Infrared (IR) spectroscopy is one of the most important analytical tools in organic chemistry. It allows chemists to identify the functional groups present in a molecule by measuring the absorption of infrared radiation, which causes characteristic bond vibrations. In A-Level Chemistry, students are expected to interpret IR spectra, justify functional group assignment, and understand how molecular structure affects absorption frequency.
1. Foundations of Infrared Spectroscopy | 红外光谱的基本原理
Molecules are not rigid; chemical bonds behave like tiny springs that can stretch and bend. When infrared radiation has the same frequency as the natural vibration of a bond, the molecule absorbs the energy and the vibration becomes more vigorous.
For a vibration to be infrared active, the molecule must undergo a change in dipole moment during the vibration. For example, the symmetric stretch of carbon dioxide (CO₂) produces no net dipole change and is infrared inactive, while the asymmetric stretch does produce a dipole change and is infrared active.
An IR spectrum is usually plotted as transmittance (%) against wavenumber (cm⁻¹). Absorption peaks appear as downward dips; a stronger or more concentrated sample gives broader and deeper dips.
There are two main classes of molecular vibrations: stretching and bending. Stretching vibrations change the bond length along the bond axis, whereas bending vibrations change the bond angle. Stretching generally requires more energy than bending, so stretching peaks appear at higher wavenumbers.
Polyatomic groups can undergo symmetric and asymmetric stretching. For example, a methylene group (CH₂) can vibrate with the two C-H bonds moving in phase (symmetric) or out of phase (asymmetric); the asymmetric stretch usually occurs at a slightly higher wavenumber.
These vibration modes are quantised. Only certain frequencies of IR radiation are absorbed, producing a spectrum with discrete absorption bands rather than a continuous absorption curve.
3. Factors Governing Vibrational Frequency | 决定振动频率的因素
The vibrational frequency of a bond depends on the strength of the bond and the masses of the two atoms. Stronger bonds vibrate faster: a carbon-carbon triple bond absorbs at a higher wavenumber than a double bond, which in turn absorbs higher than a single bond.
Heavier atoms vibrate more slowly. This explains why C-H stretching appears near 3000 cm⁻¹, C-C stretching near 1200 cm⁻¹, and C-Cl stretching near 750 cm⁻¹. The reduced mass μ of two atoms of masses m₁ and m₂ gives a quantitative relation:
In this expression, ν is the vibrational frequency and k is the bond force constant. Although A-Level students are not required to calculate frequencies, this relationship helps explain why C≡C, C=C and
Published by TutorHao | A-Level Chemistry Revision Series | aleveler.com
📚 Ion Polarisation and Its Effect on Chemical Bonding | 离子极化及其对化学键性质的影响
Ionic compounds are traditionally described as being held together by the electrostatic attraction between oppositely charged ions. However, in reality, few bonds are perfectly ionic. The concept of ion polarisation explains how ions can distort each other’s electron clouds, leading to a gradual shift from ionic toward covalent character. This idea is essential for explaining many physical and chemical properties of compounds studied in A-Level Chemistry.
Ion polarisation refers to the distortion of the electron cloud of an anion by the positively charged cation. When a small, highly charged cation approaches a large anion, it pulls the anion’s electron cloud toward itself. This distortion makes the anion no longer spherical and creates a partial sharing of electron density — a hallmark of covalent bonding.
Fajans’ rules provide a simple framework for predicting the extent of polarisation. The greater the polarising power of the cation and the greater the polarisability of the anion, the more covalent character the bond will have.
2. Factors Affecting the Polarising Power of a Cation | 影响阳离子极化力的因素
Charge density. Polarising power is directly related to charge density, which is the charge per unit volume of the ion. A small cation with a high charge, such as Al³⁺ or Mg²⁺, has a very high charge density and therefore exerts a strong attractive force on the anion’s electron cloud.
Electronic configuration. Cations with a pseudo-noble gas configuration (e.g., Zn²⁺, Cu⁺) have a greater polarising power than cations with a noble gas configuration of the same size and charge. This is because the d-orbitals are less effective at shielding the nuclear charge, allowing the nucleus to attract the anion’s electrons more strongly.
Size of the cation. For ions of the same charge, a smaller cation has greater polarising power. For example, among the Group 2 ions, Be²⁺ is the smallest and hence the most polarising.
3. Factors Affecting the Polarisability of an Anion | 影响阴离子极化率的因素
Size of the anion. Large anions have loosely held outer electrons, making them easy to distort. For example, iodide (I⁻, ionic radius ≈ 220 pm) is much more polarisable than fluoride (F⁻, ionic radius ≈ 133 pm).
Charge of the anion. A higher negative charge increases the repulsion among the anion’s own electrons, making the electron cloud more diffuse and easier to polarise. Thus, S²⁻ is more polarisable than Cl⁻.
Nature of the anion. Anions that are larger and have lower electronegativity are generally more polarisable. Oxide (O²⁻) and sulfide (S²⁻) anions are highly polarisable, which explains why many oxides and sulfides exhibit significant covalent character.
5. Effect on Bond Character: Ionic vs Covalent Continuum | 对键性质的影响：离子性—共价性连续体
Ionic and covalent bonds are not two entirely separate categories but rather two ends of a continuum. Ion polarisation is the mechanism that moves a bond from the ionic end toward the covalent end. When a cation strongly distorts the anion’s electron cloud, the electron density becomes concentrated between the two nuclei, resembling a covalent bond.
Percentage ionic character. One way to estimate bond character is to compare the experimental dipole moment or bond length with theoretical values for a perfect ionic bond. The greater the degree of polarisation, the larger the deviation from ideal ionic behaviour.
Solubility behaviour. Ionic compounds tend to dissolve in polar solvents such as water, while covalent compounds dissolve in non-polar solvents. The chlorides of Group 2 elements provide a clear illustration: BeCl₂ is covalent and soluble in organic solvents, whereas BaCl₂ is ionic and water-soluble.
6. Examples: Chlorides Across Period 3 | 实例：第三周期氯化物
Across Period 3, the chlorides show a clear trend in bonding character. NaCl and MgCl₂ are ionic solids with high melting points. AlCl₃ is borderline: it is ionic in the solid state but exists as a dimer Al₂Cl₆ in the vapour phase, showing significant covalent character. SiCl₄, PCl₅, and SCl₄ are simple covalent molecular compounds with low melting points.
The trend is explained by the increasing charge density of the cation from Na⁺ to Al³⁺, which increases polarising power. The Al³⁺ ion is small and highly charged, so it strongly polarises the large Cl⁻ ion, giving Al–Cl bonds substantial covalent character. Beyond silicon, the bonding becomes fully covalent as both atoms share electrons equally.
7. Examples: Halides of the Same Metal | 实例：同一金属的卤化物
Consider the halides of aluminium: AlF₃, AlCl₃, AlBr₃, and AlI₃. As the halide ion becomes larger, its polarisability increases. Fluoride is the smallest and least polarisable, so AlF₃ is predominantly ionic with a high melting point (1291 °C). AlCl₃ is intermediate, and AlI₃ is largely covalent, with a relatively low melting point and a layer structure.
Experimentally observable. The electrical conductivity of molten compounds decreases as covalent character increases. Molten AlF₃ conducts electricity well, while molten AlI₃ is a poor conductor because it consists of neutral molecules rather than free ions.
8. Effect on Melting and Boiling Points | 对熔点和沸点的影响
Ionic compounds generally have high melting and boiling points because the electrostatic forces between ions are strong and extend throughout the crystal lattice. When a bond acquires covalent character, the extended ionic lattice is replaced by discrete molecules or a layered structure, resulting in weaker intermolecular forces and thus lower melting points.
BeCl₂ vs BaCl₂. BeCl₂ has a melting point of 399 °C and exists as a covalent polymer in the solid state, while BaCl₂ melts at 962 °C and forms a typical ionic lattice. The small Be²⁺ ion with its very high charge density polarises Cl⁻ so strongly that the bonds become largely covalent.
Trend in Group 2 chlorides. Melting points decrease from MgCl₂ to BaCl₂? Actually, the trend is more subtle: MgCl₂ (714 °C), CaCl₂ (782 °C), SrCl₂ (874 °C), BaCl₂ (962 °C). The increase from Ca to Ba reflects the increasing ionic character as the cation becomes larger and less polarising. MgCl₂ shows some covalent character due to the small size of Mg²⁺.
9. Effect on Thermal Stability of Carbonates and Nitrates | 对碳酸盐和硝酸盐热稳定性的影响
Ion polarisation is key to understanding the thermal decomposition of metal carbonates and nitrates. The carbonate ion CO₃²⁻ is large and polarisable. When a small, highly charged cation polarises the carbonate ion, it attracts electron density away from the C–O bonds, weakening them and making the ion less stable.
Carbonates. For Group 2 carbonates, thermal stability increases down the group: MgCO₃ decomposes at about 350 °C, while BaCO₃ requires about 1360 °C. As the cation size increases down the group, its polarising power decreases, so the carbonate ion is less destabilised.
Nitrates. The same principle applies to nitrates. The nitrate ion NO₃⁻ is also polarisable. Group 1 nitrates (except LiNO₃) decompose to the nitrite and oxygen:
However, LiNO₃, with the small and highly polarising Li⁺ ion, decomposes further to Li₂O, NO₂, and O₂ because the nitrate is destabilised so much. Among Group 2 nitrates, all decompose to the oxide, NO₂, and O₂, but the temperature required increases down the group: Mg(NO₃)₂ decomposes around 330 °C, while Ba(NO₃)₂ decomposes around 590 °C.
10. Quantitative Aspect: Lattice Energy vs Polarisation | 定量方面：晶格能与极化
Ion polarisation also explains why some experimentally determined lattice energies are higher (more exothermic) than values calculated from a purely ionic model. The additional stability arises from the partial covalent character of the bond, which contributes extra enthalpy beyond simple Coulombic attraction.
For example, the lattice energy of AgCl calculated from the Born–Haber cycle is more negative than that predicted by electrostatic models. This discrepancy is attributed to the significant covalent character of Ag–Cl bonds, caused by the high polarising power of Ag⁺, which has a d¹⁰ configuration with poor shielding.
Solubility. Increased covalent character often reduces solubility in water because covalent bonds do not readily dissociate into ions. For example, AgCl is far less soluble in water than NaCl, despite both containing a Group 1-like metal cation; the Ag⁺ ion strongly polarises Cl⁻, making AgCl largely covalent in character.
Colour. Ion polarisation can also influence the colour of compounds. When an anion’s electron cloud is strongly distorted, the energy gap between molecular orbitals changes, causing absorption of visible light. Many transition metal and post-transition metal compounds are coloured due to this polarisation effect, whereas purely ionic salts such as NaCl are white.
12. Examination Tips and Common Pitfalls | 考试要点与常见误区
Key points to remember:
需要记住的要点：
Polarising power increases with increasing charge density of the cation — smaller size and higher charge mean stronger polarising power.
极化力随阳离子电荷密度增大而增强——体积越小、电荷越高，极化力越强。
Polarisability increases with increasing anion size and negative charge. Large anions such as I⁻, S²⁻, and O²⁻ are easily distorted.
极化率随阴离子体积和负电荷的增加而增大。大阴离子如I⁻、S²⁻和O²⁻容易被扭曲。
Cations with d-electrons (pseudo-noble gas) are more polarising than comparable noble-gas cations because of poor nuclear shielding.
含d电子的阳离子（伪稀有气体构型）比同等的稀有气体构型阳离子具有更强的极化力，因为核屏蔽效果差。
Greater polarisation → more covalent character → lower melting point, reduced water solubility, and altered thermal stability.
极化越强 → 共价性越强 → 熔点降低、水溶性下降、热稳定性改变。
Common mistakes:
常见误区：
Confusing polarising power (cation property) with polarisability (anion property).
混淆极化力（阳离子性质）与极化率（阴离子性质）。
Stating that all chlorides of Period 3 are ionic at room temperature; AlCl₃ is covalent in the vapour phase and shows significant covalent character in the solid state.
Forgetting that the thermal stability of carbonates and nitrates increases as the cation becomes larger because polarisation decreases.
忘记碳酸盐和硝酸盐的热稳定性随阳离子变大而增加，因为极化减弱。
13. Conclusion | 结论
Ion polarisation is a unifying concept that bridges ionic and covalent bonding. By understanding how cations distort the electron clouds of anions, we can predict and explain trends in melting points, solubility, thermal stability, lattice energies, and even colour. This is not just a theoretical curiosity — it is a core explanatory tool in A-Level Chemistry that appears across multiple topics, from periodicity to Group 2 chemistry.
Mastering the factors affecting polarisation — cation charge density, anion size, and electronic configuration — allows you to answer a wide range of exam questions with confidence. Always link the microscopic distortion of electron clouds to the macroscopic properties you observe, and you will find that many seemingly unrelated properties share the same root cause.
📚 A-Level Chemistry: Nucleophilic Addition of HCN to Aldehydes and Ketones | A-Level化学：醛酮与HCN的亲核加成反应
The nucleophilic addition of hydrogen cyanide (HCN) to aldehydes and ketones is a classic reaction in A-Level Chemistry, forming hydroxynitriles (cyanohydrins). This reaction is highly valued in examinations for its mechanism, practical conditions, and synthetic applications.
HCN adds across the carbon-oxygen double bond of an aldehyde or ketone. The product is a hydroxynitrile, also called a cyanohydrin, containing both a hydroxyl group (-OH) and a nitrile group (-C≡N) on the same carbon atom.
HCN is a very weak acid (pKa ≈ 9.2). In its molecular form, it is a poor nucleophile because the cyanide carbon is not strongly electron-rich. However, the cyanide ion CN⁻ is a strong nucleophile. Therefore, the reaction actually proceeds via the cyanide ion, not molecular HCN itself.
The equilibrium for the dissociation of HCN in solution is critical:
HCN 在溶液中的解离平衡至关重要：
HCN ⇌ H⁺ + CN⁻
Because the equilibrium lies far to the left, the concentration of CN⁻ is very low. This makes the reaction slow and inefficient if only HCN is used alone.
由于平衡强烈偏左，CN⁻ 的浓度非常低。如果仅使用 HCN，反应会非常缓慢且效率低下。
3. The Role of a Catalyst | 催化剂的作用
A small amount of potassium cyanide (KCN) or sodium cyanide (NaCN) is added as a source of cyanide ions. These salts dissociate fully in solution, providing a steady concentration of CN⁻, which then acts as the true nucleophile.
CN⁻ attacks the carbonyl carbon, forming a negatively charged intermediate.
CN⁻ 进攻羰基碳，形成带负电的中间体。
The intermediate abstracts a proton (H⁺) from HCN, regenerating CN⁻.
中间体从 HCN 夺取质子（H⁺），再生出 CN⁻。
Thus, KCN or NaCN is not consumed overall; it acts as a catalyst.
因此，KCN 或 NaCN 在总体上不被消耗，起到催化剂作用。
4. Mechanism: Step-by-Step | 反应机理：逐步解析
The mechanism is called nucleophilic addition. Let us examine each step in detail.
该机理称为亲核加成。下面逐步详细分析。
Step 1: Nucleophilic attack — The cyanide ion CN⁻ donates its lone pair of electrons to the partially positive carbonyl carbon atom. The π bond of the C=O group breaks, and the two electrons move to the oxygen atom, forming an alkoxide ion.
Step 2: Protonation — The alkoxide oxygen then abstracts a proton from HCN (or from water in acidic work-up), forming the neutral hydroxynitrile product and regenerating the cyanide catalyst.
5. Reactivity of Aldehydes vs Ketones | 醛与酮的反应活性比较
Aldehydes are generally more reactive than ketones in nucleophilic addition reactions. There are two main reasons for this difference.
在亲核加成反应中，醛通常比酮更活泼。造成这种差异的原因主要有两点。
Steric factor: In ketones, the carbonyl carbon is bonded to two alkyl or aryl groups, creating greater steric hindrance. This makes it harder for the nucleophile to approach.
电子因素： 酮中羰基碳与两个烷基或芳基相连，空间位阻较大，亲核试剂难以接近。
Electronic factor: Alkyl groups are electron-donating. In ketones, two alkyl groups donate electron density to the carbonyl carbon, making it less electrophilic. Aldehydes have only one alkyl group (or hydrogen), so the carbonyl carbon is more positive and more attractive to nucleophiles.
Methanal (HCHO) is the most reactive aldehyde because the carbonyl carbon is attached to two hydrogen atoms, offering negligible steric hindrance and no electron-donating alkyl groups.
甲醛（HCHO）是活性最高的醛，因为羰基碳连接两个氢原子，空间位阻极小且没有给电子烷基。
6. Practice: Specific Examples | 实例练习
Let us write the products for some common reactants.
The hydroxynitrile product is a versatile intermediate in organic synthesis. The nitrile group can be converted into a carboxylic acid or an amine, which allows chemists to lengthen carbon chains.
羟基腈产物是有机合成中的重要中间体。氰基可转化为羧酸或胺，从而实现碳链增长。
Hydrolysis to carboxylic acid:
水解生成羧酸：
CH₃CH(OH)CN + 2H₂O + H⁺ → CH₃CH(OH)COOH + NH₄⁺
This reaction is especially valuable because the carbonyl carbon of the original aldehyde/ketone becomes a new carboxylic acid carbon, increasing the chain length by one carbon atom.
该反应尤为珍贵，因为原醛/酮的羰基碳变为新的羧酸碳，使碳链增加一个碳原子。
Reduction to amine:
还原生成胺：
CH₃CH(OH)CN + 4[H] → CH₃CH(OH)CH₂NH₂
This route provides a method to synthesise β-amino alcohols, which are important in pharmaceutical chemistry.
这条路线提供了合成 β-氨基醇的方法，在药物化学中具有重要意义。
8. Key Points for Examinations | 考试要点
Students often lose marks in CIE A-Level Chemistry exams for the following reasons:
学生在CIE A-Level化学考试中常因以下原因失分：
Forgetting that the nucleophile is CN⁻, not HCN itself.
忘记真正的亲核试剂是 CN⁻ 而非 HCN 本身。
Omitting the negative charge on the alkoxide intermediate in the mechanism diagram.
在机理图中遗漏烷氧负离子中间体的负电荷。
Writing the product incorrectly, especially the position of the -OH and -CN groups.
产物结构书写错误，尤其是 -OH 和 -CN 的位置。
Not stating that KCN/NaCN is a catalyst, and that the reaction requires warming.
未能说明 KCN/NaCN 是催化剂以及反应需要加热。
Confusing this mechanism with nucleophilic substitution or electrophilic addition.
将此机理与亲核取代或亲电加成混淆。
Always draw curly arrows precisely: the arrow must start from the lone pair of CN⁻ and point toward the carbonyl carbon, and the other arrow must show the π bond pair moving to the oxygen.
In practice, the reaction is carried out by mixing the aldehyde or ketone with a solution of sodium or potassium cyanide, followed by addition of a slight excess of dilute hydrochloric acid or sulfuric acid. The acid is needed to protonate the alkoxide intermediate but must not be too strong, as it would convert all CN⁻ into HCN and stop the reaction.
Alternatively, the reaction can be performed with HCN gas in the presence of a basic catalyst such as KCN.
另一种方法是在 KCN 等碱性催化剂存在下使 HCN 气体参与反应。
Reagent | 试剂
Condition | 条件
Purpose | 目的
KCN or NaCN | 氰化钾或氰化钠
Aqueous solution | 水溶液
Provide CN⁻ | 提供 CN⁻
Dilute H₂SO₄ | 稀硫酸
Slight excess | 稍过量
Protonate intermediate | 质子化中间体
Heat | 加热
Gently warm | 微热
Increase reaction rate | 提高反应速率
10. Comparison with Other Nucleophilic Additions | 与其他亲核加成反应的比较
HCN addition shares the same fundamental mechanism as the addition of sodium hydrogensulfite (NaHSO₃) and Grignard reagents, but there are notable differences.
HCN 加成与亚硫酸氢钠（NaHSO₃）和格氏试剂的加成具有相同的基本机理，但也存在显著差异。
HCN addition is reversible; the product can revert to the carbonyl compound under basic conditions.
HCN 加成是可逆的；产物在碱性条件下可复原为羰基化合物。
NaHSO₃ addition is also reversible and useful for purifying aldehydes.
NaHSO₃ 加成同样可逆，常用于提纯醛。
Grignard reagent addition is irreversible and forms alcohols, not hydroxynitriles.
格氏试剂加成不可逆，生成醇而非羟基腈。
Understanding these comparisons helps in exam questions that ask you to distinguish reaction types.
理解这些比较有助于回答要求区分反应类型的考试问题。
11. Curly Arrow Notation in the Mechanism | 机理中的弯箭头表示法
In CIE exams, the mechanism is often tested as a drawn diagram. Ensure your diagram includes:
在CIE考试中，该机理常以图示形式考查。确保你的图示包含以下几点：
The correct structure of the aldehyde or ketone, showing the δ⁺ and δ− on the C=O bond.
醛或酮的正确结构，标明 C=O 的 δ⁺ 和 δ−。
The CN⁻ ion with a complete octet and a negative charge.
具有完整八隅体和负电荷的 CN⁻ 离子。
A curly arrow from the lone pair of CN⁻ to the carbonyl carbon.
一条从 CN⁻ 孤对电子指向羰基碳的弯箭头。
A curly arrow from the C=O π bond to the oxygen atom.
一条从 C=O π 键指向氧原子的弯箭头。
The alkoxide intermediate with a negative charge on oxygen.
氧上带有负电荷的烷氧负离子中间体。
A second curly arrow from the alkoxide oxygen to H⁺ of HCN.
第二条从烷氧负离子氧指向 HCN 中 H⁺ 的弯箭头。
Do not forget one arrow or you will lose marks even if the overall equation is correct.
