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.
Extended only (Supplement) · Chemistry 0620, 2026–2028.
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.