📚 Case Study Practical Exercises for Year 13 Edexcel Chemistry | Year 13 Edexcel 化学：案例分析实战演练
Mastering Year 13 Edexcel Chemistry demands more than memorising facts; it requires the ability to apply concepts to unfamiliar, real-world scenarios. This article presents a series of structured case studies that mirror the style of synoptic exam questions, covering key A2 topics such as chemical equilibrium, kinetics, organic synthesis, spectroscopic analysis, thermodynamics, and electrochemical cells. Each case study includes a scenario, guided questions, and worked solutions, helping you build the analytical thinking needed for top marks.
掌握Year 13 Edexcel化学不仅需要记忆事实，更需要将概念应用于不熟悉的现实情境。本文提供一系列结构化案例研究，模拟综合考题的风格，涵盖化学平衡、动力学、有机合成、光谱分析、热力学和电化学电池等A2核心主题。每个案例包含情境、引导性问题和详细解答，帮助你培养高分所需的分析思维。
1. Industrial Equilibrium – The Haber Process Reimagined | 工业平衡——重新思考哈伯法
The Haber process (N₂ + 3H₂ ⇌ 2NH₃, ΔH = -92 kJ mol⁻¹) is operated at 400– Use your knowledge of equilibrium, kinetics, and industrial costs to evaluate this proposal.
哈伯法（N₂ + 3H₂ ⇌ 2NH₃，ΔH = -92 kJ mol⁻¹）在400–450°C和200 atm下使用铁催化剂运行。一家初创化学公司提议使用钌基催化剂，在250°C和100 atm下便具活性。请运用平衡、动力学和工业成本知识评估此方案。
First, recall Le Chatelier's principle: since the forward reaction is exothermic, lowering the temperature shifts the equilibrium position towards more ammonia, increasing the equilibrium yield. At 250°C, the equilibrium constant Kc is larger than at 450°C, so potentially more NH₃ could be produced per pass.
首先，回忆勒夏特列原理：由于正反应放热，降低温度会使平衡位置向生成更多氨的方向移动，提高平衡产率。在250°C时，平衡常数Kc比450°C时大，因此每次转化可能产生更多NH₃。
temperature drastically reduces the rate of reaction. The iron catalyst normally requires ~ might be comparable. Rate equation analysis: the reaction is thought to follow kinetics determined by the dissociative adsorption of N₂, often described by rate = k [N₂]¹[H₂]⁰⁵. The rate constant k follows the Arrhenius equation: k = A e −Eₐ/RT . The lower Eₐ with ruthenium would increase k at a given temperature, compensating for the lower T.
然而，较低的温度会显著降低反应速率。铁催化剂通常需要约400°C才能提供足够的活化能降低。钌基催化剂据称具有更低的活化能，因此在250°C时的速率可能相当。速率方程分析：该反应动力学被认为由N₂的解离吸附控制，常表达为速率 = k [N₂]¹[H₂]⁰⁵。速率常数k遵循阿伦尼乌斯方程：k = A e −Eₐ/RT 。钌更低的Eₐ会在给定温度下增大k，从而补偿较低的温度。
oles). However, lower pressure reduces plant construction and compression costs significantly. The company must perform a cost-benefit analysis: the ruthenium catalyst is much more expensive than iron, but using less pressure could save energy. The optimum would need to balance yield, rate, and catalyst lifetime. Often, industrial choices are a compromise: iron remains popular because of its low cost and robustness despite needing higher T and P.
压力从200 atm降至100 atm会降低平衡产率，因为产物侧气体摩尔数较少（4摩尔→2摩尔）。然而，较低的压力能显著降低工厂建设和压缩成本。公司必须进行成本效益分析：钌催化剂比铁昂贵得多，但使用较低压力可节省能源。最佳条件需平衡产率、速率和催化剂寿命。工业选择常常是折中方案：铁因其成本和稳定性而沿用至今，尽管需要更高的温度和压力。
2. Kinetics in Action – Decomposition of Hydrogen Peroxide | 动力学实践——过氧化氢分解
+ O₂ ) catalysed by manganese(IV) oxide. Students vary the concentration of H₂O₂ and measure the initial rate by collecting O₂ gas. The data suggest the rate is first order with respect to H₂O₂. In an industrial setting, a new silver nanoparticle catalyst is tested. Given the kinetic data, evaluate the feasibility of scaling up.
