Nucleophilic Substitution for WJEC A-Level Chemistry | A-Level WJEC 化学：亲核取代考点精讲

📚 Nucleophilic Substitution for WJEC A-Level Chemistry | A-Level WJEC 化学：亲核取代考点精讲

Nucleophilic substitution is a fundamental reaction type where a nucleophile attacks an electron‑deficient carbon atom, displacing a leaving group. For WJEC A‑Level Chemistry, you must understand both the S_N1 and S_N2 mechanisms, the factors that influence which pathway is taken, and how to apply these ideas to reactions of halogenoalkanes with common nucleophiles such as OH⁻, CN⁻ and NH₃.

亲核取代是一种基础反应类型，在反应中亲核试剂进攻缺电子的碳原子，并将离去基团取代。在 WJEC A‑Level 化学考试中，你必须理解 S_N1 和 S_N2 两种机理，理解影响反应路径选择的各种因素，并能把这些概念应用于卤代烷与 OH⁻、CN⁻ 和 NH₃ 等常见亲核试剂的反应中。

1. What is Nucleophilic Substitution? | 什么是亲核取代？

Nucleophilic substitution is a reaction in which a nucleophile (an electron‑pair donor) replaces an atom or group of atoms that is attached to an sp³‑hybridised carbon. The carbon–halogen bond in halogenoalkanes is polar owing to the high electronegativity of the halogen, making the carbon δ+ and susceptible to attack by nucleophiles.

亲核取代是一种反应，其中亲核试剂（电子对给体）取代了连接在 sp³ 杂化碳上的一个原子或原子团。卤代烷中的碳–卤键由于卤素的高电负性而具有极性，这使得碳带部分正电荷 (δ+)，容易受到亲核试剂的进攻。

The general equation can be written as Nu⁻ + R–X → R–Nu + X⁻, where Nu⁻ is the nucleophile and X is the halogen leaving group. The reaction is classified as either S_N1 (two steps, unimolecular rate‑determining step) or S_N2 (one step, bimolecular).

反应的通式可写作 Nu⁻ + R–X → R–Nu + X⁻，其中 Nu⁻ 为亲核试剂，X 为卤素离去基团。该反应被划分为 S_N1（两步，单分子速率控制步骤）或 S_N2（一步，双分子）两种类型。

2. The Nucleophile | 亲核试剂

A nucleophile is a species that donates a lone pair of electrons to an electron‑deficient centre. Common nucleophiles in WJEC include OH⁻ (hydroxide), CN⁻ (cyanide) and NH₃ (ammonia). A strong nucleophile is usually a good base, but nucleophilicity is also influenced by charge density, polarisability and solvation. In water, CN⁻ is a stronger nucleophile than OH⁻ towards halogenoalkanes.

亲核试剂是指能将孤对电子提供给缺电子中心的物种。WJEC 考试中常见的亲核试剂包括 OH⁻（氢氧根）、CN⁻（氰根）和 NH₃（氨）。强亲核试剂通常也是强碱，但亲核性还受电荷密度、极化度和溶剂化作用的影响。在水中，CN⁻ 对卤代烷的亲核性强于 OH⁻。

Polarisable nucleophiles, such as I⁻, are particularly effective in S_N2 reactions because their electron cloud can be distorted easily to form a partial bond to carbon in the transition state.

极化度较高的亲核试剂（如 I⁻）在 S_N2 反应中特别有效，因为它们的电子云容易发生形变，从而在过渡态中与碳形成部分键。

3. The Leaving Group | 离去基团

The leaving group is the atom or group that departs with the bonding pair of electrons. In halogenoalkanes, the leaving ability follows the order I⁻ > Br⁻ > Cl⁻ > F⁻, which correlates with the weakness of the carbon–halogen bond. A good leaving group stabilises the negative charge effectively, making the cleavage of the C–X bond easier.

离去基团是带着键合电子对离去的原子或原子团。在卤代烷中，离去能力的顺序为 I⁻ > Br⁻ > Cl⁻ > F⁻，这与碳–卤键的强弱相关。一个好的离去基团能有效稳定负电荷，使 C–X 键更容易断裂。

When the leaving group departs, it takes the bonding electrons; therefore the carbon becomes more positively charged in the process. This is why tertiary halogenoalkanes that can form stable carbocations readily undergo S_N1 reactions.

