📚 A-Level Chemistry: Structure and Properties of Amino Acids | A-Level 化学：氨基酸的结构与性质
Amino acids are fundamental organic molecules that serve as the building blocks of proteins and play a central role in the CIE A-Level Chemistry syllabus. This article provides a complete and systematic review of the structure, properties, and reactions of amino acids, with every concept paired in English and Chinese to support bilingual learning.
氨基酸是构成蛋白质的基本有机分子，也是CIE A-Level 化学考纲中的核心内容。本文将系统完整地梳理氨基酸的结构、性质与反应，每个知识点均配以中英双语讲解，助力双语学习。
1. General Structure of Amino Acids | 氨基酸的通式结构
An amino acid is an organic compound that contains both an amino group (-NH₂) and a carboxyl group (-COOH). For A-Level purposes, we focus on the α-amino acids, in which both functional groups are attached to the same carbon atom, called the α-carbon. The general structural formula is H₂N-CH(R)-COOH, where R represents a variable side chain that distinguishes one amino acid from another.
氨基酸是同时含有氨基（-NH₂）和羧基（-COOH）的有机化合物。在A-Level 阶段，我们重点关注α-氨基酸，即两个官能团连接在同一个碳原子（称为α-碳）上。通式为 H₂N-CH(R)-COOH，其中R为可变的侧链基团，不同氨基酸的区别正在于R。
H₂N—CH(R)—COOH
The α-carbon is bonded to four different groups: an amino group, a carboxyl group, a hydrogen atom, and an R group. This tetrahedral arrangement is significant because, except for glycine, the α-carbon is a chiral centre, giving rise to optical isomerism, which will be discussed in Section 5.
α-碳连接四个不同基团：氨基、羧基、氢原子和R基团。这一四面体排列意义重大，因为除甘氨酸外，α-碳均为手性中心，从而产生光学异构现象，我们将在第5节详细讨论。
2. The Zwitterion | 两性离子
A distinctive property of amino acids is their existence as zwitterions in aqueous solution. At or near neutral pH, the carboxyl group loses a proton to form a carboxylate ion (-COO⁻), while the amino group gains a proton to form an ammonium ion (-NH₃⁺). The molecule thus carries both a positive and a negative charge simultaneously, yet its overall charge is zero.
氨基酸的一个显著特性是在水溶液中以两性离子形式存在。在中性pH左右，羧基失去一个质子形成羧酸根离子（-COO⁻），氨基获得一个质子形成铵离子（-NH₃⁺）。分子同时带有正电荷和负电荷，但整体净电荷为零。
H₃N⁺—CH(R)—COO⁻
The zwitterion structure explains why amino acids have relatively high melting points and are much more soluble in water than in non-polar organic solvents. The strong electrostatic interactions between zwitterions in the solid state give amino acids a salt-like, crystalline character. This is a frequently examined point in CIE structured questions.
两性离子结构解释了氨基酸熔点较高、在水中的溶解度远大于在非极性有机溶剂中溶解度的原因。固态时两性离子间的强静电相互作用使氨基酸呈现类似盐的结晶性质。这是CIE简答题中经常考查的知识点。
3. Acid-Base Behaviour | 酸碱行为
Because amino acids contain both an acidic carboxyl group and a basic amino group, they are amphoteric: they can react with both acids and bases. When treated with a strong acid such as hydrochloric acid, the carboxylate group of the zwitterion accepts a proton, producing a positively charged cation.
由于氨基酸同时含有酸性羧基和碱性氨基，因此具有两性：既能与酸反应，也能与碱反应。当与盐酸等强酸反应时，两性离子的羧酸根接受一个质子，生成带正电荷的阳离子。
H₃N⁺—CH(R)—COO⁻ + H⁺ → H₃N⁺—CH(R)—COOH
When treated with a strong base such as sodium hydroxide, the ammonium group donates a proton to the hydroxide ion, producing a negatively charged anion.
当与氢氧化钠等强碱反应时，铵基向氢氧根离子提供质子，生成带负电荷的阴离子。
H₃N⁺—CH(R)—COO⁻ + OH⁻ → H₂N—CH(R)—COO⁻ + H₂O
Students should be able to write both equations and state the ionic form present under acidic, neutral, and basic conditions. This amphoteric behaviour is analogous to that of other bifunctional molecules studied elsewhere in the syllabus.