某学校实验室研究MnO₂催化H₂O₂分解（2H₂O₂ → 2H₂O + O₂）。学生改变H₂O₂浓度并通过收集O₂测量初始速率。数据表明反应对H₂O₂为一级。在工业环境中测试了一种新型银纳米颗粒催化剂。根据动力学数据，评估放大的可行性。
MnO₂. With the silver nanocatalyst, experiments show k = 0.150 s⁻¹ at the same temperature – r, the Arrhenius pre-exponential factor A and Eₐ must be considered for temperature dependence. If the industrial process runs at a slightly elevated temperature to increase rate further, the activation energy can be determined from an Arrhenius plot (ln k vs 1/T). Suppose Eₐ (MnO₂) = 75 kJ mol⁻¹ and Eₐ (Ag) = 45 kJ mol⁻¹; the silver catalyst not only raises k but also makes the rate less sensitive to temperature fluctuations, simplifying process control.
速率方程为速率 = k [H₂O₂]¹。在MnO₂催化、298 K下速率常数k为0.015 s⁻¹。银纳米催化剂在相同温度下k = 0.150 s⁻¹——提高了10倍。这种显著增强源于纳米颗粒极高的表面积，提供了大量活性位点。然而，需考虑阿伦尼乌斯指前因子A和Eₐ对温度的依赖性。如果工业过程在稍高温度下运行以进一步提高速率，活化能可由阿伦尼乌斯图（ln k对1/T）确定。假设Eₐ(MnO₂) = 75 kJ mol⁻¹而Eₐ(Ag) = 45 kJ mol⁻¹；银催化剂不仅提升k，还使速率对温度波动不那么敏感，简化了过程控制。
Scaling up involves mass transfer considerations: in the school lab, powder catalyst is stirred in solution; industrial reactors may use a fixed bed of catalyst. The rate might become limited by diffusion of H₂O₂ to the surface. Using nanoparticle-coated beads in a flow reactor could maintain high effective k. Additionally, silver is more expensive, but the dramatic rate improvement could justify its cost if product throughput is high. A full economic assessment would compare catalyst cost, reactor size, energy consumption, and waste treatment.
放大涉及传质考量：在学校实验室，粉末催化剂被搅入溶液；工业反应器可能使用固定床催化剂。此时速率可能受限于H₂O₂扩散至表面的速度。在流动反应器中使用涂覆纳米颗粒的珠粒可保持高有效k。此外，银成本较高，但如果产品产量高，速率的重大改进可证明其合理性。完整的经济评估需比较催化剂成本、反应器尺寸、能耗及废物处理。
3. Organic Synthesis – Designing a Pathway to Ibuprofen | 有机合成——设计布洛芬合成路线
Ibuprofen, a widely used anti-inflammatory, has the structure (CH₃)₂CHCH₂C₆H₄CH(CH₃)COOH. Propose a two-step synthesis from a suitable aromatic precursor, using knowledge of electrophilic substitution and carbon–carbon bond forming reactions. Evaluate the green chemistry aspects of your route.
布洛芬是一种广泛使用的消炎药，结构为(CH₃)₂CHCH₂C₆H₄CH(CH₃)COOH。从合适的芳香前体出发，运用亲电取代和碳–碳键形成反应，提出两步合成路线。评价你设计路线的绿色化学方面。
A common starting material is isobutylbenzene, C₆H₅CH₂CH(CH₃)₂. Step 1: Friedel-Crafts acylation using acetyl chloride (CH₃COCl) and AlCl₃ catalyst to install the acetyl group, giving 4-isobutylacetophenone. The acyl group is meta-directing, but because the alkyl chain is para-directing and steric effects favour para substitution, the product is mainly the para isomer. Step 2: a haloform reaction (or better, a more atom-economical carbonylation/oxidation). However, in classic synthesis, the acetyl group is converted into the carboxylic acid via the iodoform reaction (excess I₂ and NaOH) followed by acidification, producing ibuprofen directly: CH₃CO- group becomes -COO⁻, and the -CH(CH₃)COOH side chain is formed by rearrangement? Wait, correct pathway: The traditional Boots synthesis used a Friedel-Crafts acylation with propionic anhydride, then reduction, then halogenation and Grignard reaction. Modern green synthesis by BHC uses only three steps: acylation using HF and Ac₂O, catalytic hydrogenation, and carbonylation with CO and palladium catalyst, achieving near 100% atom economy. For a simplified two-step proposal, we might propose: Step 1 – Friedel-Crafts alkylation of isobutylbenzene with methyl acrylate? Not typical. It's better to explain an idealised route: Step 1 – React isobutylbenzene with methyl 2-chloropropanoate in the presence of AlCl₃ (Friedel-Crafts) to introduce the -CH(CH₃)COOCH₃ group, then Step 2 – hydrolysis to the acid. This route has moderate atom economy due to HCl generation. Evaluate green chemistry: The BHC process is superior because it eliminates stoichiometric inorganic waste and uses recyclable catalysts, demonstrating atom economy and reduced environmental factor (E-factor).
