HSC Exam Practice

Sample response. An aldehyde is an organic compound in which the carbonyl group (C=O) is located at the terminal (end) carbon of the chain; the carbonyl carbon carries a hydrogen atom, giving the functional group −CHO. A ketone is an organic compound in which the carbonyl group is bonded to two carbon atoms within the chain; the carbonyl carbon carries no hydrogen atom, giving the general structure R–CO–R′.

Marking notes. 1 mark for aldehyde: terminal C=O + H on carbonyl C (or −CHO explicitly). 1 mark for ketone: internal C=O + no H on carbonyl C (or R–CO–R′ explicitly).

Sample response. (a) CH₃CH₂CHO: 3 carbons, C=O at chain end → propanal (aldehyde). (b) CH₃COCH₂CH₃: 4 carbons, C=O at C2 → butan-2-one (ketone). (c) HCHO: 1 carbon, C=O at chain end with two H atoms → methanal (aldehyde).

Marking notes. 1 mark per correct name + classification pair. Deduct if name correct but class wrong, or vice versa.

Sample response. Propan-1-ol contains an O–H bond, which allows it to both donate and accept hydrogen bonds between molecules. These O–H ··· O hydrogen bonds are strong IMFs requiring substantial energy to break, leading to a high boiling point of 97 °C. Propanal has no O–H bond — it contains only a polar C=O group, enabling dipole-dipole interactions between molecules, but cannot donate H-bonds between its own molecules. Dipole-dipole forces are weaker than H-bonds, so less energy is needed to separate propanal molecules, giving a lower bp of 49 °C. The difference is therefore due to the type and strength of IMF, not molecular mass.

Marking notes. 1 mark for correctly identifying O–H H-bonding (both donor and acceptor) in propan-1-ol; 1 mark for correctly stating propanal has dipole-dipole forces only (no O–H, cannot donate H-bonds); 1 mark for explicitly linking stronger IMF in propan-1-ol to its higher BP.

Sample response. Procedure: add a few drops of the unknown organic compound to freshly prepared Tollens’ reagent ([Ag(NH₃)₂]⁺, ammoniacal silver nitrate) in a clean glass test tube; warm gently in a water bath at ~50 °C for 2–3 minutes. Positive result (aldehyde): a shiny silver mirror forms on the inside of the test tube as Ag⁺ is reduced to metallic silver Ag⁰. Negative result (ketone): no silver mirror; solution remains clear/colourless. Safety note: Tollens’ reagent must be prepared immediately before use and used promptly — aged solutions can form explosive silver azide (Ag₃N).

Marking notes. 1 mark for correct reagent and heating procedure; 1 mark for positive observation (silver mirror) and negative observation (no change); 1 mark for safety note about fresh preparation / explosive azide risk.

Sample response. The structural feature is the presence of a hydrogen atom bonded to the carbonyl carbon in an aldehyde (−CHO). Oxidation of a carbonyl compound by mild oxidising agents requires removal of this H atom from the carbonyl carbon. In an aldehyde (R–CHO), the carbonyl C has one H available for the oxidising agent to abstract — the aldehyde is oxidised to a carboxylate (R–COO⁻). In a ketone (R–CO–R′), the carbonyl carbon is bonded to two carbon atoms and carries no H atom — there is nothing for the mild oxidising agent to remove — so no reaction occurs under standard laboratory conditions.

Marking notes. 1 mark for identifying that the H on the carbonyl carbon (in aldehyde) is the key structural feature; 1 mark for explaining that oxidation requires removal of this H; 1 mark for explaining that ketones lack this H on the carbonyl C, so oxidation cannot proceed.

Sample response. Short-chain aldehydes and ketones are miscible with water because the lone pairs on the C=O oxygen can accept hydrogen bonds from water’s O–H groups, enabling sufficient interactions with the polar solvent to overcome the relatively small non-polar component of the molecule. As chain length increases, the non-polar hydrocarbon portion of the molecule grows and the ratio of polar C=O groups to non-polar CH₂ units decreases. Eventually the dispersion-only interactions between the non-polar chain and water become insufficient to mix with the polar solvent network, and solubility decreases.

Marking notes. 1 mark for correctly identifying that the C=O oxygen accepts H-bonds from water enabling short-chain solubility; 1 mark for explaining that increasing chain length increases the non-polar hydrocarbon component, reducing miscibility.

Part (a) — 2 marks. The largest single gap is between propan-1-ol (97 °C) and propanone (56 °C) — a difference of 41 °C. This gap marks the transition from dipole-dipole forces (propanone, no H-bonding between molecules) to O–H hydrogen bonding (propan-1-ol, both donor and acceptor). H-bonds are significantly stronger than dipole-dipole forces, so substantially more energy is required to separate propan-1-ol molecules. (1 mark for correctly identifying the propanone/propan-1-ol gap using specific values; 1 mark for IMF explanation.)

Part (b) — 2 marks. Propanal and propanone are structural isomers — they have the same molecular formula (C₃H₆O) and therefore similar molecular masses. Both are carbonyl compounds with a polar C=O group, so both exhibit dipole-dipole interactions as their dominant IMF. Neither has an O–H bond, so neither can H-bond between its own molecules. The small difference (7 °C) reflects the slightly more polar internal C=O in propanone (flanked by two alkyl groups) compared to the terminal C=O in propanal. (1 mark for identifying same IMF type/strength: dipole-dipole only; 1 mark for explaining the small difference via internal vs terminal C=O polarity or accepting the small difference as consistent with almost equal IMF strength.)

