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IQ4Lesson 13 of 2345 minCarbonyl Chemistry
Aldehydes & Ketones
Aldehydes give almonds their bitter aroma, vanilla its warmth, and cinnamon its spice — and the single structural difference between an aldehyde and a ketone determines not just these flavour properties but everything about how these compounds react and how you detect them in the lab.
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Vanilla and Raspberry — Same Group, Different Position
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.
Silver mirror on glass · aldehyde oxidised · Ag⁺ reduced · ketone: no reaction
Fehling's Test (aldehyde only)
R–CHO + 2Cu²⁺ → R–COO⁻ + Cu₂O(s)
Blue → brick-red precipitate (Cu₂O) · ketone: stays blue · Benedict's equivalent
Learning Intentions
📖
Know
Aldehyde: terminal C=O (–CHO), H on carbonyl C, suffix –al
Ketone: internal C=O (R–CO–R'), no H on carbonyl C, suffix –one
Tollens': silver mirror = aldehyde; no mirror = ketone or alcohol
Fehling's/Benedict's: brick-red = aldehyde; stays blue = ketone
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Understand
Why aldehydes and ketones have dipole-dipole forces but cannot H-bond with each other (no O–H to donate)
Why aldehyde BP < alcohol BP < carboxylic acid BP at same carbon count
Why ketones are resistant to oxidation — no H on carbonyl C to remove
Why the aldehyde H (on C, not O) does not enable H-bonding
✏️
Can Do
Name and draw structural formulae for aldehydes and ketones up to C5
Rank C4 compounds in order of boiling point using IMF reasoning
Identify unknown carbonyl compounds from Tollens', Fehling's, and K₂Cr₂O₇ test results
Explain the oxidation resistance of ketones structurally
Key Terms — scan these before reading
HydrocarbonAn organic compound containing only carbon and hydrogen atoms.
Functional groupA specific atom arrangement determining characteristic chemical reactions.
Homologous seriesA family of compounds with the same functional group, differing by CH₂.
Addition polymerA polymer formed by monomers adding together without loss of atoms.
Condensation polymerA polymer formed with elimination of a small molecule such as water.
EsterificationA condensation reaction between a carboxylic acid and an alcohol forming an ester.
Core Concepts
01
Aldehyde & Ketone Structure — The Carbonyl in Two Contexts
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
Must Do: When drawing a structural formula to show aldehyde vs ketone, always explicitly show the H on the carbonyl carbon for an aldehyde (write –CHO, not just C=O). A ketone NEVER has H on the carbonyl carbon. In full structural formulae, draw C=O with H on the same carbon for aldehydes, and no H on the carbonyl carbon for ketones. This single feature is the marker's checkpoint.
Common Error: Confusing the aldehyde H (bonded to carbon in –CHO) with an O–H group. The H in an aldehyde is a C–H bond, not an O–H bond — there is no O–H group in an aldehyde or ketone. Students who think "aldehyde has H near O, therefore it forms H-bonds between molecules" are wrong. The C–H bond of –CHO cannot donate H-bonds. This distinction controls the boiling point comparison (see Card 2).
Exam TipFor organic chemistry questions, draw full structural formulas showing all atoms and bonds — condensed or skeletal formulas alone may lose marks in HSC extended-response questions.
02
Physical Properties — IMF Analysis & Boiling Points
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.
Intermolecular Forces
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.
Alkane
Dispersion only
No polar bonds
Aldehyde / Ketone
Dipole-dipole + dispersion
C=O dipole · cannot H-bond each other
Alcohol
H-bonding + dispersion
O–H donates AND accepts H-bonds
Boiling Point Ladder — C4 Compounds
Butane (−1°C)
−1°C
dispersion forces only
Butanal (75°C)
75°C
C=O dipole-dipole (no H-bond)
Butan-2-one (80°C)
80°C
C=O dipole-dipole, slightly more polar
Butan-1-ol (118°C)
118°C
O–H H-bonding (donor + acceptor)
Butanoic acid (164°C)
164°C
H-bond + dimerisation (2 H-bonds per pair)
Note: butan-2-one (80°C) > butanal (75°C) — internal C=O flanked by two alkyl groups is slightly more polar than terminal C=O, giving slightly stronger dipole-dipole forces.
