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Questions on structural isomerism and E/Z isomerism
Structural isomerism of three kinds, chain, position and functional group, with examples for C₄H₁₀, C₄H₉Br, C₃H₆O and C₄H₈O; stereoisomerism; E/Z isomerism from restricted rotation about the C=C bond; the Cahn–Ingold–Prelog priority rules; cis and trans as a special case; and the effect on boiling points, melting points and fats.
20 original questions · 64 marks · the structural isomerism and E/Z isomerism notes · Introduction to organic chemistry · Flashcards for this unit
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Discuss the isomerism shown by the alkenes with molecular formula C₄H₈: but-1-ene, but-2-ene and 2-methylpropene.
Mark scheme
Indicative content
- but-1-ene and but-2-ene are position isomers, the C=C starting on a different carbon of the same unbranched chain
- 2-methylpropene has a branched skeleton, so it is a chain isomer of both butenes
- but-2-ene is the only one of the three to show E/Z isomerism, existing as E-but-2-ene and Z-but-2-ene
- this is because rotation about the C=C is restricted and each carbon of the C=C carries two different groups, CH₃ and H
- but-1-ene and 2-methylpropene show no E/Z isomerism, because one carbon of the C=C carries two identical groups
- the cycloalkanes cyclobutane and methylcyclopropane are also C₄H₈ and are functional group isomers of these alkenes
Levels
- Level 3 (5–6 marks) all three kinds of structural isomerism in the set are identified with the pairs that show them, E/Z isomerism is explained for but-2-ene through restricted rotation and two different groups on each carbon, and the answer says why the other two alkenes show no E/Z isomerism
- Level 2 (3–4 marks) position and chain isomerism are identified and but-2-ene is said to show E/Z isomerism; restricted rotation and two different groups on each carbon, or why the other two alkenes do not show it, are not explained
- Level 1 (1–2 marks) one kind of isomerism is named; but-1-ene is often said to show E/Z isomerism too because it has a C=C bond
Give the names of the five structural isomers with the molecular formula C₆H₁₄.
Mark scheme
- B1 hexane, CH₃CH₂CH₂CH₂CH₂CH₃, the unbranched chain
- B1 2-methylpentane, (CH₃)₂CHCH₂CH₂CH₃, a chain of five with a methyl branch on carbon 2
- B1 3-methylpentane, CH₃CH₂CH(CH₃)CH₂CH₃, a chain of five with a methyl branch on the middle carbon
- B1 2,2-dimethylbutane, (CH₃)₃CCH₂CH₃, a chain of four with two methyl branches on carbon 2
- B1 2,3-dimethylbutane, (CH₃)₂CHCH(CH₃)₂, a chain of four with one methyl branch on each inner carbon
State what is meant by structural isomers and by stereoisomers, and state which kind of isomerism each of these pairs shows: butan-1-ol and butan-2-ol; E-but-2-ene and Z-but-2-ene.
Mark scheme
- B1 structural isomers have the same molecular formula but a different structural formula
- B1 stereoisomers have the same structural formula but a different arrangement of the atoms in space; reject 'same molecular formula' alone
- B1 butan-1-ol and butan-2-ol are structural isomers, position isomers with the OH on a different carbon
- B1 E- and Z-but-2-ene are stereoisomers, E/Z isomers about the C=C, with the same structural formula
Give the names of the four structural isomers with the molecular formula C₄H₉Br.
Mark scheme
- B1 1-bromobutane, CH₃CH₂CH₂CH₂Br, with Br at the end of the unbranched chain
- B1 2-bromobutane, CH₃CH₂CHBrCH₃, with Br on an inner carbon of the unbranched chain
- B1 1-bromo-2-methylpropane, (CH₃)₂CHCH₂Br, with Br at an end of the branched chain
- B1 2-bromo-2-methylpropane, (CH₃)₃CBr, with Br on the branching carbon
Justify the name (Z)-3-methylpent-2-ene for the isomer of CH₃CH=C(CH₃)CH₂CH₃ in which the two methyl groups attached to the carbons of the C=C are on opposite sides of the double bond.
Mark scheme
- B1 the longest chain through the C=C has five carbons, numbered from the end that gives the double bond the lower locant, so the stem is pent-2-ene with the methyl branch on carbon 3
- B1 on carbon 2 the groups are CH₃ and H, and CH₃ has the higher priority because carbon's atomic number, 6, exceeds hydrogen's, 1
- B1 on carbon 3 the groups are CH₃ and C₂H₅; the first atoms tie, so the next atoms out are compared, (C, H, H) for ethyl against (H, H, H) for methyl, and C₂H₅ has the higher priority
- B1 with the two methyls opposite, the CH₃ on carbon 2 lies on the same side as the C₂H₅ on carbon 3, so the two higher-priority groups are together and the isomer is Z; reject E from treating the methyls as the groups compared
Explain why but-2-ene, CH₃CH=CHCH₃, exists as E and Z isomers.
Mark scheme
- B1 there is restricted rotation about the C=C double bond; reject 'no rotation about the molecule'
- B1 each carbon of the C=C carries two different groups, a CH₃ and an H
- B1 so the two CH₃ groups can be on the same side (Z) or on opposite sides (E), and the two cannot interconvert
Explain, in terms of the bonding, why there is no rotation about a C=C double bond at room temperature.
Mark scheme
- B1 the double bond consists of a σ bond and a π bond
- B1 the π bond forms by the sideways overlap of a p orbital on each carbon, above and below the plane of the molecule
- B1 rotation would twist the p orbitals out of line and break the π bond, which needs far more energy than is available at room temperature
Explain why 2-methylpropane boils at a lower temperature than its chain isomer butane.
