High School Chemistry

Introduction to Organic Chemistry有机化学入门

Carbon is the atom that builds life, fuels, plastics, and pharmaceuticals. Its ability to form four covalent bonds — and to bond to itself in chains, rings, and double bonds — creates millions of distinct compounds. This guide builds the organic chemistry toolkit from scratch: why carbon is unique (tetravalency), the hydrocarbons (alkanes, alkenes, alkynes), systematic IUPAC naming, the major functional groups (alcohols, carboxylic acids, and more), structural isomers, and an introduction to organic reactions and polymers. Worked examples and KaTeX formulas throughout.碳是构建生命、燃料、塑料和药物的原子。它形成四个共价键的能力——以及与自身成键形成链、环和双键的能力——创造了数百万种不同的化合物。本指南从零开始构建有机化学（organic chemistry，有机化学）工具箱：为什么碳（carbon，碳）独特（四价性）、烃类（hydrocarbon，烃；烷烃 alkane、烯烃 alkene、炔烃 alkyne）、系统性 IUPAC 命名、主要官能团（functional group，官能团；醇、羧酸等）、同分异构体（isomer，同分异构体），以及有机反应与聚合物（polymer，聚合物）入门。全程配有例题与公式。

7 sections7 节内容 US NGSS · ON SCH4U · BC · ABUS NGSS · ON SCH4U · BC · AB Grade 12 depth (ON SCH4U B / AB Chem 30 C)12 年级深度（ON SCH4U B / AB Chem 30 C）

Where this lives in your syllabus本单元在各大纲中的位置

DIVERGENCE: organic chemistry is NOT a dedicated NGSS performance expectation. Carbon chemistry appears only indirectly through HS-PS1-2: "Construct and revise an explanation for the outcome of a simple chemical reaction based on the outermost electron states of atoms…" with the clarification "Examples of chemical reactions could include… the reaction of carbon and oxygen, or of carbon and hydrogen." Functional groups, IUPAC organic naming, and hydrocarbon classification are not assessed under NGSS. A US NGSS student encounters organic chemistry only as an application context for bonding and reaction types — not as a standalone assessed topic.差异说明：有机化学在 NGSS 中没有专属表现期望。碳化学仅通过 HS-PS1-2 间接出现："根据原子最外层电子状态、元素周期表中的趋势及化学性质规律，建立并修正对简单化学反应结果的解释。"澄清说明中举例包括"碳与氧的反应或碳与氢的反应"。官能团、IUPAC 有机命名和烃类分类均不在 NGSS 评估范围内。美国 NGSS 学生只在成键与反应类型的应用场景中接触有机化学，而非独立评估主题。

Note.提示。 Major divergence from provincial curricula. The content of this guide (hydrocarbon families, IUPAC naming, functional groups, isomers, reactions) is fully assessed in ON SCH4U, BC Chemistry 11, and AB Chemistry 30 — but lies outside the NGSS assessment boundary. A US-only student can use this guide for enrichment and AP Chemistry preparation; it is not required for NGSS assessments.与各省课纲存在重大差异。本指南内容（烃族、IUPAC 命名、官能团、同分异构体、反应）在 ON SCH4U、BC Chemistry 11 和 AB Chemistry 30 中均需全面评估，但超出 NGSS 评估范围。纯美国 NGSS 学生可将本指南用于拓展学习和 AP Chemistry 备考；NGSS 评估不作要求。

SCH4U Strand B (Organic Chemistry). Overall Expectations: B2 "investigate organic compounds and organic chemical reactions, and use various methods to represent the compounds"; B3 "demonstrate an understanding of the structure, properties, and chemical behaviour of compounds within each class of organic compounds." Specific expectations include B2.1 terminology (organic compound, functional group, saturated/unsaturated hydrocarbon, structural isomer, polymer); B2.2 IUPAC nomenclature for hydrocarbons, alcohols, aldehydes, ketones, carboxylic acids, esters, ethers, amines, amides, and simple aromatics; B3.3 organic reaction types including substitution, addition, elimination, oxidation, esterification, and hydrolysis; B3.4 addition vs condensation polymerization; B3.5 isomerism.SCH4U B 单元（有机化学）。总体期望：B2"研究有机化合物和有机化学反应，并用各种方法表示这些化合物"；B3"理解各类有机化合物的结构、性质和化学行为"。具体期望包括 B2.1 术语（有机化合物、官能团、饱和/不饱和烃、结构同分异构体、聚合物）；B2.2 烃类、醇、醛、酮、羧酸、酯、醚、胺、酰胺和简单芳香族化合物的 IUPAC 命名；B3.3 有机反应类型（取代、加成、消除、氧化、酯化、水解）；B3.4 加成聚合与缩合聚合；B3.5 同分异构现象。

Chemistry 11 Big Idea: "Organic chemistry and its applications have significant implications for human health, society, and the environment." Content: "organic compounds" — elaborated as "organic compounds: names, structures, geometry"; "applications of organic chemistry: First Peoples traditional practices (e.g., medicines), pharmaceuticals, petrochemicals, polymers, cosmetics, metabolism, agriculture, food, biotechnology." Organic chemistry is assessed in Grade 11 in BC, not Grade 12.Chemistry 11 大概念："有机化学及其应用对人类健康、社会和环境有重大影响。"内容："有机化合物"——细化为"有机化合物：名称、结构、几何形态"；"有机化学的应用：原住民传统实践（如药物）、药物化学、石化、聚合物、化妆品、代谢、农业、食品、生物技术"。在 BC，有机化学在 11 年级评估，而非 12 年级。

Note.提示。 BC places organic chemistry in Grade 11 (Chemistry 11), whereas Ontario (SCH4U) and Alberta (Chemistry 30) defer it to Grade 12. A BC Chemistry 11 student therefore meets organic compounds, naming, and applications one year earlier than their Ontario or Alberta peers. The structural content (names, structures, geometry) overlaps with §1–§4 of this guide.BC 将有机化学安排在 11 年级（Chemistry 11），而安大略（SCH4U）和阿尔伯塔（Chemistry 30）推迟到 12 年级。因此，BC Chemistry 11 学生比安大略或阿尔伯塔同学早一年接触有机化合物、命名和应用。结构内容（名称、结构、几何形态）与本指南 §1–§4 重叠。

Chemistry 30 Unit C "Chemical Changes of Organic Compounds." General Outcome 1: "explore organic compounds as a common form of matter" — knowledge outcomes include: define organic compounds as compounds containing carbon; name and draw structural, condensed structural, and line diagrams using IUPAC nomenclature for saturated and unsaturated aliphatic (including cyclic) and aromatic compounds containing up to 10 carbon atoms in the parent chain, containing only one type of functional group, including halogenated hydrocarbons, alcohols, carboxylic acids, and esters; define structural isomerism. General Outcome 2: "describe chemical reactions of organic compounds" — knowledge outcomes include: define and provide examples of simple addition, substitution, elimination, esterification, and combustion reactions; define monomers, polymers, and polymerization.Chemistry 30 C 单元"有机化合物的化学变化"。总目标 1："探索有机化合物作为物质的一种常见形式"——知识结果包括：将有机化合物定义为含碳化合物；用 IUPAC 命名法命名并绘制母链最多含 10 个碳原子、仅含一种官能团的饱和和不饱和脂肪族（含环状）及芳香族化合物的结构式、简式和线段式，包括卤代烃、醇、羧酸和酯；定义结构同分异构体。总目标 2："描述有机化合物的化学反应"——知识结果包括：定义并举例说明简单加成、取代、消除、酯化和燃烧反应；定义单体、聚合物和聚合反应。

Read me first先读这里

How to use this guide如何使用本指南

Organic chemistry is a Grade 12 topic in Ontario (SCH4U) and Alberta (Chemistry 30), a Grade 11 topic in BC (Chemistry 11), and has no dedicated NGSS performance expectation in the US. All seven sections are core for ON SCH4U and AB Chemistry 30 Diploma students; BC Chemistry 11 students cover naming, structures, and functional groups but may encounter fewer reaction types than the SCH4U list. The table below maps each row to the sections most relevant to that curriculum.有机化学在安大略（SCH4U）和阿尔伯塔（Chemistry 30）是 12 年级主题，在 BC（Chemistry 11）是 11 年级主题，在美国 NGSS 中没有专属表现期望。对于 ON SCH4U 和 AB Chemistry 30 文凭考生，全部 7 节均为核心内容；BC Chemistry 11 学生涵盖命名、结构和官能团，但接触的反应类型可能少于 SCH4U 清单。下表将每行对应到与该课纲最相关的节次。

If you are in…如果你在…	Focus on these sections重点学习	Source依据
🇺🇸 US NGSS HS Chemistry美国 NGSS 高中化学	Enrichment / AP preparation only. NGSS has no dedicated organic PE. Carbon chemistry appears under HS-PS1-2 as a reaction example. Use this guide for AP Chemistry preparation (Unit 4 covers molecular structure and functional groups).仅用于拓展/AP 备考。NGSS 无专属有机化学表现期望。碳化学以反应举例形式出现在 HS-PS1-2 中。可将本指南用于 AP Chemistry 备考（Unit 4 涵盖分子结构和官能团）。	`NGSS HS-PS1 (Chemistry)` — HS-PS1-2 Clarification Statement (C+H, C+O examples)— HS-PS1-2 澄清说明（碳氢、碳氧反应举例）
🇨🇦 ON SCH4U安大略 SCH4U	All seven sections in full. SCH4U Strand B is the broadest organic scope on this page: nine functional-group classes, six reaction types, and polymerization. §4 (IUPAC naming) is especially thorough for SCH4U B2.2.全部 7 节完整学习。SCH4U B 单元是本页范围最广的有机部分：九类官能团、六种反应类型和聚合反应。§4（IUPAC 命名）对 SCH4U B2.2 尤为详尽。	`Ontario SCH3U/4U Chemistry` — SCH4U Strand B B2.1, B2.2, B2.4, B3.1–B3.5— SCH4U B 单元 B2.1、B2.2、B2.4、B3.1–B3.5
🇨🇦 BC Chemistry 11BC Chemistry 11	§1–§5 in full (carbon bonding, hydrocarbons, naming, functional groups, isomers). BC Chemistry 11 content lists "organic compounds: names, structures, geometry" and "applications of organic chemistry." §6–§7 (reaction types, polymers) are lighter in BC 11 than in SCH4U.§1–§5 完整学习（碳的成键、烃类、命名、官能团、同分异构体）。BC Chemistry 11 内容列出"有机化合物：名称、结构、几何形态"和"有机化学的应用"。§6–§7（反应类型、聚合物）在 BC 11 中比 SCH4U 轻。	`BC Chemistry 11/12` — Chemistry 11 Big Idea "Organic chemistry and its applications…"; Content "organic compounds"; Elaboration "organic compounds: names, structures, geometry"— Chemistry 11 大概念"有机化学及其应用……"；内容"有机化合物"；细化"有机化合物：名称、结构、几何形态"
🇨🇦 AB Chemistry 30阿尔伯塔 Chemistry 30	All seven sections in full. AB Chemistry 30 Unit C caps naming at 10-carbon parent chains, one functional group — use §4 with that boundary in mind. §7 (reactions and polymers) maps to GO2 directly: addition, substitution, elimination, esterification, combustion, and polymerization are all assessed.全部 7 节完整学习。AB Chemistry 30 C 单元将命名限制在最多 10 个碳的母链、一种官能团——学习 §4 时请牢记此边界。§7（反应与聚合物）直接对应 GO2：加成、取代、消除、酯化、燃烧和聚合反应均需评估。	`Alberta Chemistry 20/30` — Chemistry 30 Unit C GO1/GO2, knowledge outcome text— Chemistry 30 C 单元 GO1/GO2，知识结果文本

Once you have located your row, use the two cards below for the speed at which you should work through the recommended sections.找到所在行后，用下面两张卡片决定推进速度。