千万不能遗漏任何一条箭头，即使总反应式正确也会因此失分。
12. Summary | 总结
The addition of HCN to aldehydes and ketones is a nucleophilic addition reaction producing hydroxynitriles. The cyanide ion CN⁻ is the active nucleophile, supplied by KCN or NaCN. Aldehydes react faster than ketones due to steric and electronic factors. The products are valuable synthetic intermediates; hydrolysis converts them to α-hydroxy carboxylic acids and reduction converts them to amino alcohols.
Mastering the mechanism, conditions, and reactivity trends will help you answer both multiple-choice and structured questions confidently in your CIE A-Level Chemistry examination.
📚 Presenting Experimental Data and Observations in Chemistry | 化学实验数据与观察结果的规范呈现
In CIE A-Level Chemistry, the ability to accurately present experimental data and observations is as crucial as the experimental procedure itself. Examiners consistently reward candidates who demonstrate meticulous recording techniques and clear data processing. This guide breaks down the essential skills for transforming raw results into polished, high-scoring scientific reports.
Raw data forms the foundation of any scientific investigation. In the exam, you must record your results directly into the answer booklet in a structured format. Never scribble results on rough paper and copy them later, as transcription errors are a leading cause of lost marks.
Ensure you record additional qualitative observations (e.g., colour changes, precipitate formation) *at the moment* they occur, not retrospectively. This timeliness ensures the authenticity and completeness of your evidence.
A well-designed results table allows the reader to interpret data intuitively. The independent variable is typically placed in the first column, followed by the dependent variable(s).
设计良好的结果表能让读者直观地解读数据。通常，自变量放在第一列，随后是因变量。
Every column must have a clear heading that includes the quantity name, followed by the symbol and its unit in brackets. For example: “Volume of HCl / cm³” or “Mass of precipitate / g”.
Units are stated in the heading, not repeated in every cell.
Values are recorded with spacing to prevent confusion.
A table is drawn with a single horizontal line for the top and bottom borders, and one below the headings.
Raw data is never processed within the same column.
制表的关键规则：
单位标注在表头中，不在每个单元格内重复。
数值之间留出适当间距，以防混淆。
表格上下仅用单横线作为边框，表头下方用单横线分隔。
原始数据从不在同一列内进行加工处理。
3. Significant Figures and Precision | 有效数字与精确度
The number of significant figures (s.f.) in a measurement reflects its precision. A measurement of 25.0 cm³ is more precise than 25 cm³. When recording data, align your significant figures with the precision of the apparatus (e.g., burette readings to 2 decimal places).
During calculations, carry extra significant figures to avoid rounding errors, but present the final answer to a consistent and appropriate number of significant figures, usually determined by the raw data (typically 2 or 3 s.f.).
Qualitative observations are vital for deducing reaction pathways. Vague terms like ‘heated’ or ‘turned blue’ are insufficient. You must describe the intensity, rate, and locality of the change.
Colour change: “White solid”, “pale blue solution”, “black precipitate formed”.
Gas evolution: “Effervescence observed” or “bubbles of gas released rapidly”.
Heat change: “Significant exothermic reaction” or “mixture becomes cold”.
Precipitate type: “Curdy white precipitate” or “gelatinous blue precipitate”.
使用精确的术语：
颜色变化：“白色固体”、“淡蓝色溶液”、“生成黑色沉淀”。
气体逸出：“观察到出现气泡”或“迅速释放大量气泡”。
热变化：“明显的放热反应”或“混合物变冷”。
沉淀类型：“絮状白色沉淀”或“胶状蓝色沉淀”。
5. Fundamentals of Graphing | 绘图基础
Graphs visually represent the relationship between variables. In A-Level chemistry, you are often required to plot data from rate experiments or titrations. Choose a scale that utilises at least half of the graph paper in both directions.
Axes must be labelled with the quantity and unit (e.g., “Time / s”). The independent variable goes on the x-axis, and the dependent variable on the y-axis. Plot each point with a sharp pencil, using a small cross (‘×’) rather than a dot, for precision.
坐标轴必须标注物理量和单位（例如“时间 / s”）。自变量位于 x 轴，因变量位于 y 轴。用削尖的铅笔绘制每个点，使用小叉号（’×’）而非圆点，以确保精确。
6. Drawing Line of Best Fit | 绘制最佳拟合线
Once all points are plotted, draw a line of best fit. It does not have to pass through every point, but should represent the general trend, with an equal number of points above and below the line.
A straight line is appropriate for first-order kinetics or calibration curves. A smooth curve is used for non-linear relationships, such as the volume of gas produced over time in a reaction reaching completion.
Treat anomalous points (outliers) carefully. If a point clearly does not fit the trend, mark it as anomalous, and exclude it from the line of best fit. Investigate potential systematic errors.
7. Interpreting Graphs and Calculating Slopes | 图表解读与斜率计算
The slope (gradient) of a linear graph often contains a key constant. For example, in a concentration-time graph, the initial rate is found by calculating the gradient of the tangent at t=0.
📚 Complete Guide to Typical Reactions of Alcohols | A-Level化学：醇的典型反应归纳
Alcohols are one of the most versatile functional groups in A-Level organic chemistry. Their reactions span substitution, elimination, oxidation, and esterification, making them a favourite topic in CIE examinations. This article systematically organises every key reaction you need to master, with conditions, mechanisms, and exam-focused observations.
The functional group of an alcohol is the hydroxyl group (–OH) attached to a saturated carbon atom. The general formula for aliphatic alcohols is CₙH₂ₙ₊₁OH. The electronegative oxygen atom creates a polar O–H bond, which underpins most of the chemical behaviour discussed below.
This classification is not merely structural – it determines whether the alcohol can be oxidised, and to what product. Examiners frequently test this distinction, so memorise it early.
Alcohols burn readily in air or oxygen, releasing carbon dioxide and water. Complete combustion is highly exothermic, which is why ethanol is used as a fuel additive.
In limited oxygen, incomplete combustion occurs, producing carbon monoxide (CO) and carbon (soot). This is important when comparing the environmental impact of alcohol fuels versus petrol.
Exam tip: When writing combustion equations for any alcohol, always balance by counting carbon atoms first, then hydrogen, then oxygen last.
考试提示：写任何醇的燃烧方程式时，先数碳原子配平，再配氢，最后配氧。
3. Reaction with Sodium – Acidic Character | 与金属钠的反应——酸性特征
Alcohols react with sodium metal, releasing hydrogen gas. This demonstrates the weakly acidic nature of the O–H bond, although alcohols are far less acidic than water or carboxylic acids.
醇与金属钠反应放出氢气。这体现了O–H键的弱酸性特征，尽管醇的酸性远弱于水和羧酸。
2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂↑
2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂↑
Sodium ethoxide (C₂H₅ONa) is an ionic compound formed when the sodium replaces the hydrogen of the hydroxyl group. The reaction of ethanol with sodium is less vigorous than that of water with sodium, because the alkyl group donates electron density, making the O–H bond less polarised.
The trend in reactivity with sodium follows the order: methanol > primary > secondary > tertiary, due to increasing electron donation from additional alkyl groups. This reaction is a useful chemical test for the presence of an –OH group, as vigorous effervescence of hydrogen gas is observed.
Primary alcohols react with hydrogen halides to form haloalkanes. However, the reaction with HCl alone is slow; a catalyst such as anhydrous zinc chloride (ZnCl₂) is required.
伯醇与氢卤酸反应生成卤代烷。但单纯与HCl反应较慢，需加入无水氯化锌（ZnCl₂）作催化剂。
C₂H₅OH + HCl → C₂H₅Cl + H₂O
C₂H₅OH + HCl → C₂H₅Cl + H₂O
Alternatively, the use of PCl₅, PCl₃, or SOCl₂ provides much faster and more complete substitution. These reagents are preferred for preparing pure haloalkanes in the laboratory.
With phosphorus pentachloride (PCl₅), misty white fumes of HCl gas are observed – a classic test for the presence of an –OH group:
与五氯化磷（PCl₅）反应时，可观察到HCl气体的白色酸雾——这是检验–OH基团的经典方法：
C₂H₅OH + PCl₅ → C₂H₅Cl + POCl₃ + HCl↑
C₂H₅OH + PCl₅ → C₂H₅Cl + POCl₃ + HCl↑
With thionyl chloride (SOCl₂), the by-products SO₂ and HCl are both gaseous, so pure chloroalkane can be isolated easily by simple distillation:
与氯化亚砜（SOCl₂）反应时，副产物SO₂和HCl均为气体，通过简单蒸馏即可获得纯净的氯代烷：
C₂H₅OH + SOCl₂ → C₂H₅Cl + SO₂↑ + HCl↑
C₂H₅OH + SOCl₂ → C₂H₅Cl + SO₂↑ + HCl↑
Exam tip: The order of reactivity of hydrogen halides is HI > HBr > HCl. For HBr, sodium bromide mixed with dilute sulfuric acid is commonly used in situ.
Alcohols undergo elimination (dehydration) when heated in the presence of an acid catalyst, producing an alkene and water. The most common conditions are concentrated sulfuric acid (H₂SO₄) at approximately 170°C, or passing alcohol vapour over heated aluminium oxide (Al₂O₃) catalyst.
The reaction proceeds via protonation of the –OH group, followed by loss of water and formation of a carbocation intermediate, then loss of a β-hydrogen to form the C=C double bond. For unsymmetrical alcohols, dehydration often gives a mixture of alkene isomers, with the more substituted alkene (Zaitsev’s rule) as the major product.
It is crucial to note the temperature dependence: at 140°C, ethanol undergoes intermolecular dehydration to form ethoxyethane (diethyl ether), whereas at 170°C, intramolecular dehydration yields ethene:
Primary alcohols are oxidised first to aldehydes and then to carboxylic acids. The typical oxidising agent is acidified potassium dichromate(VI), K₂Cr₂O₇/H₂SO₄, which changes from orange to green as Cr(VI) is reduced to Cr(III).
To stop at the aldehyde stage, the aldehyde must be distilled off immediately since it has a lower boiling point than the alcohol. For example, ethanol (b.p. 78°C) is oxidised to ethanal (b.p. 21°C), which distils over before it can be further oxidised.
If reflux conditions are used, the aldehyde is further oxidised to the carboxylic acid:
若采用回流条件，醛会进一步被氧化为羧酸：
CH₃CHO + [O] → CH₃COOH (reflux)
CH₃CHO + [O] → CH₃COOH （回流条件）
Combining these, the overall oxidation of a primary alcohol under reflux is:
综合来看，伯醇在回流条件下的总氧化反应为：
CH₃CH₂OH + 2[O] → CH₃COOH + H₂O
CH₃CH₂OH + 2[O] → CH₃COOH + H₂O
7. Oxidation of Secondary and Tertiary Alcohols | 仲醇与叔醇的氧化
Secondary alcohols are oxidised to ketones under the same conditions. Ketones are resistant to further oxidation because the carbonyl carbon has no hydrogen atom attached and the C–C bonds are not easily cleaved under mild conditions.
Tertiary alcohols do not undergo oxidation under these conditions because the carbon bearing the –OH group has no hydrogen atom available for elimination. The solution remains orange, which is a useful way to distinguish tertiary alcohols from primary and secondary alcohols.
Tertiary alcohol → No reaction (no H on the alcohol carbon) 叔醇 → 不反应（醇碳上无氢）
Colour change: orange (Cr₂O₇²⁻) → green (Cr³⁺). This is a test-tube observation you must be able to describe clearly in the exam.
颜色变化：橙色（Cr₂O₇²⁻）→ 绿色（Cr³⁺）。这是考试中你必须能清晰描述的试管观察现象。
8. Esterification | 酯化反应
Alcohols react with carboxylic acids in the presence of a strong acid catalyst (concentrated H₂SO₄) to form esters and water. This is a reversible equilibrium reaction, so reflux is used to drive the reaction towards the ester product.
Ethyl ethanoate (CH₃COOC₂H₅) is a volatile, sweet-smelling liquid. In the mechanism, the acid catalyst protonates the carboxylic acid, and the alcohol attacks the carbonyl carbon; water is eliminated. Esters are named as alkyl alkanoates: the alkyl group comes from the alcohol, and the alkanoate part comes from the carboxylic acid.
For example, methanol + butanoic acid gives methyl butanoate. Esters are widely used as flavouring agents, fragrances, and solvents, and they are the basis of biodiesel.
The iodoform test identifies compounds containing the CH₃CH(OH)– group (or CH₃CO– group). When such an alcohol is warmed with iodine in aqueous sodium hydroxide, a pale yellow precipitate of iodoform (CHI₃) forms.
Ethanol and propan-2-ol give positive iodoform tests, whereas propan-1-ol does not. This is a fast and reliable way to distinguish between isomeric alcohols.
乙醇和丙-2-醇的碘仿反应呈阳性，而丙-1-醇呈阴性。这是区分异构醇快速可靠的方法。
10. Complete Reaction Summary Table | 反应总览表
The following table consolidates all the key reactions for quick revision. You should be able to reproduce any of these equations under timed exam conditions.
下表汇总了所有关键反应，便于快速复习。你应当在限时考试条件下能够写出其中任何方程式。
Reaction Type
Reagent & Conditions
Product(s)
Observation
Combustion
O₂, ignition
CO₂ + H₂O
Blue flame, heat released
With sodium
Na metal, room temp
Alkoxide salt + H₂
Effervescence of H₂
Substitution
PCl₅ / PCl₃ / SOCl₂
Haloalkane
Misty HCl fumes (PCl₅)
Dehydration
Conc. H₂SO₄, 170°C
Alkene + H₂O
Gas bubbles of alkene
Oxidation (1°)
K₂Cr₂O₇/H⁺, reflux
Carboxylic acid
Orange → green
Oxidation (2°)
K₂Cr₂O₇/H⁺, reflux
Ketone
Orange → green
Oxidation (3°)
K₂Cr₂O₇/H⁺
No reaction
Stays orange
Esterification
RCOOH + H₂SO₄, reflux
Ester + H₂O
Sweet smell
11. Key Distinguishing Tests | 关键鉴别实验
CIE exams frequently ask how to distinguish between different alcohols or between alcohols and other functional groups. Here are the most reliable test-tube reactions.
CIE考试经常要求区分不同醇或醇与其他官能团。以下是最可靠的试管鉴别反应。
Test for –OH group: Add PCl₅ – appearance of misty white HCl fumes confirms an –OH group. Alternatively, add sodium metal – effervescence of H₂ confirms an alcoholic –OH (though carboxylic acids also react). 检验–OH基团：加入PCl₅，出现白色酸雾（HCl气体）即可确认–OH存在。或者加入金属钠，产生H₂气泡也可确认醇羟基（但羧酸也会反应）。
Distinguish 1°/2° from 3° alcohol: Warm with acidified K₂Cr₂O₇. Orange → green indicates 1° or 2°; no change indicates 3°. 区分伯/仲醇与叔醇：与酸化K₂Cr₂O₇共热。橙色变绿说明是伯醇或仲醇；不变色则为叔醇。
Distinguish 1° from 2° alcohol:Oxidation products differ – aldehyde (can be further oxidised, gives silver mirror with Tollens’ reagent) versus ketone (no silver mirror). 区分伯醇与仲醇：氧化产物不同——伯醇生成醛（可被继续氧化，与托伦试剂产生银镜），仲醇生成酮（无银镜）。
Iodoform test: Warm with I₂/NaOH – pale yellow precipitate confirms CH₃CH(OH)– group. 碘仿试验：与I₂/NaOH共热，淡黄色沉淀确认存在CH₃CH(OH)–基团。
12. Common Exam Pitfalls and Revision Checklist | 常见错误与复习清单
Students frequently lose marks on small but critical details. Here are the most common pitfalls and the correct responses.
学生常因细小但关键的问题失分。以下是常见错误及正确应对。
Pitfall: Writing the oxidation of ethanol directly to ethanoic acid without the aldehyde intermediate. Correction: Show the stepwise oxidation: ethanol → ethanal → ethanoic acid. 易错点：直接将乙醇写成氧化为乙酸而忽略醛中间体。更正：展示分步氧化：乙醇 → 乙醛 → 乙酸。
Pitfall: Using KMnO₄ when the question specifies K₂Cr₂O₇. Correction: Always match the reagent to the question. KMnO₄ (purple → colourless) is an alternative oxidant but the colour change is different. 易错点：题目指定K₂Cr₂O₇却写成KMnO₄。更正：试剂必须与题目一致。KMnO₄（紫色→无色）是替代氧化剂，但颜色变化不同。
Pitfall: Forgetting the temperature for dehydration. Correction: 170°C for alkene, 140°C for ether. 易错点：忘记脱水温度。更正：170°C生成烯烃，140°C生成醚。
Pitfall: Saying tertiary alcohols “cannot be oxidised” without mentioning the reason. Correction: The carbon bearing –OH has no hydrogen atom, so elimination to a carbonyl compound is impossible. 易错点：只说叔醇”不能被氧化”而不说明原因。更正：连接–OH的碳上没有氢原子，因此无法消除生成羰基化合物。
Pitfall: Confusing distillation and reflux conditions. Correction: Distillation removes the aldehyde product as it forms; reflux keeps everything boiling so complete oxidation to the acid occurs. 易错点：混淆蒸馏与回流条件。更正：蒸馏将生成的醛及时移出；回流使体系持续沸腾，实现完全氧化到酸。
Final checklist before the exam: (1) Can you write balanced equations for all reactions? (2) Can
Published by TutorHao | A-Level Chemistry Revision Series | aleveler.com
In the CIE A-Level Chemistry examination, practical skills are assessed through both the written Paper 3 and Paper 5, and through the practical-based questions embedded in theory papers. Mastering experimental operations, precise measurement techniques, and structured observation recording is essential for achieving high marks. This guide consolidates the key operational points, measurement protocols, and observation strategies that appear most frequently in CIE examinations.
在 CIE A-Level 化学考试中，实验技能通过 Paper 3 和 Paper 5 的实践考核，以及理论试卷中的实验类题目进行综合评估。熟练掌握实验操作、精确测量技术和结构化观察记录是获取高分的关键。本指南系统梳理了 CIE 考试中最常出现的操作要点、测量规范和观察策略。
1. Laboratory Safety and Basic Instrument Familiarisation | 实验室安全与基础仪器识别
Before conducting any experiment, candidates must be familiar with the standard laboratory apparatus and their appropriate usage. Safety goggles must always be worn when handling corrosive chemicals, heating substances, or carrying out reactions that may produce splashes. When diluting concentrated sulfuric acid, always add the acid to water slowly while stirring — never the reverse — to prevent violent exothermic splashing. A wash bottle filled with distilled water should be used for rinsing apparatus, not for adding reagent quantitatively.
2. Mass Measurement: Weighing by Difference vs. Direct Weighing | 质量测量：差量称量与直接称量
For solid reagents, the analytical balance or electronic balance is used. The weighing bottle method (weighing by difference) is preferred for hygroscopic substances or when transferring powders: weigh the bottle with the sample, transfer the required amount, then reweigh the bottle. The difference gives the exact mass transferred. This avoids the error of weighing a container on which sample residue remains. For quick approximations, weighing paper or a watch glass may be used, but never place chemicals directly on the balance pan.
3. Volume Measurement: Burette, Pipette and Measuring Cylinder | 体积测量：滴定管、移液管与量筒
A burette reads to 0.05 cm³, and for accurate titration a 50.00 cm³ burette is used. The initial reading and final reading should both be estimated to two decimal places, the second decimal being a 0 or 5. A pipette, typically 25.00 cm³, delivers a fixed volume with high accuracy and should be rinsed with the solution it will contain before use. A measuring cylinder is less accurate, with an uncertainty of approximately ±0.5 cm³, and should only be used for approximate volume measurements such as preparing dilute solutions where precision is not critical.
4. Temperature Measurement and Thermometer Techniques | 温度测量与温度计使用技术
When measuring temperature changes in enthalpy experiments, use a thermometer with a suitable range, reading to the nearest 0.5°C. The bulb of the thermometer should be fully immersed in the liquid but must not touch the bottom of the container. For cooling curve experiments, record the temperature at regular time intervals, stirring gently before each reading to ensure thermal equilibrium. Calibration of the thermometer should be checked using a melting ice-water mixture for 0°C and steam above boiling water for 100°C at standard pressure.
5. Titration Technique: The Core of Quantitative Analysis | 滴定技术：定量分析的核心
The key procedural steps for a successful acid-base titration are: (1) rinse the burette with the acid solution and the pipette with the alkali solution; (2) pipette a precise volume of alkali into a conical flask, adding a few drops of indicator; (3) run the acid from the burette until the indicator changes colour persistently, swirling the flask continuously; (4) near the endpoint, add the titrant dropwise; (5) record the burette reading and repeat. A rough titration should be performed first, followed by three accurate titrations. Concordant results should be within 0.10 cm³ of each other, and the mean of concordant titres is calculated for the final result.
6. Observation and Recording: What to Look For | 观察与记录：应关注什么
When recording observations, candidates must describe what is seen and heard, not what is inferred. Key observable changes include: colour changes (e.g., blue solution turning green then dark brown), gas evolution (bubbling rate, gas smell, colour), precipitate formation (colour, texture, whether it dissolves in excess reagent), temperature changes (exothermic or endothermic), and physical changes (melting, dissolving, crystallisation). For example, when adding aqueous ammonia dropwise to copper(II) sulfate solution, record: a pale blue precipitate forms initially, which dissolves in excess ammonia giving a deep blue solution.
