Chemistry · Year 12 · Module 7 · Lesson 4
HSC Exam Practice
Hydrocarbon Structure & Bonding
Short answer
1.Short answer — definitions and explanations
Define sigma bond (σ) and pi bond (π). In your answer, include the type of orbital overlap involved in each.
Identify the hybridisation state of the carbon atoms in each of the following molecules and state the expected bond angle at each carbon: (a) propane, (b) propene (at the C=C carbons), (c) propyne (at the C≡C carbons).
Explain why alkanes with only C–C single bonds can rotate freely about every C–C bond, whereas alkenes cannot rotate freely about the C=C double bond.
Describe the conditions necessary for geometric (cis-trans) isomerism to exist in an alkene. State whether but-1-ene (CH₂=CH–CH₂–CH₃) displays geometric isomerism and justify your answer.
Outline, with reference to bond length and bond energy data, how the bond between two carbon atoms changes as bond order increases from 1 to 3.
Distinguish between cis-but-2-ene and trans-but-2-ene with reference to the spatial arrangement of their substituents and one measurable physical property that differs between them.
Data response
2.Data response — bond energy and reactivity
The bar chart below shows the average bond energy (kJ mol−1) for C–C, C=C, and C≡C bonds. Use it to answer (a) and (b).
(a) Describe the trend shown in the bar chart and account for it using the concept of sigma and pi bonds.
(b) A student claims that because C≡C has the highest bond energy, alkynes will be the least reactive hydrocarbons towards addition reactions. Evaluate this claim using information from the bar chart.
3.Data response — boiling points of geometric isomers
The table below shows the boiling points of cis-but-2-ene and trans-but-2-ene. Use it to answer (a) and (b).
| Compound | Boiling point (°C) | Dipole moment (D) |
|---|---|---|
| cis-but-2-ene | +3.7 | 0.33 |
| trans-but-2-ene | +0.9 | 0.00 |
Source: CRC Handbook of Chemistry and Physics, 103rd edn (2022).
(a) Account for the difference in boiling points between the two isomers, with reference to their dipole moments and intermolecular forces.
(b) A student separates a mixture of cis- and trans-but-2-ene by careful fractional distillation. Explain why this separation is possible, using your knowledge of bonding and isomerism.
Extended response
4.Extended response
Assess the impact of hybridisation and bond type on the structure, physical properties, and chemical reactivity of hydrocarbons. In your response, refer to at least one example of an alkane, an alkene, and an alkyne.
Chemistry · Year 12 · Module 7 · Lesson 4
Answer Key & Marking Guidelines
Section 1 · Short answer · 3 marks · Band 3
Sample response. A sigma bond is formed by head-on (axial) overlap of atomic orbitals along the internuclear axis; it allows free rotation about the bond axis. A pi bond is formed by side-on (lateral) overlap of unhybridised p orbitals above and below the internuclear axis; rotation about a pi bond would break the overlap and is therefore restricted.
Marking notes. 1 mark for sigma = head-on overlap (or axial/direct overlap). 1 mark for pi = side-on overlap (or lateral/parallel p-orbital overlap). 1 mark for identifying that sigma allows rotation and pi does not (or equivalent distinction).
Section 1 · Short answer · 3 marks · Band 3
Sample response. (a) Propane: sp³, 109.5°. (b) Propene at C=C carbons: sp², 120°. (c) Propyne at C≡C carbons: sp, 180°.
Marking notes. 1 mark per correct hybridisation + bond angle pair. Accept either order within each pair.
Section 1 · Short answer · 3 marks · Band 4
Sample response. In alkanes, C–C single bonds consist of only a sigma bond. Because the sigma bond is formed by head-on orbital overlap along the internuclear axis, rotation about the axis does not disrupt the overlap — so rotation is free. In alkenes, the C=C double bond contains one sigma bond and one pi bond. The pi bond is formed by side-on p-orbital overlap; rotating about the C=C axis by 90° would destroy this parallel overlap, breaking the pi bond (requiring ∼270 kJ mol−1). This energy is not available at room temperature, so rotation about C=C does not occur freely.