If you are cramming the night before如果你在临阵磨枪

Memorise five things: carbon forms 4 bonds (tetravalent); alkanes are C$_n$H$_{2n+2}$, saturated; alkenes have a C=C double bond (C$_n$H$_{2n}$); IUPAC names use the longest carbon chain as the parent; and functional groups define reactivity. Read every cram-cheat box. Skip the going-deeper boxes if time is short.背熟五件事：碳形成 4 个键（四价）；烷烃通式 C$_n$H$_{2n+2}$，饱和；烯烃含 C=C 双键（C$_n$H$_{2n}$）；IUPAC 命名以最长碳链为母链；官能团决定反应性。读每个速记框。时间紧时可跳过深入框。

If you are going for the top mark如果你目标顶分

Be precise about IUPAC naming rules for branched chains; know the general formulas for alkanes (C$_n$H$_{2n+2}$), alkenes (C$_n$H$_{2n}$), and alkynes (C$_n$H$_{2n-2}$); draw all structural isomers of a given formula; identify functional groups from a structural formula; and write balanced equations for addition, esterification, and combustion reactions. ON SCH4U B2.2 expects nine functional-group classes; AB Chemistry 30 caps at one functional group per molecule but demands correct condensed structural and line formulas.精确掌握支链的 IUPAC 命名规则；熟知烷烃（C$_n$H$_{2n+2}$）、烯烃（C$_n$H$_{2n}$）和炔烃（C$_n$H$_{2n-2}$）的通式；画出给定分子式的所有结构同分异构体；从结构简式识别官能团；写出加成、酯化和燃烧反应的配平方程式。ON SCH4U B2.2 要求九类官能团；AB Chemistry 30 限制每个分子仅一种官能团，但要求正确书写简缩式和键线式。

Section 1 · Why carbon? Tetravalency and bonding第 1 节 · 为何是碳？四价性与成键

Why Carbon? Tetravalency and Bonding为何是碳？四价性与成键

Carbon's four bonds are the foundation of all organic chemistry.碳的四个键是所有有机化学的基础。

Organic compound有机化合物 — a compound containing carbon (with some inorganic exceptions: carbonates CO$_3^{2-}$, oxides CO and CO$_2$, cyanides CN$^-$, carbides). Almost all organic molecules also contain hydrogen; most contain oxygen, nitrogen, or other elements too.— 含碳化合物（少数无机例外：碳酸盐 CO$_3^{2-}$、氧化物 CO 和 CO$_2$、氰化物 CN$^-$、碳化物）。几乎所有有机分子也含有氢；大多数还含有氧、氮或其他元素。
Tetravalency四价性 — carbon has 4 valence electrons and always forms exactly 4 covalent bonds. This allows it to bond to 4 hydrogen atoms (CH$_4$), to other carbons in chains and rings, and to form double bonds (C=C) or triple bonds (C≡C). No other element has this combination of bond-forming flexibility and bond strength.— 碳有 4 个价电子，始终形成恰好 4 个共价键。这使它能与 4 个氢原子（CH$_4$）成键，在链和环中与其他碳成键，并形成双键（C=C）或三键（C≡C）。没有其他元素同时具备这种成键灵活性和键强度。
Catenation成链性 — carbon's unique ability to bond to itself in long chains, branched chains, and rings. Silicon can also catenate, but Si–Si bonds are weaker and less versatile.— 碳与自身成键形成长链、支链和环的独特能力。硅也能成链，但 Si–Si 键较弱且灵活性较差。

AB Chemistry 30 Unit C GO1 defines: "define organic compounds as compounds containing carbon, recognizing inorganic exceptions such as carbonates, cyanides, carbides and oxides of carbon." SCH4U B2.1 names "organic compound," "saturated hydrocarbon," and "unsaturated hydrocarbon" as assessed terminology.AB Chemistry 30 C 单元 GO1 定义："将有机化合物定义为含碳化合物，承认无机例外如碳酸盐、氰化物、碳化物和碳的氧化物。" SCH4U B2.1 将"有机化合物"、"饱和烃"和"不饱和烃"列为被评估术语。

Worked Example 1 · Degrees of unsaturation例题 1 · 不饱和度

The molecular formula C$_4$H$_8$ has fewer hydrogens than a fully saturated alkane C$_4$H$_{10}$. Calculate the degree of unsaturation (DoU) and state what it could mean structurally.分子式 C$_4$H$_8$ 的氢原子数少于完全饱和烷烃 C$_4$H$_{10}$。计算不饱和度（DoU）并说明其结构含义。

Formula for DoU (carbon and hydrogen only):DoU 公式（仅含碳和氢）：

$$ \text{DoU} = \frac{2C + 2 - H}{2} $$

Apply:代入：

$$ \text{DoU} = \frac{2(4) + 2 - 8}{2} = \frac{2}{2} = 1. $$

Interpretation:解释： DoU = 1 means one degree of unsaturation, which could be one C=C double bond (e.g., but-1-ene, CH$_2$=CHCH$_2$CH$_3$) or one ring (e.g., cyclobutane). ✓DoU = 1 表示一个不饱和度，可以是一个 C=C 双键（如丁-1-烯，CH$_2$=CHCH$_2$CH$_3$）或一个环（如环丁烷）。✓

How many covalent bonds does carbon always form in organic molecules?碳在有机分子中总是形成几个共价键？

§1 · Q1

Carbon has 4 valence electrons and always forms 4 covalent bonds in organic chemistry (tetravalent). These 4 bonds can be to H, other C, O, N, etc. — and can include double or triple bonds (counted as 2 or 3 bonds to the same atom).碳有 4 个价电子，在有机化学中始终形成 4 个共价键（四价）。这 4 个键可以与 H、其他 C、O、N 等成键——并可包含双键或三键（计为与同一原子的 2 个或 3 个键）。

Carbon is tetravalent: it always forms exactly 4 bonds. This is why CH$_4$ (methane) has 4 C–H bonds, and C$_2$H$_4$ (ethene) has one C=C double bond plus 2 C–H bonds on each carbon — still 4 bonds per carbon total.碳是四价的：它始终形成恰好 4 个键。这就是为什么 CH$_4$（甲烷）有 4 个 C–H 键，而 C$_2$H$_4$（乙烯）有一个 C=C 双键加上每个碳上 2 个 C–H 键——每个碳总共仍是 4 个键。

Which of the following is NOT classified as an organic compound?以下哪种物质不属于有机化合物？

§1 · Q2

Carbon dioxide (CO$_2$)二氧化碳（CO$_2$）

Methane (CH$_4$)甲烷（CH$_4$）

Ethanol (C$_2$H$_5$OH)乙醇（C$_2$H$_5$OH）

Glucose (C$_6$H$_{12}$O$_6$)葡萄糖（C$_6$H$_{12}$O$_6$）

CO$_2$ is an inorganic exception: it contains carbon but is classified as inorganic because it has no C–H bonds and is an oxide of carbon. This exception is explicitly named in AB Chemistry 30 Unit C GO1.CO$_2$ 是无机例外：它含有碳，但因为没有 C–H 键且是碳的氧化物，被归类为无机物。这一例外在 AB Chemistry 30 C 单元 GO1 中明确列出。

Organic compounds contain carbon. However, some carbon-containing compounds are classified as inorganic by convention: CO, CO$_2$, carbonates, cyanides, and carbides. All the others listed (CH$_4$, C$_2$H$_5$OH, C$_6$H$_{12}$O$_6$) contain C–H bonds and are organic.有机化合物含有碳。但按惯例，某些含碳化合物被归类为无机物：CO、CO$_2$、碳酸盐、氰化物和碳化物。列出的其他物质（CH$_4$、C$_2$H$_5$OH、C$_6$H$_{12}$O$_6$）含有 C–H 键，均为有机物。

Going deeper — why is organic chemistry so vast? Carbon's unique electronic structure深入 — 为何有机化学如此庞大？碳独特的电子结构

Carbon's electron configuration is $1s^2\, 2s^2\, 2p^2$ — with 4 valence electrons, it sits exactly in the middle of the octet. This means carbon neither strongly loses electrons (like metals) nor strongly gains electrons (like halogens): it shares. The four bonds formed are roughly equal in energy (sp$^3$ hybridization for single bonds, sp$^2$ for double, sp for triple), giving it enormous geometric flexibility. Compare to silicon (same group, same valence): Si–Si bonds are about $226\ \text{kJ/mol}$, much weaker than C–C ($346\ \text{kJ/mol}$). Weaker bonds mean silicon chains are less stable at Earth-surface temperatures and in the presence of water and oxygen. This is why Earth's biochemistry is built on carbon, not silicon.碳的电子排布为 $1s^2\, 2s^2\, 2p^2$——4 个价电子，恰好处于八隅体的中间。这意味着碳既不像金属那样强烈失去电子，也不像卤素那样强烈获得电子：它选择共用。形成的四个键能量大致相等（单键 sp$^3$ 杂化，双键 sp$^2$，三键 sp），赋予它极大的几何灵活性。与同族的硅相比：Si–Si 键约 $226\ \text{kJ/mol}$，远弱于 C–C（$346\ \text{kJ/mol}$）。键较弱意味着硅链在地表温度及水和氧的存在下不够稳定。这就是为什么地球的生物化学建立在碳而非硅上。

Section 2 · Alkanes (saturated hydrocarbons)第 2 节 · 烷烃（饱和烃）

Alkanes (Saturated Hydrocarbons)烷烃（饱和烃）

Alkanes: only C–C and C–H single bonds; general formula C$_n$H$_{2n+2}$.烷烃：仅含 C–C 和 C–H 单键；通式 C$_n$H$_{2n+2}$。

Saturated饱和 — every carbon forms single bonds only; no C=C or C≡C. The molecule is "saturated" with hydrogen: it cannot add any more H without breaking C–C bonds.— 每个碳仅形成单键；没有 C=C 或 C≡C。分子"饱和"于氢：在不断裂 C–C 键的情况下，无法再加氢。
Homologous series同系物 — each successive alkane differs by –CH$_2$–. The first ten: methane CH$_4$ ($n=1$), ethane C$_2$H$_6$, propane C$_3$H$_8$, butane C$_4$H$_{10}$, pentane, hexane, heptane, octane, nonane, decane ($n=10$).— 每个相邻烷烃相差 –CH$_2$–。前十个：甲烷 CH$_4$（$n=1$）、乙烷 C$_2$H$_6$、丙烷 C$_3$H$_8$、丁烷 C$_4$H$_{10}$、戊烷、己烷、庚烷、辛烷、壬烷、癸烷（$n=10$）。
Physical properties物理性质 — non-polar, London (dispersion) forces only; boiling point increases with chain length (more surface area → stronger London forces). Short alkanes (C$_1$–C$_4$) are gases at room temperature; C$_5$–C$_{17}$ are liquids; C$_{18}$+ are waxy solids.— 非极性，仅有色散力；沸点随链长增加（更大的接触面积 → 更强的色散力）。短链烷烃（C$_1$–C$_4$）在室温下为气体；C$_5$–C$_{17}$ 为液体；C$_{18}$+ 为蜡状固体。

$$ \text{Alkane general formula: } \text{C}_n\text{H}_{2n+2} $$ AB Chemistry 30 Unit C GO1 lists "methane, ethane, ethanol, propane, octane" as significant organic compounds in daily life. BC Chemistry 11 Content: "organic compounds: names, structures, geometry." SCH4U B3.1 requires comparing classes of organic compounds by names and structural formulas.AB Chemistry 30 C 单元 GO1 将"甲烷、乙烷、乙醇、丙烷、辛烷"列为日常生活中重要的有机化合物。BC Chemistry 11 内容："有机化合物：名称、结构、几何形态"。SCH4U B3.1 要求按名称和结构简式比较各类有机化合物。

Worked Example 2 · Molecular formula and combustion of propane例题 2 · 丙烷的分子式与燃烧

Propane is the alkane with $n = 3$. (a) Write its molecular formula and one line structural formula. (b) Write the balanced complete combustion equation.丙烷是 $n = 3$ 的烷烃。(a) 写出其分子式和一条结构简式。(b) 写出配平后的完全燃烧方程式。

(a) Molecular formula:(a) 分子式： C$_3$H$_{2(3)+2}$ = C$_3$H$_8$. Structural: CH$_3$–CH$_2$–CH$_3$ (condensed: CH$_3$CH$_2$CH$_3$).C$_3$H$_{2(3)+2}$ = C$_3$H$_8$。结构简式：CH$_3$–CH$_2$–CH$_3$（简缩式：CH$_3$CH$_2$CH$_3$）。

