📚 A-Level Chemistry: Reaction Mechanisms (Insert 3, June 2022) | A-Level 化学：反应机理（Insert 3，2022年6月）
The A-Level Chemistry exam often includes an insert booklet containing essential data and reaction mechanisms. Insert 3 from the June 2022 examination series is particularly important for organic chemistry, as it summarises key reaction mechanisms such as free radical substitution, electrophilic addition, nucleophilic substitution, elimination, and more. Mastering these mechanisms allows you to draw curly arrow diagrams correctly, predict products, and explain reaction pathways.
A-Level 化学考试通常会提供包含核心数据和反应机理的插入页。2022 年 6 月考试系列的 Insert 3 对有机化学尤为重要，它总结了自由基取代、亲电加成、亲核取代、消除等关键反应机理。熟练这些机理能帮助你正确绘制弯箭头示意图、预测产物并解释反应路径。
1. Free Radical Substitution | 自由基取代
Free radical substitution is the mechanism by which alkanes react with halogens (e.g., Cl₂ or Br₂) in the presence of ultraviolet (UV) light. The overall reaction for methane with chlorine is CH₄ + Cl₂ → CH₃Cl + HCl. This is a chain reaction involving three stages: initiation, propagation, and termination.
自由基取代是烷烃在紫外光（UV）存在下与卤素（如 Cl₂ 或 Br₂）反应的机理。甲烷与氯气的总反应为 CH₄ + Cl₂ → CH₃Cl + HCl。这是一个包含引发、增长和终止三个阶段的链反应。
Initiation: Cl₂ → 2Cl• (UV light)
UV light provides energy for homolytic fission of the Cl–Cl bond, generating two highly reactive chlorine radicals. Each chlorine radical has an unpaired electron, represented by the dot (Cl•).
紫外光提供能量使 Cl–Cl 键均裂，生成两个高活性的氯自由基。每个氯自由基带一个未成对电子，用点（Cl•）表示。
Propagation: Cl• + CH₄ → •CH₃ + HCl • •CH₃ + Cl₂ → CH₃Cl + Cl•
The chlorine radical abstracts a hydrogen atom from methane, forming a methyl radical (•CH₃) and HCl. The methyl radical then reacts with a chlorine molecule to produce chloromethane and regenerate a chlorine radical, sustaining the chain.
氯自由基从甲烷中夺取一个氢原子，形成甲基自由基（•CH₃）和 HCl。随后甲基自由基与氯分子反应，生成氯甲烷并再生一个氯自由基，使链反应持续进行。
Termination: 2Cl• → Cl₂ • Cl• + •CH₃ → CH₃Cl • 2•CH₃ → C₂H₆
Two radicals combine to form a stable molecule, ending the chain. Termination produces small amounts of ethane from methyl radical coupling.
两个自由基结合形成稳定分子，使链终止。终止步骤中甲基自由基偶联会生成少量乙烷。
2. Electrophilic Addition to Alkenes | 烯烃的亲电加成
Alkenes undergo electrophilic addition with reagents such as HBr, Br₂, and concentrated H₂SO₄. The electron-rich C=C double bond attracts the electrophile, leading to the formation of a carbocation intermediate and subsequent attack by a nucleophile.
烯烃与 HBr、Br₂ 和浓 H₂SO₄ 等试剂发生亲电加成。富电子的 C=C 双键吸引亲电试剂，形成碳正离子中间体，随后被亲核试剂进攻。
Example: CH₃CH=CH₂ + HBr → CH₃CHBrCH₃ (major product)
In the first, rate-determining step, the H⁺ from HBr attacks the double bond, generating the more stable secondary carbocation (CH₃CH⁺–CH₃) rather than the primary one. This follows Markovnikov’s rule.
在第一步决速步骤中，HBr 的 H⁺ 进攻双键，生成较稳定的二级碳正离子（CH₃CH⁺–CH₃）而非一级碳正离子，遵循马尔科夫尼科夫规则。
Step 1: CH₃CH=CH₂ + H⁺ → CH₃CH⁺–CH₃ (slow)
Step 2: CH₃CH⁺–CH₃ + Br⁻ → CH₃CHBrCH₃ (fast)
The bromide ion (Br⁻) then acts as a nucleophile and rapidly donates an electron pair to the positively charged carbon, forming the final addition product.
随后溴离子（Br⁻）作为亲核试剂快速给带正电荷的碳提供电子对，形成最终加成产物。
With bromine water (Br₂), the mechanism is similar, but the intermediate is a cyclic bromonium ion rather than a planar carbocation, leading to anti-addition.