常用起始原料为异丁基苯C₆H₅CH₂CH(CH₃)₂。第一步：弗里德-克拉夫茨酰基化反应，使用乙酰氯(CH₃COCl)和AlCl₃催化剂引入乙酰基，得到4-异丁基苯乙酮。酰基为间位定位基，但由于烷基链是对位定位且空间效应利于对位，产物主要为对位异构体。第二步：卤仿反应（更佳方案为原子经济性更高的羰基化/氧化）。经典合成中，乙酰基经碘仿反应（过量I₂和NaOH）后酸化转化为羧酸，从而直接生成布洛芬：CH₃CO-基团变为-COO⁻，并通过重排形成-CH(CH₃)COOH侧链。等一下，正确路线：最初Boots合成法采用丙酸酐进行Friedel-Crafts酰化，然后还原，卤化及格氏反应。现代BHC绿色合成仅用三步：以HF和Ac₂O酰化，催化加氢，再用CO和钯催化剂进行羰基化，达到接近100%原子经济性。为简化两步建议，可提出：第一步——异丁基苯与2-氯丙酸甲酯在AlCl₃存在下发生Friedel-Crafts反应引入-CH(CH₃)COOCH₃基团，第二步——水解成酸。此路线因产生HCl而具有中等原子经济性。绿色化学评价：BHC工艺因消除化学计量无机废物并使用可回收催化剂而更为优越，体现了原子经济性和降低的环境因子（E-factor）。
4. Spectroscopic Identification – Unknown Carbonyl Compound | 光谱鉴定——未知羰基化合物
An unknown liquid with a fruity smell is analysed. Elemental analysis gives C 62.0%, H 10.4%, O 27.6% rong absorption at 1740 cm⁻¹. NMR data: ¹H NMR (δ): 1.2 (3H, t), 2.3 (2H, q), 4.1 (2H, t), 1.6 (2H, sextet), 0.9 (3H, t). Determine the structure and justify using all evidence.
一种具有果香味的未知液体经分析：元素分析结果为C 62.0%，H 10.4%，O 27.6%。质谱显示分子离子峰在m/z = 116。IR光谱：1740 cm⁻¹处有强吸收。NMR数据：¹H NMR (δ): 1.2 (3H, t), 2.3 (2H, q), 4.1 (2H, t), 1.6 (2H, sextet), 0.9 (3H, t)。确定结构并用全部证据论证。
st (1.725): C≈3.0, H≈6.0, O=1.0, so empirical formula is C₃H₆O, with empirical mass 58. The molecular ion peak at 116 indicates the molecular formula is C₆H₁₂O₂. IR at 1740 cm⁻¹ suggests an ester C=O stretch (esters ~1735– iplet at 0.9 (3H), sextet at 1.6 (2H), and triplet at 4.1 (2H) indicate a propyl group attached to oxygen (CH₃CH₂CH₂-O-). Together, this is ethyl propanoate? H₃ is adjacent to CH₂, so it would be a triplet due to CH₂, and the CH₂ would be a quartet? hyl propanoate, the ethyl from the acid side is CH₃CH₂-CO: the CH₃ appears as a triplet at ~1.1-1.2 ppm, the CH₂ as a quartet at ~2.3 ppm. The ester alkyl (alcohol side) is CH₂CH₃: the O-CH₂ appears as a quartet at ~4.1 ppm, and the CH₃ as a triplet at ~1.3 ppm. But our NMR shows: 1.2 (3H, t) – this is the CH₃ on the acid side; 2.3 (2H, q) – CH₂ on acid side; 4.1 (2H, t) – O-CH₂; 1.6 (2H, sextet) and 0.9 (3H, t) – pears as a sextet (multiplet) at 1.6, and the terminal methyl at 0.9 (triplet). The acid part is still ethyl: CH₃CH₂- (1.2 t, 2.3 q). Therefore, the ester is ethyl butanoate? thyl propanoate, which has propanoate (ethylcarboxyl) and ethyl alcohol. Not matching. Let' ropanoate (CH₃CH₂CO-). The alcohol side signals: 4.1 (t, 2H), 1.6 (sextet, 2H), 0.9 (t, 3H). That's an n-propyl group: -OCH₂CH₂CH₃. So the ester is propyl propanoate? fect. Naming: propyl propanoate. IR: ester indeed. So the structure is CH₃CH₂COOCH₂CH₂CH₃.
