IB化学 HL 有机反应机理 SN1 SN2 加成取代
IB 化学 HL 的有机机理，是最容易看懂却拿不到分的一块内容。学生把产物写对了，箭头却反着画，中间体的电荷也标错，阅卷时这些细节照样扣分。新版大纲把这块内容放在 Reactivity 3.4，标题直译过来是"化学变化的机理是什么"，整节的落点是电子对从哪里来、到哪里去。2016 版大纲里的 Topic 10 与 Topic 20 编号已经不再使用，翻旧笔记找资料时需要对号入座。下面按考点把六类机理串起来，每一类都给判断依据、中间体和常见的错误。
Mechanisms in IB Chemistry HL look harmless until you have to draw them. Marks are lost on arrow direction, on the charge written on an intermediate, and on a final product that contradicts the mechanism you just drew. The current syllabus, first examined in May 2025, places this work in Reactivity 3.4, where the question is how electron pairs move rather than which reagent to memorise; the six families below are linked by that single question.
亲核体与亲电体 异裂和均裂的分界
判断机理的第一件事不是挑反应类型，而是先看电子对从哪来、到哪去。亲核体给出电子对，亲电体接受电子对，这条分界贯穿整个 Reactivity 3.4。卤代烃的亲核取代、烯烃的亲电加成、苯环的亲电取代都被收进"电子对共享"这个框架，Lewis 酸碱理论是它的语言：亲核体是 Lewis 碱，亲电体是 Lewis 酸。键的断裂方式决定后面写哪套机理。异裂时成键电子对整对归到一个原子上，得到正负离子，离子型机理从这里开始；均裂时两个原子各拿一个电子，得到自由基，链式反应走这条路。很多学生以为异裂和均裂只是说法差异，其实画出第一根箭头之前就得判完，因为中间体的电荷、箭头方向、产物构型全都由它决定。找亲核体看孤对电子和负电荷，OH-、CN-、NH3、H2O 都算；找亲电体看缺电子中心，碳正离子、羰基碳、被 FeBr3 极化的溴分子都属于这一类。溶剂还会改写路径：极性质子溶剂靠氢键稳住离去的阴离子与碳正离子，对 SN1 有利；极性非质子溶剂把亲核体裸露出来，进攻更直接，对 SN2 有利。
Every mechanism question opens with the same two questions: where does the electron pair come from, and where does it go. Nucleophiles donate a pair and electrophiles accept one, and Reactivity 3.4 uses that distinction as its working language, with nucleophiles acting as Lewis bases and electrophiles as Lewis acids; ionic substitution, addition and aromatic substitution all sit under it. Bond fission comes next, because heterolytic fission keeps both bonding electrons on one atom and produces ions, while homolysis splits the pair and produces radicals that propagate chains. Deciding this before drawing the first arrow pays off later, since the intermediate charge and the stereochemistry both follow from the choice.
SN2 一步协同 与构型翻转
SN2 的机理一句话就能说完：亲核体从离去基团的正对面进攻，新键形成和旧键断裂同时发生，中间没有碳正离子。因为进攻方向和离去方向相反，碳原子的构型会翻转，手性原料得到手性相反的产物，答题时写成"构型翻转"。速率方程给出第二条证据：速率 = k[卤代烃][亲核体]，两个物种都参与慢步骤，所以是双分子反应。取代基越多空间位阻越大，位阻挡住了背面进攻的路线，一级卤代烃（如溴乙烷）容易走 SN2，三级几乎走不通。离去基团的能力按共轭碱的强弱倒过来排：I- 是最弱的碱，最容易离去，C-I 反应最快，C-F 基本不动。常见的错误是学生在 SN2 里画出两步，还顺手补一个碳正离子中间体，这个中间体在这条路径上根本不存在。回到具体例子，溴乙烷与氢氧化钠水溶液反应时，真正进攻的粒子是氢氧根上的那对电子，它撞的是连溴的那个碳，而不是溴原子本身。
The SN2 pathway is one concerted step: the nucleophile attacks the carbon from the side opposite the leaving group, the new bond forms as the old one breaks, and no carbocation ever appears. Inversion of configuration follows directly from that geometry, so a chiral starting material ends up with the opposite handedness at the reacting carbon. Kinetics confirm the picture, because the rate depends on both the halogenoalkane and the nucleophile, which means both species sit in the slow step. Steric bulk slows the reaction down, which is why primary halogenoalkanes react readily and tertiary ones barely react at all; leaving-group ability runs the other way from base strength, so iodide departs easily and fluoride hardly departs at all.
