A-Level Chemistry: Identifying Structural Isomers of Organic Compounds | A-Level 化学：有机物的结构异构体辨析

📚 A-Level Chemistry: Identifying Structural Isomers of Organic Compounds | A-Level 化学：有机物的结构异构体辨析

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.

结构异构体是指具有相同分子式但结构式不同的化合物。这意味着原子以不同的顺序或方式连接，导致物理性质不同，有时化学性质也不同。

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.

例如，丁烷和 2-甲基丙烷都具有分子式 C₄H₁₀，但它们的碳骨架不同。丁烷是直链，而 2-甲基丙烷是支链。由于分子间作用力不同，它们的沸点差异显著。

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.

需要注意的是，立体异构（顺反异构和光学异构）不属于结构异构，因为原子的连接方式完全相同，只是空间排列不同。

3. Chain Isomerism | 链异构

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.

当具有相同分子式的化合物在碳链的支链方式上不同时，就产生链异构。直链异构体和支链异构体就是链异构的实例。

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.

对于烯烃，也存在链异构，但必须保持 C=C 双键不变。例如，C₄H₈ 有丁-1-烯和 2-甲基丙-1-烯等链异构体，但这些也涉及位置异构，因此分类时要小心。

4. Position Isomerism | 位置异构

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.

类似地，丁醇 (C₄H₉OH) 有四种结构异构体：丁-1-醇、丁-2-醇、2-甲基丙-1-醇和 2-甲基丙-2-醇。注意这四种异构体同时包含位置异构和链异构。

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.

对于烯烃，位置异构非常常见。例如，C₄H₈ 有丁-1-烯 (CH₂=CHCH₂CH₃) 和丁-2-烯 (CH₃CH=CHCH₃)。它们是位置异构体，因为双键从碳 1 移动到碳 2。

But-1-ene vs But-2-ene | 丁-1-烯 vs 丁-2-烯

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).

在画位置异构体时，总是给碳链编号，使官能团获得尽可能低的位次。这是一个惯例，不是异构体差异。例如，CH₃CH₂CH₂Cl 和 ClCH₂CH₂CH₃ 是同一个化合物（1-氯丙烷）。

5. Functional Group Isomerism | 官能团异构

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.

当具有相同分子式的化合物具有完全不同的官能团时，就产生官能团异构。这是最具挑战性的类型，因为化学性质可能大不相同。

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.

注意，分子式 C₃H₆O 还包括丙-2-烯-1-醇 (CH₂=CHCH₂OH)，它是一种不饱和醇——第三种官能团异构体。因此当题目要求列出所有异构体时，必须根据不饱和度考虑烯醇、醛、酮、环醚和环氧等。

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.

不饱和度 (IHD) 是一个强有力的工具。对于 C₃H₆O，IHD = (2×3 + 2 − 6)/2 = 1。这意味着存在一个双键或一个环。醛、酮、烯醇和环醚都满足这个条件。

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.

在考试中，你可能会被要求画出某一给定分子式的所有结构异构体，或者判断两个给定结构是否互为异构体。系统的方法对于避免遗漏或重复非常重要。

Here is a step-by-step strategy:

以下是一个分步策略：

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’).

步骤 1：IHD。对于 C₄H₁₀O，IHD = (2×4 + 2 − 10)/2 = 0。没有双键或环。分子是饱和的。由于氧是二价的，可能的官能团是醇 (R–OH) 和醚 (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.

步骤 2：醇异构体。画出四种碳骨架：丁烷（直链）和 2-甲基丙烷（支链）。对于丁烷，将 –OH 放在碳 1 和碳 2 上，得到丁-1-醇和丁-2-醇。将 –OH 放在碳 3 上，由于对称性与碳 2 相同。对于 2-甲基丙烷，中心碳是叔碳，得到 2-甲基丙-2-醇；末端甲基碳等价，得到 2-甲基丙-1-醇。共四种醇。

Name	Structure
Butan-1-ol	CH₃CH₂CH₂CH₂OH
Butan-2-ol	CH₃CH(OH)CH₂CH₃
2-Methylpropan-1-ol	(CH₃)₂CHCH₂OH
2-Methylpropan-2-ol	(CH₃)₃COH

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.

步骤 3：醚异构体。对于 C₄H₁₀O，醚的形式为 R–O–R’。碳原子总数为四个。烷基的可能分割为 1+3（甲氧基丙烷）和 2+2（乙氧基乙烷）。对于 1+3 分割，丙基可以是正丙基或异丙基，得到甲基正丙基醚 (CH₃OCH₂CH₂CH₃) 和甲基异丙基醚 (CH₃OCH(CH₃)₂)。对于 2+2 分割，只有乙氧基乙烷 (CH₃CH₂OCH₂CH₃)。共三种醚。

Total: 4 alcohols + 3 ethers = 7 structural isomers for C₄H₁₀O. This is a classic answer you can memorise, but you must also be able to derive it.