(b) Complete combustion (hydrocarbon + O$_2$ → CO$_2$ + H$_2$O):(b) 完全燃烧（烃 + O$_2$ → CO$_2$ + H$_2$O）：

$$ \text{C}_3\text{H}_8 + 5\,\text{O}_2 \longrightarrow 3\,\text{CO}_2 + 4\,\text{H}_2\text{O} $$

Check: C: $3=3$ ✓; H: $8=8$ ✓; O: $10=6+4=10$ ✓.核验：C: $3=3$ ✓；H: $8=8$ ✓；O: $10=6+4=10$ ✓。

What is the molecular formula of the alkane with 6 carbon atoms?含 6 个碳原子的烷烃的分子式是什么？

§2 · Q1

C$_6$H$_{12}$C$_6$H$_{12}$

C$_6$H$_{14}$C$_6$H$_{14}$

C$_6$H$_{10}$C$_6$H$_{10}$

C$_6$H$_{16}$C$_6$H$_{16}$

Alkane formula: C$_n$H$_{2n+2}$. For $n=6$: C$_6$H$_{2(6)+2}$ = C$_6$H$_{14}$. This is hexane. C$_6$H$_{12}$ would be a cycloalkane or alkene (one degree of unsaturation).烷烃通式：C$_n$H$_{2n+2}$。$n=6$：C$_6$H$_{2(6)+2}$ = C$_6$H$_{14}$。这是己烷。C$_6$H$_{12}$ 则是环烷烃或烯烃（一个不饱和度）。

Use C$_n$H$_{2n+2}$: with $n=6$, $H = 2(6)+2 = 14$. C$_6$H$_{12}$ has DoU = 1 (one double bond or ring); C$_6$H$_{10}$ has DoU = 2. C$_6$H$_{16}$ has too many H for tetravalent carbon.用 C$_n$H$_{2n+2}$：$n=6$ 时，$H = 2(6)+2 = 14$。C$_6$H$_{12}$ 的 DoU = 1（一个双键或环）；C$_6$H$_{10}$ 的 DoU = 2。C$_6$H$_{16}$ 对四价碳而言氢太多了。

Why do longer-chain alkanes have higher boiling points than shorter ones?为什么长链烷烃的沸点比短链烷烃高？

§2 · Q2

They have more C=C double bonds, increasing polarity它们有更多 C=C 双键，增加了极性

They form hydrogen bonds with water它们与水形成氢键

Their molecules have higher molar mass and are more ionic它们的分子摩尔质量更高且更具离子性

Larger molecules have more surface area, increasing London (dispersion) force strength较大的分子有更多表面积，增强了色散力

Alkanes are non-polar; their only intermolecular forces are London (dispersion) forces. Longer chains have more electrons and larger surface area, producing stronger dispersion forces and requiring more energy (higher boiling point) to separate molecules.烷烃是非极性的；它们唯一的分子间作用力是色散力。更长的链有更多电子和更大的表面积，产生更强的色散力，需要更多能量（更高的沸点）来分离分子。

Alkanes have no double bonds, no polarity, and no hydrogen bonds (no O–H or N–H). They are not ionic. The only force is London (dispersion), which increases with molecular size.烷烃没有双键、没有极性、没有氢键（没有 O–H 或 N–H）。它们不是离子性的。唯一的作用力是色散力，随分子尺寸增加而增强。

Section 3 · Alkenes and alkynes (unsaturated hydrocarbons)第 3 节 · 烯烃与炔烃（不饱和烃）

Alkenes and Alkynes (Unsaturated Hydrocarbons)烯烃与炔烃（不饱和烃）

Unsaturated hydrocarbons contain C=C or C≡C bonds and can undergo addition reactions.不饱和烃含有 C=C 或 C≡C 键，可发生加成反应。

Alkenes烯烃 — one C=C double bond; general formula C$_n$H$_{2n}$. Named with the suffix -ene: ethene C$_2$H$_4$, propene C$_3$H$_6$, but-1-ene C$_4$H$_8$. The position of the double bond is given by the lowest locant (but-1-ene vs but-2-ene).— 含一个 C=C 双键；通式 C$_n$H$_{2n}$。以 -烯后缀命名：乙烯 C$_2$H$_4$、丙烯 C$_3$H$_6$、丁-1-烯 C$_4$H$_8$。双键位置用最小定位数表示（丁-1-烯与丁-2-烯）。
Alkynes炔烃 — one C≡C triple bond; general formula C$_n$H$_{2n-2}$. Named with the suffix -yne: ethyne (acetylene) C$_2$H$_2$, propyne C$_3$H$_4$.— 含一个 C≡C 三键；通式 C$_n$H$_{2n-2}$。以 -炔后缀命名：乙炔 C$_2$H$_2$、丙炔 C$_3$H$_4$。
Addition reactions加成反应 — unsaturated molecules react with small molecules (H$_2$, Br$_2$, HCl, H$_2$O) across the double or triple bond, producing a single product with no atoms lost. Alkanes do NOT undergo addition (saturated).— 不饱和分子与小分子（H$_2$、Br$_2$、HCl、H$_2$O）在双键或三键处反应，生成一个产物，无原子损失。烷烃不发生加成反应（饱和）。

$$ \text{Alkene: C}_n\text{H}_{2n} \qquad \text{Alkyne: C}_n\text{H}_{2n-2} $$ SCH4U B3.3 includes addition reactions among the assessed reaction types. AB Chemistry 30 Unit C GO2: "define, illustrate and provide examples of simple addition … reactions." BC Chemistry 11 Content: "organic compounds: names, structures, geometry" — alkene geometry (flat around C=C) is assessed.SCH4U B3.3 将加成反应列入被评估的反应类型。AB Chemistry 30 C 单元 GO2："定义、说明并举例简单加成……反应"。BC Chemistry 11 内容："有机化合物：名称、结构、几何形态"——烯烃几何形态（C=C 周围平面）需评估。

Worked Example 3 · Addition of bromine to ethene例题 3 · 溴与乙烯的加成

Ethene (C$_2$H$_4$) reacts with bromine (Br$_2$) in an addition reaction. Write the balanced equation and name the product. State why this reaction is used as a test for unsaturation.乙烯（C$_2$H$_4$）与溴（Br$_2$）发生加成反应。写出配平方程式并命名产物。说明为何此反应用于检验不饱和度。

Balanced equation:配平方程式：

$$ \text{CH}_2\text{=CH}_2 + \text{Br}_2 \longrightarrow \text{CH}_2\text{Br--CH}_2\text{Br} $$

Product:产物： 1,2-dibromoethane (also called ethylene dibromide). The C=C double bond is broken and replaced by two C–Br single bonds.1,2-二溴乙烷（也称乙烯二溴化物）。C=C 双键断裂，被两个 C–Br 单键取代。

Bromine test:溴测试： Brown/orange Br$_2$ solution (in water or CCl$_4$) is rapidly decolourised by alkenes and alkynes. Alkanes do not react under these conditions, so no colour change. This is a standard qualitative test for unsaturation.棕/橙色 Br$_2$ 溶液（在水或 CCl$_4$ 中）被烯烃和炔烃迅速脱色。烷烃在这些条件下不反应，因此无颜色变化。这是检验不饱和度的标准定性测试。

What is the general molecular formula for alkenes?烯烃的通式分子式是什么？

§3 · Q1

C$_n$H$_{2n}$C$_n$H$_{2n}$

C$_n$H$_{2n+2}$C$_n$H$_{2n+2}$

C$_n$H$_{2n-2}$C$_n$H$_{2n-2}$

C$_n$H$_n$C$_n$H$_n$

Alkenes have one C=C double bond (DoU = 1), so they are 2H short of the alkane formula C$_n$H$_{2n+2}$: alkene = C$_n$H$_{2n}$. Alkynes (one triple bond, DoU = 2) are C$_n$H$_{2n-2}$.烯烃有一个 C=C 双键（DoU = 1），比烷烃通式 C$_n$H$_{2n+2}$ 少 2 个 H：烯烃 = C$_n$H$_{2n}$。炔烃（一个三键，DoU = 2）为 C$_n$H$_{2n-2}$。

Remember: alkane = C$_n$H$_{2n+2}$ (no degrees of unsaturation); alkene = C$_n$H$_{2n}$ (one C=C); alkyne = C$_n$H$_{2n-2}$ (one C≡C). Each degree of unsaturation removes 2 hydrogen atoms.记住：烷烃 = C$_n$H$_{2n+2}$（无不饱和度）；烯烃 = C$_n$H$_{2n}$（一个 C=C）；炔烃 = C$_n$H$_{2n-2}$（一个 C≡C）。每个不饱和度减少 2 个氢原子。

A sample decolourises bromine water rapidly. What does this indicate?一个样品使溴水迅速脱色。这说明什么？

§3 · Q2

The compound is an alkane该化合物是烷烃

The compound is an alcohol该化合物是醇

The compound contains a C=C or C≡C bond该化合物含有 C=C 或 C≡C 键

The compound is ionic该化合物是离子性的

Bromine (Br$_2$) undergoes addition across C=C and C≡C bonds, consuming the orange Br$_2$ and producing colourless dibromo products. Alkanes, alcohols, and ionic compounds do not react with Br$_2$ under these conditions.溴（Br$_2$）在 C=C 和 C≡C 键上发生加成反应，消耗橙色 Br$_2$ 并生成无色的二溴产物。烷烃、醇和离子性化合物在这些条件下不与 Br$_2$ 反应。

Alkanes are saturated — no reaction with Br$_2$ under mild conditions. Alcohols and ionic compounds also don't decolourise bromine water in this test. Only unsaturated compounds (alkenes, alkynes) react rapidly by addition.烷烃是饱和的——在温和条件下不与 Br$_2$ 反应。醇和离子性化合物在此测试中也不会使溴水脱色。只有不饱和化合物（烯烃、炔烃）通过加成快速反应。

Section 4 · IUPAC naming第 4 节 · IUPAC 命名

IUPAC Naming of Organic Compounds有机化合物的 IUPAC 命名

IUPAC naming: four steps for hydrocarbons.IUPAC 命名：烃类的四个步骤。

Step 1 — Find the parent chain第 1 步——找母链 : the longest continuous carbon chain that contains the principal functional group (or the C=C / C≡C for alkenes/alkynes). Count the carbons → meth-(1), eth-(2), prop-(3), but-(4), pent-(5), hex-(6), hept-(7), oct-(8), non-(9), dec-(10).：含主要官能团（或烯/炔的 C=C / C≡C）的最长连续碳链。数碳原子数 → 甲(1)、乙(2)、丙(3)、丁(4)、戊(5)、己(6)、庚(7)、辛(8)、壬(9)、癸(10)。
Step 2 — Number the chain第 2 步——给链编号 : start from the end that gives the principal group (or first point of difference) the lowest locant. For alkenes, the double bond gets the lowest number.：从使主要基团（或第一个差异点）具有最小定位数的一端开始编号。对于烯烃，双键获得最小编号。
Step 3 — Name substituents/multipliers第 3 步——命名取代基/倍数词 : branching alkyl groups are named with their locant + name (methyl-, ethyl-, propyl-…) listed alphabetically before the parent name. Use di-, tri-, tetra- for multiples of the same substituent.：支链烷基按其定位数 + 名称（甲基-、乙基-、丙基-……）以字母顺序列于母链名称前。相同取代基的倍数用二-、三-、四-。
Step 4 — Add suffix第 4 步——加后缀 : alkane → -ane; alkene → -ene (with locant); alkyne → -yne; alcohol → -ol; carboxylic acid → -oic acid; ester → -oate; amine → -amine; amide → -amide.：烷烃 → -烷；烯烃 → -烯（加定位数）；炔烃 → -炔；醇 → -醇；羧酸 → -酸；酯 → -酸…酯；胺 → -胺；酰胺 → -酰胺。