首先确定实验式。假定100 g：C: 62.0/12.0 = 5.167 mol; H: 10.4/1.0 = 10.4 mol; O: 27.6/16.0 = 1.725 mol。除以最小值（1.725）：C≈3.0，H≈6.0，O=1.0，实验式为C₃H₆O，实验式量58。分子离子峰116指示分子式为C₆H₁₂O₂。IR 1740 cm⁻¹表明酯羰基伸缩振动（酯类约1735–1750 cm⁻¹）。NMR信号：1.2处的三重峰（3H）和2.3处的四重峰（2H）表明一个与羰基相连的乙基（CH₃CH₂-C=O）。0.9处的三重峰（3H）、1.6处的六重峰（2H）和4.1处的三重峰（2H）表明一个丙氧基（CH₃CH₂CH₂-O-）。综合分析，此为丙酸丙酯？正确应为CH₃CH₂COOCH₂CH₂CH₃，分子式C₆H₁₂O₂，分子量116。命名为propyl propanoate（丙酸丙酯）。结构确证无误。
5. Thermodynamics & Feasibility – Fuel Cell vs Combustion | 热力学与可行性——燃料电池对比燃烧
nergy change for the fuel cell reaction 2H₂(g) + O₂(g) → 2H₂O(l) is ΔG = -474 kJ (for 2 mol H₂). The engine operates at 25% efficiency, while the fuel cell runs at 50% efficiency. Compare the maximum theoretical work and discuss entropy effects.
氢可用于内燃机或燃料电池驱动车辆。已知H₂燃烧热ΔH = -286 kJ mol⁻¹，燃料电池反应2H₂(g) + O₂(g) → 2H₂O(l) 的吉布斯自由能变ΔG = -474 kJ（对应2 mol H₂）。内燃机效率为25%，燃料电池效率为50%。比较最大理论功并讨论熵效应。
thalpy released), ΔH = -572 kJ. The entropy change ΔS for the reaction is negative because 3 moles of gas become liquid, so TΔS is negative, making ΔG = ΔH - TΔS, which will be less negative than ΔH. Given ΔG = -474 kJ, the maximum work is 474 kJ per 2 mol H₂. The remaining energy (572 - 474 = 98 kJ) is lost as heat to the surroundings because of the entropy decrease. In an engine, combustion releases all the enthalpy as heat, then a heat engine converts part of it to work limited by Carnot; here at 25% 3 kJ. The fuel cell bypasses thermal conversion and directly yields electrical work; at 50% ilises ΔG more effectively and avoids the large entropy penalty of direct heat conversion.