总计：4 种醇 + 3 种醚 = C₄H₁₀O 的 7 种结构异构体。这是一个经典答案，可以记忆，但你也必须能够推导出来。

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.

烯醇：三个碳的双键可以在 C1–C2 或 C2–C3 之间。前者给出丙-2-烯-1-醇 (CH₂=CHCH₂OH)，后者给出丙-1-烯-1-醇 (CH₃CH=CHOH)。然而，丙-1-烯-1-醇是一种烯醇，会互变异构为丙醛，因此在 A-Level 考试中通常不计为稳定可分离的异构体。你应该将其作为互变异构体提及，而不是实际意义上的结构异构体。

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.

环醚：三元环环氧化物，环氧丙烷（带甲基取代基的环氧乙烷）以甲基环氧乙烷形式存在。这是一个有效的结构异构体，但学生较少画出。

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.

因此，对于 C₃H₆O，考试主要考的结构异构体是丙醛和丙酮。一些教学大纲还包括丙-2-烯-1-醇和甲基环氧乙烷，因此请仔细检查你的考纲要求。

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.

错误 1：重复画出相同的结构。例如，将 2-氯丁烷和 3-氯丁烷画成不同的异构体。它们是同一个化合物，因为从另一端编号时，碳 3 与碳 2 相同。始终检查最长链，并从给出最低位次的一端编号。

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).

错误 2：忽略官能团异构体。例如，对于 C₂H₆O 只画出醇而忘记醚。C₂H₆O 恰好有两种异构体：乙醇 (CH₃CH₂OH) 和甲氧基甲烷 (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.

错误 3：违反化合价规则。碳必须始终有四根键，氧两根，氮三根，氢一根。碳有五根键或氧有三根键的结构是无效的。始终数一下键数。

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.

错误 4：忘记检查不饱和度。如果分子有一个双键，你不能画出一个饱和醇而不添加第二个双键。始终验证所画结构中的氢原子总数与给定分子式一致。

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.

错误 5：混淆结构异构与立体异构。顺反异构不是结构异构。它们具有相同的连接顺序。在 CIE 考试中，立体异构问题单独处理。

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.

结构异构体的物理性质不同，因为它们的形状和官能团影响分子间作用力。考试题经常要求你比较沸点或溶解性。

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.

对于烷烃，支链异构体的沸点低于直链异构体。例如，2,2-二甲基丙烷的沸点为 9.5 °C，而戊烷的沸点为 36.1 °C。支链异构体的表面积减小，导致范德华力减弱。

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.

对于醇，其沸点远高于同分异构的醚，因为醇能形成氢键。乙醇的沸点为 78.4 °C，而甲氧基甲烷的沸点为 −24 °C。两者分子式均为 C₂H₆O，但醇的沸点高得多。

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.

水溶性：少于四个碳的醇可溶于水，因为它们与水形成氢键。醚的溶解度较小。羧酸和酯也不同：羧酸由于 –OH 基团可以形成更多氢键，而酯不能作为氢键供体。

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.

质谱：醛和酮通常碎裂产生酰基离子。支链烷烃产生更丰富的小碎片离子，而直链烷烃产生更均匀的峰系列。分子离子 (M⁺) 对所有这些异构体具有相同的 m/z，但碎裂模式不同。

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.

红外光谱：在 3230–3550 cm⁻¹ 附近的宽吸收表示 O–H 伸缩振动（醇）。在 1700–1750 cm⁻¹ 附近的尖锐强吸收表示 C=O 伸缩振动（醛或酮）。醚在 1050–1150 cm⁻¹ 附近显示 C–O 伸缩振动，且没有 O–H 吸收。

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.

¹H NMR：醛在 δ 9–10 ppm 附近显示特征峰。丙醛的醛基氢呈三重峰。酮没有这样的峰。醇在 δ 1–5 ppm 附近显示可变的宽峰，这是 –OH 质子，可以通过加入 D₂O 来确认。

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.

¹³C NMR：醛基碳通常在 δ 190–200 ppm 处出现，而酮在 δ 200–210 ppm 处出现。与 –OH 相连的醇碳在 δ 50–90 ppm 处出现，醚碳的化学位移略高。

12. Exam Tips and Revision Checklist | 考试技巧与复习清单

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

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A-Level Chemistry: Identifying Structural Isomers of Organic Compounds | A-Level 化学：有机物的结构异构体辨析