SCH4U B2.2 requires IUPAC nomenclature for nine functional-group classes. AB Chemistry 30 Unit C GO1 requires naming with up to 10 carbons in the parent chain and only one functional group. BC Chemistry 11 requires "organic compounds: names, structures, geometry."SCH4U B2.2 要求九类官能团的 IUPAC 命名。AB Chemistry 30 C 单元 GO1 要求母链最多 10 个碳、仅一种官能团的命名。BC Chemistry 11 要求"有机化合物：名称、结构、几何形态"。

Worked Example 4 · Naming a branched alkane例题 4 · 支链烷烃的命名

Name the following compound: CH$_3$–CH(CH$_3$)–CH$_2$–CH$_2$–CH$_3$.命名以下化合物：CH$_3$–CH(CH$_3$)–CH$_2$–CH$_2$–CH$_3$。

Step 1 — Longest chain:第 1 步——最长链： The backbone is 5 carbons → parent name: pentane.主链为 5 个碳 → 母链名称：戊烷（pentane）。

Step 2 — Number the chain:第 2 步——编号： Number from the end nearest the branch. The branch (–CH$_3$) is at C-2 (from the left) or C-4 (from the right). Lowest locant = 2. Correct direction: CH$_3$(C1)–CH(C2)(CH$_3$)–CH$_2$(C3)–CH$_2$(C4)–CH$_3$(C5).从最近支链的一端编号。支链（–CH$_3$）在 C-2（从左）或 C-4（从右）。最小定位数 = 2。正确方向：CH$_3$(C1)–CH(C2)(CH$_3$)–CH$_2$(C3)–CH$_2$(C4)–CH$_3$(C5)。

Step 3 — Substituent:第 3 步——取代基： One methyl group at C-2 → 2-methyl.C-2 处一个甲基 → 2-甲基（2-methyl）。

Full name:完整名称： 2-methylpentane. ✓2-甲基戊烷（2-methylpentane）。✓

What is the IUPAC name for a straight-chain alkane with 4 carbon atoms?含 4 个碳原子的直链烷烃的 IUPAC 名称是什么？

§4 · Q1

Propane丙烷

Pentane戊烷

Butane丁烷

Ethane乙烷

The prefix for 4 carbons is but-; with the alkane suffix -ane, the name is butane. Memory aid: meth(1), eth(2), prop(3), but(4), pent(5), hex(6).4 个碳的前缀是 丁（but-）；加上烷烃后缀 -烷（-ane），名称为丁烷（butane）。记忆提示：甲(1)、乙(2)、丙(3)、丁(4)、戊(5)、己(6)。

Propane = 3C; pentane = 5C; ethane = 2C. For 4C: but- + -ane = butane.丙烷 = 3C；戊烷 = 5C；乙烷 = 2C。4C：丁(but-) + -烷(-ane) = 丁烷（butane）。

In IUPAC naming, from which end of the chain do you begin numbering?在 IUPAC 命名中，从链的哪一端开始编号？

§4 · Q2

Always from the left end of the structural formula as drawn始终从结构简式绘制的左端开始

From the end that gives the principal group or first branch the lowest locant number从使主要基团或第一个支链获得最小定位数的一端开始

Always from the end closest to the middle of the chain始终从最靠近链中间的一端开始

From the end with the most substituents从取代基最多的一端开始

The lowest-locant rule: number the chain so that the principal functional group (or first branch for alkanes) gets the smallest possible number. For example, 2-methylpentane, not 4-methylpentane, because 2 < 4.最小定位数规则：给链编号使主要官能团（对于烷烃则是第一个支链）获得尽可能小的数字。例如，是 2-甲基戊烷，而不是 4-甲基戊烷，因为 2 < 4。

Structural formulas can be drawn in any direction; the numbering direction depends on minimising the locant of the principal group or first point of difference — not on how the formula is drawn or where substituents cluster.结构简式可以从任何方向绘制；编号方向取决于使主要基团或第一个差异点的定位数最小，而不是公式的绘制方式或取代基的聚集位置。

Going deeper — naming alkenes with E/Z geometry深入 — 含 E/Z 几何构型的烯烃命名

The C=C double bond is planar (all four atoms around it are in one plane). Rotation around the C=C bond does NOT occur freely, unlike around C–C single bonds. This means that if each carbon of the double bond bears two different substituents, two distinct geometric isomers exist: Z (from German zusammen, "together") — priority groups on the same side; and E (from German entgegen, "opposite") — priority groups on opposite sides. Priority is assigned by the Cahn–Ingold–Prelog (CIP) rules: higher atomic number = higher priority. For example, (Z)-but-2-ene has both methyl groups on the same side; (E)-but-2-ene has them on opposite sides. SCH4U B3.5 names stereoisomers including geometric isomers as assessed content.C=C 双键是平面的（其周围四个原子都在同一平面内）。C=C 键的旋转不能自由发生，这与 C–C 单键不同。这意味着如果双键的每个碳上各连有两个不同的取代基，则存在两种不同的几何异构体：Z（德语 zusammen，"同侧"）——优先基团在同一侧；E（德语 entgegen，"对侧"）——优先基团在相对两侧。优先级按 Cahn-Ingold-Prelog（CIP）规则确定：原子序数越高优先级越高。例如，(Z)-丁-2-烯的两个甲基在同一侧；(E)-丁-2-烯的甲基在相反两侧。SCH4U B3.5 将包括几何异构体在内的立体异构体列为被评估内容。

Section 5 · Functional groups第 5 节 · 官能团

Functional Groups官能团

A functional group is the reactive part of an organic molecule — it determines its chemical behaviour.官能团是有机分子中具有反应性的部分——它决定分子的化学行为。

Class类别	Functional group官能团	Suffix后缀	Example示例
Alcohol醇	–OH (hydroxyl)	-ol	Ethanol CH$_3$CH$_2$OH乙醇 CH$_3$CH$_2$OH
Carboxylic acid羧酸	–COOH (carboxyl)	-oic acid	Ethanoic acid CH$_3$COOH乙酸 CH$_3$COOH
Ester酯	–COO– (ester linkage)	-oate	Ethyl ethanoate CH$_3$COOC$_2$H$_5$乙酸乙酯 CH$_3$COOC$_2$H$_5$
Aldehyde醛	–CHO	-al	Ethanal CH$_3$CHO乙醛 CH$_3$CHO
Ketone酮	C=O (flanked by C)	-one	Propanone CH$_3$COCH$_3$丙酮 CH$_3$COCH$_3$
Amine胺	–NH$_2$	-amine	Methylamine CH$_3$NH$_2$甲胺 CH$_3$NH$_2$
Amide酰胺	–CONH$_2$	-amide	Ethanamide CH$_3$CONH$_2$乙酰胺 CH$_3$CONH$_2$
Ether醚	–O– (between carbons)	ether / -oxy	Ethoxyethane C$_2$H$_5$OC$_2$H$_5$乙醚 C$_2$H$_5$OC$_2$H$_5$
Halogenoalkane卤代烷	–X (X = F, Cl, Br, I)	halo- prefix	Chloromethane CH$_3$Cl氯甲烷 CH$_3$Cl

SCH4U B2.2 requires IUPAC naming for all nine classes above. AB Chemistry 30 Unit C GO1 specifies hydroxyl, carboxyl, ester linkage, and halogen functional groups. BC Chemistry 11 Content: "organic compounds: names, structures, geometry."SCH4U B2.2 要求对以上九类进行 IUPAC 命名。AB Chemistry 30 C 单元 GO1 规定羟基、羧基、酯键和卤素官能团。BC Chemistry 11 内容："有机化合物：名称、结构、几何形态"。

Worked Example 5 · Esterification: making an ester from acid + alcohol例题 5 · 酯化反应：酸与醇制酯

Ethanoic acid (CH$_3$COOH) reacts with ethanol (CH$_3$CH$_2$OH) in the presence of an acid catalyst. (a) Write the balanced equation. (b) Name the ester product.乙酸（CH$_3$COOH）在酸催化剂存在下与乙醇（CH$_3$CH$_2$OH）反应。(a) 写出配平方程式。(b) 命名酯产物。

(a) Esterification equation:(a) 酯化方程式：

$$ \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \underset{\Delta}{\overset{\text{H}^+}{\rightleftharpoons}} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} $$

(b) Name the ester:(b) 酯的命名： The acid part (CH$_3$COO–) comes from ethanoic acid → ethanoate. The alcohol part (–C$_2$H$_5$) comes from ethanol → ethyl. Full name: ethyl ethanoate. Note: esterification is reversible; water must be removed to drive the equilibrium toward product.酸的部分（CH$_3$COO–）来自乙酸 → 乙酸根（ethanoate）。醇的部分（–C$_2$H$_5$）来自乙醇 → 乙基（ethyl）。完整名称：乙酸乙酯（ethyl ethanoate）。注意：酯化反应是可逆的；必须除去水以使平衡向产物方向移动。

Which functional group is present in all alcohols?所有醇中都含有哪种官能团？

§5 · Q1

–COOH (carboxyl)–COOH（羧基）

–OH (hydroxyl)–OH（羟基）

C=O (carbonyl, not attached to H or OH)C=O（羰基，不与 H 或 OH 相连）

–NH$_2$ (amino)–NH$_2$（氨基）

Alcohols are defined by the hydroxyl group –OH bonded to a carbon chain (not to a carbonyl carbon). Examples: methanol CH$_3$OH, ethanol C$_2$H$_5$OH, propan-1-ol C$_3$H$_7$OH.醇由连接到碳链（而非羰基碳）的羟基 –OH 定义。示例：甲醇 CH$_3$OH、乙醇 C$_2$H$_5$OH、丙-1-醇 C$_3$H$_7$OH。

–COOH is a carboxylic acid; C=O flanked by carbons is a ketone; –NH$_2$ is an amine. All alcohols carry the –OH (hydroxyl) group on a carbon chain.–COOH 是羧酸；两侧连碳的 C=O 是酮；–NH$_2$ 是胺。所有醇在碳链上都带有 –OH（羟基）。

What two functional classes react together in esterification?酯化反应中哪两类官能团发生反应？

§5 · Q2

Alkene + alkane烯烃 + 烷烃

Aldehyde + ketone醛 + 酮

Amine + alcohol胺 + 醇

Carboxylic acid + alcohol羧酸 + 醇

Esterification: carboxylic acid (–COOH) + alcohol (–OH) → ester (–COO–) + water. This is a condensation reaction — water is eliminated. AB Chemistry 30 Unit C GO2 and SCH4U B3.3 both name esterification as an assessed reaction type.酯化反应：羧酸（–COOH）+ 醇（–OH）→ 酯（–COO–）+ 水。这是缩合反应——水被消除。AB Chemistry 30 C 单元 GO2 和 SCH4U B3.3 均将酯化列为被评估的反应类型。

Esterification is specifically the reaction between a carboxylic acid and an alcohol, producing an ester and water. Amine + carboxylic acid produces an amide (not an ester). Alkene + alkane don't react without special conditions.酯化反应特指羧酸与醇的反应，生成酯和水。胺 + 羧酸生成酰胺（不是酯）。烯烃 + 烷烃在特殊条件之外不反应。

Going deeper — hydrogen bonding in alcohols and carboxylic acids深入 — 醇和羧酸中的氢键

The –OH group in alcohols and –COOH in carboxylic acids enable hydrogen bonding: the O–H bond is highly polar (O is electronegative), so the H bears a significant partial positive charge and is attracted to lone pairs on neighbouring O atoms. This raises boiling points dramatically compared to hydrocarbons of similar molar mass. Ethanol (M = 46 g/mol, bp 78 °C) has a much higher boiling point than propane (M = 44 g/mol, bp –42 °C) because ethanol molecules hydrogen-bond to each other while propane molecules interact only through weak London forces. Carboxylic acids form particularly strong dimers via two hydrogen bonds, explaining why acetic acid (ethanoic acid) has an unusually high bp (118 °C) for its size. SCH4U B3.2 requires comparing physical properties within and across functional-group classes, which connects directly to intermolecular force type.醇中的 –OH 和羧酸中的 –COOH 使氢键成为可能：O–H 键高度极性（O 电负性强），所以 H 带有显著的部分正电荷，并被邻近 O 原子上的孤对电子所吸引。与类似摩尔质量的烃相比，这大幅提高了沸点。乙醇（M = 46 g/mol，沸点 78 °C）的沸点远高于丙烷（M = 44 g/mol，沸点 -42 °C），因为乙醇分子之间通过氢键相互作用，而丙烷分子只有弱色散力。羧酸通过两个氢键形成特别强的二聚体，这解释了为何乙酸（乙酸）的沸点（118 °C）对其大小而言异常高。SCH4U B3.2 要求比较同类和不同类官能团的物理性质，这与分子间作用力类型直接相关。