Part (c) — 2 marks. Propanone would give a negative result with Fehling’s solution — the solution would remain blue and no brick-red precipitate would form. Propanone is a ketone: the carbonyl carbon has no H atom bonded to it, flanked instead by two CH₃ groups. Fehling’s solution oxidises only compounds that can have the H removed from the carbonyl carbon (aldehydes). Since propanone has no such H, Cu²⁺ is not reduced to Cu₂O. (1 mark for correct prediction: negative result, solution stays blue; 1 mark for structural explanation: no H on carbonyl C in ketone, so Cu²⁺ not reduced.)

Sample response. The claim is incorrect: aldehydes and ketones differ significantly in both chemical reactivity and, to a lesser extent, physical properties, despite both containing C=O. The differences arise from a single structural feature — the presence or absence of a hydrogen atom on the carbonyl carbon — which cascades into every aspect of their chemistry.

Chemical tests: Tollens’ reagent ([Ag(NH₃)₂]⁺) produces a silver mirror with an aldehyde but no reaction with a ketone. This is because the aldehyde H on the carbonyl carbon is available for the oxidising agent to abstract — the aldehyde is oxidised to a carboxylate while Ag⁺ is reduced to Ag⁰. The ketone has no H on the carbonyl carbon, so Ag⁺ cannot be reduced and no mirror forms. Similarly, Fehling’s solution (blue Cu²⁺) gives a brick-red Cu₂O precipitate with aldehydes only — ketones leave the solution blue because Cu²⁺ requires the same carbonyl H removal to be reduced. These two tests are specific to aldehydes; the claim that aldehydes and ketones behave identically is therefore clearly false for at least two major chemical tests.

Physical properties: aldehydes and ketones share the same IMF type — permanent dipole-dipole interactions from the polar C=O, without hydrogen bonding between their own molecules (neither has an O–H bond) — so their boiling points are similar at the same carbon chain length. For example, butanal (75 °C) and butan-2-one (80 °C) are close. However, the small difference (butan-2-one slightly higher) arises because the internal C=O of the ketone is flanked by two alkyl groups, making it marginally more polar than the terminal C=O of the aldehyde. This represents a small but real physical property difference arising from the structural difference. Solubility in water is similar for short chains in both classes — the C=O oxygen accepts H-bonds from water in both cases.

Overall evaluation: the claim is incorrect. Aldehydes and ketones share IMF type and broadly similar boiling points at the same carbon count, but their chemical reactivity with oxidising agents (Tollens’, Fehling’s, K₂Cr₂O₇) is fundamentally different: aldehydes are easily oxidised; ketones are resistant to oxidation under normal conditions. This difference stems entirely from the single structural feature — the presence of H on the carbonyl carbon in an aldehyde but not in a ketone.

Marking criteria. 1 mark — correctly identifies that the key structural difference is presence/absence of H on carbonyl C; 1 mark — describes Tollens’ result for aldehyde (silver mirror) and ketone (no reaction) with structural explanation; 1 mark — describes Fehling’s / K₂Cr₂O₇ result for aldehyde and ketone with structural explanation; 1 mark — correctly compares IMF type in both (dipole-dipole only, no H-bond between own molecules) and states similar boiling points at same chain length; 1 mark — identifies a physical property where they differ (e.g. butan-2-one > butanal BP due to internal vs terminal C=O polarity); 1 mark — explicitly evaluates the claim, rejecting the “identical chemical tests” component with evidence; 1 mark — response is structured (chemical tests addressed separately from physical properties, with explicit structural justifications throughout).

Sample response. The student’s error is claiming that propanone was oxidised to propanoic acid by K₂Cr₂O₇. This is incorrect: propanone is a ketone, and ketones cannot be oxidised by acidified K₂Cr₂O₇ (or any standard laboratory oxidising agent) under normal conditions. The structural reason is that the carbonyl carbon of propanone (C2 in CH₃COCH₃) carries no hydrogen atom — it is bonded to two CH₃ groups. Oxidation of a carbonyl compound requires removal of the H from the carbonyl carbon; since propanone has no such H, the oxidising agent has nothing to act on and no reaction occurs. The correct observation when acidified K₂Cr₂O₇ is added to propanone would be that the solution remains orange — there is no colour change to green, no precipitate, and no detectable reaction. The only way propanoic acid (a 3-carbon carboxylic acid) could be produced is by oxidising a primary 3-carbon alcohol (propanal as intermediate), not by oxidising propanone.

Marking notes. 1 mark for correctly identifying the error (propanone is a ketone, cannot be oxidised to propanoic acid); 1 mark for structural explanation (no H on carbonyl C of propanone); 1 mark for stating the correct observation (solution stays orange — no reaction); 1 mark for explaining why propanoic acid cannot be produced from propanone under normal conditions; 1 mark for noting the correct pathway to propanoic acid (via primary alcohol / aldehyde intermediate, not from propanone).