烯（HDPE）。支化程度不同导致物理性质大相径庭：LDPE 柔韧，用于薄膜；HDPE 刚硬，用于瓶子和管道。
Other common addition polymers include poly(propene), poly(chloroethene) (PVC), poly(tetrafluoroethene) (PTFE) and poly(phenylethene) (polystyrene). You must be able to draw a section of the polymer chain given the monomer, or deduce the monomer from a given segment of the polymer.
其他常见的加成聚合物包括聚丙烯、聚氯乙烯（PVC）、聚四氟乙烯（PTFE）和聚苯乙烯。必须做到：给定单体能画出聚合物链段，或者从给出的聚合物链段推断出原来的单体。
3. Drawing and Naming Addition Polymers | 加成聚合物的绘制与命名
To name an addition polymer, write ‘poly’ followed by the name of the monomer in brackets. If the monomer name is multi‑word, keep the brackets, e.g. poly(chloroethene). Displayed formulas must show the repeat unit inside brackets, with the subscript ‘n’ placed outside, and the continuation bonds crossing the brackets.
给加成聚合物命名时，写“poly”后接括号内的单体名称。如果单体名称由多个词组成，保留括号，例如 poly(chloroethene)。所画的显示式必须把重复单元放在方括号内，下标 n 写在外侧，并且穿过方括号的连续键要画出来。
When deducing the monomer, scan the backbone and identify the repeating unit. Remove the two single bonds at either side and restore the C=C double bond. If substituents such as Cl, CH₃ or C₆H₅ are present, they become side groups on the alkene monomer. Practice with chloroethene → PVC, propene → poly(propene) and phenylethene → polystyrene.
推断单体时，先观察主链找到重复单元。去掉两侧的单键，恢复碳碳双键。如果存在Cl、CH₃或C₆H₅等取代基，它们就成为烯烃单体上的侧基。多做练习：氯乙烯→PVC、丙烯→聚丙烯、苯乙烯→聚苯乙烯。
4. Condensation Polymerisation | 缩合聚合
Condensation polymerisation occurs between monomers that carry two reactive functional groups – one on each end. Each time a link forms, a small molecule such as H₂O or HCl is eliminated. The empirical formula of the polymer is therefore different from that of the combined monomers.
缩合聚合发生在带有两个活性官能团（每端一个）的单体之间。每形成一个链接，就会脱去一个小分子（如水或HCl）。因此聚合物的经验式与单体总和的经验式不同。
The two most important families of condensation polymers are polyesters and polyamides. In the exam you will draw the repeating unit and show the linkage explicitly: the ester link —COO— or the amide link —CONH—. Always check that the repeating unit you draw matches the total number of carbons and other atoms in the original monomers minus the small molecule lost.
最重要的两类缩合聚合物是聚酯和聚酰胺。考试时需要画出重复单元并明确标出连接键：酯基 —COO— 或酰胺基 —CONH—。一定要检查所画的重复单元总原子数是否等于原料单体扣除脱去小分子后的原子数。
5. Polyesters | 聚酯
Polyesters are formed from a diol and a dicarboxylic acid, or from a single monomer that contains both an alcohol and a carboxylic acid group. Each ester link releases one molecule of water. A classic example is Terylene (PET), made from ethane‑1,2‑diol and benzene‑1,4‑dicarboxylic acid.
聚酯由二元醇和二元羧酸反应生成，或者由同时含有醇羟基和羧基的单一单体形成。每形成一个酯键释放一分子水。经典例子是涤纶（PET），由乙二醇和对苯二甲酸制成。
The repeating unit of PET is —OCH₂CH₂OOC—C₆H₄—CO—. You must be able to draw the full displayed structure. When writing the equation for polyester formation, show the structural formulas of the two monomers and the repeating unit of the polymer, plus (2n‑1) H₂O for a chain of n repeats, or simply ‘+ n H₂O’ per mole of repeat if the chain is considered infinite. Be precise with the ester linkage: the carbonyl carbon comes from the acid, and the C–O–C oxygen from the alcohol.
PET的重复单元是 —OCH₂CH₂OOC—C₆H₄—CO—。你需要能画出完整的显示式。书写聚酯形成方程式时，展示两种单体的结构式以及聚合物的重复单元，同时标出脱去的水分子数量：对于 n 个重复单元脱 (2n‑1) H₂O，或当链长无限时，按每摩尔重复单元 n H₂O 处理，通常考试只需写 n H₂O。要精准画出酯键：羰基碳来自羧酸，C–O–C 中的氧来自醇。
6. Polyamides | 聚酰胺
Polyamides are produced either from a diamine and a dicarboxylic acid, or from an amino acid that has both amine and carboxylic acid groups. The amide (peptide) bond forms with the elimination of water. Nylon‑6,6 is made from 1,6‑diaminohexane and hexane‑1,6‑dioic acid, while Kevlar uses benzene‑1,4‑diamine and benzene‑1,4‑dicarboxylic acid for extreme strength.