7. Qualitative Analysis: Ion Tests and Confirmatory Reactions | 定性分析：离子检验与确认反应
For cation analysis, aqueous sodium hydroxide and aqueous ammonia are the two principal reagents. Record the colour of any precipitate formed and its solubility in excess reagent. For example, Al³⁺ gives a white precipitate with NaOH which dissolves in excess NaOH; Fe³⁺ gives a reddish-brown precipitate that does not dissolve; Cu²⁺ gives a pale blue precipitate insoluble in excess NaOH but soluble in excess NH₃ forming a deep blue solution. For anion tests, carbonate releases CO₂ with effervescence, sulfate gives a white precipitate with BaCl₂ acidified with HCl, and halide ions are identified using silver nitrate followed by ammonia: Cl⁻ gives white AgCl soluble in dilute NH₃, Br⁻ gives cream AgBr soluble in concentrated NH₃, I⁻ gives yellow AgI insoluble in concentrated NH₃.
8. Error Analysis and Uncertainty in Measurements | 误差分析与测量不确定度
Every measurement carries uncertainty. A balance with precision ±0.005 g used for two weighings has a total uncertainty of ±0.010 g. A burette has an uncertainty of ±0.05 cm³ per reading, so two readings give ±0.10 cm³ total uncertainty. Percentage error is calculated as the total absolute uncertainty divided by the measurement value multiplied by 100%. To reduce percentage errors, choose apparatus that gives measurements close to the full scale, increase the quantity being measured where safe, and use more precise instruments. Systematic errors cause readings to be consistently too high or too low; random errors cause unpredictable scatter. Human reaction time in timing experiments is a typical random error, while using an uncalibrated balance introduces systematic error.
9. Graphical Methods and Data Presentation | 图表方法与数据呈现
When plotting graphs from experimental data, the dependent variable is plotted on the y-axis and the independent variable on the x-axis. Axes should be labelled with quantity and unit, and a suitable scale chosen so that the plotted points occupy at least half of the graph paper in each direction. Points are marked with a sharp pencil as small crosses or dots, and a best-fit straight line or smooth curve is drawn with a ruler or flexible curve. The line of best fit does not need to pass through every point; it should balance the points above and below. For a rate-concentration graph, a straight line through the origin indicates first-order kinetics, while a horizontal line indicates zero-order. The gradient of a tangent to a concentration-time curve at a specific time gives the instantaneous rate of reaction.
在绘制实验数据图表时，因变量绘制在 y 轴，自变量绘制在 x 轴。坐标轴应标注物理量和单位，选择合适的比例使各数据点在两个方向上至少占据半张坐标纸。用削尖的铅笔以叉号或圆点标记数据点，用直尺或曲线板绘制最佳拟合直线或平滑曲线。最佳拟合线不必穿过每个点，应使点均衡分布在线条的上下两侧。对于速率-浓度图，过原点的直线表示一级反应动力学，水平线表示零级反应。浓度-时间曲线上某一点切线的斜率即为该时刻的瞬时反应速率。
10. Heating, Reflux and Filtration Operations | 加热、回流与过滤操作
When heating a liquid in a test tube, hold the tube at an angle of approximately 45°, point the mouth away from people, and ensure the liquid level is not more than one-third of the tube’s volume. Use a water bath for heating flammable or easily decomposed substances. A Bunsen burner with a blue flame (air hole open) delivers a hotter flame than the yellow luminous flame. Reflux is used for organic reactions conducted at elevated temperatures — a condenser is attached vertically to the reaction flask so vapours condense and return, preventing loss of volatile reactants. For filtration, fold the filter paper into a cone, wet it with distilled water to adhere to the funnel, and pour the mixture along a stirring rod to avoid splashing.
11. Conclusion: Linking Practical Skills to Exam Success | 结论：将实验技能与考试成功联系起来
Mastery of experimental operations, accurate measurement techniques, and systematic observation recording are indispensable skills for CIE A-Level Chemistry. In Paper 3, the examiner awards marks not only for the numerical answer but also for the method, the precision of readings, the clarity of observations and the validity of conclusions. In Paper 5, planning and analysis skills are assessed, requiring you to apply the same principles to novel scenarios. By practising these techniques and internalising the vocabulary of experimental chemistry, you will approach any practical question with confidence and precision.
熟练的实验操作、精确的测量技术和系统的观察记录是 CIE A-Level 化学不可或缺的技能。在 Paper 3 中，考官不仅为数值答案给分，还根据方法、读数精度、观察的清晰度和结论的合理性来评分。在 Paper 5 中，评估的是计划和分析能力，要求你将相同的原理应用于新的情境。通过反复练习这些技术并将实验化学词汇内化，你将能够自信且精确地应对任何实验问题。
Published by TutorHao | Chemistry Revision Series | aleveler.com
📚 A-Level Chemistry: Alcohol Homologous Series and Property Trends | A-Level化学：醇的同系物与性质规律
Alcohols are one of the most important homologous series in organic chemistry, characterised by the presence of the hydroxyl (-OH) functional group. Understanding their structural patterns and how physical and chemical properties change along the series is essential for CIE A-Level Chemistry.
A homologous series is a family of organic compounds with the same general formula, similar chemical properties, and successive members differing by a CH₂ unit. For alcohols, the general formula is CₙH₂ₙ₊₁OH, or equivalently CₙH₂ₙ₊₂O.
The first four members of the series are methanol (CH₃OH), ethanol (C₂H₅OH), propan-1-ol (C₃H₇OH) and butan-1-ol (C₄H₉OH). Each member has one more CH₂ group than the previous one, and as the chain lengthens, certain trends become apparent.
Starting from propanol, structural isomers exist. Propan-1-ol has the -OH group on the terminal carbon, while propan-2-ol has it on the middle carbon. For butanol, there are four isomers: butan-1-ol, butan-2-ol, 2-methylpropan-1-ol and 2-methylpropan-2-ol.
This classification into primary, secondary and tertiary alcohols is based on the number of carbon atoms attached to the carbon bearing the -OH group. This distinction is crucial because it determines the products of oxidation and the rate of reaction with halogenating agents.
Alcohols have significantly higher boiling points than alkanes of similar relative molecular mass. For example, ethanol (Mr = 46) boils at 78 °C, whereas propane (Mr = 44) boils at -42 °C. This dramatic difference arises from hydrogen bonding between the -OH groups of neighbouring molecules.
Within the alcohol series, boiling points increase as the carbon chain length increases. This is because the strength of London dispersion forces increases with molecular size and surface area, requiring more energy to overcome. However, branching reduces the boiling point because it decreases the surface area and weakens intermolecular forces.
Boiling point order | 沸点顺序： butan-1-ol (117 °C) > propan-1-ol (97 °C) > ethanol (78 °C) > methanol (65 °C)
4. Solubility in Water | 在水中的溶解度
Methanol, ethanol and propanol are completely miscible with water. The -OH group forms hydrogen bonds with water molecules, which is thermodynamically favourable. As the hydrocarbon chain lengthens, the non-polar character of the molecule increases, and the solubility decreases significantly. Butan-1-ol is only partially soluble, and pentan-1-ol is nearly insoluble.
A useful general rule is that a compound is likely to be water-soluble if the number of carbon atoms is not more than four per hydroxyl group. Beyond this, the hydrophobic alkyl chain dominates and water solubility is lost. This trend is important when considering the biological activity and solvent properties of alcohols.
Alcohols are weak acids, much weaker than water. This is due to the electronegativity of oxygen, which polarises the O-H bond, allowing a proton to be lost. However, the alkyl group attached to oxygen is electron-donating, which destabilises the alkoxide ion produced and reduces the acidity compared with water.
Among alcohols, acidity follows the order methanol > primary > secondary > tertiary. The electron-donating effect of additional alkyl groups stabilises the cation character less effectively and destabilises the alkoxide anion more strongly, so the conjugate base is less stable and the acid is weaker.
In practice, alcohols react with sodium metal to liberate hydrogen gas. The reaction is slower than that of water with sodium, and it confirms the weakly acidic nature of the hydroxyl proton.
在实际反应中，醇与金属钠反应释放氢气。该反应比水与钠的反应慢，这证实了羟基质子的弱酸性本质。
6. Oxidation of Alcohols | 醇的氧化反应
Primary alcohols can be oxidised to aldehydes and then to carboxylic acids. Secondary alcohols are oxidised to ketones, and tertiary alcohols are resistant to oxidation under mild conditions because the carbon bearing the -OH group has no hydrogen atom attached.
The oxidising agent commonly used is acidified potassium dichromate(VI), K₂Cr₂O₇/H₂SO₄, which changes colour from orange to green when reduced. Alternatively, acidified potassium manganate(VII), KMnO₄/H⁺, is a stronger oxidising agent and can be used for complete oxidation.
For a primary alcohol, careful distillation of the aldehyde product must be performed before it is further oxidised. If reflux conditions are used instead, the carboxylic acid is obtained directly. The key distinction between these two techniques is frequently examined in CIE questions.
Alcohols undergo elimination reactions when heated with concentrated sulfuric or phosphoric acid. A water molecule is removed from adjacent carbon atoms, producing an alkene. For example, ethanol heated with concentrated H₂SO₄ at 170 °C yields ethene.
The mechanism follows an E1 pathway for tertiary alcohols and an E2 pathway for primary alcohols. The acid protonates the hydroxyl group to form a good leaving group, water, before the elimination of a β-hydrogen and formation of the C=C double bond.
When unsymmetrical alcohols are dehydrated, the major product follows Zaitsev’s rule, which states that the more substituted alkene is formed preferentially. For example, butan-2-ol mainly gives but-2-ene rather than but-1-ene.
Alcohols react with hydrogen halides to form haloalkanes and water. The reaction proceeds via protonation of the -OH group followed by nucleophilic substitution, in which water is the leaving group and the halide ion acts as the nucleophile.
The reactivity of hydrogen halides increases in the order HCl < HBr < HI, reflecting the increasing nucleophilicity and the decreasing bond strength of the H-X bond. Tertiary alcohols react fastest because the intermediate carbocation is most stable, while primary alcohols are the slowest.
When using HCl, a catalyst such as anhydrous zinc chloride is often required, especially for primary alcohols. This is a useful practical detail that may appear in exam questions on reaction conditions and rates.
Alcohols react with carboxylic acids in the presence of concentrated sulfuric acid to form esters and water. This is a reversible condensation reaction. For example, ethanol and ethanoic acid react to give ethyl ethanoate, CH₃COOC₂H₅, which has a characteristic sweet smell.
The concentrated sulfuric acid serves dual roles: as a catalyst and as a dehydrating agent that shifts the equilibrium to the product side. The reaction is slow at room temperature, and reflux is typically used to achieve a reasonable yield.
Esters are widely used as flavourings, fragrances and solvents. The reverse reaction, hydrolysis, can occur under acidic or basic conditions. Base-catalysed hydrolysis (saponification) is irreversible because the carboxylate salt formed does not re-esterify.
The iodoform reaction is a distinctive test for the presence of a methyl carbonyl group (CH₃CO-) or a secondary alcohol with the CH₃CH(OH)- structure. Ethanol is the only primary alcohol that gives a positive iodoform test because it is oxidised to ethanal, which then reacts further.
In this test, iodine and sodium hydroxide react with the compound to produce a pale yellow precipitate of triiodomethane, CHI₃, which has an antiseptic smell. The reagent is often represented as I₂/NaOH or NaIO (sodium iodate(I)).
For ethanol, the reaction involves initial oxidation to ethanal, followed by the substitution of all three α-hydrogens of the methyl group by iodine, and then cleavage of the C-C bond. The positive result is indicated by the yellow solid and is a reliable qualitative analysis tool.
Although both alcohols and phenols contain the -OH group, their chemistry differs significantly. In phenols, the hydroxyl group is attached directly to an aromatic ring, allowing lone-pair electrons on oxygen to delocalise into the ring, which increases the acidity of the O-H bond.
Phenols are more acidic than alcohols and can react with sodium hydroxide solution, whereas alcohols generally cannot. However, phenols are still weaker acids than carboxylic acids. This pH-based distinction is a classical exam theme: phenol is a weak acid that is soluble in NaOH but not in NaHCO₃.
To achieve top marks in CIE A-Level Chemistry, you must be able to recall the general formula of alcohols, explain the trend in boiling points and solubility, classify alcohols as primary, secondary or tertiary, and predict the products of oxidation, dehydration, substitution and esterification reactions.
Key reaction conditions must be memorised: acidified K₂Cr₂O₇ for oxidation, 170 °C with concentrated H₂SO₄ for dehydration, and reflux with a carboxylic acid in acidic conditions for esterification. The iodoform test and the comparison with phenols are also high-yield areas.
Finally, always remember the underlying principle: the hydroxyl group determines the characteristic reactions of alcohols, but the alkyl chain influences physical properties and the degree of substitution controls the outcome of oxidation and elimination. Mastery of these ideas will allow you to approach unfamiliar alcohols systematically.
📚 A-Level Chemistry | Homologous Series of Aldehydes and Ketones | A-Level化学：醛与酮的同系物规律总结
In A-Level Chemistry, aldehydes and ketones are two closely related families within the same functional group class: the carbonyl compounds. Understanding their homologous series, naming conventions, physical properties, and chemical trends is essential for exam success in CIE papers.
Aldehydes contain the carbonyl group (C=O) bonded to at least one hydrogen atom, giving the functional group -CHO. The carbonyl carbon is therefore always at the end of the carbon chain.
Ketones contain the carbonyl group bonded to two carbon atoms, giving the functional group -CO-. The carbonyl carbon is therefore located inside the carbon chain, never at a terminal position.
酮的羰基碳与两个碳原子相连，官能团为 -CO-。因此酮的羰基碳位于碳链内部，绝不会出现在链端。
R-CHO (aldehyde) | R-CO-R’ (ketone)
The simplest aldehyde is methanal (HCHO), and the simplest ketone is propanone (CH₃COCH₃). Note that methanal is the only aldehyde with no alkyl group attached to the carbonyl carbon, and propanone is the simplest ketone because a ketone requires at least three carbon atoms in total.
Both aldehydes and saturated aliphatic ketones share the same general molecular formula: CₙH₂ₙO. This is because both contain one carbonyl group and no other oxygen or unsaturation.
For a straight-chain aldehyde with n carbon atoms, the molecular formula is CₙH₂ₙO. For example, ethanal is C₂H₄O, propanal is C₃H₆O, and butanal is C₄H₈O.
对于含 n 个碳原子的直链醛，分子式均为 CₙH₂ₙO。例如乙醛是 C₂H₄O，丙醛是 C₃H₆O，丁醛是 C₄H₈O。
The same formula also applies to ketones. Propanone is C₃H₆O, butanone is C₄H₈O, and pentan-2-one is C₅H₁₀O. Consequently, aldehydes and ketones with the same carbon number are structural isomers of each other.
For aldehydes, the IUPAC name uses the suffix ‘-al’. The longest chain containing the -CHO group is numbered from the carbonyl carbon, which is always carbon 1. Therefore, no locant is needed for simple aldehydes.
For ketones, the suffix ‘-one’ is used, and the position of the carbonyl group must be indicated by a locant when necessary. The chain is numbered to give the carbonyl carbon the lowest possible number.
CH₃COCH₃: propanone (no locant needed because the carbonyl must be at position 2)
CH₃COCH₂CH₃: butanone (also written as butan-2-one)
CH₃CH₂COCH₂CH₃: pentan-3-one
CH₃COCH₂CH₂CH₃: pentan-2-one
When both an aldehyde and other functional groups are present, the suffix ‘-al’ takes priority over ‘-one’ in naming. This is a common source of exam errors.
4. Structural Isomerism within the Series | 同分异构现象
Because aldehydes and ketones share the formula CₙH₂ₙO, functional group isomerism arises between them. Additionally, chain isomerism occurs within each family as the carbon skeleton branches.
For C₃H₆O, there are only two isomers: propanal (an aldehyde) and propanone (a ketone).
对于 C₃H₆O，只有两个异构体：丙醛（醛）和丙酮（酮）。
For C₄H₈O, the isomers include butanal, 2-methylpropanal, butanone, and no other simple ketone because a four-carbon ketone can only place the carbonyl at position 2 or 3, but position 3 is equivalent to position 2 by chain reversal.
Exam tip: When asked to draw all isomers of C₅H₁₀O that are aldehydes, remember to consider both straight-chain and branched aldehydes: pentanal, 2-methylbutanal, 3-methylbutanal, and 2,2-dimethylpropanal.
Aldehydes and ketones are polar molecules because the carbonyl group has a permanent dipole due to the electronegativity difference between carbon and oxygen. However, they cannot form hydrogen bonds with themselves because there is no hydrogen atom attached directly to the oxygen.
Compared to alkanes of similar molar mass, aldehydes and ketones have higher boiling points because of dipole-dipole interactions. Compared to alcohols of similar molar mass, they have lower boiling points because alcohols can form intermolecular hydrogen bonds.
As the carbon chain lengthens, boiling points increase smoothly due to greater van der Waals forces. For isomeric aldehydes and ketones, the straight-chain isomer generally has a higher boiling point than a branched isomer.
Boiling point order: alkane < aldehyde/ketone < alcohol (for similar Mr)
6. Physical Properties: Solubility | 物理性质：溶解性
The lower members of the aldehyde and ketone series, such as methanal, ethanal, and propanone, are soluble in water. This is because the carbonyl oxygen can act as a hydrogen bond acceptor with water molecules.
As the alkyl chain length increases, the non-polar hydrocarbon portion dominates, and solubility in water decreases sharply. For example, pentanal is only slightly soluble, and larger aldehydes and ketones are essentially insoluble.
This trend is analogous to that observed in alcohols, but aldehydes and ketones are generally less soluble than alcohols of similar molar mass because they cannot donate hydrogen bonds to water.
这一趋势与醇类似，但相同摩尔质量的醛酮通常比醇更难溶于水，因为它们不能向水提供氢键。
7. Reactivity of the Carbonyl Group | 羰基的反应活性
The carbonyl carbon is electron-deficient because the electronegative oxygen atom withdraws electron density. This makes the carbonyl carbon susceptible to attack by nucleophiles.
羰基碳因氧原子吸电子而缺电子，容易受到亲核试剂的进攻。
Aldehydes are generally more reactive than ketones toward nucleophilic addition. There are two main reasons: sterically, aldehydes have at most one alkyl group attached to the carbonyl carbon, so the nucleophile can approach more easily; electronically, alkyl groups are electron-donating, so ketones have two alkyl groups that stabilise the positive character of the carbonyl carbon and reduce its reactivity.
This difference explains why aldehydes are more easily oxidised and why they form more stable cyanohydrins and hemiacetals with higher equilibrium constants than ketones.
这一差异解释了为什么醛更容易被氧化，以及为什么醛形成氰醇和半缩醛时的平衡常数比酮更大。
8. Oxidation of Aldehydes and Ketones | 醛与酮的氧化
The most important chemical distinction between aldehydes and ketones is their behaviour toward oxidising agents. Aldehydes can be oxidised to carboxylic acids, whereas ketones resist oxidation under mild conditions.
醛与酮最重要的化学区别在于它们对氧化剂的反应。醛能被氧化成羧酸，而酮在温和条件下不被氧化。
With Tollens’ reagent (ammoniacal silver nitrate), aldehydes reduce Ag⁺ to metallic silver, producing a silver mirror. Ketones do not react. This forms the basis of a classic test-tube identification.
With Fehling’s or Benedict’s solution, aldehydes reduce Cu²⁺ to Cu⁺, giving a brick-red precipitate of copper(I) oxide. Ketones do not react, except α-hydroxy ketones which can give a false positive.
Acidified potassium dichromate(VI) oxidises aldehydes to carboxylic acids, with a colour change from orange to green. Ketones are not oxidised under these conditions.
酸化的重铬酸钾可将醛氧化为羧酸，颜色由橙色变为绿色；酮在此条件下不被氧化。
Reagent
Observation with aldehyde
Observation with ketone
Tollens’ reagent
Silver mirror
No change
Fehling’s solution
Brick-red precipitate
No change
Acidified K₂Cr₂O₇
Orange to green
No change
9. Reduction to Alcohols | 还原为醇
Aldehydes and ketones can be reduced to primary and secondary alcohols respectively, using reducing agents such as NaBH₄ or LiAlH₄.
醛和酮可分别被还原为伯醇和仲醇，常用还原剂为 NaBH₄ 或 LiAlH₄。
When an aldehyde is reduced, the product is a primary alcohol because the -CHO group becomes -CH₂OH. For example, ethanal is reduced to ethanol.
醛被还原时产物为伯醇，因为 -CHO 变成 -CH₂OH。例如乙醛被还原为乙醇。
When a ketone is reduced, the product is a secondary alcohol. For example, propanone is reduced to propan-2-ol.
酮被还原时产物为仲醇。例如丙酮被还原为丙-2-醇。
R-CHO + 2[H] → R-CH₂OH
R-CO-R’ + 2[H] → R-CH(OH)-R’
NaBH₄ is preferred in A-Level practical work because it is safer and can be used in aqueous or alcoholic solution. LiAlH₄ is more powerful but requires anhydrous conditions.
The carbonyl group undergoes nucleophilic addition. Two textbook reactions are particularly important: reaction with hydrogen cyanide (HCN) to form hydroxynitriles, and reaction with 2,4-dinitrophenylhydrazine (Brady’s reagent) to form orange precipitates.
With HCN, the nucleophile is CN⁻, and the carbonyl carbon forms a new C-C bond. The product is a 2-hydroxynitrile. For ethanal, the product is 2-hydroxypropanenitrile (CH₃CH(OH)CN).
This reaction is valuable because it increases the carbon chain length by one carbon, an important synthetic strategy in organic chemistry.
这个反应很有价值，因为它将碳链增加一个碳原子，是有机合成中重要的延长碳链策略。
With Brady’s reagent (2,4-DNPH), both aldehydes and ketones form orange or yellow precipitates of 2,4-dinitrophenylhydrazones. This test confirms the presence of a carbonyl group but does not distinguish between aldehydes and ketones.