离去基团离开时带走了键合电子，因此在此过程中碳会带上更多的正电荷。这就是为什么能形成稳定碳正离子的叔卤代烷容易发生 S_N1 反应。

4. The S_N2 Mechanism – One‑Step Bimolecular | S_N2 机理——一步双分子反应

In an S_N2 reaction the nucleophile attacks the carbon from the opposite side of the leaving group, forming a transition state where the carbon is simultaneously bonded to both the nucleophile and the leaving group. The process is concerted: bond‑making and bond‑breaking occur at the same time.

在 S_N2 反应中，亲核试剂从离去基团的背面进攻碳原子，形成一个过渡态，此时碳同时与亲核试剂和离去基团成键。该过程是协同的：成键和断键同时发生。

The rate equation is rate = k [R–X] [Nu⁻], hence the term bimolecular. The reaction is favoured by primary halogenoalkanes because steric hindrance around the carbon is minimal. The energy profile shows a single transition state and no intermediate.

速率方程为 rate = k [R–X] [Nu⁻]，因此称为双分子反应。该反应适于伯卤代烷，因为碳周围的空间位阻最小。能量曲线图显示一个单一的过渡态，没有中间体。

5. S_N2 Stereochemistry – Walden Inversion | S_N2 立体化学——瓦尔登翻转

If the carbon centre is chiral, an S_N2 reaction proceeds with complete inversion of configuration. This is known as Walden inversion. The nucleophile attacks from the back, so the three remaining groups are pushed to the opposite side, like an umbrella turning inside out in the wind.

如果碳中心是手性的，S_N2 反应会伴随着完全的构型翻转，这被称为瓦尔登翻转。亲核试剂从背面进攻，因此剩下的三个基团被推向相反的一侧，就像雨伞在风中被吹翻一般。

For example, optically active (R)‑2‑bromobutane reacts with OH⁻ to give (S)‑butan‑2‑ol. The product has the opposite optical rotation. Examiners may ask you to draw the transition state or predict the stereochemical outcome.

例如，具有光学活性的 (R)‑2‑溴丁烷与 OH⁻ 反应得到 (S)‑丁‑2‑醇。产物的旋光方向相反。考官可能会要求你画出过渡态或预测立体化学结果。

6. The S_N1 Mechanism – Two‑Step Unimolecular | S_N1 机理——两步单分子反应

S_N1 stands for substitution nucleophilic unimolecular. The reaction occurs in two distinct steps. First, the C–X bond breaks heterolytically to generate a planar carbocation intermediate (rate‑determining step). Second, the nucleophile attacks the carbocation rapidly from either side.

S_N1 代表单分子亲核取代。反应分两步进行。首先，C–X 键异裂生成一个平面型的碳正离子中间体（速率控制步骤）。其次，亲核试剂从平面的任一侧快速进攻碳正离子。

The rate equation is rate = k [R–X]; the nucleophile concentration does not appear. This mechanism is favoured by tertiary halogenoalkanes because the tertiary carbocation is stabilised by the inductive effect of three alkyl groups.

速率方程为 rate = k [R–X]，亲核试剂的浓度不出现。该机理适于叔卤代烷，因为叔碳正离子通过三个烷基的诱导效应得以稳定。

7. S_N1 Stereochemistry – Racemisation | S_N1 立体化学——外消旋化

Since the carbocation intermediate is planar, the nucleophile can attack from either face with equal probability. This leads to a racemic mixture if the substrate was originally a single enantiomer. The product loses its optical activity because an equal mixture of the two enantiomers is optically inactive.

由于碳正离子中间体是平面型的，亲核试剂可以等概率地从平面的任一面进攻。如果底物原本是单一对映体，这将导致外消旋混合物。产物失去光学活性，因为等量的两种对映体混合物不显示旋光性。

In practice, slight excess of inversion can sometimes be observed because the leaving group may partially shield one face before it diffuses away – this is called a solvent cage effect – but for WJEC the expected answer is racemisation.