与溴水（Br₂）反应时，机理类似，但中间体是环状溴鎓离子而非平面碳正离子，导致反式加成。
3. Nucleophilic Substitution: SN1 and SN2 | 亲核取代：SN₁ 与 SN₂
Haloalkanes can react with nucleophiles such as OH⁻, NH₃, or CN⁻ via nucleophilic substitution. The mechanism depends on the class of haloalkane: primary haloalkanes favour SN2, while tertiary haloalkanes favour SN1. Secondary haloalkanes can proceed via either pathway depending on conditions.
卤代烷可与 OH⁻、NH₃ 或 CN⁻ 等亲核试剂发生亲核取代。机理取决于卤代烷的种类：一级卤代烷倾向 SN₂，三级卤代烷倾向 SN₁。二级卤代烷根据条件可通过两种途径进行。
SN2 Mechanism: This is a one-step, concerted process in which the nucleophile attacks the carbon bearing the leaving group from the opposite side, leading to inversion of configuration. The transition state involves a pentacoordinate carbon. The rate equation is rate = k[haloalkane][nucleophile].
SN₂ 机理：这是一个一步协同过程，亲核试剂从离去基团背面进攻碳原子，导致构型翻转。过渡态涉及五配位碳。速率方程为 rate = k[卤代烷][亲核试剂]。
Example: CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻
SN1 Mechanism: This is a two-step process. The rate-determining step is the unimolecular loss of the leaving group to form a planar carbocation intermediate. The nucleophile then attacks the carbocation from either face, giving a racemic mixture. The rate equation is rate = k[haloalkane], independent of nucleophile concentration.
SN₁ 机理：这是一个两步过程。决速步骤是单分子的离去基团离去，形成平面碳正离子中间体。亲核试剂随后可从任一面进攻碳正离子，得到外消旋混合物。速率方程为 rate = k[卤代烷]，与亲核试剂浓度无关。
Example: (CH₃)₃CBr + OH⁻ → (CH₃)₃COH + Br⁻
4. Elimination Reactions | 消除反应
When a haloalkane is heated with a strong base (e.g., KOH in ethanol), elimination of HX occurs, forming an alkene. The hydroxide ion acts as a base rather than a nucleophile. This is the E2 mechanism for primary and secondary haloalkanes: a single step where the base removes a β-hydrogen while the halogen leaves, forming a double bond.
当卤代烷与强碱（如 KOH 的乙醇溶液）加热时，发生 HX 的消除，生成烯烃。氢氧根离子此时作为碱而非亲核试剂。对于一级和二级卤代烷，这是 E2 机理：一步反应中，碱夺取 β-氢，同时卤原子离去，形成双键。
E2: CH₃CH₂Br + OH⁻ → CH₂=CH₂ + H₂O + Br⁻
The E2 mechanism requires an anti-periplanar arrangement of the β-hydrogen and the leaving group for efficient overlap. Tertiary haloalkanes can undergo E1 elimination via a carbocation intermediate, analogous to SN1, but the E2 pathway is often dominant with strong bases and heat.
E2 机理要求 β-氢与离去基团呈反式共平面排列，以获得有效的轨道重叠。三级卤代烷可通过碳正离子中间体进行 E1 消除，类似于 SN₁，但在强碱和加热条件下 E2 路径常占主导。
5. Electrophilic Substitution of Benzene | 苯的亲电取代
Benzene, despite its high electron density, does not undergo addition reactions because that would destroy its aromatic stability. Instead, it undergoes electrophilic substitution, where an electrophile replaces a hydrogen atom on the ring. Classic examples include nitration, halogenation, and Friedel–Crafts alkylation/acylation.
苯尽管电子密度高，却不发生加成反应，因为这会破坏其芳香稳定性。它发生的是亲电取代，由亲电试剂取代环上的一个氢原子。经典例子包括硝化、卤化和傅-克烷基化/酰基化。
Nitration: A mixture of concentrated HNO₃ and concentrated H₂SO₄ generates the electrophile NO₂⁺ (nitronium ion). The benzene ring donates a pair of π electrons to NO₂⁺, forming an arenium ion (sigma complex). Loss of a proton restores aromaticity, giving nitrobenzene.
硝化：浓 HNO₃ 与浓 H₂SO₄ 的混合物生成亲电试剂 NO₂⁺（硝鎓离子）。苯环提供一对 π 电子给 NO₂⁺，形成芳正离子（σ 络合物）。失去一个质子后恢复芳香性，得到硝基苯。
Step 1: C₆H₆ + NO₂⁺ → [C₆H₆NO₂]⁺
Step 2: [C₆H₆NO₂]⁺ → C₆H₅NO₂ + H⁺
The overall equation is C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O, with H₂SO₄ acting as a catalyst. Similar mechanisms apply to bromination (using Br₂ with FeBr₃ catalyst to generate Br⁺) and alkylation (using a carbocation generated from a haloalkane and AlCl₃).