学生应能写出上述两个方程式，并说明在酸性、中性和碱性条件下氨基酸分别以何种离子形式存在。这种两性行为与考纲中其他双官能团分子的性质具有类比性。
4. Isoelectric Point | 等电点
The isoelectric point (pI) is the pH at which an amino acid exists predominantly as the neutral zwitterion and therefore carries no net electrical charge. At this pH, the amino acid does not migrate in an electric field, which is the principle behind electrophoretic separation.
等电点（pI）是氨基酸主要以中性两性离子形式存在、净电荷为零时的pH值。在等电点pH下，氨基酸在电场中不发生迁移，这正是电泳分离的基本原理。
At pH values below the pI, the amino acid carries a net positive charge because the carboxylate group is protonated in the acidic environment. At pH values above the pI, it carries a net negative charge because the ammonium group loses a proton in the basic environment.
当pH低于pI时，酸性环境中羧酸根被质子化，氨基酸带净正电荷。当pH高于pI时，碱性环境中铵基失去质子，氨基酸带净负电荷。
For an amino acid with a neutral side chain, the isoelectric point can be calculated from the two pKₐ values using the following formula:
对于侧链中性的氨基酸，可通过两个pKₐ值计算等电点，公式如下：
pI = (pKₐ₁ + pKₐ₂) / 2
where pKₐ₁ corresponds to the carboxyl group and pKₐ₂ corresponds to the ammonium group. The concept of pI is essential for understanding how amino acids behave in electrophoresis and ion-exchange chromatography.
其中pKₐ₁对应羧基，pKₐ₂对应铵基。等电点概念对于理解氨基酸在电泳和离子交换层析中的行为至关重要。
5. Optical Isomerism | 光学异构
With the exception of glycine, all α-amino acids contain a chiral centre at the α-carbon because it is bonded to four different groups: -NH₂, -COOH, -H, and -R. As a result, each of these amino acids exists as a pair of enantiomers, which are non-superimposable mirror images. The two enantiomers rotate plane-polarised light in opposite directions and are designated as the L- and D-forms.
除甘氨酸外，所有α-氨基酸的α-碳均连接四个不同基团：-NH₂、-COOH、-H和-R，因此是手性中心。由此，每种氨基酸都存在一对对映异构体，互为不可重叠的镜像。两种对映体使平面偏振光的旋光方向相反，分别称为L型和D型。
Naturally occurring amino acids in proteins are almost exclusively L-amino acids. This is a useful fact for exam answers discussing biological specificity. Glycine, however, is optically inactive because its R group is simply a hydrogen atom, so the α-carbon is bonded to two identical hydrogen atoms and is not chiral.
天然蛋白质中的氨基酸几乎全部为L型氨基酸。这一事实在回答生物学特异性相关题目时很有用。然而，甘氨酸无光学活性，因为其R基团只是一个氢原子，α-碳连接两个相同氢原子，不是手性中心。
6. Peptide Bond Formation | 肽键的形成
Amino acids join together through condensation reactions to form peptides and proteins. The carboxyl group of one amino acid reacts with the amino group of a second amino acid, eliminating a water molecule and forming an amide (peptide) bond, -CONH-. The product is a dipeptide; further condensations yield tripeptides, polypeptides, and proteins.
氨基酸通过缩合反应连接形成多肽和蛋白质。一个氨基酸的羧基与另一个氨基酸的氨基反应，脱去一个水分子，形成酰胺键（肽键）-CONH-。产物为二肽；继续缩合可生成三肽、多肽和蛋白质。
H₂N—CH(R₁)—COOH + H₂N—CH(R₂)—COOH → H₂N—CH(R₁)—CONH—CH(R₂)—COOH + H₂O
This reaction is an example of condensation polymerisation, and it directly connects the chemistry of amino acids to synthetic polyamides such as nylon, which appear elsewhere in the CIE syllabus. When writing exam answers, be sure to state that a water molecule is eliminated and that the linkage formed is an amide/peptide bond.
该反应是缩合聚合的实例，将氨基酸化学与考纲中其他部分涉及的合成聚酰胺（如尼龙）直接联系起来。考试作答时务必说明脱去一个水分子，并指出生成的键为酰胺键/肽键。
7. Chromatographic Separation | 层析分离
Amino acids can be separated and identified using paper chromatography or thin-layer chromatography. A sample mixture is spotted onto the chromatogram, and a suitable solvent is allowed to rise up the paper. Because different amino acids differ in polarity, they partition differently between the stationary phase (water adsorbed on the paper) and the mobile phase (the solvent), and therefore travel different distances.
