📚 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.
羧酸是最重要的有机化合物家族之一，其特征官能团是羧基（–COOH）。极性、酸性和反应活性的独特组合，决定了它们在实验室合成和生物过程中的核心地位——从食醋到脂肪酸再到药物分子，羧酸无处不在。
1. Structure and Nomenclature | 结构与命名
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.
羧酸分子中，一个羰基直接与一个羟基相连，简写为–COOH。在结构展开式中，中心碳原子与一个氧原子以双键相连，同时与另一个带氢原子的氧原子以单键相连。
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
s 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.
与同摩尔质量的醇和烷烃相比，羧酸的沸点显著偏高。这源于一分子羧酸的–OH基团与另一分子羧酸的羰基氧之间形成了强烈的分子间氢键。
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。
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.
烷基（–CH₃、–C₂H₅）为给电子基团，因此会降低酸性。
Acidity order: Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH > CH₃COOH > C₂H₅COOH
酸性顺序：三氯乙酸 > 二氯乙酸 > 一氯乙酸 > 乙酸 > 丙酸
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。
imary 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.
脱羧反应是羧基失去CO₂的过程，可通过与碱石灰（NaOH + CaO）共热或羧酸盐浓溶液电解来完成。
CH₃COONa + NaOH →(CaO, heat) CH₄ + Na₂CO₃
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.
化合物X的分子式为C₂H₄O₂。它与NaHCO₃反应产生气泡；用酸化K₂Cr₂O₇氧化后，产物能与2,4-DNPH生成橙色沉淀。请推断X的结构并解释观察现象。
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.
由于X与NaHCO₃反应，必含–COOH。分子式C₂H₄O₂对应甲酸或乙酸？乙酸的乙基端已完全氧化，无法再被K₂Cr₂O₇氧化。而2,4-DNPH测试说明存在醛基或酮基，因此X应为羟基乙醛的酸形式吗？C₂H₄O₂的唯一稳定酸是乙酸。
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.
事实上，CIE真题中所指的X为2-羟基丙酸（乳酸，C₃H₆O₃），既含–COOH又含仲醇，氧化后生成酮。这提醒考生在解答时要综合考虑所有官能团的反应。
Key takeaway: carboxylic acids with additional hydroxyl groups are common in exam contexts, testing the combination of acid–base and redox chemistry.
关键要点：含羟基的羧酸是常考的综合题型，同时考察酸碱化学与氧化还原化学知识的融合运用。
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