Section 6 · Structural isomers第 6 节 · 结构同分异构体

Structural Isomers结构同分异构体

Structural isomers: same molecular formula, different connectivity.结构同分异构体：分子式相同，连接方式不同。

Definition定义 — structural isomers have identical molecular formulas but different structural formulas (different bonding arrangements). They are distinct compounds with different physical and chemical properties.— 结构同分异构体的分子式相同，但结构简式不同（不同的成键方式）。它们是具有不同物理和化学性质的不同化合物。
Types of structural isomerism结构同分异构体的类型
- Chain isomers链异构体 — same formula, different carbon skeleton (e.g. butane vs 2-methylpropane, both C$_4$H$_{10}$).— 分子式相同，碳骨架不同（如丁烷与 2-甲基丙烷，均为 C$_4$H$_{10}$）。
- Position isomers位置异构体 — same functional group at a different position on the same carbon skeleton (e.g. propan-1-ol vs propan-2-ol).— 相同官能团在同一碳骨架的不同位置（如丙-1-醇与丙-2-醇）。
- Functional-group isomers官能团异构体 — same molecular formula but different functional groups (e.g. ethanol C$_2$H$_6$O vs methoxymethane / dimethyl ether CH$_3$OCH$_3$, same C$_2$H$_6$O).— 分子式相同但官能团不同（如乙醇 C$_2$H$_6$O 与甲醚 / 二甲醚 CH$_3$OCH$_3$，均为 C$_2$H$_6$O）。

SCH4U B3.5 names "structural isomerism in organic compounds" as an assessed concept. AB Chemistry 30 Unit C GO1: "define structural isomerism as compounds having the same empirical formulas, but with different structural formulas." BC Chemistry 11 requires identifying structural isomers from molecular formulas.SCH4U B3.5 将"有机化合物中的结构同分异构现象"列为被评估概念。AB Chemistry 30 C 单元 GO1："将结构同分异构定义为具有相同实验式但结构式不同的化合物。"BC Chemistry 11 要求从分子式中识别结构同分异构体。

Worked Example 6 · Drawing structural isomers of C$_4$H$_{10}$例题 6 · 绘制 C$_4$H$_{10}$ 的结构同分异构体

Draw and name all structural isomers of C$_4$H$_{10}$ (molecular formula of butane).画出并命名 C$_4$H$_{10}$（丁烷的分子式）的所有结构同分异构体。

Isomer 1 — straight chain:异构体 1——直链： CH$_3$–CH$_2$–CH$_2$–CH$_3$ → butane (or n-butane). 4 carbons in a row.CH$_3$–CH$_2$–CH$_2$–CH$_3$ → 丁烷（butane）（或正丁烷）。4 个碳排成一行。

Isomer 2 — branched chain:异构体 2——支链：

CH$_3$–CH(CH$_3$)–CH$_3$ → 2-methylpropane (or iso-butane). 3-carbon parent chain (propane) with a methyl group at C-2.CH$_3$–CH(CH$_3$)–CH$_3$ → 2-甲基丙烷（2-methylpropane）（或异丁烷）。3 碳母链（丙烷）在 C-2 处连有甲基。

Verification:验证： Both have the formula C$_4$H$_{10}$ ✓, but different carbon skeletons → structural (chain) isomers. No further isomers exist for C$_4$H$_{10}$. Butane (bp –1 °C) and 2-methylpropane (bp –12 °C) have different physical properties despite identical molecular formulas.两者的分子式均为 C$_4$H$_{10}$ ✓，但碳骨架不同 → 结构（链）异构体。C$_4$H$_{10}$ 没有更多异构体。丁烷（沸点 –1 °C）和 2-甲基丙烷（沸点 –12 °C）尽管分子式相同，但物理性质不同。

Propan-1-ol and propan-2-ol both have the formula C$_3$H$_8$O. What type of structural isomers are they?丙-1-醇和丙-2-醇的分子式均为 C$_3$H$_8$O。它们属于哪种类型的结构同分异构体？

§6 · Q1

Chain isomers (different carbon skeleton)链异构体（不同碳骨架）

Functional-group isomers (different functional group)官能团异构体（不同官能团）

Position isomers (same functional group, different position)位置异构体（相同官能团，不同位置）

They are not isomers; they have different molecular formulas它们不是同分异构体；它们的分子式不同

Both compounds have the same 3-carbon skeleton and the same –OH functional group, but the hydroxyl is at C-1 in propan-1-ol (CH$_3$CH$_2$CH$_2$OH) and at C-2 in propan-2-ol (CH$_3$CHOHCH$_3$). Same group, different position = position isomers.两种化合物具有相同的 3 碳骨架和相同的 –OH 官能团，但丙-1-醇（CH$_3$CH$_2$CH$_2$OH）中羟基在 C-1，丙-2-醇（CH$_3$CHOHCH$_3$）中在 C-2。相同的基团，不同的位置 = 位置异构体。

Chain isomers differ in carbon skeleton length; functional-group isomers have entirely different functional groups (e.g., alcohol vs ether). Here the skeleton and functional group are the same — only the position of the –OH differs. That is position isomerism.链异构体的碳骨架长度不同；官能团异构体的官能团完全不同（如醇与醚）。这里骨架和官能团相同——只有 –OH 的位置不同。这是位置同分异构现象。

Ethanol (CH$_3$CH$_2$OH) and dimethyl ether (CH$_3$OCH$_3$) both have the formula C$_2$H$_6$O. How do they differ?乙醇（CH$_3$CH$_2$OH）和二甲醚（CH$_3$OCH$_3$）的分子式均为 C$_2$H$_6$O。它们有何不同？

§6 · Q2

They are functional-group isomers: ethanol has an –OH group; dimethyl ether has a C–O–C linkage它们是官能团异构体：乙醇含 –OH 基团；二甲醚含 C–O–C 连接

They are position isomers: the same functional group is at a different position它们是位置异构体：相同官能团在不同位置

They are chain isomers: different carbon skeleton length它们是链异构体：不同的碳骨架长度

They are not isomers; they have different empirical formulas它们不是同分异构体；它们的实验式不同

Ethanol has the hydroxyl group (–OH), making it an alcohol; dimethyl ether has a C–O–C ether linkage. Same molecular formula C$_2$H$_6$O, completely different functional groups = functional-group isomers. They have very different properties: ethanol bp 78 °C; dimethyl ether bp –24 °C.乙醇含羟基（–OH），是一种醇；二甲醚含 C–O–C 醚键。分子式同为 C$_2$H$_6$O，但官能团完全不同 = 官能团异构体。它们的性质差异很大：乙醇沸点 78 °C；二甲醚沸点 –24 °C。

Both have formula C$_2$H$_6$O so they are isomers (same molecular formula). They have different functional groups (–OH vs –O–), making them functional-group isomers. The carbon skeleton length is the same (2 carbon atoms each).两者均为 C$_2$H$_6$O，所以它们是同分异构体（相同的分子式）。它们有不同的官能团（–OH 与 –O–），使它们成为官能团异构体。碳骨架长度相同（各 2 个碳原子）。

Section 7 · Organic reactions and polymers第 7 节 · 有机反应与聚合物

Organic Reactions and Polymers (Introduction)有机反应与聚合物（入门）

Five core reaction types + polymerization to know.需掌握的五种核心反应类型 + 聚合反应。

Addition加成 — unsaturated molecule + small molecule → single product. No atoms lost. Examples: alkene + H$_2$ → alkane (hydrogenation); alkene + Br$_2$ → dibromoalkane.— 不饱和分子 + 小分子 → 单一产物。无原子损失。示例：烯烃 + H$_2$ → 烷烃（氢化）；烯烃 + Br$_2$ → 二溴烷烃。
Substitution取代 — one atom or group replaces another. Example: CH$_4$ + Cl$_2$ → CH$_3$Cl + HCl (free-radical halogenation of alkane; requires UV light).— 一个原子或基团取代另一个。示例：CH$_4$ + Cl$_2$ → CH$_3$Cl + HCl（烷烃的自由基卤代；需要紫外光）。
Elimination消除 — removal of atoms from adjacent carbons to form a C=C double bond. Example: haloalkane + KOH → alkene + HX.— 从相邻碳上去除原子以形成 C=C 双键。示例：卤代烷 + KOH → 烯烃 + HX。
Esterification酯化 — carboxylic acid + alcohol → ester + water (condensation; reversible with acid catalyst). See §5.— 羧酸 + 醇 → 酯 + 水（缩合；酸催化可逆）。见 §5。
Combustion燃烧 — hydrocarbon + O$_2$ → CO$_2$ + H$_2$O (complete combustion). Incomplete → CO + soot.— 烃 + O$_2$ → CO$_2$ + H$_2$O（完全燃烧）。不完全 → CO + 炭黑。
Polymerization聚合 — many small monomer molecules link to form a large polymer chain. Addition polymerization: alkene monomers add across C=C (e.g. ethene → polyethylene). Condensation polymerization: monomers link with loss of a small molecule (e.g. water) — produces nylon, polyester.— 许多小的单体分子连接成大的聚合物链。加成聚合：烯烃单体通过 C=C 加成（如乙烯 → 聚乙烯）。缩合聚合：单体通过消除小分子（如水）连接——产生尼龙、聚酯。

SCH4U B3.3 lists substitution, addition, elimination, oxidation, esterification, and hydrolysis as assessed reaction types. AB Chemistry 30 Unit C GO2: "define, illustrate and provide examples of simple addition, substitution, elimination, esterification and combustion reactions" and polymerization.SCH4U B3.3 将取代、加成、消除、氧化、酯化和水解列为被评估的反应类型。AB Chemistry 30 C 单元 GO2："定义、说明并举例简单加成、取代、消除、酯化和燃烧反应"及聚合反应。

Worked Example 7 · Addition polymerization: ethene to polyethylene例题 7 · 加成聚合：乙烯到聚乙烯

Ethene (C$_2$H$_4$, monomer) undergoes addition polymerization. (a) Write the repeating unit of polyethylene. (b) Identify the type of polymerization and explain why no small molecule is released.乙烯（C$_2$H$_4$，单体）发生加成聚合。(a) 写出聚乙烯的重复单元。(b) 确定聚合类型并解释为何不释放小分子。

(a) Repeating unit:(a) 重复单元：

$$ n\,\text{CH}_2\text{=CH}_2 \longrightarrow {-}\text{[CH}_2\text{-CH}_2\text{]}_n{-} $$

(b) Type and reason:(b) 类型与原因： This is addition polymerization. The C=C double bond opens (one bond of the double bond breaks and electrons re-pair with adjacent monomers), and all atoms of the monomer are incorporated into the polymer chain. No atoms are eliminated, so no small by-product (like water) is released. Contrast with condensation polymerization, where –OH or –NH$_2$ groups react and release H$_2$O or another small molecule.这是加成聚合。C=C 双键打开（双键中的一个键断裂，电子与相邻单体重新配对），单体的所有原子都被纳入聚合物链中。没有原子被消除，因此不释放任何小的副产物（如水）。与缩合聚合对比，缩合聚合中 –OH 或 –NH$_2$ 基团反应并释放 H$_2$O 或其他小分子。

What type of reaction occurs when chloromethane (CH$_3$Cl) is formed from methane and chlorine in the presence of UV light?在紫外光存在下，甲烷与氯气反应生成氯甲烷（CH$_3$Cl），这是什么类型的反应？

