Chemistry • Year 12 • Module 7 • Lesson 13
Aldehydes & Ketones: Structure, Properties & Tests
Apply structural reasoning to real boiling-point data, a test-results summary table, and an Australian case study to practise Band 4–5 exam-style analysis.
1. Interpret the test-results summary table
The table below shows the results of three chemical tests applied to four classes of organic compound. Study the table carefully, then answer the questions that follow. 8 marks
| Compound class | Tollens’ reagent | Fehling’s/Benedict’s | Acidified K₂Cr₂O₇ |
|---|---|---|---|
| Aldehyde (e.g. ethanal) | Positive — silver mirror | Positive — brick-red ppt (Cu₂O) | Orange → green |
| Ketone (e.g. propanone) | Negative — no mirror | Negative — stays blue | Stays orange |
| Primary alcohol (e.g. ethanol) | Negative — no mirror | Negative — stays blue | Orange → green |
| Tertiary alcohol (e.g. 2-methylpropan-2-ol) | Negative — no mirror | Negative — stays blue | Stays orange |
1.1 Using the data in the table, identify which single test would allow you to definitively distinguish an aldehyde from a primary alcohol. Justify your answer. 2 marks
1.2 An unknown compound X gives a negative Tollens’ test, a negative Fehling’s test, but turns acidified K₂Cr₂O₇ from orange to green. Identify the functional group class of compound X and explain your reasoning. 3 marks
1.3 Explain why the K₂Cr₂O₇/H⁺ test is described as less specific than Tollens’ test for detecting aldehydes. Use data from the table to support your answer. 3 marks
2. Interpret the boiling-point graph — C4 carbonyl compounds
The bar chart below shows the boiling points of five C4 compounds. Study the graph, then answer the questions. 8 marks
2.1 Describe the overall trend in boiling point across the five compounds and identify the factor most responsible for the trend. 2 marks
2.2 Butanal and butan-2-one have the same molecular formula (C₄H₈O) and both contain a C=O group. Use the graph to estimate the difference in their boiling points, and explain the structural reason why butan-2-one has the slightly higher value. 3 marks
2.3 A student claims: “Butanal should have a lower boiling point than butane because the C=O group makes the molecule less stable.” Identify the error in this reasoning and provide the correct explanation. 3 marks
3. Case study — Australian context
6 marks
Stimulus. In Australia, formaldehyde (methanal, HCHO) is widely used in the resins that bind particle board and MDF used in furniture and building interiors. Safe Work Australia classifies formaldehyde as a hazardous substance and sets a workplace exposure standard of 1 ppm (ceiling). Off-gassing from new flat-pack furniture or freshly installed flooring can temporarily raise indoor formaldehyde concentrations. Acetone (propanone, CH₃COCH₃), by contrast, is used as the primary solvent in many nail polish removers sold in Australian pharmacies. Acetone is not classified as a carcinogen and is rapidly metabolised by the body. Both compounds are C≤3 carbonyl compounds, yet their toxicity, regulation, and applications differ dramatically.
3.1 Using structural chemistry, explain why formaldehyde (an aldehyde) is classified as a reducing agent, whereas acetone (a ketone) is not. Relate your answer to the result of Tollens’ test on each compound. 3 marks
3.2 Acetone is described as a versatile solvent because it “bridges polar and non-polar solubility.” Using intermolecular force reasoning, explain what structural features of acetone support this claim, and why this makes it effective for removing nail polish (a polar acrylic polymer carrying non-polar pigment particles). 3 marks
Q1.1 — Single distinguishing test for aldehyde vs primary alcohol
Sample response. Tollens’ reagent (or Fehling’s/Benedict’s) would definitively distinguish an aldehyde from a primary alcohol. Both give an orange-to-green result with acidified K₂Cr₂O₇ (both are oxidisable), but only the aldehyde gives a positive Tollens’ test (silver mirror); the primary alcohol gives no mirror. Therefore Tollens’ (or Fehling’s) alone is the single definitive test.
Marking notes. 1 mark for correctly identifying Tollens’ or Fehling’s as the distinguishing test; 1 mark for justification (aldehyde positive, primary alcohol negative, with reference to a specific observation such as silver mirror).
Q1.2 — Identify compound X
Sample response. Compound X is a primary (or secondary) alcohol — most likely a primary alcohol. The negative Tollens’ and Fehling’s results rule out an aldehyde. The positive K₂Cr₂O₇ result shows that X can be oxidised; ketones and tertiary alcohols cannot be oxidised under these conditions, so they are also eliminated. Only primary or secondary alcohols give a negative Tollens’ / Fehling’s but a positive K₂Cr₂O₇ result.
Marking notes. 1 mark for identifying primary (or secondary) alcohol; 1 mark for using Tollens’/Fehling’s negative to eliminate aldehyde; 1 mark for using K₂Cr₂O₇ positive to eliminate ketone/tertiary alcohol.