Mark scheme
- B1 2-methylpropane is branched and more compact, so its molecules have a smaller area of contact with each other
- B1 the van der Waals forces between its molecules are therefore weaker
- B1 less energy is needed to overcome these forces, so its boiling point is lower; reject any answer that breaks covalent bonds on boiling
Identify, as E or Z, the isomer of CH₃CCl=CHCH₂CH₃ in which the Cl atom and the C₂H₅ group are on opposite sides of the double bond, giving the priority on each carbon.
Mark scheme
- B1 on carbon 2, Cl has a higher priority than CH₃, because Cl has the higher atomic number, 17 against 6
- B1 on carbon 3, C₂H₅ has a higher priority than H, atomic number 6 against 1
- B1 the two higher-priority groups are on opposite sides, so the isomer is E-2-chloropent-2-ene; accept (E)-2-chloropent-2-ene
Suggest why Z-butenedioic acid, HOOCCH=CHCOOH, loses water to form a cyclic compound on heating, but E-butenedioic acid does not.
Mark scheme
- B1 in the Z isomer both COOH groups are on the same side of the C=C, close to each other
- B1 they can react with each other, losing a molecule of H₂O and closing a ring
- B1 in the E isomer the COOH groups are on opposite sides, too far apart, and restricted rotation stops them being brought together
Suggest why a fat made from fatty acids with E double bonds has a higher melting point than an oil made from the Z isomers of the same acids.
Mark scheme
- B1 a Z double bond puts a kink in the hydrocarbon chain, whereas an E chain is nearly straight
- B1 straight chains pack more closely together, as saturated chains do, with a larger area of contact
- B1 the van der Waals forces between the E chains are stronger, so more energy is needed to melt the fat
Compare the cis/trans and E/Z systems for naming the isomer of 2-chlorobut-2-ene, CH₃CCl=CHCH₃, in which the two CH₃ groups are on the same side.
Mark scheme
- B1 both systems describe the arrangement of groups about a C=C that cannot rotate
- B1 cis/trans compares the positions of the carbon chain, so this isomer, with both CH₃ groups on one side, is cis
- B1 E/Z compares the CIP highest-priority groups; Cl outranks CH₃ on carbon 2, so the priority groups are opposite and the isomer is E
Give the names of the three structural isomers with the molecular formula C₅H₁₂.
Mark scheme
- B1 pentane, CH₃CH₂CH₂CH₂CH₃, the unbranched chain of five
- B1 2-methylbutane, CH₃CH(CH₃)CH₂CH₃, a chain of four with one branch
- B1 2,2-dimethylpropane, C(CH₃)₄, a chain of three with two branches on the middle carbon
Evaluate the claim that 1,1-dichloroethene and 1,2-dichloroethene are stereoisomers of each other.
Mark scheme
- B1 both are C₂H₂Cl₂, so they are isomers
- B1 the Cl atoms are on different carbons, so the structural formulae differ and they are structural (position) isomers, not stereoisomers
- B1 1,2-dichloroethene does have stereoisomers of its own, the E and Z forms, whereas 1,1-dichloroethene has none because one carbon carries two hydrogens
Explain why 1-chloroprop-1-ene, CHCl=CHCH₃, shows E/Z isomerism but 3-chloroprop-1-ene, CH₂=CHCH₂Cl, does not.
Mark scheme
- B1 both have a C=C, about which rotation is restricted
- B1 in 1-chloroprop-1-ene carbon 1 carries Cl and H and carbon 2 carries CH₃ and H, so each carbon of the C=C has two different groups
- B1 in 3-chloroprop-1-ene carbon 1 carries two hydrogen atoms, and swapping identical groups gives the same molecule, so there are no E/Z isomers
Suggest how many isomers, including stereoisomers, have the molecular formula C₂H₂BrCl, and name each one.
Mark scheme
- B1 with Br and Cl on the same carbon there is one compound, 1-bromo-1-chloroethene, CH₂=CBrCl, which has no E/Z isomers because the other carbon carries two hydrogens
- B1 with Br and Cl on different carbons the compound is 1-bromo-2-chloroethene, CHBr=CHCl, in which each carbon carries two different groups, so it has E and Z isomers
- B1 there are three isomers in all: 1-bromo-1-chloroethene, E-1-bromo-2-chloroethene and Z-1-bromo-2-chloroethene
Name the two chain isomers with the molecular formula C₄H₁₀.
Mark scheme
- B1 butane, CH₃CH₂CH₂CH₃, the unbranched isomer
- B1 2-methylpropane, CH₃CH(CH₃)CH₃, the branched isomer
Identify the type of structural isomerism shown by propanal and propanone, and by 1-chloropropane and 2-chloropropane.
Mark scheme
- B1 propanal and propanone are functional group isomers, an aldehyde and a ketone
- B1 1-chloropropane and 2-chloropropane are position isomers, with the Cl on a different carbon
Explain why but-1-ene, CH₂=CHCH₂CH₃, does not show E/Z isomerism.
Mark scheme
- B1 carbon 1 of the C=C carries two hydrogen atoms, which are identical groups
- B1 swapping two identical groups gives the same molecule, so only one compound exists
Outline how the Cahn–Ingold–Prelog rules decide whether CH₃ or CH₂OH has the higher priority.
Mark scheme
- B1 both groups are joined through a carbon atom, so the first atoms tie and the next atoms out are compared
- B1 CH₂OH carries (O, H, H) and CH₃ carries (H, H, H); O has the higher atomic number, so CH₂OH has the higher priority
Practise structural isomerism and E/Z isomerism one question at a time