§7 · Q1

Addition加成

Substitution取代

Elimination消除

Esterification酯化

CH$_4$ + Cl$_2$ → CH$_3$Cl + HCl. One H atom on methane is replaced (substituted) by a Cl atom. This is halogenation by free-radical substitution. Addition reactions require a C=C bond to open; alkanes are saturated and do not undergo addition.CH$_4$ + Cl$_2$ → CH$_3$Cl + HCl。甲烷上的一个 H 原子被 Cl 原子取代。这是自由基取代的卤代反应。加成反应需要 C=C 键打开；烷烃是饱和的，不发生加成反应。

Addition requires a double or triple bond; methane (CH$_4$) has no C=C. Elimination removes atoms to form a double bond. Esterification involves –COOH and –OH. In this reaction, Cl substitutes for H — substitution.加成需要双键或三键；甲烷（CH$_4$）没有 C=C。消除去除原子形成双键。酯化涉及 –COOH 和 –OH。在此反应中，Cl 取代 H——是取代反应。

Polyethylene is made by addition polymerization of ethene. Which statement about this process is correct?聚乙烯由乙烯加成聚合制得。关于此过程，下列哪种说法正确？

§7 · Q2

Water is released as a by-product水作为副产物被释放

The monomer must contain a –COOH group单体必须含有 –COOH 基团

Each monomer loses one atom before joining the chain每个单体在加入链之前失去一个原子

All atoms of each monomer are incorporated into the polymer; no small molecule is released每个单体的所有原子都被纳入聚合物；不释放小分子

In addition polymerization, the C=C double bond opens and monomers link end-to-end. All atoms are retained in the polymer — no by-product is formed. This distinguishes it from condensation polymerization, which releases water (or HCl, etc.) as each monomer connects.在加成聚合中，C=C 双键打开，单体首尾相连。所有原子都保留在聚合物中——不形成副产物。这使它区别于缩合聚合，后者在每个单体连接时释放水（或 HCl 等）。

Water is released in condensation polymerization (e.g. nylon, polyester), not addition polymerization. Addition polymerization requires a C=C or C≡C bond in the monomer. All atoms join the chain; nothing is lost.水在缩合聚合（如尼龙、聚酯）中释放，而非加成聚合。加成聚合要求单体中有 C=C 或 C≡C 键。所有原子加入链中；什么都不会损失。

Exam Tactics考试策略

Exam Strategy and Common Pitfalls考试策略与常见陷阱

Naming and formula discipline命名与分子式纪律

Apply the four IUPAC steps every time.每次都应用四步 IUPAC 规则。 Longest chain containing principal group → number from the nearest end → name substituents alphabetically → add correct suffix. Skipping any step loses marks.含主要基团的最长链 → 从最近端编号 → 按字母顺序命名取代基 → 添加正确后缀。跳过任何步骤都会失分。
Check your molecular formula.核对分子式。 Draw the structural formula and count each element. Verify it matches the given molecular formula. Degree of unsaturation (DoU) $= (2C + 2 - H)/2$ is a fast check: 0 = alkane; 1 = one ring or double bond; 2 = two rings or double bonds, or one triple bond.画出结构简式，数各元素数目。核验与给定分子式相符。不饱和度 (DoU) $= (2C + 2 - H)/2$ 是快速检验：0 = 烷烃；1 = 一个环或双键；2 = 两个环或双键，或一个三键。
General formulas are exam shortcuts.通式是考试捷径。 Alkane C$_n$H$_{2n+2}$; alkene C$_n$H$_{2n}$; alkyne C$_n$H$_{2n-2}$. If a formula doesn't fit, the molecule has a ring or extra functional group.烷烃 C$_n$H$_{2n+2}$；烯烃 C$_n$H$_{2n}$；炔烃 C$_n$H$_{2n-2}$。如果分子式不符，该分子含有环或额外的官能团。

Functional groups and reactions (§5, §7)官能团与反应（§5、§7）

Identify the functional group first.先识别官能团。 The reaction type follows from the functional group: alkene → addition; carboxylic acid + alcohol → esterification; alkane + halogen (UV) → substitution; haloalkane + KOH → elimination.反应类型由官能团决定：烯烃 → 加成；羧酸 + 醇 → 酯化；烷烃 + 卤素（紫外光）→ 取代；卤代烷 + KOH → 消除。
Bromine water decolourisation = unsaturation test.溴水脱色 = 不饱和度测试。 Brown Br$_2$ solution decolourises rapidly with alkenes or alkynes (addition reaction); no change with alkanes or alcohols under mild conditions.棕色 Br$_2$ 溶液被烯烃或炔烃迅速脱色（加成反应）；在温和条件下与烷烃或醇无变化。
Addition vs condensation polymerization.加成聚合与缩合聚合。 Addition: monomer has C=C, no by-product (e.g. polyethylene). Condensation: monomers have two functional groups each, water (or HCl) is released (e.g. nylon, polyester).加成聚合：单体含 C=C，无副产物（如聚乙烯）。缩合聚合：单体各有两个官能团，释放水（或 HCl）（如尼龙、聚酯）。

Isomers (§6) — spot and draw systematically同分异构体（§6）——系统识别与绘制

Start with the longest chain, then branch.从最长链开始，再考虑支链。 For C$_5$H$_{12}$: (1) pentane (5C straight), (2) 2-methylbutane (4C + 1 methyl), (3) 2,2-dimethylpropane (3C + 2 methyls). Systematically shorten the chain by one each time until no more arrangements fit.对于 C$_5$H$_{12}$：(1) 戊烷（5C 直链），(2) 2-甲基丁烷（4C + 1 甲基），(3) 2,2-二甲基丙烷（3C + 2 甲基）。每次将链缩短一个碳，直到不能再排列为止。
Confirm each is a unique structure.确认每个结构是唯一的。 Rotate or mirror the structure before declaring it new. 2-methylbutane and 3-methylbutane are the same compound (just drawn from opposite ends); 2-methylbutane and 2-methylbutane are trivially the same.在声明为新结构之前，先旋转或镜像该结构。2-甲基丁烷和 3-甲基丁烷是同一化合物（只是从相反端画出的）；2-甲基丁烷和 2-甲基丁烷显然是同一化合物。

Answer hygiene for organic questions有机题的作答规范

Draw the structural formula, not just the name.画出结构简式，而不仅仅是名称。 In SCH4U and AB Chemistry 30 exams, condensed structural or full structural formulas are typically required for full marks. Molecular formula alone is insufficient unless specifically asked.在 SCH4U 和 AB Chemistry 30 考试中，通常需要简缩式或完整结构简式才能得满分。除非特别要求，仅写分子式是不够的。
Balance combustion equations by formula.按公式配平燃烧方程式。 For C$_x$H$_y$ + O$_2$ → CO$_2$ + H$_2$O: coefficients are $x$ (CO$_2$), $y/2$ (H$_2$O), and $(x + y/4)$ (O$_2$). Multiply through to clear fractions if needed.对于 C$_x$H$_y$ + O$_2$ → CO$_2$ + H$_2$O：系数为 $x$（CO$_2$）、$y/2$（H$_2$O）和 $(x + y/4)$（O$_2$）。如需要，乘以整数以消除分数。
Include arrow type for reversible reactions.可逆反应需注明箭头类型。 Esterification and polyester formation are reversible ($\rightleftharpoons$); combustion and addition polymerization are not ($\longrightarrow$).酯化和聚酯形成是可逆的（$\rightleftharpoons$）；燃烧和加成聚合不是（$\longrightarrow$）。

Active Recall主动复习

Flashcards闪卡

Why is carbon unique?碳为何独特？

Tetravalent (4 bonds); catenation (bonds to itself in chains and rings); strong C–C bonds (~346 kJ/mol). Creates millions of organic compounds.四价（4 个键）；成链性（与自身成键形成链和环）；C–C 键强（约 346 kJ/mol）。形成数百万种有机化合物。

Alkane general formula?烷烃通式？

$$\text{C}_n\text{H}_{2n+2}$$ Saturated; only single bonds; example: propane C$_3$H$_8$.饱和；仅单键；示例：丙烷 C$_3$H$_8$。

Alkene and alkyne formulas?烯烃和炔烃通式？

Alkene C$_n$H$_{2n}$ (one C=C). Alkyne C$_n$H$_{2n-2}$ (one C≡C). Each degree of unsaturation removes 2 H.烯烃 C$_n$H$_{2n}$（一个 C=C）。炔烃 C$_n$H$_{2n-2}$（一个 C≡C）。每个不饱和度减少 2 个 H。

IUPAC naming — 4 steps?IUPAC 命名——4 步？

1. Longest chain. 2. Number from nearest-principal-group end. 3. Name substituents (alphabetical, with locants). 4. Add suffix (-ane/-ene/-ol/-oic acid…).1. 最长链。2. 从最近主要基团端编号。3. 命名取代基（字母顺序，加定位数）。4. 添加后缀（-烷/-烯/-醇/-酸……）。

Alcohol functional group and suffix?醇的官能团与后缀？

–OH (hydroxyl group); suffix -ol. Example: ethanol CH$_3$CH$_2$OH. Hydrogen-bonds → high boiling point.–OH（羟基）；后缀 -醇。示例：乙醇 CH$_3$CH$_2$OH。氢键 → 高沸点。

Carboxylic acid group and suffix?羧酸官能团与后缀？

–COOH (carboxyl); suffix -oic acid. Example: ethanoic acid CH$_3$COOH. Acidic (donates H$^+$).–COOH（羧基）；后缀 -酸。示例：乙酸 CH$_3$COOH。酸性（提供 H$^+$）。

Esterification equation?酯化方程式？

RCOOH + R'OH ⇌ RCOOR' + H$_2$O. Acid catalyst, reversible. No atoms lost to the ester — water is the by-product.RCOOH + R'OH ⇌ RCOOR' + H$_2$O。酸催化，可逆。没有原子被酯丢失——水是副产物。

Addition reaction definition?加成反应定义？

Unsaturated molecule + small molecule → single product. Double/triple bond opens; no atoms lost. Alkenes + Br$_2$, H$_2$, HX, H$_2$O.不饱和分子 + 小分子 → 单一产物。双键/三键打开；无原子损失。烯烃 + Br$_2$、H$_2$、HX、H$_2$O。

Substitution in alkanes?烷烃中的取代？

Alkane + X$_2$ (UV light) → haloalkane + HX. One H replaced by X. Free-radical mechanism. Example: CH$_4$ + Cl$_2$ → CH$_3$Cl + HCl.烷烃 + X$_2$（紫外光）→ 卤代烷 + HX。一个 H 被 X 取代。自由基机制。示例：CH$_4$ + Cl$_2$ → CH$_3$Cl + HCl。

Structural isomers: definition?结构同分异构体：定义？

Same molecular formula, different structural (connectivity) arrangement. Types: chain, position, functional-group. Have different physical and chemical properties.分子式相同，结构（连接方式）不同。类型：链异构、位置异构、官能团异构。具有不同的物理和化学性质。

Addition polymerization?加成聚合？

Monomer has C=C; C=C bond opens; monomers link end-to-end. No by-product. Example: $n\,$CH$_2$=CH$_2$ → [–CH$_2$–CH$_2$–]$_n$ (polyethylene).单体含 C=C；C=C 键打开；单体首尾相连。无副产物。示例：$n\,$CH$_2$=CH$_2$ → [–CH$_2$–CH$_2$–]$_n$（聚乙烯）。

Bromine water test for unsaturation?溴水检验不饱和度？

Orange Br$_2$(aq) rapidly decolourises with alkenes/alkynes (addition). No colour change with alkanes. This distinguishes saturated from unsaturated hydrocarbons.橙色 Br$_2$(aq) 与烯烃/炔烃（加成）迅速脱色。与烷烃无颜色变化。用于区分饱和与不饱和烃。

Degree of unsaturation (DoU) formula?不饱和度（DoU）公式？

$$\text{DoU} = \frac{2C + 2 - H}{2}$$ 0 = saturated; 1 = one ring or C=C; 2 = two rings/double bonds or one triple bond.0 = 饱和；1 = 一个环或 C=C；2 = 两个环/双键或一个三键。

