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Isomers are molecules that share the same molecular formula but have different structural arrangements of their atoms. This seemingly simple definition conceals enormous chemical consequences: isomers can have different physical properties, different chemical reactivities, and even different biological effects.
At Edexcel A-Level, you need to understand two main categories: structural isomerism and stereoisomerism.
Structural isomers have the same molecular formula but different structural formulae — the atoms are connected in a different order. There are three types you need to know.
Chain isomers differ in the arrangement of the carbon skeleton. The carbon chain can be straight or branched.
For example, C₄H₁₀ has two chain isomers:
As the number of carbons increases, the number of possible chain isomers grows rapidly. C₅H₁₂ has three isomers; C₆H₁₄ has five; C₇H₁₆ has nine; C₁₀H₂₂ has seventy-five.
Branched isomers tend to have lower boiling points than their straight-chain counterparts because their more compact shape reduces the surface area available for van der Waals interactions.
Position isomers have the same carbon skeleton and the same functional group, but the functional group is attached at a different position on the chain.
For example, C₃H₇OH has two position isomers:
Similarly, C₄H₈ can exist as but-1-ene (double bond between C1 and C2) or but-2-ene (double bond between C2 and C3).
Functional group isomers have the same molecular formula but contain different functional groups entirely.
Key examples:
These isomers often have dramatically different physical and chemical properties because the functional group dictates reactivity.
When asked to draw all the isomers of a given molecular formula, use a systematic approach:
Worked Example: Draw all structural isomers of C₄H₉Cl.
Longest chain = 4 carbons (butane skeleton with one Cl):
Shortest chain = 3 carbons + 1 methyl branch (methylpropane skeleton):
That gives four structural isomers.
Stereoisomers have the same structural formula — the atoms are connected in the same order — but differ in the spatial arrangement of those atoms. There are two types at A-Level.
E/Z isomerism arises when there is restricted rotation around a bond, most commonly a C=C double bond, and when each carbon of the double bond carries two different groups.
Because the pi bond in a C=C prevents free rotation, groups on either side of the double bond are locked in position. If the two higher-priority groups (determined by the Cahn-Ingold-Prelog rules) are on the same side, the isomer is designated Z (from the German zusammen, meaning together). If they are on opposite sides, the isomer is E (from entgegen, meaning opposite).
For example, in but-2-ene (CH₃CH=CHCH₃):
Consider 2-chlorobut-2-ene: CH₃-CCl=CH-CH₃.
On the left carbon of C=C: -CH₃ (C, priority lower) and -Cl (Cl, atomic number 17, priority higher). On the right carbon of C=C: -CH₃ (C, priority lower) and -H (H, priority lower).
Wait — the right carbon has -CH₃ and -H. Higher priority on the right = -CH₃ (C > H).
So the two higher-priority groups are -Cl (left) and -CH₃ (right). If they are on the same side → Z. Opposite sides → E.
E/Z isomerism requires:
If either carbon of the double bond has two identical groups, E/Z isomerism is not possible. For example, CH₂=CHCl can show E/Z isomerism because the left C has H,H (wait — both H, so left C has two identical groups). Therefore CH₂=CHCl does not show E/Z isomerism. But CHCl=CHBr does (left C has H and Cl; right C has H and Br — both different pairs).
Optical isomerism arises when a molecule contains a chiral centre — a carbon atom bonded to four different groups. Such a carbon is sometimes called an asymmetric carbon and is often marked with an asterisk (*) in structural formulae.
A molecule with a chiral centre exists as two non-superimposable mirror images called enantiomers. These are related in the same way as your left and right hands — identical in every measurement except that one is the mirror image of the other.
Enantiomers have:
A racemic mixture (or racemate) is a 50:50 mixture of both enantiomers. It shows no overall optical activity because the rotations cancel out. Many chemical syntheses produce racemic mixtures because the reaction has no chiral influence to favour one enantiomer.
Exam link: SN1 reactions produce racemic mixtures because the planar carbocation intermediate is attacked equally from both sides. SN2 reactions produce inversion of configuration, so if you start with one enantiomer, you get the other. This connects stereochemistry directly to mechanism.
To identify a chiral centre, look for a carbon bonded to four different groups. For example, in butan-2-ol (CH₃CH(OH)CH₂CH₃), carbon 2 is bonded to: -OH, -H, -CH₃, and -CH₂CH₃ — four different groups, making it a chiral centre.
