Explore the carbonyl group in two contexts — discover why a single structural difference (terminal vs internal C=O) determines naming, boiling points, and every chemical test result for aldehydes and ketones.
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
Vanillin (the primary flavour compound in vanilla extract) is an aldehyde. Raspberry ketone (the compound responsible for the distinctive scent of raspberries) is, as its name suggests, a ketone. Both contain a C=O group — the carbonyl — as their key functional feature.
Before you read on: Draw what you think the structural difference between an aldehyde and a ketone is. Where is the C=O in an aldehyde — at the end or in the middle of the chain? What is the C=O bonded to in a ketone? Is there an H on the carbonyl carbon in either structure? Record your best prediction below.
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The carbonyl group (C=O) is identical in both aldehydes and ketones — what differs is its position in the molecule and what is bonded to the carbonyl carbon, and that single structural difference has cascading consequences for naming, properties, and every reaction both compounds undergo.
The C=O bond is strongly polar (δ+ on C, δ− on O, electronegativity difference ~1.0) — significantly more polar than C=C (~0.4). This polarity makes carbonyl compounds more reactive and more polar than alkanes or alkenes.
Propanal vs propanone (both C₃H₆O — functional group isomers) — the presence of H on the carbonyl carbon determines all chemical differences: naming, oxidisability, and test results
Locant not needed — always C1
Locant required for C5+ to show C=O position
Aldehydes and ketones are more polar than alkanes (due to the C=O dipole) but cannot form H-bonds with each other (no O–H or N–H) — placing them in an intermediate IMF tier that produces predictable and testable boiling point and solubility patterns.
The C=O bond is strongly polar (δ+ on C, δ− on O). This creates permanent dipole-dipole interactions between carbonyl molecules — stronger than dispersion-only (alkanes), but weaker than H-bonding (alcohols, carboxylic acids).
No H-bonding between aldehyde/ketone molecules: Neither has an O–H or N–H bond, so they cannot donate H-bonds to each other. The lone pairs on the C=O oxygen can ACCEPT H-bonds from water (one-way — relevant for solubility), but not from each other.
Because aldehydes and ketones have the same molecular formula at the same carbon count (functional group isomers), you cannot distinguish them by formula or molecular mass — you distinguish them by chemistry, specifically by reactions that exploit the single structural difference: the H on the carbonyl carbon.
Tollens' reagent is ammoniacal silver nitrate — a solution of [Ag(NH₃)₂]⁺ (diamminesilver(I) complex). It is a mild oxidising agent.
Procedure: add a few drops of the unknown compound to freshly prepared Tollens' reagent in a clean glass test tube; warm gently in a water bath (~50°C) for 2–3 minutes.
Shiny silver mirror forms on the inside of the test tube. Ag⁺ reduced to Ag⁰ (silver metal). Aldehyde oxidised to carboxylate.
R–CHO + 2[Ag(NH₃)₂]⁺ + 2OH⁻ → R–COO⁻ + 2Ag⁰ + 4NH₃ + H₂O
No silver mirror. Solution remains clear/colourless. Ag⁺ is NOT reduced. Ketone is not oxidised — no H on carbonyl C for oxidising agent to remove.
Fehling's solution contains Cu²⁺ ions complexed with sodium potassium tartrate (deep blue). Benedict's uses sodium citrate — both behave identically at HSC level.
Deep blue → brick-red/orange-red precipitate (Cu₂O). Cu²⁺ reduced to Cu⁺. Aldehyde oxidised to carboxylate.
R–CHO + 2Cu²⁺ + 5OH⁻ → R–COO⁻ + Cu₂O↓ + 3H₂O
Solution stays blue. No precipitate forms. Cu²⁺ is NOT reduced. Ketone cannot be oxidised under these mild conditions.
| Test | Aldehyde | Ketone | 1° Alcohol | 3° Alcohol |
|---|---|---|---|---|
| Tollens' | + silver mirror | − no mirror | − no mirror | − no mirror |
| Fehling's/Benedict's | + brick-red ppt | − stays blue | − stays blue | − stays blue |
| K₂Cr₂O₇/H⁺ | orange → green | stays orange | orange → green | stays orange |
The resistance of ketones to oxidation is not a coincidence or a special rule — it is a direct structural consequence of having no H on the carbonyl carbon, and understanding this structurally is more useful than memorising it as a fact.
