UNIT 2 CHEMISTRY · PEARSON AAQ · E1.4
Isomerism
FREE CHEMISTRY REVISION · TOPIC 53
Isomerism
Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · E1.4
What you will learn
Identify when compounds are isomers and explain why they occur. Structural isomers have the same molecular formula but different atom connectivity. Stereoisomers have the same molecular formula and connectivity but different spatial arrangements. This lesson covers the specified E/Z and cis–trans stereoisomerism, without extending the required content into optical isomerism.
1. Use a reliable decision sequence
| Question | If yes | If no |
|---|---|---|
| Do the molecular formulae match exactly? | Compare connectivity next | They are not isomers |
| Are any atom connections different? | They are structural isomers | Compare spatial arrangement |
| Is a distinct, constrained spatial arrangement shown? | They may be stereoisomers; apply the alkene conditions below | A rotated or redrawn version can simply be the same compound |
A molecular formula gives totals, not connections. Different drawings alone do not demonstrate different compounds. Trace the carbon skeleton, the positions of functional groups and multiple bonds, and which atoms are joined. A name difference caused by incorrect numbering does not create an isomer.
2. Structural isomerism
| Type | What differs | Original example pair | Shared formula |
|---|---|---|---|
| Chain | Carbon skeleton/branching | butane and 2-methylpropane | C₄H₁₀ |
| Position | Position of a group or multiple bond on the same skeleton | propan-1-ol and propan-2-ol | C₃H₈O |
| Position | Double-bond position | pent-1-ene and pent-2-ene | C₅H₁₀ |
| Functional-group | Atoms connect into different functional groups | CH₃CH₂OH and CH₃OCH₃ | C₂H₆O |
Functional-group isomerism means the same atom inventory forms different characteristic groups. CH₃CH₂OH contains C–O–H and is an alcohol. CH₃OCH₃ has C–O–C and is an ether. The ether is included to explain the isomer relationship; detailed ether naming and reactions are not a requirement here.
Worked identification 1 — chain isomers
Butane has CH₃CH₂CH₂CH₃: 4 C and 3+2+2+3=10 H. 2-methylpropane has CH₃CH(CH₃)CH₃: 4 C and 3+1+3+3=10 H. One has an unbranched chain; the other has a carbon bonded to three other C atoms. Same C₄H₁₀, different connections → chain structural isomers.
Worked identification 2 — position isomers
CH₃CH₂CH₂OH and CH₃CH(OH)CH₃ both have C₃H₈O. The three-carbon skeleton remains unbranched, but the C–O connection moves from an end carbon to the middle carbon. They are position structural isomers. Changing the page orientation cannot move this connection.
Worked identification 3 — functional-group isomers
CH₃CH₂OH has H=3+2+1=6. CH₃OCH₃ has H=3+3=6. Both contain two C and one O: C₂H₆O. In the first, O is joined to C and H; in the second, O is joined to two C atoms. Different connectivity and functional groups → functional-group structural isomers.
Worked identification 4 — a multiple bond changes position
CH₂=CHCH₂CH₂CH₃ and CH₃CH=CHCH₂CH₃ both have C₅H₁₀. Their unbranched five-carbon skeleton is the same, but the double bond starts at C1 or C2. They are position isomers. Pent-4-ene is not a third position: numbering that drawing from the other end gives pent-1-ene.
3. When can an alkene have E/Z stereoisomers?
The arrangement about C=C cannot interconvert by ordinary free rotation while the double bond remains intact. To show two E/Z arrangements, each carbon of the double bond must carry two different substituents. The pair at one end can match the pair at the other end: all four substituents need not be different.
| Alkene | Groups at the two C=C carbons | E/Z possible? |
|---|---|---|
| CH₂=CHCH₃ (propene) | H/H at one end; H/CH₃ at the other | No: the H/H end fails |
| CH₃CH=CHCH₃ (but-2-ene) | H/CH₃ at each end | Yes |
| (CH₃)₂C=CHCH₃ (2-methylbut-2-ene) | CH₃/CH₃ at one end; H/CH₃ at the other | No: identical methyl groups at one end |
| BrC(Cl)=CHCH₃ | Br/Cl at one end; H/CH₃ at the other | Yes |
The double-bond atoms themselves are not the two substituents being compared. Examine the two groups attached to each of those atoms apart from the other double-bond carbon. Draw all four attachments clearly before deciding. Single-bond rotations can change conformation without changing connectivity; these are not the restricted alkene arrangements being counted here.
4. Cis and trans in an appropriate alkene
Worked identification 5 — but-2-ene
Each C=C carbon has one H and one CH₃, so the condition is met. With the two methyl groups on the same side, the structure is cis-but-2-ene; opposite sides gives trans-but-2-ene. Both are C₄H₈ and have the same connectivity. They are stereoisomers, not position isomers.