Worked Example: Does 2-chlorobutane have a chiral centre?
Carbon 2 is bonded to: -Cl, -H, -CH₃, and -CH₂CH₃. All four groups are different. Yes, C2 is a chiral centre, and 2-chlorobutane exists as a pair of enantiomers.
Worked Example: Does 2-chloropropane have a chiral centre?
Carbon 2 is bonded to: -Cl, -H, -CH₃, and -CH₃. Two of the groups are the same (both -CH₃). Therefore C2 is not a chiral centre, and 2-chloropropane does not show optical isomerism.
Different isomers can have very different chemical behaviour:
| Type | Same Molecular Formula? | Same Structural Formula? | Difference |
|---|---|---|---|
| Chain isomers | Yes | No | Different carbon skeleton |
| Position isomers | Yes | No | Functional group at different position |
| Functional group isomers | Yes | No | Different functional group |
| E/Z isomers | Yes | Yes | Different spatial arrangement around C=C |
| Optical isomers (enantiomers) | Yes | Yes | Non-superimposable mirror images |
Edexcel 9CH0 specification Topic 6 — Organic Chemistry I, sub-topic 6.2 covers structural isomerism (chain, position and functional-group classes) and stereoisomerism (E/Z and optical), including the use of CIP priority rules for E/Z assignment and the identification of chiral carbons (refer to the official specification document for exact wording). Sub-topic 6.2 is examined directly on Paper 2 and synoptically on Paper 3 through NMR, IR and chromatographic identification of isomers. The Edexcel specification expects candidates to (a) draw all structural isomers of a given molecular formula up to about C5, (b) assign E/Z labels using CIP priorities, (c) recognise chiral centres (a carbon bonded to four different groups), (d) explain the optical activity of a single enantiomer and the optical inactivity of a racemic mixture. The data booklet provides no priority-rule shortcut — assignments must be done from first principles.
Question (8 marks):
(a) Draw and name all the alcohol structural isomers of C5H12O, classifying each as primary, secondary or tertiary. (5)
(b) For the compound 2-bromobut-2-ene (CH3−CBr=CH−CH3): (i) Determine whether E/Z isomerism is possible and justify. (2) (ii) Draw the E isomer. (1)
Solution with mark scheme:
(a) Step 1 — list all chain skeletons for C5H12. There are three pentyl skeletons: pentane (linear), 2-methylbutane (one branch), 2,2-dimethylpropane (two branches on a 3-C chain).
Step 2 — place -OH at every distinct position on each skeleton.
M1 — three distinct carbon skeletons identified.
M1 — pentane-derived isomers: pentan-1-ol (1°), pentan-2-ol (2°), pentan-3-ol (2°).
M1 — 2-methylbutane-derived isomers: 3-methylbutan-1-ol (1°), 3-methylbutan-2-ol (2°), 2-methylbutan-2-ol (3°), 2-methylbutan-1-ol (1°).
A1 — 2,2-dimethylpropan-1-ol (1°).
A1 — total: 8 alcohol isomers of C5H12O correctly classified (4 primary, 3 secondary, 1 tertiary).
(b)(i) Step 1 — apply the E/Z criterion. For E/Z isomerism, each carbon of the C=C must carry two different groups.
C2 of 2-bromobut-2-ene carries: -Br and -CH3. Different. C3 carries: -H and -CH3. Different.
M1 — both C=C carbons identified as carrying two distinct substituents.
A1 — therefore E/Z isomerism is possible.
(ii) Step 2 — assign CIP priorities. On C2: Br (atomic number 35) > CH3 (carbon, atomic number 6), so Br is higher priority. On C3: CH3 (carbon) > H (hydrogen), so CH3 is higher priority. The E isomer places the two higher-priority groups on opposite sides of the C=C.
A1 — correct E isomer drawn with Br and CH3 on opposite sides of the double bond:
BrH/C=C/CH3CH3Total: 8 marks (M3 A5).
Question (6 marks): Compound Z has the structural formula CH3−CH(OH)−CH(CH3)−CH2−CH3.
(a) Identify all chiral centres in Z, justifying your choice. (3)
(b) State whether Z is optically active as drawn. Explain your reasoning. (2)
(c) Describe what would be observed if a sample of pure (R)-Z were placed in a polarimeter, compared with a racemic mixture. (1)
Mark scheme decomposition by AO:
(a)
(b)
(c)
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