Oxidation of a carbonyl compound requires removing the H from the carbonyl carbon:
| Property | Aldehyde | Ketone |
|---|---|---|
| Oxidisable to COOH? | Yes — easily (H on carbonyl C) | No under normal conditions |
| Tollens' test | Positive — silver mirror | Negative |
| Fehling's/Benedict's | Positive — brick-red Cu₂O | Negative — stays blue |
| K₂Cr₂O₇/H⁺ | Orange → green | Stays orange |
| Position of C=O | Terminal (C1) | Internal (C2 or higher) |
| H on carbonyl C? | Yes | No |
| Test | A | B | C | D |
|---|---|---|---|---|
| Tollens' reagent | Silver mirror | No reaction | No reaction | Silver mirror |
| Fehling's solution | Brick-red ppt | Stays blue | Stays blue | Brick-red ppt |
| K₂Cr₂O₇/H⁺ | Orange → green | Stays orange | Orange → green | Orange → green |
Propanal (an aldehyde) gives a positive Tollens' test — it is oxidised to propanoic acid, reducing Ag⁺ to Ag(s) (the silver mirror). Propanone (a ketone) gives a negative result. The difference: aldehydes have a hydrogen atom on the carbonyl carbon (R–CHO) that can be removed during oxidation. Ketones (R–CO–R) have no such hydrogen — both R groups are carbon chains, so there is no available H for the oxidation pathway. Ketones require very harsh conditions to oxidise and do not react with mild oxidising agents like Tollens' or Fehling's.
Complete the Learn phase to unlock Practice.
For each compound below, predict the result of each test (positive/negative and describe the observation). Justify each prediction.
Rank each set of C3 compounds in order of increasing boiling point, identify the IMF responsible for each compound's position, and explain the largest single jump in the ranking.
Set: Propane · Propanal · Propanone · Propan-1-ol · Propanoic acid
Actual BPs (use to verify your ranking): propane −42°C · propanal 49°C · propanone 56°C · propan-1-ol 97°C · propanoic acid 141°C
Question 1. A student adds Fehling's solution to pentan-3-one and warms it in a water bath. What does the student observe?
Question 2. Which correctly explains why propanal has a lower boiling point than propan-1-ol, despite both being C3 compounds containing oxygen?
Question 3. A student treats an unknown compound with Tollens' reagent and observes a silver mirror. What can the student conclude?
Question 4. Which statement correctly explains why ketones cannot be further oxidised to carboxylic acids under normal laboratory conditions?
Question 5. Propanone (acetone) is described as a "universal solvent." Which combination of properties best explains this?
Question 6 4 marks
Propanal (CH₃CH₂CHO) and propanone (CH₃COCH₃) are functional group isomers. (a) Describe the structural difference between these two compounds. (b) For each compound, predict and explain the result when treated with Tollens' reagent. (c) Explain why propanone cannot be oxidised to propanoic acid under normal conditions, while propanal can.
Question 7 5 marks
Arrange the following C4 compounds in order of increasing boiling point and explain the ranking in terms of intermolecular forces: butane, butanal, butan-2-one, butan-1-ol, butanoic acid. In your explanation, identify the specific IMF for each compound and explain why butan-2-one has a higher boiling point than butanal despite both being C4 carbonyl compounds with the same molecular formula.
Question 8 6 marks
A forensic scientist performs Tollens' and Fehling's tests on three unknown organic compounds (X, Y, Z), all with molecular formula C₃H₆O. Results: X gives a silver mirror with Tollens' and a brick-red precipitate with Fehling's. Y gives no reaction with Tollens' and no reaction with Fehling's. Z gives a silver mirror with Tollens'. (a) Classify X, Y, and Z as aldehyde or ketone. (b) Identify the IUPAC name and structural formula of each compound. (c) Explain the molecular basis for X giving a positive Tollens' test while Y does not. (d) Write the equation for oxidation of X with excess K₂Cr₂O₇/H⁺ under reflux conditions.
Q1 — C: Pentan-3-one is a ketone (C=O at C3, flanked by two ethyl groups, no H on carbonyl C). Fehling's solution is specific for aldehydes — it requires the H on the carbonyl carbon for the Cu²⁺ reduction to occur. Ketones are non-reducing; Cu²⁺ stays blue.
Q2 — B: Propan-1-ol (BP 97°C) has an O–H group — it both donates H-bonds and accepts H-bonds. Propanal (BP 49°C) has only the C=O — lone pairs can accept H-bonds from water, but propanal cannot donate H-bonds between its own molecules. Net IMF between propanal molecules is dipole-dipole only.
Q3 — B: A silver mirror with Tollens' reagent confirms the compound is an aldehyde. Only aldehydes (which have H on the carbonyl carbon) reduce Ag⁺ to Ag⁰. Ketones give no silver mirror. Primary alcohols do NOT give a positive Tollens' test.
Q4 — A: Oxidation of a carbonyl compound requires removing the H from the carbonyl carbon. In a ketone, the carbonyl C is bonded to two other C atoms and has NO H atom. Without that H, the oxidising agent has nothing to remove.