在恒温恒压下，从化学反应可获得的最大功即为吉布斯自由能变ΔG。2 mol H₂燃烧释放焓ΔH = -572 kJ。反应熵变ΔS为负值（3摩尔气体变为液体），故TΔS为负，使得ΔG = ΔH - TΔS，其绝对值小于ΔH。已知ΔG = -474 kJ，故最大功为每2 mol H₂ 474 kJ。剩余能量（572 - 474 = 98 kJ）因熵减少而以热的形式散失。内燃机将全部焓以热释放，再由热机受卡诺限制转化为功：25%效率下输出功 = 0.25 × 572 = 143 kJ。燃料电池绕过热能转换，直接产生电功；50%效率下输出功 = 0.50 × 474 = 237 kJ。因此燃料电池产生更多功，因为它更有效地利用ΔG，避免了直接热转化带来的巨大熵损失。
6. Electrochemistry – Corrosion Prevention in Ship Hulls | 电化学——船体腐蚀防护
principle, calculate the mass of zinc required to protect a 100 kg iron hull component for one year, given that the average corrosion current is 0.5 A. (Faraday constant = 96,500 C mol⁻¹, Zn²⁺/Zn E° = -0.76 V, Fe²⁺/Fe E° = -0.44 V)
某邮轮船体为钢制（主要含铁）。为防止生锈，安装了锌制牺牲阳极。解释电化学原理，计算保护一个100 kg铁制船体部件一年所需的锌质量，设平均腐蚀电流为0.5 A。（法拉第常数=96,500 C mol⁻¹，Zn²⁺/Zn E° = -0.76 V，Fe²⁺/Fe E° = -0.44 V）
as a sacrificial anode because its standard electrode potential is more negative than iron' 2H₂O + 4e⁻ → 4OH⁻), preventing Fe → Fe²⁺ + 2e⁻ from taking place. This is cathodic protection. The current of 0.5 A corresponds to a charge flow of Q = It = 0.5 A × (365×24×3600 s) = 0.5 × 31,536,000 = 15,768,000 C. Moles of electrons = Q / F = 15,768,000 / 96,500 ≈ 163.4 mol. Since Zn → Zn²⁺ + 2e⁻, each mole of Zn liberates 2 moles of electrons, so moles of Zn required = 163.4 / 2 = 81.7 mol. Mass of Zn = 81.7 mol × 65.4 g mol⁻¹ = 5,343 g ≈ 5.34 kg. Therefore, to protect that section, about 5.34 kg of zinc would be consumed in one year. In practice, multiple anodes are distributed and the mass of zinc used is larger to ensure continued protection even after some consumption.
锌作为牺牲阳极因其标准电极电势比铁更负，因而优先氧化：Zn(s) → Zn²⁺(aq) + 2e⁻。铁船体作为阴极，发生氧气还原（中性水中：O₂ + 2H₂O + 4e⁻ → 4OH⁻），从而阻止Fe → Fe²⁺ + 2e⁻发生。此为阴极保护。电流0.5 A对应电荷量 Q = It = 0.5 A × (365×24×3600 s) = 15,768,000 C。电子摩尔数 = Q / F = 15,768,000 / 96,500 ≈ 163.4 mol。由于每个Zn原子释放2个电子，需Zn摩尔数 = 163.4 / 2 = 81.7 mol。Zn质量 = 81.7 mol × 65.4 g mol⁻¹ = 5,343 g ≈ 5.34 kg。因此，保护该部件一年约需消耗5.34 kg锌。实际操作中会分散布置多个阳极，且会使用更多锌以确保即使部分消耗后仍能持续保护。
7. Acid–Base Equilibria – Buffer System in Blood | 酸碱平衡——血液缓冲系统
Blood pH is maintained around 7.40 by the carbonic acid– , calculate the pH. (Kₐ of H₂CO₃ = 4.3 × 10⁻⁷ mol dm⁻³). Predict the effect of holding one's breath on blood pH.
血液pH通过碳酸-碳酸氢盐缓冲系统维持在7.40左右。若HCO₃⁻浓度为0.024 mol dm⁻³，溶解的CO₂（以H₂CO₃形式）浓度为0.0012 mol dm⁻³，计算pH。（H₂CO₃的Kₐ = 4.3 × 10⁻⁷ mol dm⁻³）。预测屏息对血液pH的影响。
The buffer equilibrium is H₂CO₃ ⇌ ol dm⁻³. pH = -log₁₀(2.15 × 10⁻⁸) ≈ 7.67. This is slightly higher than normal 7.40, indicating the given concentrations might differ from real arterial blood values (normally [HCO₃⁻]/[H₂CO₃] ≈ 20:1, giving pH 7.4). Our ratio here is 0.024/0.0012 = 20, which should give pH = pKₐ + log(20) = 6.37 + 1.30 = 7.67. Indeed, using pKₐ = -log(4.3×10⁻⁷) = 6.37, pH = 6.37 + log(0.024/0.0012) = 6.37 + log(20) = 6.37 + 1.30 = 7.67. So the calculated value is 7.67. When holding breath, CO₂ accumulates, so [CO₂] (hence [H₂CO₃]) increases. The ratio [HCO₃⁻]/[H₂CO₃] decreases, so pH drops, potentially below 7.35 (acidosis). The body responds by increasing breathing rate to expel CO₂.