Marking notes. 1 mark for free rotation in alkanes attributed to sigma-only bond (rotation does not disrupt overlap). 1 mark for identifying that C=C contains a pi bond formed by side-on p-orbital overlap. 1 mark for explaining that rotation would break the pi bond (energy required is too high at room temperature).
Section 1 · Short answer · 3 marks · Band 4
Sample response. Conditions for geometric isomerism: (1) the molecule must contain a C=C double bond (rotation is restricted); (2) each carbon of the double bond must have two different substituents. But-1-ene (CH₂=CH–CH₂–CH₃) does not display geometric isomerism because the terminal =CH₂ carbon has two identical hydrogen substituents (H and H), so swapping them does not produce a new isomer.
Marking notes. 1 mark for stating C=C (restricted rotation) is required. 1 mark for stating each double-bond carbon must carry two different substituents. 1 mark for correctly identifying that but-1-ene does not display geometric isomerism and correctly justifying why (terminal CH₂ group has two identical H atoms).
Section 1 · Short answer · 2 marks · Band 3
Sample response. As bond order increases from 1 (C–C, 154 pm, 347 kJ mol−1) to 2 (C=C, 134 pm, 614 kJ mol−1) to 3 (C≡C, 120 pm, 839 kJ mol−1), bond length decreases and bond energy increases. Each additional bond adds electron density between the nuclei, pulling them closer (shorter bond) and requiring more energy to separate them (higher bond energy).
Marking notes. 1 mark for identifying bond length decreases with increasing bond order (with at least one data value). 1 mark for identifying bond energy increases with increasing bond order (with at least one data value).
Section 1 · Short answer · 3 marks · Band 4
Sample response. In cis-but-2-ene the two methyl (CH₃) groups are on the same side of the C=C double bond; in trans-but-2-ene they are on opposite sides. A measurable property that differs: cis-but-2-ene has a higher boiling point (+3.7 °C) than trans-but-2-ene (+0.9 °C), because the cis isomer has a net dipole moment (the bond dipoles do not cancel) leading to stronger permanent dipole–dipole intermolecular forces; the trans isomer is nonpolar (dipole moment = 0 D).
Marking notes. 1 mark for correctly describing the spatial arrangement of substituents for each isomer. 1 mark for naming a physical property that differs (boiling point or melting point accepted). 1 mark for a correct explanation linked to polarity / dipole moment or intermolecular forces.
Section 2 · Data response · 6 marks · Band 4–5
Sample response (a). Bond energy increases with bond order: C–C (347) < C=C (614) < C≡C (839 kJ mol−1). Each successive bond adds a pi bond (via side-on p-orbital overlap) to the existing sigma bond, introducing more electron density between the nuclei and requiring more energy to break the bond completely.
Sample response (b). The student's claim is partially incorrect. While C≡C has the highest total bond energy, the relevant bond energy for addition reactivity is the energy of the pi bond specifically, not the total bond energy. The pi bond of the C≡C (first pi: approx. 839 − 347 = 492 kJ mol−1 for both pi bonds) and C=C (pi bond: approx. 614 − 347 = 267 kJ mol−1) are both available for addition reactions. Alkenes and alkynes are both much more reactive than alkanes towards addition because they possess pi bonds that are more accessible and weaker than sigma bonds; alkanes (C–C sigma only) are essentially unreactive towards electrophilic addition. The claim that alkynes are "least reactive" is wrong if it compares them to alkanes — both alkynes and alkenes are far more reactive than alkanes towards addition.
Marking notes. Part (a): 1 mark for describing the increasing trend with quoted data values. 1 mark for attributing the increase to additional pi bonds / increased electron density. 1 mark for correctly linking bond order to number of pi bonds. Part (b): 1 mark for identifying that the claim conflates total bond energy with pi bond availability. 1 mark for explaining that it is the pi bond energy, not total bond energy, that determines reactivity toward addition. 1 mark for identifying that alkanes are actually least reactive (sigma-only bonds) and so the student's claim is wrong in the context of comparing all three classes.