Complete combustion of any hydrocarbon?任何烃的完全燃烧？

C$_x$H$_y$ + O$_2$ → $x$CO$_2$ + $\frac{y}{2}$H$_2$O. Multiply through to clear fractions. Incomplete combustion → CO and C (soot).C$_x$H$_y$ + O$_2$ → $x$CO$_2$ + $\frac{y}{2}$H$_2$O。乘以整数以消除分数。不完全燃烧 → CO 和 C（炭黑）。

Mixed Practice综合练习

Practice Quiz综合测验

Which of the following is a saturated hydrocarbon? 🇨🇦 SCH4U B3.1 / AB Chem 30 C以下哪种是饱和烃？🇨🇦 SCH4U B3.1 / AB Chem 30 C

Ethene (C$_2$H$_4$)乙烯（C$_2$H$_4$）

Ethyne (C$_2$H$_2$)乙炔（C$_2$H$_2$）

Benzene (C$_6$H$_6$)苯（C$_6$H$_6$）

Propane (C$_3$H$_8$)丙烷（C$_3$H$_8$）

Propane (C$_3$H$_8$) is an alkane — no double or triple bonds; all carbons are saturated (maximum H). Check: C$_3$H$_{2(3)+2}$ = C$_3$H$_8$ ✓. Ethene has C=C (DoU = 1); ethyne has C≡C (DoU = 2); benzene has a ring + double bonds (DoU = 4).丙烷（C$_3$H$_8$）是烷烃——没有双键或三键；所有碳都是饱和的（氢最多）。核验：C$_3$H$_{2(3)+2}$ = C$_3$H$_8$ ✓。乙烯有 C=C（DoU = 1）；乙炔有 C≡C（DoU = 2）；苯有一个环 + 双键（DoU = 4）。

Saturated = only single bonds (alkane formula C$_n$H$_{2n+2}$). Ethene (C$_2$H$_4$) and ethyne (C$_2$H$_2$) have double and triple bonds; benzene is aromatic. Only propane fits C$_n$H$_{2n+2}$.饱和 = 仅单键（烷烃通式 C$_n$H$_{2n+2}$）。乙烯（C$_2$H$_4$）和乙炔（C$_2$H$_2$）有双键和三键；苯是芳香族的。只有丙烷符合 C$_n$H$_{2n+2}$。

What is the IUPAC name of CH$_3$CH$_2$CH$_2$CH$_2$CH$_3$? 🇨🇦 SCH4U B2.2 / AB Chem 30 C GO1CH$_3$CH$_2$CH$_2$CH$_2$CH$_3$ 的 IUPAC 名称是什么？🇨🇦 SCH4U B2.2 / AB Chem 30 C GO1

Pentane戊烷

Butane丁烷

Hexane己烷

Propane丙烷

5 carbon atoms in a straight chain + alkane suffix -ane → pentane. Prefix: pent- (5). Formula: C$_5$H$_{12}$ = C$_5$H$_{2(5)+2}$ ✓.5 个碳原子的直链 + 烷烃后缀 -烷 → 戊烷。前缀：戊（pent-，5）。分子式：C$_5$H$_{12}$ = C$_5$H$_{2(5)+2}$ ✓。

Count the carbons in the chain: 5. Prefix for 5 = pent-. Suffix for saturated (no double/triple bonds) = -ane. Answer: pentane. Butane = 4C; hexane = 6C; propane = 3C.数链中的碳：5 个。5 的前缀 = 戊（pent-）。饱和（无双键/三键）的后缀 = -烷（-ane）。答案：戊烷（pentane）。丁烷 = 4C；己烷 = 6C；丙烷 = 3C。

Which functional group is characteristic of a carboxylic acid? 🇨🇦 SCH4U B2.1 / AB Chem 30 C GO1羧酸的特征官能团是什么？🇨🇦 SCH4U B2.1 / AB Chem 30 C GO1

–OH (hydroxyl)–OH（羟基）

–NH$_2$ (amino)–NH$_2$（氨基）

–COOH (carboxyl)–COOH（羧基）

–CHO (aldehyde)–CHO（醛基）

The carboxyl group (–COOH) contains both a carbonyl (C=O) and a hydroxyl (–OH) attached to the same carbon. It makes the compound acidic (proton donor). Example: ethanoic acid CH$_3$COOH (acetic acid). SCH4U B2.1 and AB Chemistry 30 GO1 both list this group.羧基（–COOH）包含连接在同一碳上的羰基（C=O）和羟基（–OH）。它使化合物具有酸性（质子供体）。示例：乙酸 CH$_3$COOH。SCH4U B2.1 和 AB Chemistry 30 GO1 均列出此基团。

–OH (hydroxyl) = alcohol; –NH$_2$ = amine; –CHO = aldehyde. Carboxylic acids are defined by the –COOH (carboxyl) group — C=O plus –OH on the same carbon.–OH（羟基）= 醇；–NH$_2$ = 胺；–CHO = 醛。羧酸由 –COOH（羧基）定义——同一碳上的 C=O 加 –OH。

Two compounds have the formula C$_5$H$_{12}$: pentane and 2-methylbutane. They are examples of: 🇨🇦 SCH4U B3.5 / AB Chem 30 C GO1两种化合物的分子式为 C$_5$H$_{12}$：戊烷和 2-甲基丁烷。它们是以下哪种情况的例子：🇨🇦 SCH4U B3.5 / AB Chem 30 C GO1

Geometric isomers (E/Z)几何异构体（E/Z）

Chain isomers (structural isomers with different carbon skeletons)链异构体（碳骨架不同的结构同分异构体）

Position isomers (same skeleton, functional group at different position)位置异构体（相同骨架，官能团在不同位置）

Allotropes of carbon碳的同素异形体

Pentane has a 5-carbon straight chain; 2-methylbutane has a 4-carbon parent chain plus a methyl branch — different carbon skeletons, same formula C$_5$H$_{12}$ → chain isomers (a type of structural isomer). Geometric (E/Z) isomers require a C=C with different substituents; these are both alkanes.戊烷有 5 碳直链；2-甲基丁烷有 4 碳母链加甲基支链——不同的碳骨架，相同的分子式 C$_5$H$_{12}$ → 链异构体（结构同分异构体的一种）。几何（E/Z）异构体需要 C=C 且具有不同的取代基；这两个都是烷烃。

Geometric isomers require a C=C double bond (absent here). Position isomers have the same skeleton but functional group at different positions. Allotropes are different forms of the same element. Here: same formula C$_5$H$_{12}$, different carbon skeleton = chain isomers.几何异构体需要 C=C 双键（此处没有）。位置异构体的骨架相同但官能团在不同位置。同素异形体是同一元素的不同形态。这里：相同的分子式 C$_5$H$_{12}$，不同的碳骨架 = 链异构体。

What product forms when propene (CH$_2$=CHCH$_3$) reacts with hydrogen (H$_2$) by addition? 🇨🇦 SCH4U B3.3 / AB Chem 30 C GO2丙烯（CH$_2$=CHCH$_3$）与氢气（H$_2$）发生加成反应，生成什么产物？🇨🇦 SCH4U B3.3 / AB Chem 30 C GO2

Propane (CH$_3$CH$_2$CH$_3$)丙烷（CH$_3$CH$_2$CH$_3$）

Propanol (CH$_3$CH$_2$CH$_2$OH)丙醇（CH$_3$CH$_2$CH$_2$OH）

Propanal (CH$_3$CH$_2$CHO)丙醛（CH$_3$CH$_2$CHO）

Carbon dioxide and water二氧化碳和水

Addition of H$_2$ across C=C (hydrogenation): CH$_2$=CHCH$_3$ + H$_2$ → CH$_3$CH$_2$CH$_3$ (propane). The double bond is broken; each carbon gains one H. All atoms of both reactants are in the product.H$_2$ 跨越 C=C 加成（氢化）：CH$_2$=CHCH$_3$ + H$_2$ → CH$_3$CH$_2$CH$_3$（丙烷）。双键断裂；每个碳获得一个 H。两种反应物的所有原子都在产物中。

Propanol forms when water (H$_2$O) adds across the double bond (hydration). Propanal is an oxidation product of propanol. Combustion yields CO$_2$ + H$_2$O. H$_2$ addition (hydrogenation) converts the alkene to the corresponding alkane.丙醇是水（H$_2$O）跨越双键加成（水化）的产物。丙醛是丙醇的氧化产物。燃烧生成 CO$_2$ + H$_2$O。H$_2$ 加成（氢化）将烯烃转化为相应的烷烃。

What is the degree of unsaturation (DoU) of benzene C$_6$H$_6$? 🇨🇦 BC Chemistry 11 / SCH4U B苯 C$_6$H$_6$ 的不饱和度（DoU）是多少？🇨🇦 BC Chemistry 11 / SCH4U B

DoU $= (2C + 2 - H)/2 = (2 \times 6 + 2 - 6)/2 = 8/2 = 4$. This reflects benzene's ring (1 DoU) + 3 degrees from its delocalized double bonds (or 3 equivalent C=C). High DoU is characteristic of aromatic compounds.DoU $= (2C + 2 - H)/2 = (2 \times 6 + 2 - 6)/2 = 8/2 = 4$。这反映了苯的环（1 个 DoU）+ 3 个来自其离域双键（或 3 个等价 C=C）的不饱和度。高 DoU 是芳香族化合物的特征。

Apply DoU = (2C + 2 − H)/2. For C$_6$H$_6$: (12 + 2 − 6)/2 = 8/2 = 4. Benzene has a 6-membered ring plus three alternating double bonds (represented by the ring + 3 C=C or the delocalized model) — 4 degrees of unsaturation total.应用 DoU = (2C + 2 − H)/2。C$_6$H$_6$：(12 + 2 − 6)/2 = 8/2 = 4。苯有一个 6 元环加三个交替双键（用环 + 3 个 C=C 或离域模型表示）——总共 4 个不饱和度。

In polyester formation (a condensation polymer), what small molecule is released as each monomer joins the chain? 🇨🇦 SCH4U B3.4 / AB Chem 30 C GO2在聚酯形成（缩合聚合物）中，每个单体加入链时释放什么小分子？🇨🇦 SCH4U B3.4 / AB Chem 30 C GO2

Carbon dioxide (CO$_2$)二氧化碳（CO$_2$）

No small molecule — all atoms are incorporated没有小分子——所有原子都被纳入

Water (H$_2$O)水（H$_2$O）

Hydrogen (H$_2$)氢气（H$_2$）

Condensation polymerization: each monomer linkage forms by the reaction of two functional groups (e.g. –OH + –COOH → ester bond + H$_2$O). Water is eliminated each time. This is why condensation polymers have lower molecular weight per monomer than addition polymers.缩合聚合：每个单体连接由两个官能团的反应形成（如 –OH + –COOH → 酯键 + H$_2$O）。每次都消除水。这就是为什么缩合聚合物每个单体的分子量比加成聚合物低。

CO$_2$ is released in combustion. "No small molecule" describes addition polymerization (alkene monomers). H$_2$ is not a by-product of polymerization. In condensation polymerization, –OH + –COOH → ester bond + H$_2$O eliminates water each time.CO$_2$ 在燃烧中释放。"没有小分子"描述加成聚合（烯烃单体）。H$_2$ 不是聚合反应的副产物。在缩合聚合中，–OH + –COOH → 酯键 + H$_2$O，每次都消除水。

Which reaction converts a haloalkane to an alkene? 🇨🇦 SCH4U B3.3 / AB Chem 30 C GO2哪种反应将卤代烷转化为烯烃？🇨🇦 SCH4U B3.3 / AB Chem 30 C GO2

Addition加成

Elimination消除

Esterification酯化

Combustion燃烧

Elimination: CH$_3$CHBrCH$_3$ + KOH → CH$_3$CH=CH$_2$ + KBr + H$_2$O. HBr is eliminated from adjacent carbons, forming a C=C double bond. Elimination is the reverse of addition.消除：CH$_3$CHBrCH$_3$ + KOH → CH$_3$CH=CH$_2$ + KBr + H$_2$O。HBr 从相邻碳上消除，形成 C=C 双键。消除是加成的逆过程。

Addition adds atoms across a double bond (alkene → haloalkane). Esterification makes esters. Combustion burns the compound. To make an alkene from a haloalkane, you remove HX — that is elimination.加成向双键加原子（烯烃 → 卤代烷）。酯化制备酯。燃烧使化合物燃烧。要从卤代烷制备烯烃，需要去除 HX——这是消除反应。

