Chemistry · Year 12 · Module 7 · Lesson 1
HSC Exam Practice
Introduction to Organic Chemistry & IUPAC Nomenclature I
Short answer
1.Short answer
Define organic chemistry and explain why the classification of a compound as “organic” does not require it to be produced by a living organism.
A carbon atom is described as sp2-hybridised. State the geometry around that carbon, the bond angle between its sigma bonds, and identify the type of carbon–carbon bond that must be present.
Write the correct IUPAC name for the compound with condensed formula CH3CH(CH3)CH2CH2CH3. Show your working (longest chain, numbering direction, substituent name and locant).
Distinguish between the terms homologous series and isomers. Include in your answer a named example of each concept drawn from the alkane family.
Outline the three rules used to assign the correct IUPAC name to an alkene. Apply these rules to name the compound CH3CH2CH2CH=CHCH3.
A student claims that a compound with molecular formula C6H10 must be an alkene. Assess this claim.
Data response
2.Data response — boiling points and Australian natural gas
The graph below shows the normal boiling points of the first six straight-chain alkanes.
(a) Describe the trend shown in Figure 2.1 and explain the trend using intermolecular forces.
(b) Karratha LNG (Woodside Energy, Western Australia) liquefies and exports natural gas, which is approximately 90% methane and 7% ethane. Using Figure 2.1, explain why the gas must be cooled to approximately −162°C for export as a liquid.
(c) At a storage facility in Sydney, LNG warms to −80°C during a brief power outage. Identify which component(s) of the LNG would vaporise and which would remain liquid. Justify your answer using the graph.
3.Multi-step calculation — verifying molecular formulas
A chemist at CSL (Commonwealth Serum Laboratories, Melbourne) synthesises three compounds labelled P, Q, and R during early-stage pharmaceutical research. The structural information recorded for each is:
- P: IUPAC name: 2,3-dimethylpentane
- Q: Condensed formula: CH3CH2C≡CCH2CH3
- R: Molecular formula stated as C7H12; identified as an alkyne
(a) For compound P: write the molecular formula and verify it against the appropriate general formula. State the hybridisation of every carbon in the chain.
(b) For compound Q: write the IUPAC name and state the molecular formula. Verify the formula using the appropriate general formula.
(c) For compound R: determine whether C7H12 is consistent with a straight-chain alkyne. If it is inconsistent, identify and explain the error.
Extended response
4.Extended response
Evaluate the claim that carbon’s structural diversity is entirely determined by its tetravalency. In your response, refer to hybridisation states, functional groups, homologous series, and at least one real-world application drawn from the Australian chemical industry.
Chemistry · Year 12 · Module 7 · Lesson 1
Answer Key & Marking Guidelines
Section 1 · Short answer · 2 marks · Band 3
Sample response. Organic chemistry is the study of carbon-containing compounds, regardless of their origin. A compound is classified as organic based on its structural feature (a carbon backbone), not on whether it was produced by a living organism. For example, methane is formed by geological processes but is classified as organic because it is a carbon-containing compound.
Marking notes. 1 mark for defining organic chemistry as the study of carbon-containing compounds; 1 mark for explaining that classification is based on molecular structure (carbon content), not biological origin.
Section 1 · Short answer · 3 marks · Band 3
Sample response. An sp2-hybridised carbon has trigonal planar geometry. The bond angle between its three sigma bonds is approximately 120°. The presence of sp2 hybridisation indicates a C=C double bond (a pi bond formed by sideways overlap of unhybridised p orbitals).
Marking notes. 1 mark for trigonal planar geometry; 1 mark for ~120° bond angle; 1 mark for identifying a C=C double bond (or equivalent: “double bond present”).
Section 1 · Short answer · 3 marks · Band 3–4
Sample response. Longest chain: CH3–CH–CH2–CH2–CH3 = 5 carbons (pentane). Branch: –CH3 (methyl). Number from the end nearer the branch (left end): CH3(C1)–CH(C2)–CH2(C3)–CH2(C4)–CH3(C5); methyl branch at C2. IUPAC name: 2-methylpentane.
Marking notes. 1 mark for identifying 5C parent chain (pentane); 1 mark for correct numbering direction giving locant 2; 1 mark for correct full name (2-methylpentane). Penalise 1 mark for “4-methylpentane” (wrong end of numbering) but accept working shown.
Section 1 · Short answer · 4 marks · Band 4
Sample response. A homologous series is a family of compounds that share the same functional group and general formula, with consecutive members differing by one –CH2– unit and therefore having different molecular formulas. Example: propane (C3H8) and butane (C4H10) are consecutive members of the alkane homologous series. Isomers are compounds that have the same molecular formula but different structural arrangements. Example: butane and 2-methylpropane both have the formula C4H10 but different structures.
Marking notes. 1 mark for homologous series definition (same functional group, differ by CH2); 1 mark for a correct named alkane example (propane/butane as consecutive members); 1 mark for isomers definition (same molecular formula, different structure); 1 mark for correct named isomer example (butane/2-methylpropane, C4H10).
Section 1 · Short answer · 4 marks · Band 4
Sample response. Three alkene naming rules: (1) Find the longest chain that includes the C=C double bond — this is the parent chain; change suffix to –ene. (2) Number the chain from the end nearer to the double bond to give the lowest locant to the C=C. (3) Insert the locant of the lower-numbered C=C carbon immediately before “ene” (e.g. but-1-ene). Application: CH3CH2CH2CH=CHCH3. Chain including C=C: 6 carbons (hexene). Numbering from the CH3 end at right gives C=C at C2; from the left end gives C=C at C4. Choose C2 (lower locant). Name: hex-2-ene.
