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Isomerism is one of the defining features of organic chemistry. Two compounds are isomers if they share the same molecular formula but differ in how their atoms are arranged. A single formula such as C₄H₁₀O hides seven distinct compounds; C₅H₁₂O conceals at least seventeen. The differences between them range from trivial (a barely-different boiling point) to profound (one enantiomer is medicine, the other is teratogen). This lesson works through the AQA taxonomy in detail. We treat the three sub-types of structural isomerism — chain, position, and functional-group — and then the two sub-types of stereoisomerism, namely E/Z (geometric) isomerism about a C=C double bond and optical isomerism about a chiral centre. Optical isomerism is signposted here and developed in full in lesson 2. By the end you will be able to draw every isomer of a given formula systematically, assign E or Z by the Cahn-Ingold-Prelog rules, and identify stereocentres at a glance.
Spec mapping (AQA 7405): This lesson maps to §3.3.1 (isomerism), the foundational section that underwrites all subsequent organic content. It cross-references §3.3.2 (nomenclature, treated in lesson 0 of this course), §3.3.7 (organic synthesis, where stereocontrol becomes a design constraint), and §3.3.15 (NMR spectroscopy, where ¹H and ¹³C signals distinguish isomers that have identical molecular formulae). Optical isomerism is introduced here and developed in lesson 2 of this course (§3.3.7 in the spec wording). Refer to the official AQA specification document for the exact wording of each section.
Assessment objectives: Definitions of each isomer sub-type are AO1 recall items and appear on virtually every Paper 2. AO2 questions ask you to draw all isomers of a given molecular formula, assign E or Z using CIP priority, and identify stereocentres in a given structure. AO3 questions ask you to predict the number of isomers from a formula, rationalise observed reaction outcomes in terms of stereochemistry (e.g. why an SN1 product is racemic), and connect spectroscopic data (NMR, IR, mass spec) to the isomer actually present.
Two compounds are isomers if they share the same molecular formula but differ in the arrangement of their atoms. The umbrella divides into two branches:
Isomers are not conformers. Rotation about a C–C single bond produces an infinite family of conformations (eclipsed, staggered, gauche), but these inter-convert at room temperature and are not separable. Restricted rotation about a C=C double bond, by contrast, locks atoms in place and gives rise to genuine E/Z isomers that can be bottled, melted, and characterised separately.
Isomerism
├── Structural Isomerism (different connectivity)
│ ├── Chain isomerism
│ ├── Position isomerism
│ └── Functional-group isomerism
└── Stereoisomerism (same connectivity, different 3D arrangement)
├── E/Z (geometric) isomerism
└── Optical isomerism (enantiomers)
Structural isomers have the same molecular formula but differ in how the atoms are connected. The bond list is different. Three sub-types are recognised at A-Level.
Chain isomers have the same functional group(s) but different carbon-skeleton shapes — different straight-chain lengths, branching patterns, or ring sizes.
Example 1 — C₄H₁₀. Two isomers:
Example 2 — C₅H₁₂. Three chain isomers:
The chemistry is virtually identical (same C–H and C–C bonds, same combustion enthalpy per mole within ~1%). The physical properties differ subtly: branched isomers pack less efficiently in the liquid state and have weaker London dispersion forces because their surface-area-to-volume ratio is lower. Boiling point therefore falls as branching increases.
Position isomers share the same carbon skeleton and the same functional group(s) but differ in the position of the functional group on the chain.
Example 1 — C₃H₇Br. Two position isomers:
Example 2 — C₃H₇OH. Two position isomers:
Position isomers can have dramatically different reactivity even though their bulk properties are similar. Propan-1-ol oxidises sequentially to propanal then propanoic acid; propan-2-ol oxidises only to propanone (a ketone) and cannot proceed further with mild oxidants because there is no α-H on the carbonyl carbon. Likewise, 1-bromopropane reacts predominantly via SN2 (primary, less steric hindrance), whereas 2-bromopropane shows a greater proportion of SN1 character.