Solubility in Water
Short-chain aldehydes and ketones dissolve in water because the C=O oxygen (lone pairs) accepts H-bonds from water's O–H. This one-directional H-bonding is sufficient for short-chain compounds.
Methanal, ethanal, propanal: fully miscible with water
Propanone (acetone): fully miscible — excellent general-purpose solvent (polar C=O + non-polar CH₃ groups dissolves both polar and non-polar compounds)
Must Do — IMF Explanation for BP: "Alcohols can both DONATE and ACCEPT H-bonds via the O–H group; aldehydes/ketones can only ACCEPT H-bonds via the C=O lone pairs — they cannot donate H-bonds between their own molecules. H-bonds in alcohols are therefore stronger and more numerous per molecule, requiring more energy to overcome → higher BP." The donor/acceptor distinction is what separates Band 5 from Band 6 responses.
Common Error: "Aldehydes form H-bonds because the C=O has oxygen in it." There is no O–H bond in an aldehyde or ketone. The oxygen in C=O has lone pairs but is not bonded to H. H-bonds require O–H (or N–H or F–H) as the donor — none exist in the aldehyde or ketone functional group. The C=O oxygen only ACCEPTS H-bonds from other donors (e.g. water) — it cannot donate.
Insight: Propanone (acetone) is the most widely used organic solvent in school labs for a reason that follows directly from its IMF: a polar C=O (accepts H-bonds → miscible with water and polar solvents) + two non-polar CH₃ groups (compatible with organic solvents) + low BP (56°C, evaporates quickly) + relatively low toxicity. It dissolves nail polish (polar acrylic polymer + non-polar pigments) for exactly these reasons — it bridges polar and non-polar solubility simultaneously.
03
Distinguishing Aldehydes from Ketones — Chemical Tests
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.
Test 1 — Tollens' Reagent (Silver Mirror Test)
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.
POSITIVE (Aldehyde)
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
NEGATIVE (Ketone)
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.
Test 2 — Fehling's Solution (or Benedict's Solution)
Fehling's solution contains Cu²⁺ ions complexed with sodium potassium tartrate (deep blue). Benedict's uses sodium citrate — both behave identically at HSC level.
Procedure: add unknown compound to Fehling's solution; heat in a water bath.
POSITIVE (Aldehyde)
Deep blue → brick-red/orange-red precipitate (Cu₂O). Cu²⁺ reduced to Cu⁺. Aldehyde oxidised to carboxylate. R–CHO + 2Cu²⁺ + 5OH⁻ → R–COO⁻ + Cu₂O↓ + 3H₂O
NEGATIVE (Ketone)
Solution stays blue. No precipitate forms. Cu²⁺ is NOT reduced. Ketone cannot be oxidised under these mild conditions.
Complete Test Results Summary
Tollens' reagent
Aldehyde:+ silver mirror
Ketone:− no mirror
1° Alcohol:− no mirror
3° Alcohol:− no mirror
Fehling's/Benedict's
Aldehyde:+ brick-red ppt
Ketone:− stays blue
1° Alcohol:− stays blue
3° Alcohol:− stays blue
K₂Cr₂O₇/H⁺
Aldehyde:orange → green
Ketone:stays orange
1° Alcohol:orange → green
3° Alcohol:stays orange
Key: Tollens' and Fehling's are SPECIFIC for aldehydes — they do not react with alcohols. K₂Cr₂O₇ is less specific — it reacts with aldehydes AND primary/secondary alcohols.
Must Do — Safety: Tollens' reagent must be freshly prepared — aged solutions can form explosive silver azide (Ag₃N). In any HSC practical investigation answer about Tollens' test, note: "Tollens' reagent is prepared immediately before use and the test is carried out promptly." This safety point is specifically tested in practical investigation questions.
