Chemistry • Year 12 • Module 7 • Lesson 13
Aldehydes & Ketones: Structure, Properties & Tests
Build HSC Band 5–6 extended-response technique on carbonyl structure, chemical tests, and the structural basis of oxidation resistance.
1. Data & scenario evaluation — identifying unknown carbonyl compounds (Band 5–6)
8 marks Band 5–6
Scenario. A forensic chemistry trainee in a NSW Police laboratory receives four unlabelled organic liquids, each with molecular formula C₄H₈O. A preliminary IR scan confirms a strong C=O stretch for all four. The trainee performs three confirmatory tests and records the results in the table below.
| Sample | Tollens’ reagent | Fehling’s solution | Acidified K₂Cr₂O₇ |
|---|---|---|---|
| P | Silver mirror | Brick-red ppt | Orange → green |
| Q | No mirror | Stays blue | Stays orange |
| R | Silver mirror | Brick-red ppt | Orange → green |
| S | No mirror | Stays blue | Stays orange |
Q1. Analyse and evaluate the test data to fully characterise samples P, Q, R, and S. In your response you must:
- Classify each sample as aldehyde or ketone and justify each classification using all three test results.
- State the IUPAC name of each sample (note: P and R are different C₄H₈O aldehydes; Q and S are both C₄H₈O ketones — at HSC level, butan-2-one is the only non-cyclic C₄H₈O ketone, so Q and S may be different samples of the same compound).
- Explain the structural reason why Q and S give identical negative results for all three tests.
- Write a half-equation showing the oxidation of P (butanal) by Tollens’ reagent, identifying oxidised and reduced species.
- Evaluate which single test from the table is most useful for a forensic chemist wishing to rapidly screen for the presence of an aldehyde in an unknown sample, and justify your choice.
2. Source critique — detect the scientific flaw (Band 5–6)
7 marks Band 5–6
“Aldehydes and alcohols both contain oxygen. In fact, the reason aldehydes smell fruity and dissolve in water is because both functional groups share the same O–H hydrogen-bonding unit. This means aldehydes and alcohols should have similar boiling points for the same carbon chain length — and the small observed difference is simply due to molecular mass. Similarly, Fehling’s test can detect any organic compound containing a C=O or O–H group, since both groups involve oxygen.”
— Adapted from a fictional student-blog chemistry summary, 2024
Q2. This passage contains three distinct scientific errors. For each error:
- Identify the specific claim that is wrong.
- Explain the correct chemistry, using appropriate terminology from this lesson.
- For one of the three errors, describe how you would design a simple laboratory test or observation to experimentally demonstrate that the claim is wrong.
Q1 — Marking criteria (8 marks)
Classification (2 marks). P and R: positive Tollens’ (silver mirror) + positive Fehling’s (brick-red) + positive K₂Cr₂O₇ (orange → green) → aldehydes (1 mark for both correctly classified). Q and S: all three tests negative → ketones (1 mark for both).
IUPAC names (2 marks). C₄H₈O aldehydes: P = butanal (CH₃CH₂CH₂CHO, straight chain) and R = 2-methylpropanal ((CH₃)₂CHCHO, branched). C₄H₈O ketone(s): Q = butan-2-one (CH₃COCH₂CH₃). At HSC level, butan-2-one is the only open-chain C₄H₈O ketone. S is therefore also butan-2-one (a duplicate sample demonstrating reproducible negative results) or cyclobutanone (a cyclic C₄H₈O ketone, if cyclic structures are included). Marker note: award 1 mark for correctly naming Q as butan-2-one; award 1 mark for naming S as butan-2-one or cyclobutanone and explaining both are ketones.
Structural explanation for Q and S identical results (1 mark). Despite being structural isomers, both Q and S are ketones — their C=O groups are positioned internally within the carbon chain and neither has an H atom bonded to the carbonyl carbon. Tollens’ and Fehling’s require removal of the carbonyl H; K₂Cr₂O₇ also cannot oxidise a ketone without a carbonyl H. The absence of H on the carbonyl carbon in any ketone — regardless of where the C=O sits — means all ketones are chemically equivalent in this respect and give identical negative results.
Half-equation for oxidation of butanal by Tollens’ (1 mark). CH₃CH₂CH₂CHO → CH₃CH₂CH₂COO⁻ + H⁺ + 2e⁻ (butanal is the oxidised species / reducing agent; [Ag(NH₃)₂]⁺ + e⁻ → Ag⁰ + 2NH₃ is the reduction half; silver ion is the reduced species / oxidising agent). Award 1 mark for a correct oxidation half-equation or correctly identifying both oxidised and reduced species with justification.
Evaluation of best single test (2 marks). Tollens’ (or Fehling’s) is the most useful for rapid screening because it is specific for aldehydes — only aldehydes produce a silver mirror / brick-red precipitate; no other compound class in this lesson gives a positive result. K₂Cr₂O₇ is less useful for screening because it gives a positive result for primary/secondary alcohols as well as aldehydes, making it ambiguous. Award 1 mark for naming Tollens’ or Fehling’s as the best test; 1 mark for justification (specificity for aldehyde only vs ambiguity of K₂Cr₂O₇).
Q2 — Source critique marking criteria (7 marks)
Error 1 (2 marks): “both functional groups share the same O–H hydrogen-bonding unit.” Identify (1 mark): aldehydes do not have an O–H group. The H in −CHO is bonded to carbon (a C–H bond), not to oxygen. Correct chemistry (1 mark): because there is no O–H (or N–H) bond in an aldehyde, aldehyde molecules cannot donate H-bonds to each other. They can only accept H-bonds via the C=O lone pairs from external H-bond donors (e.g. water). This is fundamentally different from an alcohol’s O–H, which both donates and accepts H-bonds.
Error 2 (2 marks): “the small observed difference [in BP] is simply due to molecular mass.” Identify (1 mark): for C4 compounds, butanal (75 °C) and butan-1-ol (118 °C) have a 43 °C difference in BP despite very similar molecular masses (72 vs 74 g/mol) — molecular mass alone cannot explain this. Correct chemistry (1 mark): the large BP gap is due to the type of IMF: butanal has dipole-dipole forces only (no H-bonding between molecules); butan-1-ol has O–H hydrogen bonds (both donor and acceptor). H-bonds are significantly stronger than dipole-dipole forces, requiring more energy to overcome — this, not molecular mass, explains the large BP difference.
Error 3 (2 marks): “Fehling’s test can detect any organic compound containing a C=O or O–H group.” Identify (1 mark): Fehling’s solution is specific for aldehydes only. Ketones (which have C=O) and alcohols (which have O–H) give negative results — the solution remains blue. Correct chemistry (1 mark): Fehling’s test works by mild oxidation of the aldehyde H on the carbonyl carbon. Ketones have no such H and cannot reduce Cu²⁺ to Cu₂O. Alcohols also cannot — Fehling’s is not an oxidant strong enough to oxidise an alcohol under the mild conditions used.
Experimental demonstration (1 mark — one of the three errors). Example for Error 3: Add equal volumes of ethanal (aldehyde), propanone (ketone), and ethanol (alcohol) each to separate test tubes of freshly prepared Fehling’s solution; warm in a water bath at 60 °C for 3 minutes. If the claim were correct, all three would show a colour change. In practice, only ethanal produces a brick-red precipitate; propanone and ethanol leave the solution blue. This directly falsifies the claim that Fehling’s detects any C=O or O–H compound.