Functional-group isomers share the molecular formula but contain different functional groups entirely. Their chemistry is correspondingly very different.
Example 1 — C₂H₆O. Two functional-group isomers:
Example 2 — C₃H₆O. Two functional-group isomers (ignoring less stable enol and cyclic forms):
Example 3 — C₃H₆O₂. Three common functional-group isomers:
Functional-group isomers are useful diagnostically — IR spectroscopy (next-but-one lesson, §3.3.15) immediately distinguishes a carboxylic acid (broad O–H around 2500–3300 cm⁻¹, strong C=O at ~1710 cm⁻¹) from an ester (no O–H, sharp C=O at ~1735 cm⁻¹).
The classic AO2 question asks: "draw all the structural isomers of [molecular formula]." Students who guess miss isomers; students who proceed systematically do not. The procedure:
DoU = (2·5 + 2 − 12) / 2 = 0. No rings or double bonds — only saturated open-chain structures.
Pentanol position isomers (C–C–C–C–C–OH skeleton, OH at each non-equivalent position):
Branched-chain alcohols on the C₄ skeleton + methyl branch:
Doubly-branched / neopentyl alcohol:
Ether functional-group isomers (R–O–R'):
Counting alcohols only, C₅H₁₂O has eight structural isomers. Adding ethers brings the total to fourteen-ish (depending on how one counts methyl-tert-butyl-ether-like names). The AQA mark scheme typically asks for the eight alcohols and is generous about whether ethers are also listed.
DoU = (2·4 + 2 − 8) / 2 = 1. One degree — either one C=C double bond or one ring.
Open-chain alkenes:
Cyclic isomers (saturated rings): 4. Cyclobutane: a four-membered carbocycle. 5. Methylcyclopropane: a three-membered ring with a methyl substituent.
So C₄H₈ has six distinct compounds: three open-chain alkenes (with but-2-ene split into E and Z) plus two cycloalkanes. Strict counting of structural isomers (ignoring stereoisomers) gives five; counting E and Z separately gives six. Be careful which the question asks for.
DoU = (2·4 + 2 − 8) / 2 = 1. One double bond (C=O or C=C) or a ring.
Aldehyde:
Ketone: 3. Butanone: CH₃COCH₂CH₃.
Alcohols with C=C (unsaturated alcohols, "enols"): 4. But-3-en-1-ol: CH₂=CHCH₂CH₂OH. 5. But-2-en-1-ol: CH₃CH=CHCH₂OH (E and Z forms). 6. But-3-en-2-ol: CH₂=CHCH(OH)CH₃ (chiral). 7. 2-Methylprop-2-en-1-ol: CH₂=C(CH₃)CH₂OH.
Ethers with C=C: Methoxypropene and ethoxyethene also exist. The mark scheme will normally signpost whether unsaturated ethers are required.
The point of these worked examples is the systematic procedure, not memorisation of the lists. A student who has internalised "longest chain first; reposition; branch; switch functional group" will reconstruct any of these on the fly.
Stereoisomers share their bond list but differ in spatial arrangement. Two sub-types arise at A-Level.
E/Z isomerism arises when:
The double bond is composed of a sigma bond (axial overlap of sp² hybrid orbitals) and a pi bond (sideways overlap of perpendicular p-orbitals). Rotating one sp² carbon relative to the other would break the pi bond — an energetically prohibitive ~270 kJ mol⁻¹ at room temperature. The geometry is therefore locked, and atoms on either side of the C=C are pinned in place.
If both groups on one sp² carbon are identical (e.g. CH₂= in but-1-ene or 2-methylpropene), no E/Z isomerism arises — flipping the molecule produces the same compound.
To assign E or Z:
CH₃–CH=CH–CH₃. On the left sp² carbon, the substituents are CH₃ and H — CH₃ wins (C > H). On the right sp² carbon, the substituents are again CH₃ and H — CH₃ wins.
BrClC=CHCH₃. On the left sp² carbon: Br (Z = 35) versus Cl (Z = 17) — Br wins. On the right sp² carbon: CH₃ (C, Z = 6) versus H (Z = 1) — CH₃ wins.