Common Error: "Fehling's test detects both aldehydes and ketones because both have a C=O group." Wrong — Fehling's test ONLY detects aldehydes. Ketones give a negative result (stays blue). The test exploits the oxidisability of the aldehyde H — the presence of a C=O alone is not sufficient. Ketones are classified as non-reducing compounds (they do not reduce Cu²⁺ under Fehling's conditions).
Insight: Benedict's solution is used in urine glucose testing in medicine — glucose is an aldehyde sugar (open-chain form has an aldehyde group) and gives a positive Benedict's test. Diabetic patients with elevated blood glucose show a colour change from blue toward brick-red. Ketone bodies (acetone, acetoacetate) that accumulate in diabetic ketoacidosis are — as their name states — ketones, and they do NOT react with Benedict's solution. The chemistry distinction between aldehydes and ketones has direct medical diagnostic significance.
04
Why Ketones Resist Further Oxidation — The Structural Basis
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.
Structural Argument
Oxidation of a carbonyl compound requires removing the H from the carbonyl carbon:
Aldehyde (R–CHO): the carbonyl C has one H → that H can be removed by an oxidising agent → aldehyde oxidised to carboxylic acid (R–COOH)
Ketone (R–CO–R'): the carbonyl C has NO H → there is nothing to remove → oxidising agent cannot act → no reaction under normal conditions
To further oxidise a ketone would require breaking a C–C bond — a much more demanding process requiring concentrated oxidising acid at high temperature, far outside HSC Module 7 scope.
Oxidation Hierarchy
Primary path: Alkane → Alcohol → Aldehyde → Carboxylic acid Secondary path: Alkane → Alcohol → Ketone (STOPS HERE) Tertiary path: Tertiary alcohol → NO REACTION (no H on C–OH) Each step removes H atoms. Steps only possible when H is available on the relevant carbon.
Practical Consequence — Product Control
Ketone oxidation resistance makes ketones valuable as stable synthetic targets. If you need a stable, non-further-oxidisable carbonyl compound → choose a secondary alcohol and oxidise it → get the ketone → it stays there.
If you need an aldehyde → must use distillation from a primary alcohol to prevent over-oxidation to acid. The ketone equivalent does not require this precaution.
Aldehyde vs Ketone — Complete Chemical Comparison
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
Stability to oxidants
Low — easily oxidised
High — resistant
Position of C=O
Terminal (C1)
Internal (C2 or higher)
H on carbonyl C?
Yes
No
Must Do: "Ketones cannot be oxidised under normal lab conditions" is the correct HSC statement. The phrase "under normal conditions" is important — it qualifies that this applies to K₂Cr₂O₇/H⁺, KMnO₄/H⁺, Tollens', and Fehling's, not to extreme conditions outside HSC scope.
Common Error: Writing "propan-2-ol → propanone → propanoic acid" as an oxidation pathway. This is wrong — propanone is a ketone and CANNOT be oxidised to propanoic acid under standard conditions. Propanoic acid (3-carbon carboxylic acid) can only be produced from a primary 3-carbon alcohol (propanal as intermediate). The pathway "secondary alcohol → ketone" is a dead end for further oxidation — there is no arrow from ketone to carboxylic acid in normal synthesis.
⚠️ Common Misconceptions — Aldehydes & Ketones
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"The H in an aldehyde (–CHO) is an O–H hydrogen, so aldehydes can H-bond with each other." The H in –CHO is bonded to CARBON, not oxygen — it is a C–H bond. There is no O–H group in an aldehyde. Aldehydes cannot donate H-bonds between their own molecules. The carbonyl oxygen has lone pairs but is not bonded to H.
✗
"Fehling's test detects all carbonyl compounds (aldehydes and ketones)." Fehling's solution is specific for aldehydes only. Ketones give a negative result — the solution stays blue. The test exploits the oxidisability of the aldehyde H; ketones have no equivalent H.