Section 2 · Data response · 5 marks · Band 4–5
Sample response (a). cis-but-2-ene has a higher boiling point (+3.7 °C) than trans-but-2-ene (+0.9 °C) despite having the same molecular formula. The cis isomer has a net dipole moment of 0.33 D because the two C–CH₃ bond dipoles point in the same general direction (both methyl groups on the same side) and do not cancel. This net dipole creates permanent dipole–dipole attractions between molecules in addition to London dispersion forces, requiring more energy (higher boiling point) to overcome these forces. The trans isomer (dipole moment = 0 D) has its methyl groups on opposite sides so bond dipoles cancel; it relies only on London dispersion forces, which are weaker, giving a lower boiling point.
Sample response (b). Fractional distillation is possible because the two isomers have different boiling points (+3.7 vs +0.9 °C). The difference in boiling points arises because the compounds are genuinely different molecules (different spatial arrangements of atoms are locked in place by the restricted rotation around the C=C double bond). When the mixture is heated, the more volatile trans isomer (lower boiling point) preferentially distils over first, allowing separation.
Marking notes. Part (a): 1 mark for identifying that cis has a higher boiling point and linking to a net dipole moment; 1 mark for explaining why cis is polar (dipoles do not cancel) and trans is non-polar (dipoles cancel); 1 mark for correctly identifying the additional intermolecular force (permanent dipole–dipole) in the cis isomer that accounts for the higher boiling point. Part (b): 1 mark for stating that the different boiling points enable separation by distillation; 1 mark for linking the different boiling points back to restricted rotation around the C=C (pi bond) that locks the substituents in distinct spatial arrangements, making them different compounds.
Section 3 · Extended response · 7 marks · Band 5–6
Sample response. Hybridisation determines the electron geometry around each carbon atom and the types of bonds available, which in turn govern structure, physical properties, and reactivity. In alkanes (e.g. methane, CH₄), carbon is sp³ hybridised, forming four equivalent sigma bonds arranged tetrahedrally at 109.5°. Because only sigma bonds are present, free rotation occurs around every C–C bond, giving long-chain alkanes a flexible structure. Alkanes are generally non-polar and relatively unreactive — they undergo substitution (halogenation, e.g. CH₄ + Cl₂ → CH₃Cl + HCl under UV light) rather than addition, because there are no pi bonds available for electrophilic attack. In alkenes (e.g. ethene, C₂H₄), carbon at the double bond is sp² hybridised, producing a planar trigonal arrangement at 120°. The unhybridised p orbital on each sp² carbon forms a pi bond by side-on overlap; this restricts rotation about C=C. Restricted rotation means that substituents locked on the same side (cis) or opposite sides (trans) of the double bond cannot interconvert at room temperature, giving rise to geometric isomerism. The accessible pi bond makes alkenes highly reactive towards electrophilic addition reactions (e.g. ethene + HBr → bromoethane). In alkynes (e.g. ethyne, C₂H₂), carbon is sp hybridised, forming a linear molecule at 180°. Two pi bonds give ethyne the highest bond energy (839 kJ mol−1) but also the greatest electron density along the bond, making it reactive towards sequential addition. Overall, hybridisation state is the primary structural determinant: as hybridisation changes from sp³ → sp² → sp, bond angles decrease from 109.5° to 120° to 180° (in terms of the geometry of the hybrid bonds), bonds shorten and strengthen, pi-bond character increases, free rotation is progressively replaced by restricted rotation, geometric isomerism becomes possible in sp² systems, and reactivity towards addition increases relative to alkanes.
Marking notes. 1 mark — sp³ alkane correctly described (sigma only, free rotation, tetrahedral 109.5°, named example). 1 mark — sp² alkene correctly described (sigma + pi, 120°, trigonal planar, named example). 1 mark — sp alkyne correctly described (sigma + 2 pi, linear, 180°, named example). 1 mark — pi bond restriction explicitly linked to geometric isomerism in alkenes. 1 mark — reactivity towards addition correctly linked to presence and accessibility of pi bonds (alkanes unreactive, alkenes/alkynes reactive). 1 mark — at least one trend in bond length or bond energy cited with supporting values (e.g. single 154 pm/347 kJ mol−1, double 134/614, triple 120/839). 1 mark — an evaluative statement assessing the overall impact of hybridisation, reaching a conclusion about how hybridisation state determines a hydrocarbon's physical and chemical behaviour.