Marking notes. 1 mark each for the three rules stated; 1 mark for correct application (hex-2-ene). Also accept “2-hexene” as an older but still acceptable NESA format.
Section 1 · Short answer · 3 marks · Band 4
Sample response. The claim is partially correct but incomplete. Alkenes have the general formula CnH2n; for n = 6: C6H12 ≠ C6H10, so C6H10 does not fit the alkene series. Checking alkynes (CnH2n−2): n = 6 → 2(6) − 2 = 10 H → C6H10 — a match. Therefore C6H10 is consistent with an alkyne (e.g. hex-1-yne), not an alkene. The student’s claim is incorrect.
Marking notes. 1 mark for testing and ruling out the alkene formula; 1 mark for correctly applying the alkyne formula and finding C6H10 consistent; 1 mark for a clear, justified conclusion that the compound is an alkyne, not an alkene.
Section 2 · Data response · 7 marks · Band 4–5
(a) Trend + explanation (3 marks). Boiling point increases from methane (−162°C) to hexane (69°C) as chain length increases [1]. The increase occurs because longer chains have more electrons and greater molecular surface area, strengthening London (dispersion) forces between molecules [1]. Stronger intermolecular forces require more energy to overcome during vaporisation, raising the boiling point [1].
(b) LNG cooling (2 marks). To exist as a liquid, a substance must be below its boiling point. Methane has a boiling point of −162°C — above this temperature it is a gas. Ethane’s boiling point is −89°C. Since the mixture is ~90% methane, it must be kept below −162°C (the boiling point of the most volatile component) to maintain the entire mixture as a liquid for storage and tanker transport [2].
(c) Which vaporises at −80°C (2 marks). At −80°C, any component with a boiling point below −80°C will have vaporised. From the graph: methane (bp −162°C) and ethane (bp −89°C) both have boiling points below −80°C, so both would vaporise [1]. Propane (bp −42°C) would also vaporise. Butane (bp −1°C), pentane, and hexane would remain liquid because their boiling points are above −80°C. Since LNG is ~90% methane + ~7% ethane, almost all of the gas would have boiled off [1].
Section 2 · Multi-step calculation · 8 marks · Band 4–5
(a) Compound P — 2,3-dimethylpentane (3 marks). Parent chain = 5C; two methyl branches (at C2 and C3) = 2 × 1C = 2C extra. Total C = 7. Using CnH2n+2 with n = 7: H = 16. Molecular formula = C7H16 [1]. Verification: C7H2(7)+2 = C7H16 — confirmed [1]. Hybridisation: all 7 carbons are sp3 because the compound contains only C–C and C–H single bonds; no double or triple bonds are present [1].
(b) Compound Q — CH3CH2C≡CCH2CH3 (3 marks). Functional group: C≡C triple bond → alkyne. Chain: 6 carbons including both C≡C carbons → parent = hex. Number from the end nearer the triple bond: from the left, C1–CH3, C2–CH2, C3≡C4, C5–CH2, C6–CH3; triple bond at C3 from either end (symmetric). Name: hex-3-yne [1]. Molecular formula: 6 carbons. CnH2n−2, n = 6: H = 10. Formula = C6H10 [1]. Verification: count H from condensed formula: CH3(3) + CH2(2) + 0 + 0 + CH2(2) + CH3(3) = 10 H — confirmed [1].
(c) Compound R — C7H12 as alkyne check (2 marks). Test general formula CnH2n−2 with n = 7: expected H = 2(7) − 2 = 12. C7H12 gives exactly 12 H — this IS consistent with an alkyne [1]. Therefore the molecular formula C7H12 is not an error; it matches a hept-X-yne. The student’s formula is correct. Full marks for correctly verifying and confirming consistency (do not penalise if student states “consistent, no error”) [1].
Section 3 · Extended response · 9 marks · Band 5–6
Marking criteria (9 marks).
1. Tetravalency defined correctly: carbon forms exactly 4 covalent bonds with itself and with H, O, N, halogens, enabling chains, branches, rings.
2. Evaluates the claim honestly: tetravalency is necessary but not sufficient for all of carbon’s diversity — the three hybridisation states (sp3, sp2, sp) add another dimension of structural variety.
3. Explains sp3 (tetrahedral, 109.5°, single bonds only → alkanes), sp2 (trigonal planar, 120°, C=C → alkenes, carbonyl groups), and sp (linear, 180°, C≡C → alkynes) with correct geometry and bond angle for each.
4. Explains how different hybridisation states create different functional groups (–OH, C=O, C=C etc.), each with distinct reactivity patterns.
5. Links functional groups to homologous series: each series (alkanes, alkenes, alcohols etc.) shares a functional group and general formula; boiling points and reactivity vary predictably within each series.
6. Provides a relevant Australian chemical industry example, such as: Karratha LNG (methane/ethane — alkane homologous series, sp3); Ampol Lytton refinery (separation of petroleum based on chain-length boiling point trend); CSL pharmaceuticals (complex organic molecules with multiple functional groups, diverse hybridisation).
7. Reaches a justified conclusion: tetravalency alone explains chain-forming ability (structural backbone diversity), but hybridisation variation and functional group chemistry are additional features that extend carbon’s structural and chemical diversity beyond what tetravalency alone predicts.
8. Response is coherently structured (introduction → analysis → examples → evaluation → conclusion).
9. Correct chemical terminology and notation throughout (no errors in hybridisation symbols, general formulas, or IUPAC examples cited).
Band 5 (7–8 marks): addresses criteria 1–7 with minor gaps. Band 6 (9 marks): all criteria addressed cohesively with specific Australian examples and a nuanced evaluative conclusion.