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"Ketones can be oxidised to carboxylic acids by using KMnO₄ instead of K₂Cr₂O₇." Neither KMnO₄/H⁺ nor K₂Cr₂O₇/H⁺ oxidises ketones under normal conditions. The resistance is structural (no H on carbonyl C) — it is independent of which oxidising agent is used.
✗
"Propanal and propanone are the same compound — they have the same formula C₃H₆O." They are functional group isomers — same molecular formula (C₃H₆O) but different structural formulae and completely different chemical properties. Propanal gives a positive Tollens' test; propanone does not. Same formula does not mean same compound.
Worked Examples
Example 1Straightforward
Identifying and Naming Aldehydes and Ketones
Problem: Identify whether each compound is an aldehyde or ketone, name it using IUPAC rules, and predict its Tollens' test result: (a) CH₃CH₂CH₂CHO; (b) CH₃COCH₂CH₂CH₃; (c) HCHO; (d) CH₃CH₂COCH₂CH₃
1
(a) CH₃CH₂CH₂CHO: carbonyl at chain end with H on same carbon (–CHO). Aldehyde. 4 carbons → butanal. Tollens': positive (silver mirror).
2
(b) CH₃COCH₂CH₂CH₃: C=O between C2 (bonded to CH₃) and C3 (bonded to CH₂CH₂CH₃). Internal carbonyl, no H on carbonyl C. Ketone. 5 carbons, C=O at C2 → pentan-2-one. Tollens': negative.
3
(c) HCHO: H₂C=O — the carbonyl C has two H atoms and one =O. Terminal by definition (C1 of 1-carbon chain). Aldehyde. → methanal (formaldehyde). Tollens': positive.
4
(d) CH₃CH₂COCH₂CH₃: C=O at C3, flanked by CH₂CH₃ on each side. Internal carbonyl, no H on carbonyl C. Ketone. 5 carbons, C=O at C3 → pentan-3-one. Tollens': negative.
Ranking Boiling Points of C4 Compounds — IMF Analysis
Problem: Arrange butane, butanal, butan-2-one, butan-1-ol, and butanoic acid in order of increasing boiling point. Explain your reasoning.
1
Identify IMF type for each:
Butane (C₄H₁₀): non-polar alkane → dispersion forces only
Butanal (C₄H₈O): aldehyde → C=O dipole-dipole + dispersion; cannot H-bond between molecules
Butan-2-one (C₄H₈O): ketone → C=O dipole-dipole + dispersion; cannot H-bond between molecules
Butan-1-ol (C₄H₁₀O): alcohol → O–H H-bonding (donor + acceptor) + dispersion
Butanoic acid (C₄H₈O₂): carboxylic acid → O–H H-bonding + dimerisation (2 simultaneous H-bonds per pair)
2
Rank by IMF strength (weakest → strongest):
Dispersion only < Dipole-dipole < H-bonding < H-bonding + dimerisation
Within dipole-dipole: butan-2-one slightly above butanal — internal C=O flanked by two alkyl groups is slightly more polar than terminal C=O, giving slightly stronger dipole-dipole forces.
3
Full order with actual BPs: Butane (−1°C) < Butanal (75°C) < Butan-2-one (80°C) < Butan-1-ol (118°C) < Butanoic acid (164°C)
Answer: Butane < Butanal < Butan-2-one < Butan-1-ol < Butanoic acid. Explanation: Butane has only dispersion forces (lowest BP). Butanal and butan-2-one have C=O dipole-dipole forces — stronger than dispersion alone but weaker than H-bonding; neither can donate H-bonds between its own molecules. Butan-2-one > butanal due to slightly more polar internal C=O. Butan-1-ol has O–H H-bonds (stronger, donor + acceptor). Butanoic acid has O–H H-bonds plus dimerisation (two simultaneous H-bonds per pair) — highest BP.