在实践中，偶尔可以观察到略微过量的翻转产物，这是因为离去基团在完全扩散离开前可能部分遮挡了一面——这被称为溶剂笼效应——但对于 WJEC，预期的答案仍是外消旋化。

8. Factors Influencing S_N1 versus S_N2 | 影响 S_N1 与 S_N2 的因素

Nature of the halogenoalkane: Primary halogenoalkanes prefer S_N2 because the α‑carbon is sterically accessible. Tertiary halogenoalkanes prefer S_N1 because the tertiary carbocation is stable and the α‑carbon is too hindered for S_N2. Secondary halogenoalkanes can proceed by either mechanism depending on the conditions.

卤代烷的结构：伯卤代烷倾向于 S_N2，因为 α‑碳位阻小。叔卤代烷倾向于 S_N1，因为叔碳正离子稳定且 α‑碳位阻过大不利于 S_N2。仲卤代烷根据反应条件可以走任一机理。

Strength and concentration of nucleophile: A high concentration of a strong nucleophile favours S_N2 because the nucleophile participates in the rate‑determining step. S_N1 rates are independent of nucleophile concentration.

亲核试剂的浓度与强度：高浓度的强亲核试剂有利于 S_N2，因为亲核试剂参与决速步骤。S_N1 速率与亲核试剂浓度无关。

Solvent: Polar protic solvents (water, alcohols) stabilise carbocations via solvation, thus favouring S_N1. Polar aprotic solvents (propanone, ethanenitrile) favour S_N2 because they solvate the cation but leave the nucleophile relatively unsolvated and more reactive.

溶剂：极性质子性溶剂（水、醇）通过溶剂化稳定碳正离子，从而有利于 S_N1。极性非质子性溶剂（丙酮、乙腈）有利于 S_N2，因为它们溶剂化阳离子却使亲核试剂较少被溶剂化，反应活性更高。

9. Key Reactions of Halogenoalkanes with Common Nucleophiles | 卤代烷与常见亲核试剂的反应

Reaction with hydroxide ions (aqueous NaOH): Halogenoalkanes are heated under reflux with aqueous sodium hydroxide to produce alcohols. For a primary halogenoalkane the mechanism is S_N2; for tertiary it is S_N1. The ionic equation is R–X + OH⁻ → R–OH + X⁻.

与氢氧根离子的反应（NaOH 水溶液）：卤代烷与氢氧化钠水溶液在加热回流下反应生成醇。伯卤代烷按 S_N2 机理进行，叔卤代烷按 S_N1 机理。离子方程式为 R–X + OH⁻ → R–OH + X⁻。

Reaction with cyanide ions (KCN in ethanol): Heating a halogenoalkane with potassium cyanide in ethanol under reflux yields a nitrile (R–CN), which increases the carbon chain length by one. The nucleophile is CN⁻ and the mechanism is typically S_N2 for primary substrates.

与氰根离子的反应（KCN 的乙醇溶液）：卤代烷与氰化钾的乙醇溶液在回流下加热得到腈 (R–CN)，使碳链增加一个碳原子。亲核试剂是 CN⁻，对伯卤代烷通常为 S_N2 机理。

Reaction with ammonia (excess NH₃ in ethanol): Halogenoalkanes react with excess ammonia to form primary amines. The initial step is nucleophilic substitution giving an alkylammonium ion, followed by deprotonation by a second NH₃ molecule. Overall: R–X + 2NH₃ → R–NH₂ + NH₄⁺X⁻.

与氨的反应（过量 NH₃ 的乙醇溶液）：卤代烷与过量氨反应生成伯胺。第一步是亲核取代生成烷基铵离子，随后被另一分子 NH₃ 去质子化。总反应为 R–X + 2NH₃ → R–NH₂ + NH₄⁺X⁻。

10. Exam Tips and Common Pitfalls | 考试技巧与常见误区

Always specify the mechanism with curly arrows that show the movement of electron pairs. For S_N2, draw the nucleophile attacking from the back and the leaving group departing simultaneously. For S_N1, show the heterolytic fission of the C–X bond to form a carbocation, followed by nucleophilic attack.

