Structural formulas and structural isomers
Describe unambiguous connections and compare the two required C₄H₁₀ and C₄H₈ examples without mistaking a rotated drawing for a new isomer.
Exam scope and assessment notes
Extended only (Supplement) · Chemistry 0971, 2026–2028. Cambridge states that IGCSE 0971 has the same subject content as IGCSE 0620, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Read displayed formulas and total molecular atom counts.
- Interpret brackets as attached groups rather than missing atoms.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does a molecular formula provide directly?
What you will learn
- Interpret the required structural formulas as unambiguous descriptions of arrangement.
- Define structural isomers and identify the specified same-formula/different-connectivity pairs.
Structural notation states which groups connect
A structural formula gives an unambiguous description of how atoms are arranged. Required examples include CH₂=CH₂ for ethene, CH₃CH₂OH for ethanol and CH₃COOCH₃ for methyl ethanoate. In CH₃COOCH₃ the carbonyl-containing acid part connects through oxygen to the final methyl group; the displayed structure makes that ester linkage explicit in the later naming/reaction lessons.
A molecular formula such as C₂H₆O counts atoms but does not specify every connection. Structural notation can group hydrogens, while a displayed drawing shows each atom/bond. Read parentheses by connectivity: CH₃CH(CH₃)CH₃ has a CH₃ group attached to the central CH, not four carbon atoms in one unbranched row. A whole-diagram rotation leaves those connections unchanged.
Compare totals first, then connections
Structural isomers are compounds with the same molecular formula but different structural formulas. C₄H₁₀ examples are CH₃CH₂CH₂CH₃ and CH₃CH(CH₃)CH₃: the first is unbranched, while the second has a branch. Both contain four C and ten H, but the carbon connections differ.
The required C₄H₈ pair is CH₃CH₂CH=CH₂ and CH₃CH=CHCH₃, corresponding to different positions of the double bond in an unbranched four-carbon chain. They are structural isomers because connection/bond-order placement differs. Reading a chain from the opposite end can write the same compound differently, so confirm actual connectivity instead of counting every rewritten or rotated formula as a new isomer.
Pause and explain
Why are the two required C₄H₁₀ examples structural isomers?
Worked example
Compare CH₃CH₂CH₂CH₃ and CH₃CH(CH₃)CH₃ for molecular totals and carbon connectivity.
Show the worked solution
- The unbranched chain contains C₄H₁₀.
- The branched formula contains three CH₃ groups and one CH, also C₄H₁₀.
- The central CH in the second connects to three carbons, unlike the unbranched chain's internal carbons; they are structural isomers.
Answer Same C₄H₁₀ molecular formula, different unbranched/branched connectivity: structural isomers.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Interpret CH₂=CH₂, CH₃CH₂OH and CH₃COOCH₃, giving atom totals while retaining their distinct connections.
Give me a hint
Group notation still needs every atom counted.
Compare my reasoning
- CH₂=CH₂ gives C₂H₄ with a carbon double bond.
- CH₃CH₂OH gives C₂H₆O with a C–O–H end.
- CH₃COOCH₃ gives C₃H₆O₂ with the ester connection, rather than another ethanol formula.
C₂H₄ ethene; C₂H₆O ethanol; C₃H₆O₂ methyl ethanoate, with different explicitly described connections.
Look for these in your work
- All three structural formulas are interpreted.
- Grouped atoms and bond connections are not erased.
Compare CH₃CH₂CH=CH₂ and CH₃CH=CHCH₃ for totals and double-bond position.
Give me a hint
Count H in each written group before deciding whether the molecular formulas match.
Compare my reasoning
- The first totals C₄H₈ and has the double bond at an end position.
- The second also totals C₄H₈ but its double bond is between the middle carbons.
- Same totals and different double-bond placement give structural isomers.
Both C₄H₈; end versus middle double-bond position makes them structural isomers.
Look for these in your work
- Both hydrogen totals are checked.
- The positional structural difference is identified.
A diagram of unbranched butane is turned upside down and called a new structural isomer. Compare that operation with making the required branched C₄H₁₀ example.
Give me a hint
Check bonds rather than page orientation.
Compare my reasoning
- Whole rotation preserves which atoms are connected in unbranched butane.
- The branched example instead changes carbon connectivity while keeping C₄H₁₀ totals.
- Therefore rotation alone is not a new structural isomer; a changed arrangement with the same formula is required.
Rotation keeps the same structure; branching changes connectivity at the same C₄H₁₀ total and gives the required different isomer.
Look for these in your work
- Orientation and connectivity are separated.
- The same-formula condition is retained.
Common mistakes
- Calling compounds with different atom totals structural isomers.
- Treating brackets as a separate unconnected molecule.
- Calling every rotated whole drawing a new structural isomer.
- Ignoring double-bond position when comparing the required C₄H₈ pair.
What can you explain now?
Define structural isomers.
Compare with the explanation
Compounds with the same molecular formula but different structural formulas.
Both matching totals and different arrangement are required.
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Make it stick
Interpret the three required structural formulas, draw both C₄H₁₀ arrangements and locate the two C₄H₈ double-bond positions.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.