Identify the correct IUPAC name for CH$_3$CH(CH$_3$)CH$_2$OH. 🇨🇦 SCH4U B2.2 / AB Chem 30 C GO1确定 CH$_3$CH(CH$_3$)CH$_2$OH 的正确 IUPAC 名称。🇨🇦 SCH4U B2.2 / AB Chem 30 C GO1

2-methylpropan-1-ol2-甲基丙-1-醇

1-methylpropan-2-ol1-甲基丙-2-醇

2-methylpropanol2-甲基丙醇

Isobutanol异丁醇

Parent chain: 3 carbons containing the –OH group → propan. Number from the OH end: C1 = CH$_2$OH, C2 = CH(CH$_3$), C3 = CH$_3$. The –OH is at C-1 → propan-1-ol. Methyl substituent at C-2 → 2-methylpropan-1-ol. ✓母链：含 –OH 基团的 3 个碳 → 丙烷（propan）。从 OH 端编号：C1 = CH$_2$OH，C2 = CH(CH$_3$)，C3 = CH$_3$。–OH 在 C-1 → 丙-1-醇（propan-1-ol）。C-2 处的甲基取代基 → 2-甲基丙-1-醇（2-methylpropan-1-ol）。✓

Find the parent chain containing the OH group (3C = propane base). Number from the OH end to give it the lowest locant (C-1). Methyl branch is at C-2. The suffix -ol confirms alcohol. Full name: 2-methylpropan-1-ol. "Isobutanol" is a common name, not IUPAC.找含 OH 基团的母链（3C = 丙烷基）。从 OH 端编号以给它最低定位数（C-1）。甲基支链在 C-2。后缀 -醇（-ol）确认是醇。完整名称：2-甲基丙-1-醇（2-methylpropan-1-ol）。"异丁醇"是通俗名称，不是 IUPAC。

Carbon has the ability to bond to itself in long chains. What is this property called? 🇨🇦 SCH4U B / BC Chemistry 11碳具有与自身成键形成长链的能力。这种性质叫什么？🇨🇦 SCH4U B / BC Chemistry 11

Q10

Tetravalency四价性

Hybridisation杂化

Electronegativity电负性

Catenation成链性

Catenation is the ability of an atom to form bonds with atoms of the same element, producing chains or rings. Carbon excels at catenation because C–C bonds are strong (~346 kJ/mol) and carbon can bond to itself in straight chains, branched chains, rings, and even cages (like buckminsterfullerene C$_{60}$).成链性是原子与同种元素原子成键形成链或环的能力。碳在成链方面表现突出，因为 C–C 键较强（约 346 kJ/mol），碳可以与自身成键形成直链、支链、环，甚至笼状结构（如富勒烯 C$_{60}$）。

Tetravalency = 4 bonds per carbon atom. Hybridisation describes orbital mixing (sp$^3$, sp$^2$, sp). Electronegativity = tendency to attract bonding electrons. Catenation specifically means self-bonding into chains/rings — the unique property that creates the vast diversity of organic compounds.四价性 = 每个碳原子 4 个键。杂化描述轨道混合（sp$^3$、sp$^2$、sp）。电负性 = 吸引成键电子的趋势。成链性特指自身成键形成链/环——创造有机化合物巨大多样性的独特性质。

Self-check自查

Readiness Checklist准备就绪清单

Tick each item when you can do it cold, without notes, on a first attempt.能在无笔记、首次尝试下完成，再勾选每一项。

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✓Define an organic compound, state the inorganic exceptions (CO, CO$_2$, carbonates, cyanides, carbides), and explain why carbon's tetravalency and catenation ability make it uniquely suited for building millions of compounds. 🇨🇦 SCH4U B2.1 / AB Chem 30 C GO1定义有机化合物，列举无机例外（CO、CO$_2$、碳酸盐、氰化物、碳化物），并解释为何碳的四价性和成链性使其独特地适合构建数百万种化合物。🇨🇦 SCH4U B2.1 / AB Chem 30 C GO1
✓State the general formulas for alkanes (C$_n$H$_{2n+2}$), alkenes (C$_n$H$_{2n}$), and alkynes (C$_n$H$_{2n-2}$); name the first 10 straight-chain alkanes; and explain why alkenes/alkynes are called unsaturated. 🇨🇦 SCH4U B3.1 / BC Chem 11 / AB Chem 30 C GO1列出烷烃（C$_n$H$_{2n+2}$）、烯烃（C$_n$H$_{2n}$）和炔烃（C$_n$H$_{2n-2}$）的通式；命名前 10 个直链烷烃；并解释为何烯烃/炔烃被称为不饱和烃。🇨🇦 SCH4U B3.1 / BC Chem 11 / AB Chem 30 C GO1
✓Apply the 4-step IUPAC naming procedure to name a branched-chain alkane or alkene with up to 10 carbons; and given an IUPAC name, draw the structural formula. 🇨🇦 SCH4U B2.2 / AB Chem 30 C GO1应用 4 步 IUPAC 命名程序命名最多 10 个碳的支链烷烃或烯烃；并根据 IUPAC 名称画出结构简式。🇨🇦 SCH4U B2.2 / AB Chem 30 C GO1
✓Identify and name the major functional groups (hydroxyl, carboxyl, ester linkage, carbonyl/aldehyde/ketone, amino, amide, ether, haloalkane) from a structural formula, and state the IUPAC suffix for each. 🇨🇦 SCH4U B2.2 / AB Chem 30 C GO1从结构简式中识别并命名主要官能团（羟基、羧基、酯键、羰基/醛基/酮基、氨基、酰胺、醚、卤代烷），并说明每种的 IUPAC 后缀。🇨🇦 SCH4U B2.2 / AB Chem 30 C GO1
✓Write balanced equations for esterification (carboxylic acid + alcohol → ester + water) and explain the equilibrium sign (reversible) and the role of the acid catalyst. 🇨🇦 SCH4U B3.3 / AB Chem 30 C GO2写出酯化反应（羧酸 + 醇 → 酯 + 水）的配平方程式，并解释平衡符号（可逆）和酸催化剂的作用。🇨🇦 SCH4U B3.3 / AB Chem 30 C GO2
✓Define structural isomerism, distinguish chain, position, and functional-group isomers, and draw all structural isomers of a given molecular formula (e.g., C$_4$H$_{10}$, C$_3$H$_8$O). 🇨🇦 SCH4U B3.5 / AB Chem 30 C GO1定义结构同分异构现象，区分链异构、位置异构和官能团异构体，并画出给定分子式（如 C$_4$H$_{10}$、C$_3$H$_8$O）的所有结构同分异构体。🇨🇦 SCH4U B3.5 / AB Chem 30 C GO1
✓Use the bromine water test to distinguish saturated from unsaturated hydrocarbons, and explain the chemistry of the addition reaction responsible for the decolourisation. 🇨🇦 SCH4U B2.4 / BC Chem 11用溴水测试区分饱和与不饱和烃，并解释导致脱色的加成反应的化学原理。🇨🇦 SCH4U B2.4 / BC Chem 11
✓Name and describe five organic reaction types: addition, substitution, elimination, esterification, and combustion; write balanced equations for each with a specific example. 🇨🇦 SCH4U B3.3 / AB Chem 30 C GO2命名并描述五种有机反应类型：加成、取代、消除、酯化和燃烧；为每种类型用具体示例写出配平方程式。🇨🇦 SCH4U B3.3 / AB Chem 30 C GO2
✓Distinguish addition polymerization (alkene monomer, no by-product, e.g. polyethylene) from condensation polymerization (bifunctional monomers, H$_2$O released, e.g. nylon/polyester), and define monomer, polymer, and repeating unit. 🇨🇦 SCH4U B3.4 / AB Chem 30 C GO2区分加成聚合（烯烃单体，无副产物，如聚乙烯）与缩合聚合（双官能团单体，释放 H$_2$O，如尼龙/聚酯），并定义单体、聚合物和重复单元。🇨🇦 SCH4U B3.4 / AB Chem 30 C GO2
✓Calculate the degree of unsaturation (DoU) for a given molecular formula, interpret the result (0 = alkane; 1 = one ring or double bond; 4 = benzene), and use it to determine what structural features must be present. 🇨🇦 SCH4U B / BC Chem 11计算给定分子式的不饱和度（DoU），解释结果（0 = 烷烃；1 = 一个环或双键；4 = 苯），并利用它确定必须存在的结构特征。🇨🇦 SCH4U B / BC Chem 11
✓Compare the physical properties (boiling points, solubility) of alkanes, alcohols, and carboxylic acids with similar molar mass, explaining the difference in terms of intermolecular forces (London forces, hydrogen bonds). 🇨🇦 SCH4U B3.2 / AB Chem 30 C GO1比较摩尔质量相近的烷烃、醇和羧酸的物理性质（沸点、溶解性），从分子间作用力（色散力、氢键）角度解释差异。🇨🇦 SCH4U B3.2 / AB Chem 30 C GO1

Next Steps后续单元

What This Feeds Into本单元的去向

Organic chemistry is the final unit in the High School Chemistry series — it draws on bonding (Unit 3), nomenclature (Unit 4), reactions and equations (Unit 6), and thermochemistry (Unit 10), and applies them to the largest class of compounds in chemistry. The cross-references below point to what comes next in the Dingrui Scholars curriculum.有机化学是高中化学系列的最后一个单元——它综合运用成键（第 3 单元）、命名（第 4 单元）、化学反应与方程式（第 6 单元）和热化学（第 10 单元），并将其应用于化学中最大的化合物类别。以下链接指向 Dingrui Scholars 课程中的后续内容。

Within High School Chemistry.在 HS Chemistry 内部。

This is the last unit in the High School Chemistry series. Organic chemistry draws directly on Unit 3 (Chemical Bonding: why carbon forms 4 covalent bonds), Unit 4 (Nomenclature: the IUPAC naming system is extended here to organic molecules), Unit 6 (Chemical Reactions: combustion and substitution reaction types), and Unit 10 (Thermochemistry: energy changes in organic reactions, especially combustion of hydrocarbons). Reviewing those units reinforces the organic content here.这是高中化学系列的最后一个单元。有机化学直接依赖第 3 单元（化学键：为何碳形成 4 个共价键）、第 4 单元（命名：IUPAC 命名系统在此延伸至有机分子）、第 6 单元（化学反应：燃烧和取代反应类型）和第 10 单元（热化学：有机反应中的能量变化，尤其是烃的燃烧）。复习这些单元可强化此处的有机内容。

College-credit feeder: IB Chemistry HL (Organic).大学学分衔接：IB Chemistry HL（有机化学）。

No dedicated IB Chemistry HL organic study guide currently exists in this series. IB Chemistry HL covers organic chemistry in Reactivity 3 (Organic Chemistry), which extends the content here with more functional groups (amides, aromatic compounds), reaction mechanisms (S$_N$1, S$_N$2, E1, E2), spectroscopic identification (IR, NMR, mass spec), and green chemistry principles. If you are targeting IB Chemistry HL, the IUPAC naming, functional groups, and reaction types in this guide form the required foundation — build on them by consulting your IB subject guide and a Reactivity 3 resource. AP Chemistry (Unit 4 and beyond) similarly assumes this foundation and extends to molecular orbital theory, acid-base properties of organic molecules, and biochemistry.本系列目前没有专属的 IB Chemistry HL 有机化学学习指南。IB Chemistry HL 在 Reactivity 3（有机化学）中涵盖有机化学内容，扩展了更多官能团（酰胺、芳香族化合物）、反应机制（S$_N$1、S$_N$2、E1、E2）、光谱鉴定（IR、NMR、质谱）和绿色化学原则。如果你目标是 IB Chemistry HL，本指南中的 IUPAC 命名、官能团和反应类型构成所需基础——在此基础上通过参考 IB 科目指南和 Reactivity 3 资源加以深化。AP Chemistry（第 4 单元及以后）同样假设这一基础，并延伸至分子轨道理论、有机分子的酸碱性质和生物化学。