一定要用弯箭头标出电子对的移动方向。对于 S_N2，画出亲核试剂从背面进攻与离去基团同步离去。对于 S_N1，画出 C–X 键异裂形成碳正离子，随后亲核试剂进攻。

Do not forget to include the relevant lone pairs and charges. WJEC examiners often deduct marks for missing partial charges (δ+/δ−) on the C–X bond or for an incomplete transition state. If a question asks why a tertiary halogenoalkane does not undergo S_N2, mention steric hindrance and the inability to achieve the required back‑side attack.

不要忘记标出相关的孤对电子和电荷。WJEC 考官常因为 C–X 键上缺少部分电荷 (δ+/δ−) 或过渡态不完整而扣分。如果题目问为什么叔卤代烷不能发生 S_N2，要提到空间位阻以及无法实现所需的背面进攻。

For synthesis questions, recall that CN⁻ extends the carbon chain, while OH⁻ introduces an alcohol group and excess NH₃ gives a primary amine. Be aware of the conditions: aqueous solvents favour alcohols, alcoholic solvents favour nitriles and amines.

在合成题中，记住 CN⁻ 可增加碳链，OH⁻ 引入醇羟基，过量 NH₃ 得到伯胺。注意反应条件：水溶剂有利于生成醇，乙醇溶剂有利于生成腈和胺。

11. Comparing S_N1 and S_N2 – Summary Table | S_N1 与 S_N2 对比总结表

Feature	S_N1	S_N2
Kinetics	Unimolecular (first order)	Bimolecular (second order)
Steps	Two (carbocation intermediate)	One (concerted transition state)
Preferred substrate	Tertiary > secondary	Primary > secondary
Stereochemistry	Racemisation	Walden inversion
Nucleophile requirement	Weak nucleophile acceptable	Strong nucleophile required
Solvent effect	Polar protic favours	Polar aprotic favours

This table highlights the critical distinctions you must be able to recall under timed conditions. Be ready to explain each point with examples, such as using (CH₃)₃CBr for S_N1 and CH₃CH₂Br for S_N2.

这张表格突出了你必须能在限时条件下回忆起来的关键区别。做好用实例解释每一点的准备，比如用 (CH₃)₃CBr 说明 S_N1，用 CH₃CH₂Br 说明 S_N2。

12. Linking Nucleophilic Substitution to Mechanism Questions in WJEC Papers | 联系 WJEC 试卷中的机理题

WJEC frequently asks students to give the mechanism for a specific nucleophilic substitution, naming the type of mechanism and drawing the relevant curly arrow diagram. You should be confident in drawing the approach of the nucleophile, the structure of the transition state (S_N2) or carbocation (S_N1), and the final product.

WJEC 经常要求考生写出特定亲核取代反应的机理，说明机理类型并画出相应的弯箭头图示。你应自信地画出亲核试剂的接近方式、过渡态 (S_N2) 或碳正离子 (S_N1) 的结构，以及最终产物。

When given the formula of an optically active halogenoalkane, you may be asked to predict the stereochemistry of the product. If the mechanism is S_N2 with a chiral centre, show inversion; if S_N1, indicate loss of optical activity.

当给出有光学活性的卤代烷分子式时，你可能会被要求预测产物的立体化学。如果机理是 S_N2 且有手性中心，就要画出翻转；如果是 S_N1，则要说明旋光性消失。

Also, remember that nucleophilic substitution competes with elimination (E1/E2). WJEC papers sometimes ask you to state the conditions that favour substitution over elimination – using a strong nucleophile that is a weak base (e.g. CN⁻) and moderate temperature helps maximise substitution yield.

同时要记住，亲核取代与消除反应 (E1/E2) 是竞争关系。WJEC 试卷有时要求你陈述有利于取代而非消除的条件——使用强亲核性弱碱性的试剂（如 CN⁻）及温和的温度有助于最大化取代产率。

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Nucleophilic Substitution for WJEC A-Level Chemistry | A-Level WJEC 化学：亲核取代 考点精讲