SN1 两步机理 碳正离子与外消旋
SN1 分成两步，慢的那一步只有一个分子参与。卤代烃先异裂，键上的电子对留给卤素，碳上出现一个带正电的碳正离子，这一步是决速步，速率方程只写成速率 = k[卤代烃]，不含亲核体的浓度。碳正离子是平面三角形，两侧都有空间，亲核体从哪一侧进攻的概率接近相同，手性底物得到两种构型数量近似相等的混合物，也就是外消旋。IBO 官方化学大纲（syllabus）对底物的处理很明确：一级卤代烃按协同的 SN2 处理，三级按两步 SN1 处理，二级两种都可能发生，要靠实验数据判断。这里最容易踩的坑，是把碳正离子的稳定性当成唯一标准。叔碳正离子被三个烷基推电子稳定，但三级卤代烃在碱或高温条件下，还可能被夺走 β-H 生成烯烃，取代与消除在同一条链上竞争。副反应也不难解释：碳正离子还有另一个选择，让碱夺走相邻碳上的一个氢，电子对退回去形成双键，烯烃和取代产物就一起出现。
SN1 takes two steps, and the slow one involves only the halogenoalkane, which is why its rate law contains no nucleophile. Ionisation releases the halide first and leaves a planar carbocation, so attack from either face becomes equally likely and a chiral substrate ends up as a racemic mixture rather than a single inverted product. The official guide treats primary halogenoalkanes through the concerted SN2 route and tertiary ones through SN1, and accepts either pathway for secondary substrates when the evidence supports it. Carbocation stability is a useful guide but not the whole story, since a tertiary halide heated with a base can also lose a beta hydrogen and give an alkene instead.
烯烃亲电加成 马氏规则与碳正离子稳定性
烯烃的双键是一个富电子区域，遇到被极化的卤素、水或卤化氢时，它会主动进攻缺电子的一端。以溴化氢加成为例，H+ 先被双键的 π 电子抓住，生成碳正离子，溴离子随后进攻带正电的碳。反应的走向由碳正离子的稳定性决定。不对称烯烃里两个可能的碳正离子稳定性不同，更稳定的那个更容易生成，也主导产物，二级优先于一级，三级优先于二级，这就是马氏规则背后的原因。它描述的是"氢加到含氢较多的碳上"这个结果，本质却是中间体的稳定性对比。很多学生把马氏规则当成要背下来的口诀，遇到共轭体系或重排就失效：碳正离子会通过 1,2-迁移变成更稳定的一个，产物随之改变。水的加成走的是同一条路线，第一步同样生成碳正离子，随后由水分子给出电子对成键，最后失掉一个质子，得到醇。
An alkene is an electron-rich region, so halogens, water and hydrogen halides must be polarised before addition can begin. In the addition of HBr, the pi electrons take the proton, a carbocation appears, and the bromide ion then closes the new bond; the more stable carbocation decides which product dominates, with secondary beating primary and tertiary beating secondary. Markovnikov's rule summarises that outcome instead of replacing it, which is why rearrangements catch people out. A carbocation will shift a hydride or an alkyl group to become more stable, and the product then no longer matches the textbook wording.