聚酰胺可由二元胺与二元羧酸反应生成，也可由同时含有氨基和羧基的氨基酸生成。酰胺（肽）键形成时脱去水。尼龙-6,6由1,6-己二胺和1,6-己二酸制成，而为了获得超高强度，凯夫拉使用1,4-苯二胺和1,4-苯二甲酸。
In a polyamide the repeating unit contains the —CONH— group. Nylon‑6,6’s repeat is —NH(CH₂)₆NHCO(CH₂)₄CO—. Again, you need to connect the amine nitrogen to the carbonyl carbon, and make it clear that the nitrogen still carries a hydrogen atom in the polymer backbone. Don’t forget that polymerisation of amino acids gives polypeptides, which are natural polyamides; they are covered in the next section.
聚酰胺的重复单元含有 —CONH— 基团。尼龙-6,6 的重复单元是 —NH(CH₂)₆NHCO(CH₂)₄CO—。同样，你必须把胺基氮连到羰基碳上，并且明确示意主链上的氮仍带有一个氢原子。别忘了氨基酸聚合生成多肽，也就是天然的聚酰胺；下一节将详细介绍。
7. Natural Polymers: Proteins | 天然聚合物：蛋白质
Proteins are condensation polymers of α‑amino acids. Each amino acid contains an amine group (–NH₂) and a carboxylic acid group (–COOH) attached to the same α‑carbon. When they polymerise, peptide bonds (–CONH–) form and a water molecule is eliminated for each link.
蛋白质是α-氨基酸的缩合聚合物。每个氨基酸在同一个α-碳上连接着一个氨基（–NH₂）和一个羧基（–COOH）。它们聚合时形成肽键（–CONH–），每生成一个连接即脱去一分子水。
The primary structure of a protein is the sequence of amino acids in the chain. Secondary structure (α‑helix, β‑pleated sheet) arises from hydrogen bonding between the N–H and C=O groups along the backbone. Tertiary structure is the overall 3D folding controlled by hydrogen bonds, ionic attractions, disulfide bridges and hydrophobic interactions. Even a single change in the amino acid sequence (as in sickle‑cell anaemia) can radically alter the shape and function of the protein. IB and OCR specifications expect you to interpret diagrams showing these levels of structure and explain how hydrogen bonding stabilises secondary structures.
蛋白质的一级结构是链中氨基酸的排列顺序。二级结构（α-螺旋、β-折叠）源于主链上 N–H 和 C=O 基团之间的氢键。三级结构是整体的三维折叠，受氢键、离子吸附、二硫桥和疏水作用共同控制。哪怕氨基酸序列只有一个位点发生变化（如镰状细胞贫血），也会大幅改变蛋白质的形状与功能。IB 和 OCR 大纲要求能解读展示这些结构层次的图示，并解释氢键如何稳定二级结构。
8. Natural Polymers: Polysaccharides | 天然聚合物：多糖
Polysaccharides are polymers of monosaccharide units joined by glycosidic bonds formed through condensation reactions that eliminate water. Starch, glycogen and cellulose are all glucose polymers, yet they have strikingly different properties because of the type of glucose and the linkage geometry.
多糖是单糖单元通过糖苷键连接而成的聚合物，这种缩合反应脱去水。淀粉、糖原和纤维素都由葡萄糖聚合而成，但由于葡萄糖构型和连接方式不同，它们的性质截然不同。
Starch is a mixture of amylose (α‑1,4‑linked, helical, compact) and amylopectin (α‑1,4‑linked with α‑1,6‑branches). Glycogen has even more branching. Both serve as energy stores. Cellulose, on the other hand, uses β‑glucose units linked by β‑1,4‑glycosidic bonds; every second glucose is flipped, allowing long straight chains to pack tightly with strong inter‑chain hydrogen bonds. This gives cellulose enormous tensile strength, making it ideal for plant cell walls. You should be able to sketch the arrangement of β‑1,4 links in cellulose and explain why it is insoluble and resistant to hydrolysis.
