Isomerism Exam Questions

(a) Structural isomers are compounds that have the same molecular formula [1] but different structural formulae / different arrangements of atoms [1]

(b) Any valid pair, e.g.: butane (CH3CH2CH2CH3) and methylpropane (CH3CH(CH3)CH3) — both have the formula C4H10 [1]

Examiner note: “Different structures” alone is insufficient — you must say “same molecular formula AND different structural formulae.” Isomers are different compounds with different properties, not just different ways of drawing the same molecule. Rotating or flipping a structure does not create a new isomer.

Isomer 1: Butane — CH3CH2CH2CH3 (a straight chain of 4 carbon atoms) [2] (1 mark for correct structure, 1 mark for name)

Isomer 2: Methylpropane (2-methylpropane) — CH3CH(CH3)CH3 (a chain of 3 carbon atoms with a CH3 branch on the middle carbon) [2] (1 mark for correct structure, 1 mark for name)

Examiner note: C4H10 has only two isomers. A common error is to draw “1-methylpropane” — this is just butane drawn from a different starting point, not a new isomer. Always check: is the longest chain the same length? Are the branches in different positions? If the answer to both is no, you have the same molecule.

Isomer 1: Pentane — CH3CH2CH2CH2CH3 (straight chain of 5 carbons) [2]

Isomer 2: Methylbutane (2-methylbutane) — CH3CH(CH3)CH2CH3 (chain of 4 carbons with a CH3 branch on the second carbon) [2]

Isomer 3: Dimethylpropane (2,2-dimethylpropane) — C(CH3)4 / CH3C(CH3)2CH3 (chain of 3 carbons with two CH3 branches on the central carbon) [2]

Examiner note: There are exactly three isomers of C5H12. Approach systematically: start with the longest chain (5C), then try 4C + 1 branch, then 3C + 2 branches. Placing the branch on carbon 1 of the 4C chain gives pentane again (just start counting from the other end), so the only unique position is carbon 2. For the 3C chain, both branches must go on the central carbon.

• Butane (straight chain) has a higher boiling point than methylpropane (branched) [1]
• The straight-chain molecule has a larger surface area, allowing more contact between molecules [1]
• This results in stronger intermolecular forces (van der Waals forces), so more energy is needed to separate the molecules [1]

Examiner note: Branching makes a molecule more compact/spherical, reducing the surface area available for intermolecular contact. Fewer points of contact mean weaker intermolecular forces and a lower boiling point. Both isomers have the same molecular formula and the same relative molecular mass, so the only variable is shape.

(a) Isomer 1: Propan-1-ol — CH3CH2CH2OH [1] Isomer 2: Methoxyethane — CH3OCH2CH3 [1]

(b) Propan-1-ol is an alcohol; methoxyethane is an ether [1]

(c) Add sodium metal to each: propan-1-ol reacts (fizzes, producing hydrogen gas) while methoxyethane does not react with sodium [1]

OR: Add acidified potassium dichromate and warm: propan-1-ol causes the solution to change from orange to green, while methoxyethane shows no colour change [1]

Examiner note: Isomers can belong to different homologous series if the atoms are arranged to give different functional groups. Propan-1-ol has an -OH group, making it an alcohol. Methoxyethane has an -O- group (ether linkage) between two carbon chains. They have the same molecular formula but very different chemical properties.

(a) Isomer 1: But-1-ene — CH2=CHCH2CH3 (double bond between carbon 1 and carbon 2) [1] Isomer 2: But-2-ene — CH3CH=CHCH3 (double bond between carbon 2 and carbon 3) [1]

(b) A cycloalkane with 4 carbon atoms (cyclobutane) also has the formula C4H8 because the ring removes two hydrogen atoms compared to the open-chain alkane C4H10, giving the same formula as the alkene [1]

(c) Add bromine water: the alkene decolourises bromine water (addition reaction across the C=C), but cyclobutane does not (it has no C=C double bond, only single bonds in a ring) [1]

Examiner note: Cycloalkanes have the same general formula as alkenes (CnH2n) but are saturated — they contain no double bonds. The ring structure accounts for the “missing” two hydrogen atoms. At IGCSE, you need to recognise that isomers can include ring structures as well as chain isomers and positional isomers.