苯环亲电取代 硝化与带电中间体
苯的六个 π 电子离域在整个环上，环是富电子的，所以苯并不排斥亲电体，它比烯烃沉稳的原因是加成会毁掉离域体系。苯环遇到带电的亲电体 E+ 时走取代：π 电子先伸出去形成一根新键，环内的离域被打断，得到一个带正电的中间体，环上那个碳暂时只有四个成键电子。随后连在同一个碳上的氢被碱夺走，C-H 键的电子对回到环上，离域重新建立，净结果是一个氢被取代。硝化是标准例子，亲电体是浓硝酸与浓硫酸混合生成的 NO2+，产物是硝基苯。R3.4.13 的指导意见写明，亲电体本身的生成过程不在考核范围，需要做的是认得出是哪种亲电体，并画对它进攻苯环的过程。阅卷时最常扣分的地方是漏掉收尾那一步：学生把 E 挂到环上就停笔，没有画回那根恢复离域的箭头。苯不能让溴水褪色，需要 FeBr3 这类 Lewis 酸先把溴分子极化到足够缺电子，反应才启动，这个对比本身就是取代优于加成的证据。
Benzene is electron rich but unusually stable, so an electrophile does not push it into addition; substitution keeps the delocalised ring intact, and that is the whole reason it is preferred. A charged electrophile such as NO2+, generated in situ from concentrated nitric and sulfuric acids, is attacked by the pi electrons, giving a positively charged intermediate in which delocalisation is temporarily broken. A base then removes the hydrogen from that same carbon and its electron pair returns to the ring, restoring aromaticity. The assessable skill is recognising the electrophile and drawing its attack, not the formation of the electrophile itself; the mark most often lost is the final arrow that rebuilds the ring.
自由基取代 均裂引发的链式三步
自由基反应的起点是光。以甲烷和氯气在紫外光下的反应为例，光照提供足够的能量让 Cl-Cl 均裂，两个氯原子各拿一个电子，形成两个氯自由基，这一步叫引发。自由基很活泼，氯自由基从甲烷上夺走一个氢，生成氯化氢和一个甲基自由基；甲基自由基再去撞氯分子，拿走一个氯并放出新的氯自由基，这样循环下来就是链增长，也是产物里会混有一氯、二氯甚至更多取代物的原因。两个自由基相遇时配成一根键，链就终止。这里需要记牢的是箭头：自由基机理里用半箭头，表示单个电子的移动；离子型机理用整箭头，表示一对电子的移动。链一旦启动就难以控制停在哪一步，所以实验室里常让甲烷过量，用来提高一氯产物的比例。
Radical substitution needs light, because UV photons supply enough energy for Cl-Cl to break homolytically and give two chlorine atoms, each holding a single electron. Those radicals are reactive enough to pull a hydrogen off methane, producing HCl and a methyl radical, which then attacks another chlorine molecule and regenerates the chain carrier; that cycle explains why the chlorination of methane rarely stops at one substitution. Pairing two radicals ends the sequence. One drawing detail separates this family from the rest: radicals move one electron at a time and use half-headed arrows, while ionic mechanisms move pairs and use full arrows.
机理推断 速率方程与分子数的证据链
画出机理不等于证明机理，考卷给的证据通常是速率方程。Reactivity 2.2 的 HL 部分要求把提出的机理与实验速率、计量关系放在一起核对。反应由若干基元步骤组成，最慢的一步决定整体速率，速率方程里出现的物种必须出现在决速步，或者出现在通向决速步的快速平衡里。分子数说的是决速步里参加碰撞的粒子个数，一级是单分子，二级是双分子，三个粒子同时碰撞的假设基本被排除。机理推断题的常见形式是给你一个速率方程和两个候选机理，让你选一个并说明理由。判断方法先是数决速步里的粒子，再看总反应方程式加总后能不能对上。这类题丢分大多不是化学问题，而是没有写清哪一步慢。中间体和过渡态也要分开：中间体停在势能曲线的谷底，存在的时间够长，可以被检测到；过渡态站在峰顶，只是瞬间的排布，写机理时不能把它当成一个能独立存在的离子。
A mechanism stays a proposal until kinetics back it up, which is why Reactivity 2.2 asks you to test a sequence of elementary steps against the experimental rate law and the overall stoichiometry. The slowest step sets the rate of the whole reaction, so any species in the rate equation must appear either in the rate-determining step or in a fast equilibrium leading into it. Molecularity counts the particles colliding in that slow step, and three-particle collisions are treated as too unlikely to propose. Questions of this kind typically hand you a rate equation and two candidate mechanisms and ask for a choice with justification; count the particles in the slow step first, then check that the steps sum to the balanced equation.
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