淀粉是直链淀粉（α-1,4-糖苷键，螺旋状，紧密）和支链淀粉（α-1,4-键并带有α-1,6-支链）的混合物。糖原的支化程度更高。两者都是能量储存形式。而纤维素使用β-葡萄糖单元，以β-1,4-糖苷键连接；每隔一个葡萄糖翻转一次，使长直链紧密堆积，链间形成大量氢键。这赋予纤维素极高的抗张强度，非常适合植物细胞壁。你应能画出纤维素中β-1,4-连接的示意图，并解释其为何不溶于水且难以水解。
9. Polymer Properties and Structure | 聚合物的性质与结构
The physical properties of polymers are governed by the strength of intermolecular forces between chains, chain length, branching and whether the polymer is amorphous or crystalline. Thermoplastic polymers, such as poly(ethene) and PET, soften on heating because the weak intermolecular forces weaken reversibly. Thermosetting polymers form covalent cross‑links during curing and cannot be remoulded.
聚合物的物理性质取决于链间分子间作用力的强度、链长、支化程度以及聚合物是无定形还是结晶态。热塑性聚合物（如聚乙烯和PET）受热软化，因为较弱的分子间作用力可逆减弱。热固性聚合物在固化过程中形成共价交联，无法再重塑。
Polyamides like nylon and Kevlar have hydrogen bonds between the C=O and N–H groups on adjacent chains. These strong intermolecular forces raise the melting point and mechanical toughness dramatically. In Kevlar, the rigid benzene rings and extensive hydrogen bonding produce a material five times stronger than steel on an equal‑weight basis. Polyesters also have permanent dipole–dipole interactions from the polar C=O groups, but no hydrogen bonding, so their strength is generally lower than polyamides.
像尼龙和凯夫拉这样的聚酰胺，在相邻链的 C=O 和 N–H 基团之间形成氢键。这些强大的分子间作用力大幅提升了熔点和力学韧性。在凯夫拉中，刚性的苯环加上广泛氢键共同造就了同等重量下强度是钢铁五倍的材料。聚酯也有来自极性 C=O 基团的永久偶极‑偶极作用，但没有氢键，因此强度通常低于聚酰胺。
Plasticisers are small, non‑volatile molecules that insert between polymer chains, pushing them apart and reducing the strength of intermolecular attractions. This makes the plastic more flexible and soft – for instance, PVC used for wire insulation contains plasticisers, while unplasticised PVC (uPVC) is rigid and used for window frames.
增塑剂是微小的不挥发分子，插入聚合物链之间将它们推开，减弱分子间作用力。这使塑料变得更加柔软和有弹性——例如，用于电线绝缘层的 PVC 含有增塑剂，而未增塑的 PVC（uPVC）则坚硬，用于窗框。
10. Biodegradable and Sustainable Polymers | 可生物降解与可持续聚合物
Most addition polymers such as poly(ethene) and poly(propene) have a saturated carbon–carbon backbone that is chemically inert and non‑biodegradable. They persist in the environment, causing serious waste management issues. Condensation polymers, however, contain ester or amide bonds that can be hydrolysed, making them susceptible to biodegradation.
大多数加成聚合物，例如聚乙烯和聚丙烯，具有化学惰性且不可生物降解的饱和碳‑碳骨架。它们长期存在于环境中，造成严重的废弃物管理问题。而缩合聚合物含有可被水解的酯键或酰胺键，因此容易被生物降解。
Polylactic acid (PLA) is a prominent biodegradable polyester derived from renewable resources such as corn starch. It is used for disposable cups, food packaging and medical sutures that dissolve in the body. Poly(glycolic acid) and poly(3‑hydroxybutyrate) (PHB) are other examples. You should recognise the structure of PLA and be able to write its hydrolysis equation showing the formation of lactic acid monomer.
聚乳酸（PLA）是一种重要的可生物降解聚酯，来自玉米淀粉等可再生资源。它用于一次性杯子、食品包装以及可被人体吸收的医疗缝合线。聚乙醇酸和聚3‑羟基丁酸酯（PHB）是另外的例子。你应能认出 PLA 的结构，并写出其水解生成乳酸单体的方程式。
Recycling and disposal methods are also tested. Mechanical recycling re‑grinds and remoulds plastics, but can degrade properties. Feedstock recycling breaks polymers back into monomers or fuels. Incineration for energy recovery must be balanced against toxic emissions. Life‑cycle assessment (LCA) is used to compare the environmental impact of polymer production, use and disposal with alternatives such as paper or glass. Both IB and OCR ask you to evaluate the sustainability of given materials based on data provided.
回收和处理方法也是考点。机械回收将塑料重新粉碎成型，但可能降低性能。原料回收将聚合物分解回单体或燃料。以焚烧方式回收能量则必须权衡有毒气体的排放。生命周期评估（LCA）用于比较聚合物在生产、使用和处置过程中的环境影响，并与纸张或玻璃等替代品进行对照。IB 和 OCR 都要求根据给出的数据评价某种材料的可持续性。
Published by TutorHao | Chemistry 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