缓冲平衡为H₂CO₃ ⇌ H⁺ + HCO₃⁻，Kₐ = [H⁺][HCO₃⁻] / [H₂CO₃]。变形得[H⁺] = Kₐ × [H₂CO₃] / [HCO₃⁻] = (4.3 × 10⁻⁷) × (0.0012) / (0.024) = 2.15 × 10⁻⁸ mol dm⁻³。pH = -log₁₀(2.15 × 10⁻⁸) ≈ 7.67。此值略高于正常值7.40，表明给定浓度可能与实际动脉血值不同（通常[HCO₃⁻]/[H₂CO₃] ≈ 20:1，得pH 7.4）。本文比值为20，应得pH = pKₐ + log(20) = 6.37 + 1.30 = 7.67。确认为7.67。当屏息时，CO₂积聚，导致[CO₂]（亦即[H₂CO₃]）上升。[HCO₃⁻]/[H₂CO₃]比值下降，pH随之下降，可能低于7.35（酸中毒）。身体以加快呼吸排出CO₂应对。
8. Transition Metals – Catalytic Converters | 过渡金属——催化转化器
mistry, explain how these metals catalyse the reaction 2CO + 2NO → 2CO₂ + N₂. Include the concepts of variable oxidation states and surface adsorption.
汽车催化转化器使用铂、钯和铑来减少NO和CO排放。运用过渡金属化学知识，解释这些金属如何催化反应2CO + 2NO → 2CO₂ + N₂。包含变氧化态和表面吸附概念。
Transition metals display variable oxidation states and have partially filled d-orbitals, enabling them to chemisorb reactant molecules onto their surface. In the catalytic converter, NO and CO adsorb onto the metal surface. Electron density can shift from the metal to the adsorbed NO, weakening the N–O bond; simultaneously, the metal can oxidise CO to CO₂ by abstracting oxygen from NO. For example, rhodium cycles between Rh(0) and Rh(III) ox states: NO dissociates on the surface, leaving adsorbed N and O atoms; CO reacts with adsorbed O to form CO₂, leaving N atoms to combine to N₂. Platinum and palladium facilitate the oxidation steps. The reaction proceeds via a Langmuir–Hinshelwood mechanism where both reactants are adsorbed, surface diffusion brings them together, and product desorbs. This lowers activation energy by providing an alternative pathway with a surface-stabilised transition state. The efficiency relies on high dispersion of metals on a ceramic honeycomb support to maximise surface area.
过渡金属具有变氧化态和部分填充的d轨道，能将反应物分子化学吸附至表面。在催化转化器中，NO和CO吸附于金属表面。电子密度可从金属转移至吸附的NO，削弱N–O键；同时，金属可通过从NO夺取氧将CO氧化为CO₂。例如，铑在Rh(0)和Rh(III)氧化态间循环：NO在表面解离留下吸附的N和O原子；CO与吸附O反应生成CO₂，N原子结合为N₂。铂和钯促进氧化步骤。反应通过Langmuir–Hinshelwood机理进行：两个反应物均被吸附，表面扩散使其相遇，产物脱附。这通过提供一条表面稳定过渡态的替代路径降低了活化能。效率依赖于将金属高度分散在陶瓷蜂窝载体上以最大化表面积。
9. Organic Mechanisms – Nucleophilic Substitution Puzzle | 有机机理——亲核取代之谜
A student reacts 1-bromobutane with NaOH(aq) under reflux, obtaining butan-1-ol. When they use 2-bromo-2-methylpropane under the same conditions, 2-methylpropan-2-ol is formed, but the reaction is significantly faster even though the C–Br bond is stronger. Explain the mechanism for each, including rate equations, and deduce why the tertiary substrate reacts faster.
某学生将1-溴丁烷与NaOH水溶液回流反应，得到丁-1-醇。当在相同条件下使用2-溴-2-甲基丙烷时，生成2-甲基丙-2-醇，且反应显著更快，尽管C–Br键更强。解释各自的机理，包括速率方程，并推断为何叔卤代烷反应更快。
e, displacing Br⁻ in one step. Rate = k [1-bromobutane][OH⁻], second order overall. The transition state involves partial C–O and C– teric hindrance around the carbon affects the rate; primary carbons are easily accessible.
1-溴丁烷为伯卤代烷；反应通过Sₙ2机理进行：OH⁻从背面进攻碳，协同一步取代Br⁻。速率 = k [1-溴丁烷][OH⁻]，总二级反应。过渡态涉及部分形成的C–O和C–Br键，碳经瓦尔登翻转。中心碳周围的空间位阻影响速率；伯位碳易接近。
2-bromo-2-methylpropane is a tertiary haloalkane. Sₙ2 is impossible because of extreme steric hindrance from three methyl groups blocking back
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