Hydrogen atoms attached to the carbon adjacent to the carbonyl group (α-hydrogens) are weakly acidic. This is because the conjugate base (enolate ion) is stabilised by resonance delocalisation of the negative charge onto the oxygen atom.
This acidity, though weak (pKa ≈ 20), enables condensation reactions. In the iodoform test, the α-methyl group of ethanol or a methyl ketone is halogenated and then cleaved to give iodoform (CHI₃), a yellow solid with a distinctive antiseptic smell.
The iodoform test is positive for ethanal and for all ketones of the form R-CO-CH₃. Propanone gives a positive result, while propanal gives a negative result because it lacks an α-methyl group adjacent to the carbonyl.
In the homologous series of aldehydes and ketones, the following trends are central to the CIE specification: both families share the general formula CₙH₂ₙO; boiling points increase with molar mass; solubility decreases with chain length; aldehydes are more reactive than ketones toward nucleophiles; and aldehydes are readily oxidised while ketones are not.
For identification tests, remember: Tollens’ and Fehling’s tests distinguish aldehydes from ketones, while Brady’s reagent detects any carbonyl group, and the iodoform test specifically detects methyl ketones and ethanal.
Mastering these patterns allows you to predict properties, draw isomers, and design simple synthetic routes involving carbonyl compounds with confidence.
掌握这些规律，你就能自信地预测性质、画出异构体，并设计涉及羰基化合物的简单合成路线。
Published by TutorHao | Chemistry Revision Series | aleveler.com
📚 Carboxylic Acids: Properties and Key Reactions | A-Level化学：羧酸的性质与常见反应
Carboxylic acids are one of the most important families of organic compounds, featuring the carboxyl functional group (–COOH). Their unique combination of polarity, acidity, and reactivity makes them essential in both laboratory synthesis and biological processes, from vinegar to fatty acids and drug molecules.
A carboxylic acid contains a carbonyl group attached directly to a hydroxyl group, written as –COOH. In displayed formula, the carbon is double-bonded to one oxygen and single-bonded to another oxygen bearing a hydrogen atom.
The carboxyl carbon is sp² hybridised, giving bond angles of approximately 120° around the carbon.
羧基碳为sp²杂化，碳原子周围的三键角接近120°。
IUPAC naming uses the suffix –oic acid; the carboxyl carbon is always number 1.
IUPAC命名采用后缀“–oic acid”（某酸）；羧基碳始终编号为1。
Common examples: ethanoic acid (CH₃COOH), benzoic acid (C₆H₅COOH).
常见实例：乙酸（CH₃COOH）、苯甲酸（C₆H₅COOH）。
2. Physical Properties | 物理性质
Carboxylic acids exhibit notably high boiling points compared to alcohols and alkanes of similar molar mass. This arises from strong intermolecular hydrogen bonds between the –OH group of one molecule and the C=O oxygen of another.
They can form stable cyclic dimers, held together by two hydrogen bonds, which effectively doubles their molecular mass in the vapour phase.
羧酸能通过两重氢键形成稳定的环状二聚体，相当于在蒸气相中将其分子量“翻倍”。
Small-chain acids (up to C₄) are miscible with water because of hydrogen bonding with H₂O.
短链羧酸（至C₄）能与水任意比互溶，原因是与H₂O形成氢键。
As the hydrocarbon chain lengthens, solubility in water decreases and melting/boiling points gradually rise.
随着碳链增长，水溶性下降，熔点与沸点逐渐升高。
3. Acidity and Ionisation | 酸性来源与电离
Carboxylic acids are weak acids. In aqueous solution, they partially dissociate to form a carboxylate ion and a hydrogen ion:
羧酸是弱酸。在水中部分电离产生羧酸根离子和氢离子：
CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺
The high stability of the carboxylate ion is the key reason for the acid character. The negative charge is delocalised across both oxygen atoms through resonance.
羧酸具有酸性的关键在于羧酸根离子高度稳定。其负电荷通过共振离域分布在两个氧原子之间。
Resonance stabilisation lowers the energy of the conjugate base, making dissociation more favourable.
共振稳定降低了共轭碱的能量，使电离更加容易发生。
pKa values typically range from 3 to 5, stronger than simple alcohols (pKa ≈ 16) but weaker than mineral acids.
pKa值通常在3–5之间，强于普通醇（pKa≈16），但弱于无机强酸。
4. Factors Affecting Acid Strength | 影响酸性强弱的因素
Electron-withdrawing groups attached to the carbon chain stabilise the carboxylate ion by spreading the negative charge, thus increasing acidity.
与碳链相连的吸电子基团能使羧酸根离子的负电荷分散而使其更稳定，由此增强酸性。
Electronegative atoms such as Cl, F, and NO₂ withdraw electron density inductively through σ bonds.
Cl、F、NO₂等电负性大的原子通过σ键以诱导效应吸走电子密度。
The closer the electronegative group is to the carboxyl group, the stronger the effect.
吸电子基团越靠近羧基，其对酸性的影响越显著。
Alkyl groups (–CH₃, –C₂H₅) are electron-donating and therefore decrease acidity.
5. Salt Formation with Metals and Bases | 与金属和碱生成盐
Carboxylic acids react with metals, alkalis, and carbonates to produce carboxylate salts, showing typical acidic behaviour.
羧酸可与金属、碱及碳酸盐反应生成羧酸盐，表现出典型的酸性行为。
Reactant
Equation
Observation
Mg
2CH₃COOH + Mg → (CH₃COO)₂Mg + H₂
Effervescence, H₂ gas
NaOH
CH₃COOH + NaOH → CH₃COONa + H₂O
Neutralisation, heat evolved
NaHCO₃
CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂
Bubbling of CO₂
CaCO₃
2CH₃COOH + CaCO₃ → (CH₃COO)₂Ca + H₂O + CO₂
Slow dissolution, CO₂ gas
The reaction with sodium carbonate is often used as a chemical test to distinguish carboxylic acids from neutral organic compounds such as phenols and alcohols.
与碳酸钠的反应常用来区分羧酸和苯酚、醇等中性有机化合物（苯酚酸性太弱，不与NaHCO₃反应）。
6. Esterification | 酯化反应
The most characteristic reaction of carboxylic acids is condensation with an alcohol to form an ester and water. This is catalysed by concentrated H₂SO₄, which also acts as a dehydrating agent.
羧酸最特征的反应是与醇缩合生成酯和水。该反应以浓H₂SO₄为催化剂，同时浓硫酸还担当脱水剂。
CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O
The reaction is reversible and proceeds slowly even without a catalyst at room temperature.
该反应可逆，室温下无催化剂时进行得很慢。
Replacement of the –OH group comes from the acid; the alcohol supplies only the –H. This is confirmed by isotope labelling with ¹⁸O.
酸提供–OH基团而醇仅提供H原子，该机理通过¹⁸O同位素标记实验得到证实。
Ester names end in –oate: ethyl ethanoate from ethanol and ethanoic acid.
酯的命名以“–oate”结尾：由乙醇和乙酸生成的产物叫乙酸乙酯。
Esters are volatile, sweet-smelling compounds used as flavourings and fragrances.
酯为芳香甜味、易挥发的化合物，广泛用于香料和食用香精。
7. Amide Formation | 酰胺的生成
Carboxylic acids react with ammonia or amines to form ammonium salts, which can be dehydrated to primary or secondary amides upon heating.
羧酸与氨或胺反应先形成铵盐，加热脱水后生成伯酰胺或仲酰胺。
CH₃COOH + NH₃ → CH₃COONH₄ →(heat) CH₃CONH₂ + H₂O
Amides are much less reactive than acids and have very high melting points due to strong hydrogen bonding.
酰胺的反应活性远低于羧酸，且因强氢键而具有很高的熔点。
Primary amides can be further dehydrated with P₂O₅ to form nitriles: CH₃CONH₂ → CH₃CN + H₂O.
伯酰胺在P₂O₅作用下可进一步脱水生成腈：CH₃CONH₂ → CH₃CN + H₂O。
This pathway is important in synthetic chemistry for building compounds with C–N bonds.
该路线在构建含C–N键化合物的合成化学中很重要。
8. Reduction to Alcohols | 还原为醇
Carboxylic acids can be reduced to primary alcohols using powerful reducing agents such as lithium aluminium hydride (LiAlH₄) in dry ether.
羧酸可被强还原剂如氢化铝锂（LiAlH₄）在无水乙醚中还原为伯醇。
CH₃COOH + 4[H] → C₂H₅OH + H₂O
NaBH₄ is too weak to reduce carboxylic acids directly, only working on aldehydes and ketones.
NaBH₄太弱，不能直接还原羧酸，只能还原醛和酮。
The reaction proceeds through an aldehyde intermediate, which is reduced further because LiAlH₄ is so reactive.
还原经醛作为中间体，因LiAlH₄反应活性极强，醛会被继续还原至醇。
Water or dilute acid is added after the reaction to destroy excess hydride and release the alcohol.
反应结束后需加入水或稀酸以分解多余的氢化物并释出醇。
9. Decarboxylation | 脱羧反应
Decarboxylation involves the removal of CO₂ from the carboxyl group. It can be accomplished by heating with soda-lime (NaOH + CaO) or by electrolysis of concentrated carboxylate solutions.
This method produces an alkane with one fewer carbon atom than the original acid.
该方法生成比原羧酸少一个碳原子的烷烃。
In biological systems, decarboxylation is catalysed by enzymes and proceeds at body temperature, e.g. in the Krebs cycle.
在生物体系中，脱羧由酶催化并在体温下进行，例如在三羧酸循环中扮演重要角色。
Care must be taken not to confuse this with the Kolbe electrolysis which also gives alkanes but via radical intermediates.
注意不要与科尔贝电解混淆，后者虽也生成烷烃，但机理经历自由基中间体。
10. Formation of Acyl Chlorides | 酰氯的生成
Reaction with thionyl chloride (SOCl₂) converts carboxylic acids into acyl chlorides, a very reactive class of acid derivatives.
羧酸与氯化亚砜（SOCl₂）反应生成酰氯——一类反应活性极高的羧酸衍生物。
CH₃COOH + SOCl₂ → CH₃COCl + SO₂ + HCl
Because SO₂ and HCl are both gases, they escape from the reaction mixture, driving the equilibrium to completion and leaving a pure product.
由于SO₂和HCl均为气体，会逸出反应体系，推动平衡完全右移，得到纯净的酰氯产物。
Acyl chlorides react rapidly with water, alcohols, and amines to form acids, esters, and amides respectively.
酰氯能与水、醇、胺猛烈反应，分别生成酸、酯和酰胺。
They are much more reactive than the parent carboxylic acids because chlorine is a better leaving group than –OH.
酰氯的活性远高于原始羧酸，因为氯离子是比–OH更好的离去基团。
11. Test to Identify Carboxylic Acids | 羧酸的鉴别
At A-Level, it is essential to know a sequence of qualitative tests for identifying a carboxylic acid within an unknown organic sample.
A-Level阶段，必须掌握一套定性的测试顺序来鉴别未知有机样品中是否含有羧酸。
Add NaHCO₃: only carboxylic acids produce CO₂ effervescence among common oxygen-containing organics.
加入NaHCO₃：常见的含氧有机物中，只有羧酸会产生CO₂气泡。
Blue litmus turns red, indicating an acidic solution.
蓝色石蕊变红，说明溶液呈酸性。
React with ethanol and concentrated H₂SO₄: the production of a sweet-smelling ester confirms the presence of the –COOH group.
与乙醇在浓H₂SO₄催化下反应，若产生水果香味酯类，进一步确证–COOH的存在。
FeCl₃ solution gives a yellow or buff-coloured precipitate with carboxylate ions, whereas phenol gives a violet colour.
羧酸根离子与FeCl₃溶液生成黄色或淡棕色沉淀，而苯酚则显紫红色。
12. Exam-Standard Worked Example | 真题标准例题
Compound X has a molecular formula C₂H₄O₂. It shows effervescence with NaHCO₃ and gives an orange precipitate with 2,4-DNPH after oxidation with acidified K₂Cr₂O₇. Suggest the structural formula of X and explain the observations.
Since X reacts with NaHCO₃, it must contain a –COOH group. The molecular formula corresponds to methanoic acid or ethanoic acid? C₂H₄O₂ is ethanoic acid. Oxidation with K₂Cr₂O₇ gives ethanedioic acid? Wait—ethanoic acid is already maximally oxidised at the carboxyl carbon. The oxidation product must come from a different functional group.
Correct approach: the oxidation step uses acidified K₂Cr₂O₇ to convert a secondary alcohol to a ketone, but no alcohol is present. For C₂H₄O₂, the only structure is CH₃COOH. Ethanoic acid does not react with K₂Cr₂O₇; the observation contradicts, so X must be HOOC–COOH? That is C₂H₂O₄, not suitable.
In fact, the exact question in CIE past papers refers to X being 2-hydroxypropanoic acid (lactic acid, C₃H₆O₃), which contains –COOH and a secondary alcohol that oxidises to a ketone. This demonstrates the importance of considering both functional groups.
Understanding organic reaction mechanisms is central to A-Level Chemistry. A mechanism is a step-by-step description of how covalent bonds break and form during a reaction, showing the movement of electrons and the structures of any intermediates or transition states.
A mechanism is not just a theoretical drawing; it explains how and why a reaction proceeds. Once you can read curly arrows, you can work out which bonds break, which bonds form, and which species act as nucleophiles or electrophiles. This skill turns memorising hundreds of reactions into understanding a few logical patterns.
📚 Haloalkanes: Properties, Reactions, and Applications | A-Level化学：卤代烷烃的用途解析
Haloalkanes are organic compounds derived from alkanes by replacing one or more hydrogen atoms with halogen atoms. Their general formula for monohaloalkanes is R–X, where R is an alkyl group and X is F, Cl, Br or I. The polarisation of the carbon–halogen bond gives haloalkanes a unique reactivity that makes them indispensable in both laboratory synthesis and industrial applications. In this article, we will examine their classification, physical properties, reaction pathways, and above all their practical uses in everyday life and advanced technology.
卤代烷烃是由烷烃中一个或多个氢原子被卤素原子取代而得到的有机化合物。单卤代烷烃的通式为 R–X，其中 R 为烷基，X 为氟、氯、溴或碘。碳–卤素键的极化使卤代烷烃具有独特的反应活性，使其在实验室合成和工业应用中不可或缺。本文将探讨其分类、物理性质、反应路径，更重要的是其在日常生活和先进技术中的实际用途。
1. Classification and Structure | 分类与结构
Haloalkanes are classified according to the number of halogen atoms and the degree of substitution at the carbon bearing the halogen. A primary (1°) haloalkane has the halogen attached to a carbon bonded to one other alkyl group; a secondary (2°) haloalkane has it bonded to two alkyl groups; a tertiary (3°) haloalkane has it bonded to three alkyl groups. For example, 1-chlorobutane is primary, 2-chlorobutane is secondary, and 2-chloro-2-methylpropane is tertiary.
The carbon–halogen bond is polar because the halogen is more electronegative than carbon. Fluorine creates the most polar bond, followed by chlorine, bromine and iodine. However, bond enthalpy decreases down the group: C–F ≈ 467 kJ mol⁻¹, C–Cl ≈ 346 kJ mol⁻¹, C–Br ≈ 290 kJ mol⁻¹, C–I ≈ 228 kJ mol⁻¹. This trend explains why iodoalkanes are generally the most reactive in nucleophilic substitution.
The physical properties of haloalkanes directly influence their applications. Boiling points increase with molecular mass and with halogen size, because stronger van der Waals forces require more energy to overcome. For example, chloromethane is a gas at room temperature, while bromoethane and iodoethane are volatile liquids.
Haloalkanes are polar but not hydrogen-bond donors, so they dissolve in organic solvents but are insoluble in water. Densities vary: chloroalkanes are usually less dense than water, whereas bromoalkanes and iodoalkanes are denser. These properties make haloalkanes useful as solvents, anaesthetics and intermediates, as discussed below.
Two reaction types dominate haloalkane chemistry: nucleophilic substitution and elimination. In nucleophilic substitution, an electron-rich species attacks the electron-deficient carbon and replaces the halogen. For example, with aqueous sodium hydroxide, a haloalkane forms an alcohol:
Structural isomerism is one of the most frequently tested topics in CIE A-Level Chemistry Paper 2 and Paper 4. Students are often asked to identify, draw, or count structural isomers of a given molecular formula. Mastering this topic requires a clear understanding of the three main types: chain isomerism, position isomerism, and functional group isomerism.
结构异构是 CIE A-Level 化学 Paper 2 和 Paper 4 中最常考的知识点之一。学生经常被要求识别、画出或数出某一分子式对应的结构异构体。掌握这一专题需要清晰理解三种主要类型：链异构、位置异构和官能团异构。
1. What Are Structural Isomers? | 什么是结构异构体？
Structural isomers are compounds that have the same molecular formula but different structural formulas. This means the atoms are connected in a different order or arrangement, leading to different physical and sometimes different chemical properties.
For example, both butane and 2-methylpropane share the molecular formula C₄H₁₀, yet they have different carbon skeletons. Butane is a straight chain, while 2-methylpropane has a branched chain. Their boiling points differ significantly because of different intermolecular forces.
Same molecular formula → Different structural formula → Structural isomers
相同分子式 → 不同结构式 → 结构异构体
2. Classification of Structural Isomerism | 结构异构的分类
Structural isomerism is divided into three main categories. You must be able to recognise each type and understand which functional groups can give rise to functional group isomerism.
结构异构分为三大类。你必须能够识别每种类型，并理解哪些官能团会产生官能团异构。
Chain isomerism – different branching of the carbon skeleton
Chain (链异构) – 碳骨架的支链不同
Position isomerism – same skeleton, functional group at different positions
Position (位置异构) – 骨架相同，官能团位于不同位置
Functional group isomerism – different functional groups with same molecular formula
Functional group (官能团异构) – 分子式相同但官能团不同
It is important to note that stereoisomerism (cis-trans and optical isomerism) is NOT considered structural isomerism because the connectivity of atoms is identical; only the spatial arrangement differs.
Chain isomerism arises when compounds with the same molecular formula differ in the branching of the carbon chain. Straight-chain isomers and branched-chain isomers are examples of chain isomerism.
Consider the molecular formula C₅H₁₂ (pentane). It has three chain isomers:
以分子式 C₅H₁₂（戊烷）为例，它有三种链异构体：
Pentane (CH₃CH₂CH₂CH₂CH₃) – a straight chain
戊烷 (CH₃CH₂CH₂CH₂CH₃) – 直链
2-Methylbutane (CH₃CH(CH₃)CH₂CH₃) – a branched chain
2-甲基丁烷 (CH₃CH(CH₃)CH₂CH₃) – 支链
2,2-Dimethylpropane (C(CH₃)₄) – a highly branched chain
2,2-二甲基丙烷 (C(CH₃)₄) – 高度支链
As branching increases, the boiling point decreases due to reduced surface area and weaker van der Waals forces. This is a classic exam point: the more branched the isomer, the lower the boiling point.
随着支链增多，沸点降低，因为表面积减小，范德华力减弱。这是一个经典考点：异构体支链越多，沸点越低。
C₅H₁₂ has 3 chain isomers | C₅H₁₂ 有 3 种链异构体
For alkenes, chain isomerism also exists, but you must keep the C=C double bond intact. For example, C₄H₈ has chain isomers such as but-1-ene and 2-methylprop-1-ene, but these also involve position isomerism, so be careful when classifying.
Position isomerism occurs when the same carbon skeleton and the same functional group exist, but the functional group is attached to different carbon atoms in the chain.
当碳骨架和官能团相同，但官能团连接在碳链的不同碳原子上时，就产生位置异构。
Classic examples include halogenoalkanes and alcohols. For the formula C₃H₇Cl, two position isomers are possible:
典型的例子包括卤代烷和醇。对于分子式 C₃H₇Cl，可能存在两种位置异构体：
1-Chloropropane (CH₃CH₂CH₂Cl) – chlorine on terminal carbon
1-氯丙烷 (CH₃CH₂CH₂Cl) – 氯在末端碳上
2-Chloropropane (CH₃CH(Cl)CH₃) – chlorine on middle carbon
2-氯丙烷 (CH₃CH(Cl)CH₃) – 氯在中间碳上
Similarly, butanol (C₄H₉OH) has four structural isomers: butan-1-ol, butan-2-ol, 2-methylpropan-1-ol, and 2-methylpropan-2-ol. Note that these four include both position isomerism and chain isomerism.
For alkenes, positional isomerism is very common. For example, C₄H₈ has but-1-ene (CH₂=CHCH₂CH₃) and but-2-ene (CH₃CH=CHCH₃). These are position isomers because the double bond moves from carbon 1 to carbon 2.
When drawing position isomers, always number the chain so that the functional group gets the lowest possible number. This is a convention, not an isomer difference. For example, CH₃CH₂CH₂Cl and ClCH₂CH₂CH₃ are the same compound (1-chloropropane).
Functional group isomerism arises when compounds with the same molecular formula have entirely different functional groups. This is the most challenging type because the chemical properties can be dramatically different.
Key pairs of functional group isomers you must know for CIE A-Level Chemistry include:
在 CIE A-Level 化学中，你必须知道的关键官能团异构体对包括：
Molecular formula
Functional group isomer A
Functional group isomer B
C₂H₆O
Ethanol (alcohol) CH₃CH₂OH
Methoxymethane (ether) CH₃OCH₃
C₃H₆O
Propanal (aldehyde) CH₃CH₂CHO
Propanone (ketone) CH₃COCH₃
C₂H₄O₂
Ethanoic acid (carboxylic acid) CH₃COOH
Methyl methanoate (ester) HCOOCH₃
C₃H₇NO₂
Alanine (amino acid) CH₃CH(NH₂)COOH
Methyl glycinate (ester) NH₂CH₂COOCH₃
Note that the molecular formula C₃H₆O also includes prop-2-en-1-ol (CH₂=CHCH₂OH), which is an unsaturated alcohol — a third functional group isomer. So when a question asks for all isomers, you must consider alkenols, aldehydes, ketones, cyclic ethers, and epoxides depending on the degree of unsaturation.