Q1.3 — Specificity of K₂Cr₂O₇ vs Tollens’
Sample response. K₂Cr₂O₇/H⁺ is less specific because it reacts with three compound classes (aldehydes and primary/secondary alcohols — all three rows show orange → green), whereas Tollens’ reagent reacts with only one class (aldehydes). The data show that both an aldehyde and a primary alcohol give an orange-to-green result with K₂Cr₂O₇, so a positive result cannot pinpoint aldehyde alone, making it impossible to distinguish an aldehyde from a primary alcohol with K₂Cr₂O₇ alone.
Marking notes. 1 mark for identifying that K₂Cr₂O₇ reacts with more than one class; 1 mark for citing the specific classes (aldehyde + primary alcohol, or including secondary); 1 mark for contrasting with Tollens’ as specific to aldehyde only, using table data as support.
Q2.1 — Trend in boiling point
Sample response. Boiling point increases from butane through butanal, butan-2-one, butan-1-ol, to butanoic acid. The most responsible factor is the type and strength of intermolecular forces: as IMF strength increases from dispersion only (butane) → dipole-dipole (carbonyl compounds) → hydrogen bonding (alcohol, carboxylic acid) → hydrogen bonding + dimerisation (carboxylic acid), more energy is required to separate molecules, so boiling point increases.
Marking notes. 1 mark for correct trend (increasing BP across the series); 1 mark for identifying IMF type/strength as the responsible factor.
Q2.2 — Butanal vs butan-2-one boiling point difference
Sample response. The graph shows butanal at 75 °C and butan-2-one at 80 °C — a difference of approximately 5 °C. Butan-2-one’s C=O group is flanked by two alkyl (CH₃ and CH₂CH₃) groups on the carbonyl carbon, which are slightly electron-donating. This makes the internal C=O slightly more polar than the terminal C=O in butanal (which is flanked by only one alkyl group on that side), producing marginally stronger dipole-dipole interactions and hence a slightly higher boiling point.
Marking notes. 1 mark for reading the difference as ~5 °C from the graph; 1 mark for identifying that the internal C=O of butan-2-one is flanked by two alkyl groups (vs one for butanal); 1 mark for linking this to slightly stronger dipole-dipole forces / higher polarity of the internal C=O.
Q2.3 — Student error correction
Sample response. The student has confused internal bond stability with intermolecular forces. Boiling point is determined by the strength of intermolecular (between-molecule) forces, not by intramolecular stability. The correct explanation is: butanal has a strongly polar C=O group that creates dipole-dipole intermolecular forces between adjacent molecules — these are much stronger than the dispersion forces that exist between non-polar butane molecules. Therefore butanal has a significantly higher boiling point (75 °C vs −1 °C) than butane, not a lower one.
Marking notes. 1 mark for identifying the error (confusing intramolecular stability with intermolecular forces); 1 mark for stating butanal’s C=O creates dipole-dipole IMF; 1 mark for explicitly stating that dipole-dipole forces are stronger than dispersion forces (butane), correctly predicting higher BP for butanal.
Q3.1 — Formaldehyde as reducing agent
Sample response. Formaldehyde is an aldehyde — its carbonyl carbon (HCHO, where the carbonyl C carries two H atoms) has an H atom available. When Tollens’ reagent ([Ag(NH₃)₂]⁺) is added, formaldehyde is oxidised (loses the H) to form formate (HCOO⁻), while Ag⁺ is reduced to metallic silver (Ag⁰), depositing as a silver mirror. Formaldehyde acts as the reducing agent because it donates electrons to the silver ions. Acetone (propanone) has no H on its carbonyl carbon (it is flanked by two CH₃ groups), so it cannot be oxidised — Tollens’ gives no reaction, and acetone is not a reducing agent.
Marking notes. 1 mark for identifying that formaldehyde has H on the carbonyl carbon (enables oxidation); 1 mark for Tollens’ test result: silver mirror (positive) for formaldehyde vs no reaction (negative) for acetone; 1 mark for explicitly linking the structural feature (presence/absence of carbonyl H) to reducing agent ability.
Q3.2 — Acetone as a bridging solvent
Sample response. Acetone has a polar C=O group and two non-polar CH₃ groups. The C=O oxygen’s lone pairs can accept H-bonds from water molecules (O–H ··· O=C), making acetone fully miscible with water and able to dissolve polar substances like acrylic polymer. The non-polar CH₃ groups interact via dispersion forces with non-polar pigment particles and hydrocarbon chains in the nail polish. This dual nature — polar C=O + non-polar methyl groups — allows acetone to solvate both the polar polymer matrix and the non-polar pigment components simultaneously, dissolving nail polish effectively.
Marking notes. 1 mark for identifying that the C=O oxygen accepts H-bonds from water (polar solubility); 1 mark for identifying CH₃ groups interact via dispersion forces with non-polar components; 1 mark for linking both features together to explain why acetone dissolves a composite polar/non-polar substance like nail polish.