Q5 — D: Propanone's versatility as a solvent comes from its dual nature: the polar C=O (accepts H-bonds from water → miscible with polar/aqueous solvents) + two non-polar CH₃ groups (compatible with non-polar organic solvents).
Q6 (4 marks):
(a) Both have formula C₃H₆O, but propanal has the C=O at the terminal carbon (C1) — the carbonyl carbon has one H atom bonded to it (–CHO group). Propanone has the C=O at the internal carbon (C2) — the carbonyl carbon is bonded to two CH₃ groups and has NO H atom. [1 mark]
(b) Propanal → silver mirror (positive). The H on the carbonyl carbon of propanal is oxidised; Ag⁺ is reduced to Ag⁰. Propanone → no reaction (negative). Propanone has no H on its carbonyl carbon for Tollens' to oxidise. [1 mark each = 2 marks]
(c) Propanal can be oxidised because the carbonyl carbon has one H atom that can be removed, forming propanoic acid. Propanone cannot be oxidised because the carbonyl carbon has no H atom — the oxidising agent has nothing to remove at that position. [1 mark]
Q7 (5 marks):
Increasing BP: butane (−1°C) < butanal (75°C) < butan-2-one (80°C) < butan-1-ol (118°C) < butanoic acid (164°C) [1 mark for correct order]
Butane: non-polar, dispersion forces only — weakest IMF, lowest BP [½ mark]
Butanal and butan-2-one: both have C=O dipole-dipole forces + dispersion, stronger than dispersion alone but unable to donate H-bonds between their own molecules [1 mark]
Butan-2-one slightly above butanal: internal C=O flanked by two alkyl groups is slightly more polar than terminal C=O [1 mark]
Butan-1-ol: O–H H-bonds — much stronger IMF than dipole-dipole → large jump in BP [½ mark]
Butanoic acid: O–H H-bonds PLUS dimerisation (two simultaneous H-bonds per pair) → strongest IMF → highest BP [1 mark]
Q8 (6 marks):
(a) X: positive Tollens' and Fehling's → aldehyde. Y: negative both → ketone. Z: positive Tollens' → aldehyde. [1 mark]
(b) C₃H₆O aldehyde: propanal (CH₃CH₂CHO). C₃H₆O ketone: propanone (CH₃COCH₃). X = Z = propanal; Y = propanone. [2 marks]
(c) X is propanal — carbonyl carbon has one H atom. Tollens' reagent ([Ag(NH₃)₂]⁺) oxidises this H, converting the aldehyde to carboxylate; Ag⁺ is reduced to Ag⁰ (silver mirror). Y is propanone — carbonyl carbon has NO H atom. Tollens' cannot act; Ag⁺ is not reduced, no silver mirror forms. [2 marks]
(d) CH₃CH₂CHO + [O] → CH₃CH₂COOH. Product: propanoic acid. Conditions: excess K₂Cr₂O₇/H₂SO₄, reflux. Colour change: orange → green. [1 mark]
Return to your structural prediction. In an aldehyde the C=O is at the END of the chain (terminal). The carbonyl carbon has one H bonded to it — written as –CHO. Vanillin has this –CHO group. The H on the carbonyl carbon is what makes vanillin (and all aldehydes) easily oxidisable and reactive in tests like Tollens' and Fehling's.
In a ketone the C=O is in the MIDDLE of the chain (internal). The carbonyl carbon has NO H — it is flanked by two other carbon groups. Raspberry ketone has this internal C=O. The absence of the H makes it resistant to mild oxidising agents and gives it a negative Tollens' test.
Look back at what you wrote before reading this lesson. How has your understanding changed?
Back at the start you were asked why vanilla and raspberry flavourings — both carbonyl compounds — smell so different despite sharing a C=O group. Now you know: the specific carbon chain length, degree of branching, and functional group position determine the molecular shape and how it fits into olfactory receptors. Vanillin is an aromatic aldehyde (–CHO terminal on a benzene-ring framework); raspberry ketone has an internal C=O flanked by two carbon groups. Same carbonyl family, completely different three-dimensional shapes and molecular interactions — which is exactly why one triggers the vanilla receptor pathway and the other does not.
Coming up in Lesson 14: Carboxylic acids — the next step up in oxidation from aldehydes — are covered in detail, including their reactions with bases (neutralisation) and alcohols (esterification), and why their acidic properties differ from mineral acids.
What is the structural difference between an aldehyde and a ketone?
What observation indicates a positive Tollens' test, and what does it tell you about the compound?
Why can't ketones be oxidised to carboxylic acids under normal conditions?
Arrange in order of increasing boiling point (C4): butane, butanal, butan-2-one, butan-1-ol, butanoic acid.
A compound with molecular formula C₄H₈O gives a brick-red precipitate with Fehling's solution and turns K₂Cr₂O₇/H⁺ green. What is the functional group and a possible IUPAC name?
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