The degree of unsaturation (IHD) is a powerful tool. For C₃H₆O, IHD = (2×3 + 2 − 6)/2 = 1. This means there is either one double bond or one ring. Aldehydes, ketones, alkenols, and cyclic ethers all satisfy this condition.
6. How to Systematically Identify Structural Isomers | 如何系统识别结构异构体
In exams, you may be asked to draw all structural isomers for a given formula, or to determine whether two given structures are isomers. A systematic approach is essential to avoid missing or duplicating isomers.
Calculate the degree of unsaturation (IHD) from the molecular formula.
计算分子的不饱和度 (IHD)。
Identify the possible functional groups based on the IHD and the atoms present.
根据不饱和度和所含原子，确定可能的官能团。
Draw the longest carbon chain first, then reduce the chain length and add branches.
先画最长碳链，然后缩短碳链并添加支链。
For each skeleton, place the functional group at every distinct position, using symmetry to eliminate duplicates.
对每个骨架，将官能团放在所有不等价的位置上，利用对称性排除重复项。
Check for other functional group possibilities (e.g., aldehydes vs ketones, alcohols vs ethers).
检查其他官能团可能性（如醛 vs 酮，醇 vs 醚）。
Finally, verify that every drawn structure has the correct molecular formula and is not a duplicate.
最后，验证每个画出的结构式具有正确的分子式且不重复。
This method works well for small molecules (up to about six carbon atoms), which are the ones typically tested at A-Level.
这个方法适用于小分子（最多约六个碳原子），而 A-Level 考试通常只考这些。
7. Worked Example: All Isomers of C₄H₁₀O | 例题：C₄H₁₀O 的所有异构体
Let us apply the systematic method to find all structural isomers of C₄H₁₀O, a formula that commonly appears in exam questions.
让我们用系统方法找出 C₄H₁₀O 的所有结构异构体，这个分子式在考试中经常出现。
Step 1: IHD. For C₄H₁₀O, IHD = (2×4 + 2 − 10)/2 = 0. No double bonds or rings. The molecule is saturated. Since oxygen is divalent, possible functional groups are alcohols (R–OH) and ethers (R–O–R’).
Step 2: Alcohol isomers. Draw the four carbon skeletons: butane (straight), 2-methylpropane (branched). For butane, place –OH on carbon 1 and carbon 2, giving butan-1-ol and butan-2-ol. Placing –OH on carbon 3 is identical to carbon 2 by symmetry. For 2-methylpropane, the central carbon is tertiary, giving 2-methylpropan-2-ol; the terminal methyl carbons are equivalent, giving 2-methylpropan-1-ol. That gives four alcohols.
Step 3: Ether isomers. For C₄H₁₀O, ethers have the form R–O–R’. The total number of carbon atoms is four. Possible splits of the alkyl groups are 1+3 (methoxypropane) and 2+2 (ethoxyethane). For the 1+3 split, the propyl group can be n-propyl or isopropyl, giving methyl propyl ether (CH₃OCH₂CH₂CH₃) and methyl isopropyl ether (CH₃OCH(CH₃)₂). For the 2+2 split, only ethoxyethane (CH₃CH₂OCH₂CH₃). So three ethers.
8. Worked Example: Isomers of C₃H₆O | 例题：C₃H₆O 的异构体
Let us examine C₃H₆O, which has IHD = 1. The possible functional groups include aldehydes, ketones, alkenols, and cyclic ethers.
让我们考察 C₃H₆O，其 IHD = 1。可能的官能团包括醛、酮、烯醇和环醚。
Aldehyde: propanal (CH₃CH₂CHO). Ketone: propanone (CH₃COCH₃). These are the two most commonly mentioned isomers.
醛：丙醛 (CH₃CH₂CHO)。酮：丙酮 (CH₃COCH₃)。这是最常提到的两种异构体。
Alkenols: with three carbons, the double bond can be between C1–C2 or C2–C3. The former gives prop-2-en-1-ol (CH₂=CHCH₂OH), and the latter gives prop-1-en-1-ol (CH₃CH=CHOH). However, prop-1-en-1-ol is an enol that tautomerises to propanal, so it is usually not counted as a stable isolatable isomer in A-Level exams. You should mention it as a tautomer, not a structural isomer for practical purposes.
Cyclic ethers: the three-membered ring epoxide, epoxypropane (oxirane with a methyl substituent) exists as methyloxirane. This is a valid structural isomer, though it is less commonly drawn by students.
So for C₃H₆O, the main structural isomers tested are propanal and propanone. Some syllabi also include prop-2-en-1-ol and methyloxirane, so check your syllabus specification carefully.
9. Common Mistakes and How to Avoid Them | 常见错误及如何避免
Students often lose marks in isomer questions due to several repeatable errors. Being aware of these pitfalls will significantly improve your exam performance.
学生在异构体题目中常因几个可重复的错误而失分。注意这些陷阱将显著提高你的考试表现。
Mistake 1: Drawing the same structure twice. For example, drawing 2-chlorobutane and 3-chlorobutane as different isomers. They are the same because carbon 3 is identical to carbon 2 when numbered from the other end. Always check the longest chain and number from the end that gives the lowest locant.
Mistake 2: Ignoring functional group isomers. For example, only drawing alcohols for C₂H₆O and forgetting the ether. C₂H₆O has exactly two isomers: ethanol (CH₃CH₂OH) and methoxymethane (CH₃OCH₃).
Mistake 3: Breaking valence rules. Carbon must always have four bonds, oxygen two, nitrogen three, and hydrogen one. A structure where a carbon has five bonds or an oxygen has three bonds is invalid. Always count the bonds.
Mistake 4: Forgetting to check the IHD. If the molecule has one double bond, you cannot draw a saturated alcohol without adding a second double bond. Always verify that the total number of hydrogens in your drawn structure matches the given formula.
Mistake 5: Confusing structural isomerism with stereoisomerism. Cis-trans isomers are not structural isomers. They have the same connectivity. In CIE exams, questions about stereoisomers are treated separately.
10. Comparing Physical Properties of Structural Isomers | 结构异构体的物理性质比较
Structural isomers have different physical properties because their shapes and functional groups affect intermolecular forces. Exam questions often ask you to compare boiling points or solubility.
For alkanes, branched isomers have lower boiling points than straight-chain isomers. For example, 2,2-dimethylpropane boils at 9.5 °C, while pentane boils at 36.1 °C. The reduced surface area of the branched isomer leads to weaker van der Waals forces.
For alcohols, boiling points are much higher than isomeric ethers because alcohols form hydrogen bonds. Ethanol boils at 78.4 °C, while methoxymethane boils at −24 °C. Both have the formula C₂H₆O, but the alcohol has a much higher boiling point.
Solubility in water: alcohols with fewer than four carbon atoms are soluble in water because they form hydrogen bonds with water. Ethers are slightly soluble but less so than comparable alcohols. Carboxylic acids and esters also differ: carboxylic acids can form hydrogen bonds more extensively due to the –OH group, while esters cannot donate hydrogen bonds.
Hydrogen bonding > dipole-dipole > van der Waals | 氢键 > 偶极-偶极 > 范德华力
11. Spectroscopic Identification of Isomers | 用波谱鉴别异构体
In Paper 4, you may be given mass spectra, IR spectra, or NMR data to deduce which structural isomer is present. Here is how you distinguish common isomer pairs.
在 Paper 4 中，你可能会得到质谱、红外光谱或核磁共振数据来推断存在哪种结构异构体。以下是如何区分常见的异构体对。
Mass spectrometry: Aldehydes and ketones often fragment to produce acylium ions. A branched alkane gives a more abundant smaller fragment ion, while a straight chain gives a more uniform series of peaks. The molecular ion (M⁺) has the same m/z for all isomers, but the fragmentation pattern differs.
IR spectroscopy: A broad absorption around 3230–3550 cm⁻¹ indicates an O–H stretch (alcohol). A sharp, strong absorption near 1700–1750 cm⁻¹ indicates a C=O stretch (aldehyde or ketone). An ether shows a C–O stretch near 1050–1150 cm⁻¹ and no O–H stretch.
¹H NMR: Aldehydes show a characteristic peak around δ 9–10 ppm. For propanal, the aldehyde proton appears as a triplet. Ketones have no such peak. Alcohols show a variable broad peak around δ 1–5 ppm for the –OH proton, which can be confirmed by adding D₂O.
¹³C NMR: The carbon of an aldehyde group typically appears at δ 190–200 ppm, while ketones appear at δ 200–210 ppm. Alcohol carbons bearing –OH appear at δ 50–90 ppm, and ether carbons at slightly higher chemical shifts.
Here is a concise checklist that will help you solve structural isomer questions confidently in the exam.
以下是一个简洁的清单，可以帮助你在考试中自信地解决结构异构问题。
Always calculate IHD first — it tells you if rings or double bonds are possible.
总是先计算不饱和度——它告诉你是否可能存在环或双键。
Draw the longest chain first, then shorten it step by step.
先画最长链，然后逐步缩短。
Number the chain to give the functional group the lowest locant, but remember the reverse orientation is the same compound.
给碳链编号使官能团位次最低，但记住反向编号是同一个化合物。
Use symmetry to eliminate duplicates.
利用对称性排除重复。
Check for functional group isomers: alcohols/ethers, aldehydes/ketones, carboxylic acids/esters, alkenes/cycloalkanes.
检查官能团异构体：醇/醚、醛/酮、羧酸/酯、烯烃/环烷烃。
For molecules with IHD = 1, consider both the double-bond isomers and ring isomers.
对于 IHD = 1 的分子，同时考虑双键异构体和环异构体。
When comparing physical properties, rank by strength of intermolecular forces.
比较物理性质时，按分子间作用力强弱排序。
In spectroscopy questions, look for the presence or absence of O–H, C=O, and aldehyde protons.
在波谱题中，检查是否存在 O–H、C=O 和醛基质子。
Finally, practise drawing isomers of common formulas: C₄H₁₀, C₅H₁₂, C₃H₇Cl, C₄H₉Br, C₂H₆O, C₃H₈O, C₃H₆O, C₄H₈O, C₄H₈, and C₄H₄. These cover all three types of structural isomerism and are the most likely to appear in your exam.
最后，练习画常见分子式的异构体：C₄H₁₀、C₅H₁₂、C₃H₇Cl、C₄H₉Br、C₂H
Published by TutorHao | A-Level Chemistry Revision Series | aleveler.com
📚 Elimination Reactions in A-Level Chemistry | A-Level化学：消除反应考点详解
Elimination reactions are one of the most frequently tested organic reaction mechanisms in CIE A-Level Chemistry. They involve the removal of two atoms or groups from adjacent carbon atoms, resulting in the formation of a double bond (C=C). Understanding the mechanism, the conditions required, and the competition with substitution reactions is essential for scoring well in Paper 4 and Paper 5.
An elimination reaction is a type of organic reaction in which two substituents are removed from a molecule, forming a new π bond in the process. The most common type studied at A-Level is the elimination of hydrogen halide (HX) from a halogenoalkane, producing an alkene.
It is important to note that the reagent used is NaOH dissolved in ethanol (alcoholic NaOH), not aqueous NaOH. The solvent plays a decisive role in determining whether elimination or substitution occurs.
2. Elimination vs Substitution: The Role of Solvent | 消除与取代的竞争：溶剂的作用
Halogenoalkanes can undergo both nucleophilic substitution and elimination reactions with NaOH. Which pathway dominates depends primarily on the solvent used.
With aqueous NaOH: OH⁻ acts as a nucleophile and attacks the electron-deficient carbon atom, resulting in substitution to form an alcohol.
使用水性NaOH时：OH⁻ 作为亲核试剂进攻缺电子的碳原子，发生取代反应生成醇。
With alcoholic NaOH: OH⁻ favours acting as a base, abstracting a proton from the β-carbon atom, leading to elimination and formation of an alkene.
使用醇性NaOH时：OH⁻ 更倾向于作为碱，从β-碳原子上夺取质子，发生消除反应生成烯烃。
Condition
Major Product
Role of OH⁻
Aqueous NaOH (H₂O)
Alcohol (substitution)
Nucleophile
Alcoholic NaOH (C₂H₅OH)
Alkene (elimination)
Base
3. The E2 Mechanism (Bimolecular Elimination) | E2机理（双分子消除）
The E2 mechanism is a one-step, concerted process in which the base abstracts a proton from the β-carbon while simultaneously the halide ion leaves from the α-carbon. Both processes occur in a single transition state, meaning the reaction is bimolecular — the rate depends on the concentrations of both the halogenoalkane and the base.
The reaction is concerted — the C-H bond breaking, C=C bond formation, and C-X bond breaking all happen in one step.
反应是协同的——C-H键断裂、C=C键形成和C-X键断裂同时发生。
The rate equation is rate = k[halogenoalkane][base], showing second-order kinetics overall.
速率方程为 rate = k[卤代烷][碱]，总反应级数为二级。
Primary and secondary halogenoalkanes typically undergo elimination via the E2 pathway when treated with a strong base such as alcoholic NaOH.
伯卤代烷和仲卤代烷在使用强碱（如醇性NaOH）处理时，通常通过E2途径发生消除。
The transition state involves an anti-periplanar arrangement, where the H and X are on opposite sides of the carbon-carbon bond.
过渡态中H和X处于碳碳键的反式共平面（anti-periplanar）排列。
Rate = k [R-X] [OH⁻]
4. The E1 Mechanism (Unimolecular Elimination) | E1机理（单分子消除）
The E1 mechanism is a two-step process. In the first, much slower step, the halogenoalkane undergoes heterolytic bond fission to form a carbocation intermediate. In the second, faster step, the base removes a proton from the β-carbon, forming the alkene. Because the slow step involves only the halogenoalkane, the rate depends solely on the concentration of the halogenoalkane.
The rate equation is rate = k[halogenoalkane], showing first-order kinetics.
速率方程为 rate = k[卤代烷]，表现为一级动力学。
E1 is favoured by tertiary halogenoalkanes because they form stable tertiary carbocations.
E1反应更有利于叔卤代烷，因为叔碳正离子更稳定。
E1 often competes with SN1 since both share the same carbocation intermediate.
E1常与SN1竞争，因为两者共享相同的碳正离子中间体。
Feature
E1
E2
Number of steps
Two
One
Kinetics
First-order
Second-order
Intermediate
Carbocation
None (concerted)
Favoured by
Tertiary halogenoalkanes
Primary/secondary halogenoalkanes
Stereochemistry
No special requirement
Anti-periplanar required
5. Zaitsev’s Rule | 查依采夫规则
When a halogenoalkane can eliminate HX to form more than one alkene, Zaitsev’s rule predicts which product will be the major product. According to this rule, the most stable alkene is the most substituted alkene, i.e., the alkene with the greatest number of alkyl groups attached to the C=C double bond.
This is because alkyl groups are electron-donating through the inductive effect, stabilising the π bond. More substituted alkenes have a more stable double bond due to hyperconjugation.
For halogenoalkanes, the standard conditions for elimination are alcoholic NaOH (or KOH) and heat. The hydroxide ion acts as a base rather than a nucleophile because the ethanol solvent does not solvate the OH⁻ ion effectively, making it “naked” and more basic.
The general reaction for a primary halogenoalkane is:
伯卤代烷的一般反应式为：
CH₃CH₂CH₂Br + NaOH (alc) → CH₃CH=CH₂ + NaBr + H₂O
In this reaction:
在该反应中：
The OH⁻ removes a hydrogen atom from the β-carbon.
OH⁻ 从β-碳上夺取一个氢原子。
The electron pair from the C-H bond forms the C=C π bond.
C-H键中的电子对形成了C=C π键。
The bromine leaves as Br⁻.
溴以Br⁻的形式离去。
It is worth remembering that in the CIE examination, you may be asked to draw the mechanism for this elimination reaction using curly arrows. Make sure you practise drawing the E2 mechanism clearly and unambiguously.
Alcohols can also undergo elimination, specifically dehydration, to form alkenes. This requires an acid catalyst such as concentrated H₂SO₄ or H₃PO₄, along with heat.
CH₃CH₂OH → CH₂=CH₂ + H₂O (catalysed by conc. H₂SO₄, ~170°C)
CH₃CH₂OH → CH₂=CH₂ + H₂O（浓H₂SO₄催化，约170°C）
Important exam points for alcohol dehydration:
关于醇脱水的考试要点：
Butan-2-ol can dehydrate to form but-1-ene and but-2-ene, with but-2-ene as the major product according to Zaitsev’s rule.
2-丁醇脱水可生成1-丁烯和2-丁烯，根据查依采夫规则，2-丁烯是主要产物。
The temperature matters: at lower temperatures (around 140°C for ethanol), substitution or ether formation may occur instead of elimination.
温度很关键：在较低温度下（乙醇约140°C），可能发生取代或生成醚的反应而非消除。
Different alkenes can be obtained by choosing the right alcohol as the starting material — this is a useful synthetic strategy.
选择不同的醇作为原料可以获得不同的烯烃——这是一种有用的合成策略。
The dehydration mechanism involves protonation of the -OH group by the acid, formation of a carbocation, and loss of H⁺ from the β-carbon to form the alkene.
脱水机理包括：酸使-OH基团质子化、生成碳正离子、从β-碳上失去H⁺形成烯烃。
8. Stereochemistry of Elimination | 消除的立体化学
Elimination reactions can exhibit stereospecificity, particularly in E2 reactions. For the E2 mechanism to proceed efficiently, the hydrogen and halogen must be anti-periplanar, meaning they lie in the same plane but on opposite sides of the C-C bond.
In cyclic systems, trans-1,2-dibromocyclohexane cannot undergo E2 elimination as readily as the cis isomer because the anti-periplanar geometry is harder to achieve.
The E2 elimination of a stereoisomer can produce a specific alkene isomer (E or Z). For example, the elimination of HBr from one stereoisomer of 2-bromobutane may preferentially form (E)-but-2-ene over (Z)-but-2-ene.
Exam questions sometimes ask you to determine the structure of the alkene product based on the stereochemistry of the starting material.
考题有时会要求你根据起始原料的立体化学判断烯烃产物的结构。
Note: In E1 reactions, the formation of a planar carbocation intermediate means that stereochemistry is generally not preserved; a mixture of E and Z isomers is often obtained.
9. Elimination vs Oxidation: Knowing the Difference | 消除与氧化的区别
Students often confuse elimination with oxidation, especially when alcohols are involved. It is essential to distinguish these processes clearly.
学生常常将消除反应与氧化反应混淆，特别是涉及醇的反应时。清晰地辨别这两种过程非常重要。
Elimination (dehydration): loses a small molecule (H₂O or HX) and forms a C=C bond. No change in the oxidation state of carbon occurs — the carbon atoms remain at the same oxidation level.
Oxidation: involves an increase in the oxidation state of carbon, typically by adding oxygen or removing hydrogen along with losing electrons. For alcohols, oxidation gives carbonyl compounds (aldehydes, ketones, carboxylic acids).
For example, ethanol can be dehydrated to ethene (elimination) or oxidised to ethanal (oxidation). These are different reactions requiring different reagents and conditions.
例如，乙醇可以脱水生成乙烯（消除反应），也可以被氧化为乙醛（氧化反应）。两者需要不同的试剂和条件。
Reaction Type
Reagents/Conditions
Product
Dehydration of ethanol
Conc. H₂SO₄, 170°C
Ethene (C₂H₄)
Oxidation of ethanol
K₂Cr₂O₇/H₂SO₄, reflux
Ethanoic acid (CH₃COOH)
Oxidation of ethanol (distillation)
K₂Cr₂O₇/H₂SO₄, distill
Ethanal (CH₃CHO)
10. How to Identify Elimination Products in Exam Questions | 如何在考题中判断消除产物
When determining the products of an elimination reaction in an exam, follow this systematic approach:
在考试中判断消除反应的产物时，可以按照以下系统性的方法进行分析：
Identify the α-carbon and β-carbon atoms. The α-carbon carries the leaving group (X or OH). The β-carbon is the adjacent carbon bearing the hydrogen.
识别α-碳和β-碳。α-碳上带有离去基团（X或OH），β-碳是相邻的、带有氢原子的碳。
Remove the leaving group and a hydrogen from the β-carbon.
移除离去基团和β-碳上的一个氢。
Connect the α and β carbons with a double bond.
将α和β碳用双键连接。
If there are two different β-carbons, apply Zaitsev’s rule: the major product is the more substituted alkene.
如果有两个不同的β-碳，应用查依采夫规则：主要产物是取代程度更高的烯烃。
Finally, determine whether the product has E/Z isomerism, and if asked, predict the more stable isomer (usually E).
Forgetting to heat the mixture: Elimination reactions generally require heating. Substitution often occurs at room temperature.
忘记加热：消除反应通常需要加热。取代反应通常在室温下便可进行。
Drawing incorrect mechanisms: In E2, the base attacks the β-hydrogen, not the α-carbon. Curly arrows must show electron movement from the C-H bond to form the C=C bond, and from the C-X bond to the halogen.
Not applying Zaitsev’s rule: When more than one alkene is possible, always identify the major product.
没有应用查依采夫规则：当可能生成不止一种烯烃时，一定要指出主要产物。
Ignoring by-products: For halogenoalkane elimination, the by-products are NaBr (or KBr) and H₂O. For alcohol dehydration, the only by-product is H₂O.