氨基酸可通过纸层析或薄层层析进行分离和鉴定。将混合样品点在层析纸上，让合适的溶剂沿纸上行。由于不同氨基酸的极性不同，它们在固定相（吸附于纸上的水）和流动相（溶剂）之间的分配不同，因此移动距离不同。
The retention factor (Rf) is defined as the distance moved by the solute divided by the distance moved by the solvent front:
比较值（Rf）定义为溶质移动距离除以溶剂前沿移动距离：
Rf = distance moved by solute / distance moved by solvent front
After development, the chromatogram is sprayed with ninhydrin solution and heated; purple-violet spots reveal the positions of the amino acids. Rf values can be compared with those of known standards to identify specific amino acids.
层析展开后，喷洒茚三酮溶液并加热，紫色斑点即可显示氨基酸的位置。将Rf值与已知标准样品对比，可鉴定具体氨基酸。
8. Chemical Tests | 化学检验
Two chemical tests are essential in the CIE specification for amino acids and proteins. The ninhydrin test: when an amino acid is warmed with ninhydrin solution, a deep blue-purple colour develops. Ninhydrin reacts with the free amino group, so the test is positive for amino acids, peptides, and proteins, and is widely used to detect amino acids on chromatograms.
CIE考纲中涉及两个重要的化学检验。茚三酮检验：氨基酸与茚三酮溶液共热时，产生深蓝紫色。茚三酮与游离氨基反应，因此氨基酸、多肽和蛋白质均呈阳性，广泛应用于层析图谱上的氨基酸检测。
The biuret test: a small amount of sodium hydroxide solution is added to the sample, followed by a few drops of dilute copper(II) sulfate solution. A purple-violet colour indicates the presence of peptide bonds. Free amino acids give a negative result because no peptide bond is present; a dipeptide gives a positive result because it contains one peptide bond. In practice, at least two peptide bonds are required for a clearly visible colour.
双缩脲检验：向样品中加入少量氢氧化钠溶液，再加几滴稀硫酸铜溶液。出现紫色表明存在肽键。游离氨基酸不含肽键，呈阴性；二肽含一个肽键，呈阳性。实际操作中通常需要至少两个肽键才能观察到明显的紫色。
9. Hydrolysis of Peptides | 肽的水解
Peptides and proteins can be hydrolysed to release their constituent amino acids. Acid hydrolysis uses 6 mol dm⁻³ hydrochloric acid under reflux for several hours. This method gives complete hydrolysis, but it destroys tryptophan and converts asparagine to aspartic acid and glutamine to glutamic acid. Therefore, the analysis of acid hydrolysates is not fully representative of the original protein.
多肽和蛋白质可水解为组成氨基酸。酸水解使用6 mol dm⁻³盐酸回流数小时。该法水解完全，但会破坏色氨酸，并将天冬酰胺转化为天冬氨酸、谷氨酰胺转化为谷氨酸。因此，酸水解产物的分析结果不能完全代表原始蛋白质。
Enzyme-catalysed hydrolysis provides a milder alternative. Protease enzymes hydrolyse specific peptide bonds under gentle conditions, producing mixtures of smaller peptides and amino acids. Because enzymes are highly specific, the products retain the true amino acid composition of the protein and can be analysed more reliably.
酶催化水解提供了一种较温和的选择。蛋白酶在温和条件下特异性切断肽键，产生较小的肽段和氨基酸混合物。由于酶具有高度专一性，产物能保留蛋白质真实的氨基酸组成，分析结果更为可靠。
10. Key Exam Points | 核心考点总结
To handle this topic confidently in the CIE A-Level Chemistry examination, students should be able to: (1) draw and interpret the general structure of an α-amino acid; (2) explain zwitterion formation and why amino acids are amphoteric; (3) state the ionic form of an amino acid in acidic, neutral, and basic conditions and relate this to the isoelectric point; (4) identify the chiral centre in an amino acid and explain why glycine does not show optical isomerism; (5) write the equation for dipeptide formation as a condensation reaction; and (6) describe the ninhydrin and biuret tests, including expected colour changes.
要在CIE A-Level 化学考试中从容应对本节内容，学生应能够：(1) 画出并解释α-氨基酸的通式结构；(2) 解释两性离子的形成及氨基酸为何具有两性；(3) 说明氨基酸在酸性、中性、碱性条件下的离子形态并将其与等电点联系起来；(4) 识别氨基酸中的手性中心并解释甘氨酸为何没有光学异构；(5) 书写二肽生成的缩合反应方程式；(6) 描述茚三酮检验和双缩脲检验的预期颜色变化。
A common exam trap is confusing the species present under different pH conditions. Remember: in acid → cation; in neutral solution / at pI → zwitterion; in base → anion. Another frequent error is claiming that the biuret test is positive for all
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