总反应方程式为 C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O，H₂SO₄ 作为催化剂。类似的机理适用于溴化（用 Br₂ 和 FeBr₃ 催化剂生成 Br⁺）和烷基化（用卤代烷与 AlCl₃ 生成的碳正离子）。
6. Nucleophilic Addition to Carbonyls | 羰基的亲核加成
Carbonyl compounds (aldehydes and ketones) contain a polar C=O bond. The carbon is electron-deficient and susceptible to attack by nucleophiles. Potassium cyanide (KCN) in acidified solution provides CN⁻ ions, which add to the carbonyl carbon, forming a hydroxynitrile.
羰基化合物（醛和酮）含有极性的 C=O 键。碳原子缺电子，易受亲核试剂进攻。酸化溶液中的氰化钾（KCN）提供 CN⁻ 离子，加至羰基碳上，形成羟腈。
CH₃COCH₃ + HCN ⇌ CH₃C(OH)(CN)CH₃
The mechanism involves the CN⁻ nucleophile attacking the carbonyl carbon, pushing the π electrons onto the oxygen, yielding an alkoxide intermediate. Subsequent protonation (e.g., by H⁺ from HCN or dilute acid) gives the hydroxynitrile. This reaction increases the carbon chain length, useful in synthesis.
机理包括 CN⁻ 亲核试剂进攻羰基碳，将 π 电子推向氧，得到烷氧负离子中间体。随后质子化（例如来自 HCN 或稀酸的 H⁺）生成羟腈。该反应可增长碳链，在合成中非常有用。
Reducing agents like NaBH₄ (in water) or LiAlH₄ (in dry ether) deliver a hydride nucleophile (H⁻), which similarly adds to the carbonyl. NaBH₄ reduces aldehydes to primary alcohols and ketones to secondary alcohols.
还原剂如 NaBH₄（在水中）或 LiAlH₄（在干燥醚中）提供氢负离子亲核试剂（H⁻），同样加到羰基上。NaBH₄ 将醛还原为伯醇，将酮还原为仲醇。
7. Understanding Curly Arrows | 理解弯箭头
Curly arrows in reaction mechanisms show the movement of an electron pair. The arrow starts from an electron-rich site (a lone pair, a π bond, or a negative charge) and points to an electron-deficient site (a positive charge or a polarised bond). Drawing accurate curly arrows is essential for scoring full marks in mechanism questions.
反应机理中的弯箭头表示电子对的移动。箭头从富电子位点（孤对电子、π 键或负电荷）出发，指向缺电子位点（正电荷或极性键）。准确绘制弯箭头是在机理题中获得满分的关键。
For example, in nucleophilic substitution, the curly arrow originates from the nucleophile’s lone pair and points towards the partially positive carbon attached to the halogen. A second arrow shows the C–X bond breaking, with the electron pair moving onto the halogen.
例如，在亲核取代中，弯箭头源自亲核试剂的孤对电子，指向与卤素相连且带部分正电荷的碳。另一支箭头表示 C–X 键断裂，电子对移向卤素。
Always ensure arrows are drawn precisely, showing electron flow from bond to atom or from atom to bond. Never use double-headed arrows to show the movement of single electrons – those are fishhook arrows used for radicals.
务必精确绘制箭头，展示电子从键到原子或从原子到键的流动。切勿用双头箭头表示单电子移动——这种鱼钩箭头用于自由基。
8. Summary of Key Conditions | 关键条件总结
The insert summarises reaction conditions for common organic transformations. The table below consolidates key mechanisms with the necessary reagents and conditions, providing a quick revision guide analogous to Insert 3.
插入页总结了常见有机转化的反应条件。下表汇总了关键机理及其所需试剂与条件，提供与 Insert 3 类似的快速复习指南。
| Reaction / 反应 | Reagents / 试剂 | Conditions / 条件 |
|---|---|---|
| Free radical substitution (alkane → haloalkane) / 自由基取代 | Cl₂ or Br₂ | UV light |
| Electrophilic addition (alkene → haloalkane) / 亲电加成 | HBr or HCl (gaseous) | Room temperature |
| Electrophilic addition (alkene → dibromoalkane) / 亲电加成 | Br₂ | Room temperature, absence of UV |
| Nucleophilic substitution (SN2, primary haloalkane) / 亲核取代 SN₂ | NaOH (aq) or KCN (aq/alc) | Warm, aqueous or ethanolic |
| Nucleophilic substitution (SN1, tertiary haloalkane) / 亲核取代 SN₁ | NaOH (aq) | Room temperature, aqueous |
| Elimination (haloalkane → alkene) / 消除反应 | KOH (ethanolic) | Heat under reflux |
| Nitration of benzene / 苯的硝化 | Conc. HNO₃ + conc. H₂SO₄ | 50–55 °C |
| Nucleophilic addition (carbonyl → hydroxynitrile) /
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