忽略副产物：卤代烷消除的副产物是NaBr（或KBr）和H₂O。醇脱水的唯一副产物是H₂O。
Confusing E1 and E2: Remember that E1 involves a carbocation intermediate and shows first-order kinetics; E2 is concerted and second-order.
混淆E1和E2：记住E1涉及碳正离子中间体且表现为一级动力学；E2是协同反应且为二级。
12. Summary: Key Points to Remember | 总结：必须记住的关键点
Here is a concise summary of the most important information regarding elimination reactions for the CIE A-Level Chemistry examination:
以下是与CIE A-Level化学考试相关的消除反应最重要信息的简明总结：
Elimination removes two atoms/groups and forms a C=C double bond.
消除反应脱去两个原子/基团并形成C=C双键。
Alcoholic NaOH + heat causes halogenoalkanes to undergo elimination to form alkenes.
醇性NaOH + 加热使卤代烷发生消除生成烯烃。
Conc. H₂SO₄ + heat causes alcohols to dehydrate to alkenes.
浓H₂SO₄ + 加热使醇脱水生成烯烃。
E2 is a one-step, bimolecular, concerted process; E1 is a two-step, unimolecular process via a carbocation.
E2是一步完成的双分子协同过程；E1是通过碳正离子的两步单分子过程。
Zaitsev’s rule: the major product is the most substituted alkene.
查依采夫规则：主要产物是取代程度最高的烯烃。
E2 requires an anti-periplanar geometry between the leaving group and the β-hydrogen.
E2要求离去基团和β-氢处于反式共平面几何构型。
Check the solvent carefully in exam questions: aqueous vs alcoholic determines the reaction pathway.
仔细审题中溶剂：水性还是醇性决定了反应路径。
Always include by-products in your equations and practise drawing curly arrow mechanisms.
在方程式中始终写出副产物，并练习画出弯箭头机理。
Mastering elimination reactions requires a clear understanding of the mechanistic differences, the conditions, and the factors that govern product selection. Practice with past paper questions to become confident in applying these concepts. Good luck with your revision!
📚 A-Level Chemistry: Nucleophilic Substitution Mechanisms of Haloalkanes | A-Level化学：卤代烷烃亲核取代反应机理
Nucleophilic substitution is one of the most important reaction types in organic chemistry. Haloalkanes are ideal models for studying this mechanism because their carbon–halogen bond is polarised, making the carbon atom electron-deficient and open to attack by electron-rich species.
Haloalkanes are alkanes in which one or more hydrogen atoms have been replaced by halogen atoms (F, Cl, Br, I). They are represented generally as R–X, where R is an alkyl group and X is a halogen atom.
The carbon–halogen bond is polar because the halogen is more electronegative than carbon. This creates a partial positive charge (δ+) on carbon and a partial negative charge (δ−) on the halogen.
This polarised bond is the key to the reactivity of haloalkanes: it allows nucleophiles to attack the carbon atom and displace the halogen as a leaving group.
这种极性键是卤代烷烃反应活性的关键：它允许亲核试剂进攻碳原子，并将卤素作为离去基团取代。
2. Nucleophiles and Leaving Groups | 亲核试剂与离去基团
A nucleophile is an electron-rich species that donates a pair of electrons to form a new covalent bond. Common nucleophiles include OH⁻, CN⁻, NH₃, and H₂O.
The leaving group is the species that departs with the pair of bond electrons. In haloalkanes, the halide ion X⁻ is the leaving group.
离去基团是带着成键电子对离开的物种。在卤代烷烃中，卤离子X⁻是离去基团。
A good leaving group must be stable when carrying a negative charge. The order of leaving group ability among halides is I⁻ > Br⁻ > Cl⁻ > F⁻, because larger halide ions are more polarisable and weaker bases.
Better leaving group = weaker base. 离去基团越好 = 碱性越弱。
Iodide is the best leaving group among halides. 碘离子是卤素中最好的离去基团。
Fluoride is a poor leaving group and C–F bonds are very strong. 氟离子是较差的离去基团，且C–F键很强。
3. Overview of Two Mechanisms | 两种机理概览
Nucleophilic substitution in haloalkanes can occur by two distinct mechanisms: S_N1 and S_N2. The symbol S_N stands for substitution, nucleophilic; the number 1 or 2 indicates whether the rate-determining step is unimolecular or bimolecular.
The mechanism chosen depends on the structure of the haloalkane, the nature of the nucleophile, the solvent, and the leaving group.
选择哪种机理取决于卤代烷烃的结构、亲核试剂的性质、溶剂以及离去基团。
S_N1: Rate = k[R–X]
S_N2: Rate = k[R–X][Nu⁻]
In S_N1, only the haloalkane appears in the rate equation; in S_N2, both the haloalkane and the nucleophile appear.
在S_N1中，速率方程中只出现卤代烷烃；在S_N2中，卤代烷烃和亲核试剂都出现。
4. The S_N1 Mechanism | S_N1机理
S_N1 is a two-step mechanism that occurs mainly with tertiary haloalkanes. In the first, slow step, the carbon–halogen bond breaks heterolytically to form a carbocation and a halide ion.
In the second, fast step, the nucleophile attacks the carbocation to form the substitution product. If the nucleophile is neutral, for example water, a quick deprotonation step follows.
Because the slow step only involves one molecule, the rate depends only on the concentration of the haloalkane.
由于慢步骤只涉及一个分子，反应速率只取决于卤代烷烃的浓度。
The carbocation intermediate is planar and trigonal, so the nucleophile can attack from either face. If the starting haloalkane is optically active, S_N1 produces a racemic mixture.
S_N2 is a one-step mechanism in which the nucleophile attacks the carbon atom at the same time that the leaving group departs. No intermediate is formed; instead, a transition state exists.
The nucleophile donates its electron pair to form a new bond while the leaving group accepts the bond electrons. This means the leaving group acts as a Lewis base, and the electron pair moves from the leaving group? No, the bond electrons shift toward the leaving group.
The reaction is bimolecular, so the rate depends on both the concentration of the haloalkane and the concentration of the nucleophile.
该反应是双分子的，因此速率同时取决于卤代烷烃和亲核试剂的浓度。
Attack occurs at the side opposite the leaving group, so the configuration of a chiral centre is inverted. This is called Walden inversion.
进攻发生在离去基团的相反一侧，因此手性中心的构型会发生翻转。这称为瓦尔登反转。
S_N2 is favoured by primary haloalkanes and by strong nucleophiles in polar aprotic solvents.
S_N2更有利于伯卤代烷烃、强亲核试剂和极性非质子溶剂。
6. Effect of Substrate Structure | 底物结构的影响
Alkyl group structure is the most important factor in deciding which mechanism operates.
烷基结构是决定哪种机理起主导作用的最重要因素。
For S_N2, methyl and primary haloalkanes react fastest because the carbon atom is less sterically hindered. As the number of alkyl groups increases, attack by the nucleophile becomes more difficult.
For S_N1, tertiary haloalkanes react fastest because the tertiary carbocation is most stable. Carbocation stability increases with more alkyl groups due to the electron-donating inductive effect and hyperconjugation.
Thus, tertiary haloalkanes tend to react by S_N1, while primary haloalkanes tend to react by S_N2. Secondary haloalkanes can react by either pathway, depending on conditions.
A common laboratory test distinguishes different haloalkanes by the precipitate formed with silver nitrate.
一个常见实验测试通过硝酸银产生的沉淀来区分不同的卤代烷烃。
The haloalkane is warmed with aqueous ethanolic silver nitrate. The halide ion is displaced by water or ethanol, and then reacts with Ag⁺ to form a silver halide precipitate.
将卤代烷烃与含水硝酸银的乙醇溶液加热。卤离子被水或乙醇置换出来，然后与Ag⁺反应生成卤化银沉淀。
Chloroalkanes give a white precipitate of AgCl.
氯代烷烃生成白色AgCl沉淀。
Bromoalkanes give a cream precipitate of AgBr.
溴代烷烃生成奶油色AgBr沉淀。
Iodoalkanes give a yellow precipitate of AgI.
碘代烷烃生成黄色AgI沉淀。
Because C–I bonds are weakest, iodoalkanes react fastest; chloroalkanes react slowest and may need more heating.
由于C–I键最弱，碘代烷烃反应最快；氯代烷烃反应最慢，可能需要更长时间加热。
Order of reactivity: RI > RBr > RCl
10. Reactions with Specific Nucleophiles | 与特定亲核试剂的反应
Different nucleophiles convert haloalkanes into different organic products. These reactions are frequently tested in A-Level exams.
不同的亲核试剂将卤代烷烃转化为不同的有机产物。这些反应在A-Level考试中经常出现。
KOH (aqueous) or NaOH (aqueous) → alcohol. 水相KOH或NaOH → 醇。
KCN (ethanolic) → nitrile, which can be hydrolysed to carboxylic acid or reduced to amine. 乙醇KCN → 腈，可水解为羧酸或还原为胺。
Watch out: aqueous conditions favour replacement by OH⁻, while ethanolic conditions favour elimination to form alkenes, especially with tertiary haloalkanes.
注意：水相条件有利于OH⁻取代，而乙醇条件有利于消除生成烯烃，尤其是叔卤代烷烃。
11. Common Exam Mistakes | 常见考试误区
Students often lose marks by confusing the rate equations, the intermediates, or the stereochemical outcomes.
学生常因混淆速率方程、中间体或立体化学结果而失分。
S_N1: unimolecular, carbocation, racemisation.
S_N1：单分子，碳正离子，外消旋化。
S_N2: bimolecular, transition state, inversion.
S_N2：双分子，过渡态，构型翻转。
Do not write that S_N1 has no intermediate, and do not draw S_N2 with a carbocation. Also, do not say that primary haloalkanes react fastest in S_N1; this is incorrect.
In mechanism drawings, show the curly arrow from the nucleophile toward the carbon atom. For S_N2, draw the nucleophile attacking from the side opposite the leaving group.
Haloalkanes undergo nucleophilic substitution through S_N1 or S_N2 mechanisms, governed by the structure of the alkyl group and the reaction conditions.
卤代烷烃通过S_N1或S_N2机理发生亲核取代反应，具体由烷基结构和反应条件决定。
S_N1 is favoured by tertiary haloalkanes, involves a carbocation intermediate, and can lead to racemisation. S_N2 is favoured by primary haloalkanes, occurs in one step, and leads to inversion of configuration.
Nucleophilic substitution is a cornerstone of A-Level Organic Chemistry. It describes how a nucleophile, an electron-rich species, replaces a leaving group attached to a carbon atom. Understanding this mechanism is essential for predicting products, interpreting reaction rates, and explaining stereochemical outcomes.
A nucleophile is a species that donates a pair of electrons to form a new covalent bond. It can be negatively charged or neutral, but it always contains a lone pair of electrons. Common nucleophiles include hydroxide (OH⁻), cyanide (CN⁻), ammonia (NH₃), and water (H₂O).
The word ‘nucleophile’ means ‘nucleus-loving’. Because the carbon atom in a halogenoalkane is electron-deficient (slightly positive), it attracts nucleophiles. The carbon-halogen bond is polarised, making the carbon a site for attack.
For A-Level, you should be able to identify nucleophilic centres and write balanced equations for nucleophilic substitution reactions. The lone pair on the nucleophile attacks the electrophilic carbon, and the halogen leaves as a halide ion.
A good leaving group must be able to stabilise the negative charge it acquires after departure. Halide ions are excellent leaving groups because their negative charge is spread over a large electron cloud. The order of leaving group ability for halogens is I⁻ > Br⁻ > Cl⁻ > F⁻.
In CIE A-Level, you are normally concerned with halogenoalkanes (also called haloalkanes or alkyl halides). The carbon-halogen bond strength determines how easily the bond breaks. Weaker bonds, such as C-I, break more readily, making iodoalkanes most reactive in nucleophilic substitution.
Although fluorine is more electronegative than chlorine, the C-F bond is so strong that fluoroalkanes are generally unreactive. This is a common exam trap: electronegativity alone does not dictate reactivity; bond strength and leaving group stability are more important.
The SN2 mechanism is a single concerted step. The nucleophile attacks the carbon from one side while the leaving group leaves from the opposite side. This backside attack leads to inversion of configuration at the carbon, like an umbrella turning inside out.
The rate equation for an SN2 reaction is: rate = k [halogenoalkane] [nucleophile]. This means the reaction is second order overall: both the substrate and the nucleophile appear in the rate-determining step. Any factor that increases the concentration of either reactant will speed up the reaction.
SN2反应的速率方程为：rate = k [卤代烷] [亲核试剂]。这意味着反应总体为二级：底物和亲核试剂都出现在决速步中。任何增加任一反应物浓度的因素都会加快反应速率。
Rate = k [R-X] [Nu⁻]
SN2 reactions are favoured by primary halogenoalkanes because the carbon is sterically unhindered. In contrast, tertiary substrates are too crowded for backside attack to occur efficiently, so SN2 is unlikely for tertiary halogenoalkanes.
The SN1 mechanism involves two distinct steps. First, the C-X bond breaks heterolytically to form a carbocation and a halide ion. Second, the nucleophile attacks the carbocation to form the product. The carbocation is planar, so attack can occur from either face, giving a racemic mixture if the carbon was chiral.
The rate equation for SN1 is: rate = k [halogenoalkane]. The concentration of the nucleophile does not appear because the slow step is only the ionisation of the C-X bond. This is first order kinetics.
SN1的速率方程为：rate = k [卤代烷]。亲核试剂的浓度不出现在方程中，因为慢步骤仅是C-X键的电离。这是一级动力学。
Rate = k [R-X]
SN1 is favoured by tertiary halogenoalkanes because tertiary carbocations are stabilised by electron-donating alkyl groups via inductive effects. The more substituted the carbocation, the more stable it is. The order of carbocation stability is: 3° > 2° > 1° > methyl.
Three main factors determine which mechanism operates: the structure of the substrate, the nature of the nucleophile, and the solvent. For CIE, you must be able to explain why a primary halogenoalkane follows SN2 while a tertiary halogenoalkane follows SN1.
Substrate structure: Primary carbocations are highly unstable, so SN1 is very slow for primary halogenoalkanes. Instead, primary substrates undergo SN2 because the carbon is accessible. Tertiary carbocations are stable, so tertiary substrates favour SN1.
Nucleophile strength: Strong, negatively charged nucleophiles such as OH⁻ and CN⁻ tend to promote SN2 because they can attack directly. Weak, neutral nucleophiles such as water or alcohols are more likely to participate in SN1, where the nucleophile only attacks after the carbocation has formed.
Solvent: Polar protic solvents (e.g., water, ethanol) stabilise carbocations and halide ions, favouring SN1. Polar aprotic solvents (e.g., propanone) leave nucleophiles less solvated, making them more reactive and favouring SN2. This detail is often tested in multiple-choice questions.
Nucleophilic substitution with water or hydroxide converts a halogenoalkane into an alcohol. Aqueous sodium hydroxide (NaOH) under reflux provides OH⁻ ions, which attack the carbon and displace the halide. This is a key reaction for preparing alcohols from halogenoalkanes.
For a primary halogenoalkane, the reaction is SN2 and the product is a primary alcohol. For a tertiary halogenoalkane, the reaction proceeds via SN1 and the product is a tertiary alcohol. The mechanism can be traced using curly arrows in exam questions.
Example equation: CH₃CH₂Br + NaOH → CH₃CH₂OH + NaBr. In ionic form: CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻. Remember to include all charges and lone pairs when drawing mechanisms.
Halogenoalkanes react with potassium cyanide (KCN) in ethanol under reflux to form nitriles. For example, CH₃CH₂Br + KCN → CH₃CH₂CN + KBr. The nitrile group (-CN) is valuable because it can be reduced to an amine or hydrolysed to a carboxylic acid, extending the carbon chain.
This reaction is an excellent example of nucleophilic substitution where the nucleophile is CN⁻. It is typically used to increase the length of a carbon chain by one carbon atom, which is an important synthetic route in organic chemistry.
Ammonia reacts with halogenoalkanes to form amines. Excess ammonia is used to avoid further substitution. For example, CH₃Br + 2NH₃ → CH₃NH₂ + NH₄Br. The first NH₃ acts as a nucleophile, and the second NH₃ removes a proton from the intermediate ammonium salt.
These reactions are often examined in synthesis questions. You should be able to identify the product formed when a halogenoalkane reacts with KCN or NH₃, and explain the role of ethanol or excess ammonia.
8. Stereochemistry: Inversion and Racemisation | 立体化学：翻转与外消旋化
Stereochemistry is a key distinguishing feature between SN1 and SN2. In SN2, the nucleophile attacks from the back side, so the configuration of a chiral centre is inverted. If the starting material is a single enantiomer, the product will be the opposite enantiomer.
In SN1, the intermediate carbocation is planar (sp² hybridised), so the nucleophile can attack from either side with equal probability. If the starting halogenoalkane is chiral, the product will be a racemic mixture — a 50:50 mixture of both enantiomers. This results in no optical activity.
Exam questions often ask you to explain why SN1 gives a racemic product while SN2 gives inversion. You should mention the planar carbocation for SN1 and the backside attack for SN2. Drawing 3D representations using wedges and dashes is essential for full marks.
Below is a summary table that you can use for quick revision. It compares the key features of SN1 and SN2 mechanisms. Knowing this table well will help you answer both short-answer and multiple-choice questions.
Notice that these two mechanisms are not mutually exclusive. Some secondary halogenoalkanes can undergo both SN1 and SN2 depending on the conditions. In CIE questions, you may need to predict which mechanism dominates based on the given reagent.
When drawing nucleophilic substitution mechanisms, always use curly arrows to show the movement of electron pairs. The arrow starts from the lone pair or bond and points where the electrons are going. Do not use half-headed arrows for heterolytic bond breaking.
Common mistakes include: showing the leaving group departing before the nucleophile attacks in SN2, forgetting the positive charge on the carbocation in SN1, and using CH₃ instead of CH₃CH₂ when writing products. Always check that the molecular formula is balanced.
Another common error is confusing nucleophilic substitution with elimination. Strong bases such as OH⁻ can also cause elimination to form alkenes, especially at high temperatures. In CIE, you should know that ethanolic NaOH favours elimination, while aqueous NaOH favours substitution.
Finally, remember that rate equations in nucleophilic substitution are determined experimentally, not from the stoichiometric equation. If given kinetic data, determine the order with respect to each reactant before deciding whether the mechanism is SN1 or SN2.
Nucleophilic substitution is a central reaction type in halogenoalkane chemistry. SN1 and SN2 differ in the number of steps, kinetics, substrate preference, and stereochemistry. Recognising these differences allows you to predict products, mechanisms, and reaction rates with confidence.
For CIE A-Level, make sure you can write both mechanisms using curly arrows, classify halogenoalkanes as primary, secondary, or tertiary, and explain experimental evidence such as rate equations and optical activity. Practice past exam questions to strengthen your understanding.
Keep this guide handy during revision. Nucleophilic substitution is not just about memorising reactions; it is about understanding electron flow and molecular structure. Master it, and you will score well in organic chemistry questions.
📚 Applications of Ammonia and Ammonium Compounds in A-Level Chemistry | A-Level化学：氨及铵化合物的应用
Ammonia and its related ammonium compounds occupy a central position in both industrial chemistry and laboratory practice. From the Haber process to fertiliser production, from the Ostwald process to qualitative analysis, these nitrogen-containing species exemplify how fundamental chemical principles translate into real-world applications. This article systematically reviews their uses within the Cambridge International A-Level Chemistry syllabus, highlighting the underlying chemistry at each step.
1. Industrial Synthesis by the Haber Process | 工业合成：哈伯法
Ammonia is manufactured on an industrial scale using the Haber process, in which nitrogen and hydrogen react reversibly under high pressure and moderate temperature. The balanced equation is N₂(g) + 3H₂(g) ⇌ 2NH₃(g), with ΔH = −92 kJ mol⁻¹. The process operates at around 450 °C and 200 atm, using a finely divided iron catalyst with promoters such as Al₂O₃ and K₂O.
The choice of conditions reflects a compromise. Although the forward reaction is exothermic, lower temperatures would slow the rate unacceptably. Higher pressures would shift the equilibrium toward ammonia but raise energy and safety costs. The ammonia is removed continuously by liquefaction, allowing unreacted gases to be recycled.
The largest single use of ammonia is in the manufacture of nitrogenous fertilisers. Plants require nitrogen for protein and nucleic acid synthesis, but most cannot absorb atmospheric N₂ directly. Fertiliser production therefore converts ammonia into forms that deliver nitrogen to crops efficiently.
Ammonium nitrate (NH₄NO₃): produced by reacting ammonia with nitric acid, NH₃ + HNO₃ → NH₄NO₃. It has a high nitrogen content (35%) and is widely used as a top dressing.
Urea (CO(NH₂)₂): synthesised from ammonia and carbon dioxide, 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O. Urea has the highest nitrogen content (46%) among common solid fertilisers and is slowly hydrolysed in soil.
Ammonium salts are particularly suitable because the NH₄⁺ ion is retained by negatively charged clay particles in soil, reducing leaching. However, over-application leads to environmental problems such as eutrophication, which will be discussed in Section 9.
3. Nitric Acid Production: The Ostwald Process | 硝酸生产：奥斯特瓦尔德法
Ammonia is the starting material for nitric acid, which is itself a vital industrial chemical. In the Ostwald process, ammonia is oxidised by air over a platinum-rhodium catalyst:
氨是硝酸的原料，而硝酸本身也是重要的工业化学品。在奥斯特瓦尔德法中，氨在铂铑催化剂上被空气氧化：
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)
The nitrogen monoxide is further oxidised to nitrogen dioxide, which is then absorbed in water to give nitric acid:
一氧化氮进一步氧化为二氧化氮，然后被水吸收得到硝酸：
2NO(g) + O₂(g) → 2NO₂(g)
3NO₂(g) + H₂O(l) → 2HNO₃(aq) + NO(g)
This three-stage route demonstrates the catalytic oxidation of ammonia, a key reaction type that appears frequently in examination questions. The nitric acid produced is used to make ammonium nitrate fertilisers, explosives, and various organic intermediates such as TNT and nylon precursors.