Cis/trans compares an appropriate identical pair across the double bond. In but-2-ene, CH₃ has higher priority than H on both ends, so cis is (2Z)-but-2-ene and trans is (2E)-but-2-ene. E/Z remains the more general naming system.
5. Assign E and Z using priority
Apply the Cahn–Ingold–Prelog (CIP) priority rules separately at each double-bond carbon. Compare the atomic numbers of the atoms directly attached: higher atomic number has higher priority. For the common atoms here: I > Br > Cl > F > O > C > H. Atomic number, rather than total group mass or visual size, controls this first comparison.
If directly attached atoms tie, compare the next attached atoms in decreasing atomic-number order, excluding the atom just followed back towards the double bond. Continue outward to the first difference. For example, an ethyl carbon presents [C,H,H], whereas a methyl carbon presents [H,H,H], so ethyl has priority over methyl. These examples do not require advanced isotope or multiple-bond tie-breaking rules.
- Select one higher-priority group on the left-hand double-bond carbon.
- Select one higher-priority group on the right-hand double-bond carbon independently.
- Higher-priority groups on the same side → Z (zusammen, together).
- Higher-priority groups on opposite sides → E (entgegen, opposite).
Worked identification 6 — four different substituents
For BrC(Cl)=CHCH₃, the left end has Br and Cl, and the right end has CH₃ and H. Br outranks Cl (35 > 17), while carbon in CH₃ outranks H (6 > 1). In the upper diagram, Br and CH₃ are together, giving Z; in the lower, they are opposite, giving E. Each formula is C₃H₄BrCl: the right alkene carbon carries one H and its methyl carries three.
Cis/trans is not a useful unambiguous label for this four-different-group example. Do not compare Br with the group at the other end to decide which end “wins”; rank the two choices within each end.
Worked identification 7 — break a carbon/carbon priority tie
Consider CH₃CH=C(CH₃)CH₂CH₃, 3-methylpent-2-ene, C₆H₁₂. At C2, CH₃ outranks H. At C3, both substituents begin with C, so the first comparison ties. Ethyl’s next-atom list [C,H,H] outranks methyl’s [H,H,H]. Therefore compare the C2 methyl with the C3 ethyl: same side gives Z; opposite gives E.
If the two methyl groups happen to be on the same side in this molecule, that does not make it Z: the higher-priority group on C3 is ethyl. This is why “cis always means Z” is an unsafe generalisation.
6. Reject false isomer pairs
Worked identification 8 — two drawings can be the same compound
CH₃CH₂CH₂CH₃ and CH₃CH₂CH₂CH₃ drawn in reverse or bent into a zigzag both have the same carbon connections. Every carbon retains the same neighbours. There is no new structural isomer; drawing shape alone is insufficient. Likewise, propan-3-ol is a misnumbered drawing of propan-1-ol, not a third position isomer.
To enumerate structural isomers, draw a candidate, complete valencies, count its formula, choose its proper name, then compare its connectivity against every structure already listed. For E/Z, first test the two-different-groups condition, then show the two distinct arrangements. Avoid counting a whole-molecule rotation as an additional arrangement.
Practice with answers
| Task | Answer |
|---|---|
| Are C₂H₆ and C₂H₆O isomers? | No: molecular formulae differ. |
| Can CH₂=C(CH₃)₂ have E/Z isomers? | No: one end has H/H and the other CH₃/CH₃. |
| Compare butan-1-ol and butan-2-ol | C₄H₁₀O in both; OH position differs → position structural isomers. |
| Which has higher priority, Br or Cl? | Br: atomic number 35 exceeds 17. |
| Which outranks the other: CH₂CH₃ or CH₃? | Ethyl: [C,H,H] exceeds [H,H,H] after the directly attached C/C tie. |
| A but-2-ene drawing has methyl groups on opposite sides | trans, equivalently (2E)-but-2-ene. |
Common misconceptions
- Any two different drawings are isomers: compare formula, connectivity and constrained arrangement.
- Same molecular formula means same compound: formula does not show connections.
- Moving OH means stereoisomerism: that changes connectivity and gives structural isomers.
- Every alkene shows E/Z: each double-bond carbon needs two different groups.
- All four substituents must be different: they only need to differ within each end.
- Z means the biggest-looking groups are together: apply atomic-number priority.
- Cis always equals Z: this correspondence holds for the simple but-2-ene pair, not every alkene.
- Ordinary rotation around a single bond produces another required alkene stereoisomer: the C=C restriction is essential.
Examination tips
- Write both molecular formulae before classifying an isomer pair.
- For structural isomers, state the actual connectivity difference.
- For E/Z, identify both substituent pairs before assigning priorities.
- Show the two higher-priority groups clearly and state same side or opposite sides.
- Explain a rejected E/Z case by naming the identical pair on one double-bond carbon.
- Keep the required stereoisomer content to E/Z and cis–trans.
Check your understanding
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