Notice that "Z" does not always mean "the two carbons are on the same side". The CIP system asks where the priority-1 group on each sp² centre sits — and priority-1 may or may not be carbon.
Common misconception: E/Z is not synonymous with cis/trans. Cis/trans uses an intuitive "are the matching groups on the same side?" — fine for but-2-ene, but ambiguous for trisubstituted alkenes such as 1-bromo-1-chloropropene. CIP priority always gives a single, unambiguous answer; that is why A-Level chemistry uses E/Z exclusively from 2017 onwards.
Optical isomerism arises when a carbon atom is bonded to four different groups. Such a carbon is called a chiral centre (or asymmetric carbon, stereocentre); the molecule and its mirror image are then non-superimposable, like a pair of hands. The two mirror-image forms are enantiomers. Enantiomers have identical scalar physical properties (boiling point, melting point, density) and react identically with achiral reagents, but rotate the plane of plane-polarised light by equal and opposite amounts. The (+)/(−) prefix denotes which way light is rotated.
A racemic mixture (50:50 of the two enantiomers) is optically inactive because the rotations cancel. Reactions that create a new chiral centre from a planar (sp²) intermediate — such as SN1 substitution or nucleophilic addition to a carbonyl — produce racemic products because attack occurs equally from both faces of the planar species.
The biological importance of chirality cannot be overstated. Enzymes are themselves chiral, so they discriminate between enantiomers. Thalidomide is the textbook cautionary tale: one enantiomer is an effective sedative, the other a teratogen. Naproxen, ibuprofen, and L-DOPA are all chiral drugs whose two enantiomers have radically different pharmacology.
This is the headline of lesson 2 of this course. Worked examples on chirality, drawing wedge-and-dash 3D structures, identifying multiple stereocentres in larger molecules, and computing 2ⁿ as an upper bound on stereoisomer count are all deferred to lesson 2. The remainder of this lesson focuses on identifying stereocentres, since AO2 questions on lesson 1 commonly ask "circle the chiral centres" without demanding 3D drawings.
A stereocentre (chiral centre) is a tetrahedral carbon bonded to four different groups. To find them:
The maximum possible number of stereoisomers is 2ⁿ, where n is the number of stereocentres. For a single stereocentre (n = 1), there are 2 enantiomers. For two stereocentres (n = 2), there are up to 4 stereoisomers — two pairs of enantiomers, which are diastereomers of each other. Internal symmetry (a meso form) can reduce the count below 2ⁿ; this is undergraduate territory and not assessed at A-Level.
CH₃–CHBr–CHBr–CH₃. Both internal carbons (C-2 and C-3) bear H, Br, CH₃, and CH(Br)CH₃. Both are stereocentres. Maximum stereoisomer count = 2² = 4. (In fact only three exist — one is a meso compound — but that is for lesson 2.)
CH₃–CH(OH)–COOH. The central carbon is bonded to H, OH, CH₃, and COOH — four different groups, so one stereocentre. Two enantiomers exist: (+)-lactic acid (D-lactic acid) and (−)-lactic acid (L-lactic acid). Mammalian muscle produces (+)-lactic acid; bacterial fermentation typically produces (−)-lactic acid.
Glucose (C₆H₁₂O₆) in its open-chain form has four stereocentres (C-2, C-3, C-4, C-5). Maximum stereoisomers = 2⁴ = 16. These sixteen are the eight pairs of enantiomers known as the aldohexoses — glucose, mannose, galactose, allose, altrose, gulose, idose, and talose. (The cyclic hemiacetal form adds one further stereocentre at the anomeric carbon, doubling the count to 32.) This is wildly out of scope for AQA A-Level but illustrates why biological molecules with many stereocentres are so diverse.
A formula such as C₃H₆O does not uniquely identify a compound — propanal and propanone share it. The standard A-Level toolkit (signposted to lesson §3.3.15) for distinguishing isomers:
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