Although the Solvay process is traditionally associated with sodium carbonate production, ammonia plays an essential catalytic role. In this process, saturated brine is treated with ammonia and carbon dioxide. The key reaction sequence involves ammonium bicarbonate formation:
The sodium bicarbonate precipitates due to its limited solubility and is heated to produce sodium carbonate:
碳酸氢钠因溶解度有限而沉淀析出，经加热分解生成碳酸钠：
2NaHCO₃ → Na₂CO₃ + CO₂ + H₂O
The ammonia is regenerated when the ammonium chloride by-product is treated with calcium hydroxide, and the CO₂ from the decomposition step is recycled. Thus, ammonia is not consumed overall; it acts as a carrier that facilitates the conversion of sodium chloride into sodium carbonate.
In aqueous solution, ammonia behaves as a weak base. Its basicity arises from the lone pair of electrons on the nitrogen atom, which can accept a proton from water:
This equilibrium explains the alkaline pH of ammonia solutions and its ability to neutralise acids. It also underlies the use of ammonium salts as buffer components.
这一平衡解释了氨溶液的碱性pH值及其中和酸的能力，也说明了铵盐作为缓冲组分的用途。
Ammonia is also a classic ligand in coordination chemistry. Because nitrogen carries a lone pair, NH₃ can donate electrons to transition metal ions, forming complex ions:
氨也是配位化学中的经典配体。由于氮原子带孤对电子，NH₃可以向过渡金属离子提供电子，形成配合离子：
[Cu(NH₃)₄]²⁺: deep blue solution, formed when excess ammonia is added to copper(II) salts.
[Cu(NH₃)₄]²⁺：深蓝色溶液，由过量氨加入铜(II)盐时生成。
[Ag(NH₃)₂]⁺: the diamminesilver(I) ion used in Tollens’ reagent for testing aldehydes.
[Ag(NH₃)₂]⁺：二氨合银(I)离子，用于检验醛的托伦斯试剂。
[Fe(H₂O)₅(NH₃)]³⁺: illustrates ligand substitution in aquated iron(III) complexes.
[Fe(H₂O)₅(NH₃)]³⁺：展示了水合铁(III)配合物中的配体取代反应。
The formation of these complexes is accompanied by characteristic colour changes, making ammonia a valuable reagent in qualitative analysis.
这些配合物的形成伴随着特征颜色变化，使氨成为定性分析中的重要试剂。
6. Reducing Properties of Ammonia | 氨的还原性
Ammonia can act as a reducing agent. In the presence of catalysts such as copper(II) oxide or platinum, ammonia reduces hot metal oxides to the metals while being oxidised to nitrogen or nitrogen monoxide. This property is exploited both in the laboratory and in industrial pollution control (e.g., the selective catalytic reduction of NOₓ in exhaust gases).
The oxidation state of nitrogen in ammonia is −3. In this reaction, nitrogen is oxidised to 0 in N₂, confirming that ammonia has been oxidised. Conversely, in the catalytic oxidation to NO (Ostwald process), nitrogen is oxidised to +2. The ability of nitrogen to adopt multiple oxidation states is a recurring theme in nitrogen chemistry.
7. Thermal Decomposition of Ammonium Salts | 铵盐的热分解
Ammonium salts decompose on heating, and the nature of the decomposition depends on the anion. This is a classic examination topic that tests both chemical knowledge and the ability to predict reactions.
N₂O + 2H₂O (at 200–260 °C; explosive at higher T 在200–260 °C；高温下爆炸)
(NH₄)₂CO₃ (carbonate 碳酸铵)
2NH₃ + CO₂ + H₂O
The decomposition of ammonium nitrate deserves special attention. Under controlled heating it gives dinitrogen oxide (laughing gas), but at higher temperatures or under confinement it can decompose explosively. This dual behaviour explains both its use as a fertiliser and its misuse as an explosive.
A key laboratory application of ammonium compounds is their detection. The standard test involves adding a strong alkali such as sodium hydroxide to the test sample, followed by gentle warming. Ammonia gas is evolved and can be identified by:
Turning moist red litmus paper blue, indicating its basic nature.
使湿润的红色石蕊试纸变蓝，表明其碱性。
Producing white fumes of ammonium chloride when a glass rod dipped in concentrated HCl is held nearby: NH₃(g) + HCl(g) → NH₄Cl(s).
用蘸有浓盐酸的玻璃棒靠近时产生氯化铵白烟：NH₃(g) + HCl(g) → NH₄Cl(s)。
NH₄⁺(aq) + OH⁻(aq) → NH₃(g) + H₂O(l)
This test is required practical work in many A-Level syllabuses. The underlying principle is that hydroxide ions deprotonate the ammonium ion, converting it into molecular ammonia, which is volatile and escapes from the solution.
The widespread use of ammonia and ammonium compounds carries significant environmental consequences. Ammonium nitrate and other nitrogenous fertilisers can leach into waterways, causing eutrophication. This process stimulates excessive algal growth; subsequent decomposition depletes dissolved oxygen, leading to fish kills and loss of aquatic biodiversity.
Additionally, ammonia itself is a pollutant. Released into the atmosphere, it reacts with acidic species to form fine particulate matter, affecting air quality and human health. Agricultural emissions of NH₃ are a growing concern in environmental policy.
From a safety perspective, ammonia is toxic and corrosive. Its threshold limit value (TLV) is 25 ppm over an eight-hour working day. Concentrated ammonia solutions require careful handling with appropriate personal protective equipment. Ammonium nitrate poses fire and explosion hazards when stored improperly, particularly when contaminated with organic materials.
The applications of ammonia and ammonium compounds illustrate the connections between bonding, equilibria, acid-base chemistry, redox behaviour, and industrial processes. When revising this topic, focus on the following:
Be able to write and balance the Haber and Ostwald process equations, and explain the choice of conditions using Le Chatelier’s principle and kinetic arguments.
能够书写并配平哈伯法和奥斯特瓦尔德法的方程式，并运用勒夏特列原理和动力学论据解释条件的选择。
Know the thermal decomposition products of common ammonium salts, especially NH₄NO₃.
掌握常见铵盐的热分解产物，尤其是NH₄NO₃。
Understand the role of ammonia as both a weak base and a ligand, and be prepared to write equations for complex ion formation.
理解氨作为弱碱和配体的双重角色，并能够书写配合离子生成的方程式。
Remember that NH₄⁺ is detected by warming with alkali and identifying the evolved NH₃ — this appears in nearly every practical paper.
牢记NH₄⁺的检验方法：加碱加热并用适当方法鉴别产生的NH₃——这几乎出现在每份实验试卷中。
The table below summarises the key industrial reactions involving ammonia and their products.
下表总结了涉及氨的关键工业反应及其产物。
Process 工艺
Reactants 反应物
Products 产物
Key conditions 关键条件
Haber 哈伯法
N₂ + H₂
NH₃
450 °C, 200 atm, Fe catalyst
Ostwald 奥斯特瓦尔德法
NH₃ + O₂
NO → NO₂ → HNO₃
Pt-Rh catalyst, ~900 °C
Solvay 索尔维法
NaCl + CO₂ + NH₃ + H₂O
Na₂CO₃ (NH₃ regenerated 氨再生)
Ammonia as catalyst 氨作催化剂
Neutralisation 中和反应
NH₃ + HNO₃
NH₄NO₃ fertiliser 硝酸铵肥料
Acid-base reaction 酸碱反应
By mastering these applications, you not only prepare for direct recall questions but also develop the analytical skills needed for applying unfamiliar contexts — a hallmark of higher-mark examination questions. Always connect the chemistry back to structure, equilibria and energetics; this is what examiners reward.
Alkenes are hydrocarbons that contain a carbon-carbon double bond (C=C), which consists of one sigma (σ) bond and one pi (π) bond. The presence of this π bond makes alkenes significantly more reactive than alkanes, and their most characteristic chemical behaviour is the addition reaction. In this article, we will explore the mechanism, regiochemistry, and key examples of electrophilic addition reactions of alkenes, tailored specifically for the CIE A-Level Chemistry syllabus.
The carbon-carbon double bond in alkenes is formed by the overlap of atomic orbitals in two distinct ways. The sigma bond arises from the end-on overlap of two sp² hybrid orbitals, while the pi bond is formed by the side-on overlap of two unhybridised p orbitals. The electron density in the π bond is located above and below the plane of the molecule, making it exposed and readily available for interaction with electrophiles.
烯烃中的碳碳双键通过原子轨道的两种不同重叠方式形成。sigma 键由两个 sp² 杂化轨道的头碰头重叠形成，而 pi 键则由两个未杂化 p 轨道的肩并肩重叠形成。π 键的电子云分布在分子平面的上方和下方，使其暴露在外，易于与亲电试剂发生相互作用。
The π bond is weaker than the σ bond, with a bond energy of approximately 264 kJ mol⁻¹, compared to about 348 kJ mol⁻¹ for a C–C σ bond. However, the total bond energy of the C=C double bond (about 612 kJ mol⁻¹) is greater than that of a single bond, which explains why addition reactions are favoured over direct cleavage of the double bond.
2. The General Mechanism: Electrophilic Addition | 通用机理：亲电加成
The general mechanism of electrophilic addition to alkenes proceeds in two main steps. In the first step, the π electrons of the double bond attack an electrophile (E⁺), forming a new σ bond between the electrophile and one of the alkene carbon atoms. The other carbon atom acquires a positive charge, generating a carbocation intermediate.
In the second step, a nucleophile (Nu⁻) attacks the positively charged carbocation, forming a second σ bond. This completes the addition process, and both carbon atoms of the former double bond now have four single bonds, consistent with sp³ hybridisation.
For the CIE syllabus, typical electrophiles include H⁺ (from hydrogen halides or acids), Br⁺/Cl⁺ (from halogens), and H⁺ from water in acid-catalysed hydration. It is essential to draw the full curly arrow mechanism in examinations, showing the movement of electron pairs clearly.
When an alkene reacts with a hydrogen halide (HCl, HBr, or HI), the product is a haloalkane. The reaction can be carried out by bubbling the gaseous hydrogen halide through the liquid alkene, or by mixing them in an inert solvent. The reactivity of hydrogen halides increases in the order HCl < HBr < HI, which correlates with the increasing bond polarity and decreasing bond strength down the group.
当烯烃与卤化氢（HCl、HBr 或 HI）反应时，产物为卤代烷烃。该反应可以将气态卤化氢通入液态烯烃中进行，或在惰性溶剂中混合反应。卤化氢的反应活性按 HCl < HBr < HI 的顺序递增，这与键的极性增强和键能减小相关。
For example, ethene reacts with hydrogen bromide to form bromoethane:
例如，乙烯与溴化氢反应生成溴乙烷：
CH₂=CH₂ + HBr → CH₃–CH₂Br
The reaction proceeds via the attack of H⁺ on the double bond, forming a carbocation, followed by attack of Br⁻ on the carbocation. In the case of symmetrical alkenes like ethene, only one product is possible because the two carbon atoms are equivalent.
4. Markovnikov’s Rule and Regioselectivity | 马尔可夫尼科夫规则与区域选择性
When an unsymmetrical alkene such as propene reacts with a hydrogen halide, two possible products could form. However, in practice, one product dominates. Markovnikov’s rule states that when HX adds to an unsymmetrical alkene, the hydrogen atom attaches to the carbon atom of the double bond that already has the greater number of hydrogen atoms, and the halogen attaches to the carbon with fewer hydrogen atoms.
The reason for this regioselectivity lies in the stability of the carbocation intermediate. The addition of H⁺ to propene can generate either a primary carbocation (CH₃–CH₂–CH₂⁺) or a secondary carbocation (CH₃–CH⁺–CH₃). Secondary carbocations are more stable than primary carbocations due to the electron-donating inductive effect of alkyl groups, which helps to delocalise and stabilise the positive charge.
Carbocation stability follows the order: benzylic > allylic > tertiary > secondary > primary > methyl. This ordering is crucial for predicting the major product in electrophilic addition reactions.
Alkenes react readily with halogens at room temperature. For example, ethene decolourises bromine water (an orange-brown solution), forming 1,2-dibromoethane, a colourless liquid. This reaction is commonly used as a qualitative test for the presence of a C=C double bond in an unknown compound.
The mechanism involves the polarisation of the Br–Br bond as the π electrons approach the bromine molecule. The bromine molecule becomes polarised, and the electrophilic Br⁺ is attracted to the double bond. A cyclic bromonium ion intermediate is formed, which is more stable than an open carbocation because the positive charge is shared between the two carbon atoms.
The Br⁻ ion then attacks the bromonium ion from the opposite side, leading to anti addition. This means the two bromine atoms add to opposite faces of the double bond, an important stereochemical consequence of halogen addition.
Alkenes react with steam in the presence of an acid catalyst, typically concentrated phosphoric acid (H₃PO₄) or sulfuric acid (H₂SO₄), to form alcohols. This is an industrial method for producing ethanol from ethene, and the reaction is reversible and exothermic.
Industrial conditions for this process are typically a temperature of 300 °C, a pressure of 60–70 atm, and phosphoric acid supported on silica as the catalyst. A high yield of ethanol is obtained by continuously removing ethanol from the reaction mixture and recycling unreacted ethene.
The mechanism is similar to hydrogen halide addition: H⁺ protonates the double bond to form a carbocation, then water acts as a nucleophile to attack the carbocation, and finally deprotonation yields the alcohol. For unsymmetrical alkenes, Markovnikov’s rule applies, giving the more substituted alcohol as the major product.
Concentrated sulfuric acid adds to alkenes to form hydrogen sulfate esters (alkyl hydrogensulfates). For example, ethene reacts with cold concentrated H₂SO₄ to form ethyl hydrogensulfate (CH₃–CH₂–OSO₃H). This compound can be hydrolysed by warming with water to produce ethanol, providing an alternative route to alcohols.
This indirect hydration method was traditionally used in industry before the development of the direct catalytic hydration process. It also exemplifies the acid-catalysed addition of water, where the acid is regenerated at the end of the overall process. In the CIE examination, students should be able to write the equations for both steps and explain the role of sulfuric acid.
8. Addition of Hydrogen (Catalytic Hydrogenation) | 氢气的加成（催化加氢）
Alkenes react with hydrogen gas in the presence of a nickel, platinum, or palladium catalyst to form alkanes. This process is called catalytic hydrogenation. The reaction requires gentle heating (around 150 °C) with a finely divided nickel catalyst, although platinum and palladium are more active and can catalyse the reaction at room temperature.
烯烃在镍、铂或钯催化剂存在下与氢气反应生成烷烃，该过程称为催化加氢。反应通常需要在约 150 °C 下使用细粉状镍催化剂进行，而铂和钯的活性更高，可在室温下催化该反应。
CH₂=CH₂ + H₂ → CH₃–CH₃ (Ni catalyst, 150 °C)
Catalytic hydrogenation proceeds via a heterogeneous mechanism on the metal surface. The alkene and hydrogen molecules adsorb onto the catalyst surface, where the H–H bond breaks and hydrogen atoms are transferred to the double bond. Both hydrogen atoms add to the same face of the double bond, resulting in syn addition.
This reaction has significant industrial importance. For example, vegetable oils (unsaturated fats) are hydrogenated to produce margarine and solid cooking fats. Partial hydrogenation controls the degree of unsaturation and thus the physical properties of the final product.
Alkenes can undergo addition polymerisation, where thousands of monomer units join together to form a long-chain polymer. This reaction is initiated by free radicals or catalysts and involves the breaking of the π bond without the elimination of any small molecule.
For example, ethene polymerises to form poly(ethene), commonly known as polyethene or polythene:
例如，乙烯聚合生成聚（乙烯），俗称聚乙烯：
n CH₂=CH₂ → [–CH₂–CH₂–]ₙ
Substituted alkenes such as propene, chloroethene (vinyl chloride), and phenylethene (styrene) polymerise similarly to form poly(propene), poly(chloroethene) or PVC, and poly(phenylethene) or polystyrene, respectively. These polymers have a range of applications, from packaging materials to pipes and insulating foams.
The structure of the repeating unit must be identified correctly in examinations. For addition polymers, the repeating unit is the same as the monomer with the double bond opened. Students should be able to draw the repeating unit, deduce the monomer from a given polymer, and recognise that addition polymers are generally non-biodegradable due to their inert C–C and C–H bonds.
The table below summarises the key electrophilic addition reactions of alkenes that you need to know for the CIE A-Level Chemistry examination. Be sure to memorise the reagents, conditions, and products, and practise drawing curly arrow mechanisms for each reaction.
Note that the addition of bromine to alkenes is also used as a test for unsaturation. The orange-brown bromine water is decolourised when shaken with an alkene, producing a colourless dibromo compound. Alkanes, in contrast, do not decolourise bromine water under these conditions, which allows the two classes of hydrocarbons to be distinguished.
Students often lose marks in examinations due to careless errors in drawing mechanisms or misapplying Markovnikov’s rule. Here are the most common pitfalls and how to avoid them:
学生在考试中经常因绘制机理粗心或误用马尔可夫尼科夫规则而失分。以下是最常见的误区及避免方法：
Always show full curly arrows from the electron-rich centre to the electron-poor centre. A curly arrow starts at a bond or lone pair and points to the atom that accepts the electrons.
Do not write ‘Br⁺’ as a free ion in the mechanism for bromine addition. Instead, show the polarisation of the Br–Br bond and the formation of the bromonium ion.
在溴加成机理中不要将 Br⁺ 写为自由离子。应展示 Br–Br 键的极化以及溴鎓离子的形成。
Apply Markovnikov’s rule only to the addition of unsymmetrical electrophiles (HX, H₂O) to unsymmetrical alkenes. Halogen addition and hydrogenation are not governed by this rule.
In hydration, do not forget that the acid catalyst is regenerated. The overall stoichiometric equation shows only alkene + water → alcohol.
在水合反应中，不要忘记酸催化剂会再生。总化学计量方程式仅显示烯烃 + 水 → 醇。
When drawing polymer structures, show the repeating unit in square brackets with an ‘n’ subscript, and ensure that the bonds on either side extend beyond the brackets.
绘制聚合物结构时，用方括号展示重复单元并标注下标 ‘n’，且方括号两侧的键必须向外延伸。
Additionally, always consider the stereochemistry of the addition. Anti addition occurs in halogen addition via the bromonium ion pathway, while syn addition occurs in catalytic hydrogenation. Although the CIE syllabus does not always require detailed stereochemical analysis, understanding this concept can help you answer extension questions confidently.
Electrophilic addition reactions are one of the most important topics in A-Level organic chemistry. Understanding the structure of the double bond, the stepwise mechanism, and the factors that determine product distribution will allow you to tackle a wide range of examination questions with confidence. Remember to practise drawing mechanisms repeatedly, as muscle memory and accuracy are key to earning full marks in the written papers.
The reactions covered in this article — addition of hydrogen halides, halogens, steam, sulfuric acid, hydrogen, and polymerisation — are all recurring themes in CIE examination papers. Master them, understand the underlying principles, and you will be well-prepared for your examinations.
📚 A-Level Chemistry: Sources and Industrial Production of Alkanes | A-Level 化学：烷烃的来源与工业获取
Alkanes are saturated hydrocarbons that form the fundamental backbone of organic chemistry. In the CIE A-Level syllabus, understanding where alkanes come from and how industry obtains them on a massive scale is not just a theoretical exercise; it is a core application of thermodynamics, physical separation techniques, and reaction kinetics. This article systematically explores the natural reservoirs of alkanes, the engineering principles behind crude oil fractionation, and the chemical transformations—cracking and reforming—that reshape these molecules to meet global energy and chemical demands.
1. Natural Sources: Natural Gas and Crude Oil | 天然来源：天然气与原油
The primary industrial sources of alkanes are natural gas and crude oil (petroleum). Natural gas consists predominantly of methane (CH₄), typically accounting for 70-90% of its composition, with smaller amounts of ethane (C₂H₆), propane (C₃H₈), and butane (C₄H₁₀). It is formed over millions of years from the anaerobic decomposition of marine organisms buried deep beneath sedimentary rocks. Because methane has a very low boiling point (−161 °C), it remains a gas at ambient conditions and is extracted via drilling, then transported through pipelines or liquefied under high pressure for shipping.
Crude oil, in contrast, is a thick, dark-brown liquid containing a complex mixture of hundreds of different hydrocarbons, including straight-chain alkanes (CnH₂n₊₂), branched alkanes, cycloalkanes, and aromatic hydrocarbons. The exact composition varies drastically depending on the geographical source. For example, Nigerian crude oil is rich in light, short-chain hydrocarbons, while Venezuelan crude oil contains a higher proportion of heavy, long-chain residues. This variability makes the refining process both challenging and economically fascinating, as refineries must be highly adaptable to different feedstock qualities.
2. Fractional Distillation: The Fundamental Separation Principle | 分馏：基础的分离原理
Fractional distillation exploits the difference in boiling points of hydrocarbons to separate crude oil into simpler, more commercially useful fractions. When crude oil is heated to approximately 400 °C, all hydrocarbons with boiling points below this temperature vaporise. The resulting vapour is fed into a tall fractional distillation column, where a temperature gradient is maintained—hot at the bottom (around 360 °C) and cool at the top (around 20 °C). As the vapour ascends, cooler hydrocarbons with higher boiling points condense on trays at specific heights, while lighter hydrocarbons continue to rise as gases.
Each tray in the column is fitted with bubble caps that force the rising vapour to bubble through the liquid on the tray. This ensures intimate gas-liquid contact, allowing volatile components to re-vaporise and less volatile components to remain condensed. This repeated condensation and re-vaporisation process, known as equilibrium staging, dramatically improves the purity of each separated fraction compared to simple distillation. The fractions are drawn off as liquids from the side of the column, while the very lightest gases such as methane (CH₄) and ethane (C₂H₆) are collected from the top.
3. Physical vs Chemical: The Nature of Fractional Distillation | 物理与化学：分馏的本质
A common conceptual error among students is to confuse fractional distillation with a chemical reaction. Fractional distillation is purely a physical separation process. No covalent bonds are broken or formed during the operation; the hydrocarbon molecules retain their original identity. The only intermolecular forces at play are London dispersion forces (instantaneous induced dipole-dipole interactions). The strength of these forces increases with the size of the molecule, which directly explains why longer-chain alkanes have higher boiling points.
For a molecule like octane (C₈H₁₈), the larger electron cloud and greater surface area allow for more significant temporary dipoles, resulting in stronger attractions between adjacent molecules. In contrast, propane (C₃H₈) has a smaller electron cloud, weaker dispersion forces, and thus a much lower boiling point (−42 °C). This fundamental principle allows the fractionating column to act as a giant filtering device, sorting hydrocarbons by molar mass and molecular size.
4. Industrial Fractions and Their Applications | 工业馏分及其应用
The crude oil fractionating column yields several distinct fractions, each defined by a specific carbon-chain length range and boiling point range. Understanding these fractions and their primary industrial applications is a crucial exam requirement. The table below summarises the key fractions in order of increasing boiling point from the top to the bottom of the column.
Domestic cooking gas (LPG), heating, chemical feedstock for hydrogen production
Gasoline / Petrol (汽油)
C₄ to C₁₂
40 to 100
Fuel for internal combustion engines; rich in octane isomers
Naphtha (石脑油)
C₆ to C₁₂
90 to 150
Chemical feedstock for catalytic reforming to produce aromatic hydrocarbons and hydrogen
Kerosene / Paraffin (煤油)
C₁₀ to C₁₆
150 to 250
Aviation jet fuel, heating oil, solvent for paraffin wax
Diesel Oil / Gas Oil (柴油)
C₁₄ to C₂₀
250 to 350
Fuel for diesel engines, central heating systems
Residual Fuel Oil / Bitumen (重油 / 沥青)
C₂₀ and above
above 350
Heavy fuel for ships, roofing materials, road surfacing (asphalt)
The demand for these fractions is highly imbalanced. In modern economies, there is an enormous demand for the lighter fractions, particularly gasoline (C₄-C₁₂) and naphtha (C₆-C₁₂), as transportation fuels and as raw materials for the petrochemical industry. However, natural crude oil typically yields a much higher proportion of heavier, longer-chain hydrocarbons. This mismatch between supply and demand creates the economic imperative for the conversion processes discussed below.
5. The Need for Cracking: Adjusting the Hydrocarbon Chain Length | 裂化的必要性：调整烃链长度
Cracking is the process of breaking larger hydrocarbon molecules into smaller, more useful ones. Since the demand for shorter-chain alkanes (like petrol) far exceeds the natural supply from fractional distillation, refineries use cracking to convert surplus long-chain alkanes (C₁₅-C₅₀) into a mixture of shorter alkanes (C₅-C₁₂) and alkenes (such as ethene and propene). This process is critically important because it not only balances the fuel market but also supplies essential monomers for the plastics and polymers industry.
There are two principal industrial methods of cracking: thermal cracking (thermal cracking) and catalytic cracking. Each has distinct mechanisms, operating conditions, and product distributions that are directly examined in the A-Level syllabus. We will explore these methods in the next two sections, paying particular attention to the free-radical mechanism of thermal cracking and the ionic/carbocation mechanism promoted by acidic zeolite catalysts.
6. Thermal Cracking: Conditions and Free-Radical Mechanism | 热裂化：条件与自由基机理
Thermal cracking is conducted at extremely high temperatures of approximately 700 to 1200 K (about 400 to 900 °C) and high pressures of 70 atmospheres or more. Under these drastic conditions, carbon-carbon bonds in large alkane molecules undergo homolytic fission, generating highly reactive free radicals. The severe conditions are required to overcome the very high activation energy (Ea) associated with breaking strong C–C single bonds (about 350 kJ/mol). Because this process requires substantial energy input, it tends to be more expensive and less selective than catalytic cracking.
The overall equation for a typical thermal cracking reaction can be represented as follows, where a long-chain alkane (decane) decomposes into a shorter alkane and an alkene:
典型热裂化反应的总方程式可表示如下，其中长链烷烃（癸烷）分解为较短的烷烃和烯烃：
C₁₀H₂₂ → C₈H₁₈ + C₂H₄
癸烷 → 辛烷 + 乙烯
The mechanism proceeds via several steps: initiation, where a C–C bond breaks homolytically to form two alkyl radicals (R•); propagation, where a radical abstracts a hydrogen atom from another alkane molecule to form a new molecule and a new radical, or undergoes β-scission to split into a smaller alkene and a smaller radical; and termination, where two radicals combine to form a stable molecule. This mechanism produces a wide distribution of products, including both straight-chain and branched hydrocarbons.
7. Catalytic Cracking: The Modern Industrial Standard | 催化裂化：现代工业标准
Catalytic cracking operates at milder conditions compared to thermal cracking, typically around 450 °C and near-atmospheric pressure. The catalyst used is an acidic zeolite, often based on silica-alumina (SiO₂-Al₂O₃). The zeolite catalyst provides a highly porous, acidic environment that promotes cleavage of C–C bonds via an ionic mechanism involving carbocation intermediates. The use of a catalyst significantly lowers the activation energy, reducing the energy cost and improving the process efficiency.
A typical catalytic cracking reaction can be represented by the equation below, where dodecane is converted into hexane and hexene:
典型的催化裂化反应可由下面的方程式表示，其中十二烷转化为己烷和己烯：
C₁₂H₂₆ → C₆H₁₄ + C₆H₁₂
十二烷 → 己烷 + 己烯
The carbocation mechanism allows for isomerisation of the carbon skeleton prior to fragmentation, which results in a higher proportion of branched-chain alkanes in the gasoline fraction. Branched alkanes have higher octane numbers than their straight-chain counterparts, meaning they combust more smoothly in petrol engines without premature detonation (engine knocking). Therefore, catalytic cracking produces higher-quality gasoline than thermal cracking, in addition to being more energy-efficient and producing fewer gaseous by-products.
While cracking breaks large molecules into smaller ones, catalytic reforming is a process that upgrades straight-chain alkanes (typically C₆-C₁₀ naphtha) into branched-chain alkanes, cycloalkanes (cycloalkanes), and aromatic hydrocarbons (arenes). This is achieved by passing the naphtha fraction over a bifunctional catalyst containing platinum on an alumina support at high temperatures (approximately 500 °C) and moderate pressures. The key chemical transformations include isomerisation (straight chain → branched chain), cyclisation (straight chain → cycloalkane) and aromatisation (cycloalkane → arene).
For example, the reforming of n-hexane (C₆H₁₄) can produce cyclohexane (C₆H₁₂), which can be further dehydrogenated to form benzene (C₆H₆). Similarly, n-heptane can be converted into toluene (C₇H₈). These reforming reactions are crucial because both branched alkanes and aromatic hydrocarbons have much higher octane numbers than straight-chain alkanes. Aromatics are also valuable feedstocks for the production of polymers, dyes, and pharmaceuticals.
9. Key Industrial Applications of Alkane Processing | 烷烃加工的主要工业应用
The industrial processing of alkanes is not solely concerned with producing fuels; it is the foundational source of many essential chemical feedstocks. The alkenes generated from cracking, particularly ethene (C₂H₄) and propene (C₃H₆), are the most important organic building blocks in the petrochemical industry. Ethene is polymerised to form poly(ethene) (polythene), used extensively in packaging and containers, while propene is polymerised to poly(propene) (polypropylene). These processes link the oil refinery directly to the plastics manufacturing sector.
Furthermore, the hydrogen gas generated simultaneously during catalytic reforming is a valuable industrial commodity. It is extensively used in the Haber process for the synthesis of ammonia (NH₃), in the hydrogenation of unsaturated vegetable oils (e.g., converting margarine oils into solid fats), and in hydrodesulfurisation—a process that removes sulfur impurities from crude oil fractions to reduce atmospheric SO₂ pollution. Thus, the refinery functions as a complex, interconnected network where every product stream has downstream economic value.
10. Exam Focus: Conditions, Equations, and Common Pitfalls | 考试焦点：条件、方程式和常见误区
When answering CIE A-Level questions on this topic, students must be precise about conditions. Thermal cracking requires a temperature of 700-1200 K and a high pressure of up to 70 atm. Catalytic cracking requires a lower temperature of about 450 °C, near-atmospheric pressure, and a zeolite catalyst. Catalytic reforming requires a platinum catalyst at approximately 500 °C. Mixing up these specific conditions is a frequent cause of lost marks. Always state both temperature and pressure, and name the catalyst when one is used.
📚 A-Level Chemistry: Typical Chemical Reactions of Alkanes | A-Level 化学：烷烃的典型化学反应
Alkanes are saturated hydrocarbons with the general formula CₙH₂ₙ₊₂. They are often described as ‘chemically inert’ because they react with very few reagents under normal conditions. However, their typical reactions—combustion, free radical substitution, cracking, and reforming—are central to both industrial chemistry and A-Level examinations. This article provides a comprehensive and exam-focused review of these characteristic reactions, complete with mechanisms, conditions, and common pitfalls.
Each carbon atom in an alkane is sp³ hybridised and forms four sigma (σ) bonds arranged tetrahedrally around the carbon centre. The C–C and C–H bonds are both single sigma bonds. The electronegativity difference between carbon (2.5) and hydrogen (2.1) is very small (about 0.4), so the C–H bond has very low polarity.
The C–C bond enthalpy is about 348 kJ mol⁻¹ and the C–H bond enthalpy is about 412 kJ mol⁻¹. These strong, non-polar, localised sigma bonds mean that alkanes are not attacked by nucleophiles, electrophiles, acids, or bases under normal conditions. No π electrons are available to interact with approaching reagents.
2. General Unreactivity: Why Alkanes Are ‘Inert’ | 为何烷烃”惰性”
Strong C–C and C–H sigma bonds are difficult to break; the activation energy for bond fission is very high.
C–C 和 C–H 强 σ 键难以断裂；键断裂所需的活化能非常高。
Low polarity: there are no obvious electron-rich or electron-poor sites on the molecule to attract attacking species.
极性低：分子上没有明显的电子富集区或电子贫乏区来吸引进攻物种。
No π bonds are present, so addition reactions (typical of alkenes) are impossible.
不存在 π 键，因此无法发生烯烃特有的加成反应。
Alkanes have no lone pairs and no vacant orbitals, so they cannot act as ligands or Lewis acids/bases.
烷烃既无孤对电子，也无空轨道，因此不能充当配体或路易斯酸碱。
As a result, the reactions that alkanes do undergo typically require extreme conditions—high temperatures, UV light, or a catalyst. The most important of these are combustion and free radical substitution.
Alkanes burn readily in excess oxygen to produce carbon dioxide and water. This is a highly exothermic process, which is exactly why alkanes are important fuels in domestic and industrial settings.
📚 A-Level Chemistry: Structure and Properties of the Alkane Homologous Series | A-Level 化学：烷烃同系物的结构与性质
Alkanes are the simplest family of organic compounds and form the foundation of the homologous series in organic chemistry. Understanding their structure and properties is essential for tackling CIE A-Level Chemistry questions on bonding, isomerism, reaction mechanisms, and industrial applications.
1. The Homologous Series and General Formula | 同系物与通式
An homologous series is a family of compounds with the same general formula, similar chemical properties, and successive members differing by a CH₂ unit. Alkanes fit this definition perfectly, with the general formula CₙH₂ₙ₊₂ for open-chain (acyclic) alkanes.
For example: CH₄ (methane, n = 1), C₂H₆ (ethane, n = 2), C₃H₈ (propane, n = 3), and C₄H₁₀ (butane, n = 4). Each member can be derived from the previous one by replacing a hydrogen atom with a methyl group, adding one carbon and two hydrogen atoms.
A common exam trap is confusing alkanes with cycloalkanes. Cycloalkanes have the general formula CₙH₂ₙ and are not part of the alkane homologous series in the strict CIE sense, although they share some similar reactions.
Each carbon atom in an alkane is sp³ hybridised, forming four sigma (σ) bonds directed towards the corners of a regular tetrahedron. The bond angle is approximately 109.5°, which minimises electron-pair repulsion according to VSEPR theory.
The C–C and C–H bonds are both sigma bonds, formed by the head-on overlap of atomic orbitals. The electron density in a σ bond lies directly between the two nuclei, making the bond relatively strong and freely rotating around the C–C axis.
Because the electronegativity difference between carbon (2.55) and hydrogen (2.20) is small, the C–H bond is essentially non-polar. As a result, alkane molecules have no permanent dipole and are non-polar overall.
Isomerism appears from C₄H₁₀ onwards. Butane has two isomers: n-butane (CH₃CH₂CH₂CH₃) and isobutane ((CH₃)₂CHCH₃). The number of possible constitutional isomers increases rapidly with carbon number.
According to IUPAC rules, the name of an alkane is derived from the longest continuous carbon chain, with the suffix “-ane”. Substituents (alkyl groups) are named by replacing “-ane” with “-yl”, such as methyl (-CH₃) and ethyl (-C₂H₅).
Select the longest chain containing the maximum number of carbon atoms as the parent chain.
选择含碳原子数最多的最长碳链作为主链。
Number the chain from the end giving the lowest locant to the first substituent.
从离第一个取代基最近的一端开始编号，使取代基的位次最小。
List substituents alphabetically (not by size), using di-, tri-, etc. for identical groups, and separate numbers by commas and numbers from names by hyphens.
For example, CH₃CH(CH₃)CH₂CH₃ is named 2-methylbutane, not 3-methylbutane. A common mark scheme requirement is to write the locant and substituent in the correct order.
4. Physical Properties: Trends Down the Series | 物理性质：沿同系物的递变规律
The physical properties of alkanes change gradually with increasing molecular mass, reflecting the increasing strength of intermolecular forces (London dispersion forces). These forces arise from temporary induced dipoles due to electron movement.
Boiling point: As the number of carbon atoms increases, the molecular surface area grows, leading to stronger dispersion forces and a higher boiling point. For isomeric alkanes, the straight-chain isomer has a higher boiling point than its branched counterpart because branching reduces the surface area for contact.
Melting point: Similar to boiling point, but a zig-zag “sawtooth” pattern is observed: alkanes with an even number of carbon atoms pack more efficiently in the solid state, giving higher melting points.
熔点：与沸点相似，但呈现”锯齿形”规律：含偶数个碳原子的烷烃在固态中堆积更紧密，因此熔点相对较高。
Alkane
Formula
Boiling point / °C
Methane
CH₄
-164
Ethane
C₂H₆
-89
Propane
C₃H₈
-42
Butane
C₄H₁₀
-0.5
Pentane
C₅H₁₂
36
Alkanes are insoluble in water because water is polar and highly hydrogen-bonded, whereas alkanes are non-polar and cannot form hydrogen bonds with water. They dissolve readily in non-polar organic solvents and are less dense than water.
5. Bond Energies and Chemical Stability | 键能与化学稳定性
The C–C bond energy is about 348 kJ mol⁻¹ and the C–H bond energy is about 412 kJ mol⁻¹. These are relatively strong covalent bonds, which explains the general unreactivity of alkanes at room temperature towards acids, bases, and oxidising agents.
Alkanes are often described as “saturated” hydrocarbons because they contain only single C–C and C–H bonds. This saturation, combined with the non-polar nature of the bonds, means alkanes do not undergo electrophilic addition reactions like alkenes do.
However, alkanes are not completely inert. Under suitable conditions — high temperature, ultraviolet light, or in the presence of oxygen — they react readily, especially through combustion and free-radical substitution.
Complete combustion of alkanes in excess oxygen produces carbon dioxide and water, releasing a large amount of energy. This is the basis for their use as fuels.
烷烃在过量氧气中完全燃烧生成二氧化碳和水，并释放大量能量。这是它们用作燃料的基础。
CH₄ + 2O₂ → CO₂ + 2H₂O
CH₄ + 2O₂ → CO₂ + 2H₂O
Incomplete combustion occurs when the oxygen supply is limited, producing carbon monoxide (CO) and/or carbon (soot) instead of CO₂. Carbon monoxide is toxic because it binds to haemoglobin more strongly than oxygen.
The general equation for complete combustion of an alkane is CₙH₂ₙ₊₂ + (3n+1)/2 O₂ → nCO₂ + (n+1)H₂O. Exam questions often ask you to balance such equations working from the general formula.
Under ultraviolet light, alkanes react with halogens — especially chlorine and bromine — in a free-radical substitution reaction. This is a key mechanism tested in CIE A-Level Chemistry.
The product is not a single chloromethane; further substitution leads to a mixture: CH₃Cl, CH₂Cl₂, CHCl₃, and CCl₄. In the exam, you must be able to identify the initiation step (homolytic fission) and explain why a mixture forms.
Relative reactivity of halogens decreases in the order F₂ > Cl₂ > Br₂ > I₂. Fluorine is dangerously explosive; iodine does not react under normal conditions. When alkanes have more than one type of hydrogen, substitution occurs preferentially at tertiary (3°) C–H bonds, then secondary (2°), then primary (1°), due to radical stability.
Alkanes are obtained primarily from crude oil and natural gas. Large alkane molecules are often converted into more valuable smaller molecules through cracking, which involves breaking C–C bonds at high temperatures and/or with a catalyst.
Thermal cracking uses high temperature (about 700–1200 K) and high pressure to produce mostly alkenes, especially ethene. Catalytic cracking uses a zeolite catalyst at lower temperatures (about 700 K) to produce branched-chain alkanes and aromatic hydrocarbons for high-octane petrol.
Cracking also produces hydrogen gas, which is used in the Haber process and hydrogenation of unsaturated oils. Understanding the purpose of cracking — matching supply of fractions to demand — is a common AO1/AO2 question.
In addition to natural sources, alkanes can be prepared in the laboratory and in industry by hydrogenation of alkenes. Ethene reacts with hydrogen over a nickel catalyst at about 150 °C to form ethane.
除了天然来源，实验室和工业上还可以通过烯烃的加氢反应制备烷烃。乙烯在镍催化剂作用下，于约 150 °C 与氢气反应生成乙烷。
CH₂=CH₂ + H₂ → CH₃CH₃ (Ni catalyst, heat)
CH₂=CH₂ + H₂ → CH₃CH₃（镍催化剂，加热）
Reduction of haloalkanes with zinc and hydrochloric acid also yields alkanes: R–X + 2[H] → R–H + HX. This method is sometimes shown in synthetic routes, though it is less important at A-Level than catalytic hydrogenation.
In a synthetic pathway question, you should recognise that converting an alkene to an alkane requires H₂/Ni, converting a haloalkane to an alkane may require reduction, and converting an alcohol to an alkane is not straightforward at A-Level.
10. Environmental and Safety Considerations | 环境与安全考量
Methane, the simplest alkane, is a potent greenhouse gas. Its release from agriculture, landfill sites, and fossil-fuel extraction contributes to global warming. Carbon dioxide from alkane combustion is also a major greenhouse gas.
Incomplete combustion produces toxic carbon monoxide, and nitrogen oxides are formed in vehicle engines at high temperatures. Alkanes are also volatile organic compounds (VOCs) that can contribute to photochemical smog in urban areas.
The octane number of a fuel relates to its resistance to knocking in a petrol engine. Branched-chain alkanes and aromatic compounds have higher octane numbers than straight-chain alkanes, which is why catalytic cracking is valuable for petrol production.
11. Common Exam Traps and Revision Tips | 常见考试陷阱与复习要点
Many students lose marks by making avoidable errors in alkane questions. Here are the most frequent pitfalls:
许多学生在烷烃题目中因可避免的错误而失分。以下是最常见的陷阱：
Using the wrong general formula: remember CₙH₂ₙ₊₂ for acyclic alkanes, not CₙH₂ₙ.
使用错误的通式：记住开链烷烃是 CₙH₂ₙ₊₂，不是 CₙH₂ₙ。
Forgetting that alkanes are non-polar and insoluble in water; do not attribute solubility to hydrogen bonding.
忘记烷烃是非极性且不溶于水；不要将其溶解性归因于氢键。
Writing the propagation steps with arrows that do not show electron movement or radicals correctly — radicals always have an unpaired electron (Cl•, •CH₃).
Claiming that alkanes undergo addition reactions because they are hydrocarbons: saturation means no π bond, so addition is impossible.
声称烷烃因是烃类而能发生加成反应：饱和意味着没有 π 键，因此不可能发生加成反应。
Confusing the order of boiling points during isomerism: branching lowers the boiling point, not raises it.
比较异构体沸点时混淆顺序：支链化降低沸点，而不是升高。
When revising, draw the displayed formulae and consider the 3D shape at each carbon atom. Practise naming compounds from structural formulae and drawing isomers from molecular formulae.
The non-polar, saturated, tetrahedral structure of alkanes determines their physical inertness and their chemical behaviour. Weak intermolecular forces explain low boiling points for small alkanes, while strong C–H and C–C bonds explain their low reactivity.
Under suitable activation — heat, light, or catalysts — alkanes undergo combustion, free-radical substitution, and cracking. These reactions underpin their role as fuels and as feedstocks for the petrochemical industry.
For CIE A-Level exams, master the general formula, IUPAC naming, the three-step free-radical mechanism, and the trends in physical properties. These core ideas will allow you to solve both structured and multiple-choice questions with confidence.