Example 3Extended Response
Identifying Unknown Compounds Using Chemical Tests (7 marks)
Problem: Four organic compounds A (C₄H₈O), B (C₄H₈O), C (C₄H₁₀O), D (C₄H₈O) gave the following test results:
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
(a) Classify A, B, C, D by functional group. (b) Identify the IUPAC name of each compound. (c) Explain why B gives negative Tollens' and negative Fehling's despite having a C=O group. (d) Write the equation for oxidation of A under reflux with excess K₂Cr₂O₇/H⁺.
1
(a) Classify by test results:
A (C₄H₈O): Tollens' + and Fehling's + → aldehyde
B (C₄H₈O): Tollens' −, Fehling's −, K₂Cr₂O₇ stays orange → no oxidation → ketone (C₄H₈O with internal C=O, no H on carbonyl C)
C (C₄H₁₀O): Tollens' −, Fehling's −, but K₂Cr₂O₇ → green → oxidisable but not aldehyde → secondary alcohol (C₄H₁₀O = saturated alcohol; secondary gives + K₂Cr₂O₇ but − Tollens')
D (C₄H₈O): Tollens' + and Fehling's + → aldehyde (isomer of A)
2
(b) Identify structures:
A (C₄H₈O aldehyde, straight chain): butanal (CH₃CH₂CH₂CHO)
B (C₄H₈O ketone): only one C₄ ketone with this formula → butan-2-one (CH₃COCH₂CH₃)
C (C₄H₁₀O secondary alcohol): butan-2-ol (CH₃CHOHCH₂CH₃)
D (C₄H₈O aldehyde, branched isomer): 2-methylpropanal ((CH₃)₂CHCHO)
3
(c) Why B (butan-2-one) gives negative Tollens' and Fehling's:
Butan-2-one is a ketone — the carbonyl carbon (C2) is bonded to CH₃ (C1) and CH₂CH₃ (C3) and has NO H atom. Tollens' reagent and Fehling's solution are mild oxidising agents that work by removing the H from the carbonyl carbon of an aldehyde. Since butan-2-one has no H on its carbonyl carbon, these oxidising agents cannot act on it. Ag⁺ in Tollens' and Cu²⁺ in Fehling's are NOT reduced → no silver mirror → no brick-red precipitate. Ketones are non-reducing compounds.
4
(d) Oxidation of A (butanal) under reflux:
Butanal is an aldehyde → excess K₂Cr₂O₇/H⁺ + reflux → full oxidation to carboxylic acid. CH₃CH₂CH₂CHO + [O] → CH₃CH₂CH₂COOH
Product: butanoic acid. Conditions: excess K₂Cr₂O₇/H₂SO₄, reflux. Orange → green.
Answer:
(a) A = aldehyde; B = ketone; C = secondary alcohol; D = aldehyde.
(b) A = butanal; B = butan-2-one; C = butan-2-ol; D = 2-methylpropanal.
(c) Butan-2-one has no H on the carbonyl C — Tollens' and Fehling's cannot oxidise it; Ag⁺ and Cu²⁺ are not reduced → negative results.
(d) CH₃CH₂CH₂CHO + [O] → CH₃CH₂CH₂COOH (butanoic acid). Excess K₂Cr₂O₇/H₂SO₄, reflux. Orange → green.
ALDEHYDE vs KETONE:
Aldehyde: R–CHO · terminal C=O · H on carbonyl C · suffix –al
Ketone: R–CO–R' · internal C=O · NO H on carbonyl C · suffix –one
Both: CₙH₂ₙO (functional group isomers)
CHEMICAL TESTS:
Tollens': aldehyde → silver mirror · ketone → no reaction
Fehling's/Benedict's: aldehyde → brick-red Cu₂O · ketone → stays blue
K₂Cr₂O₇/H⁺: aldehyde → orange→green · ketone → stays orange
(K₂Cr₂O₇ less specific — also reacts with 1°/2° alcohols)
OXIDATION RESISTANCE:
Ketone has NO H on carbonyl C → cannot be oxidised (normal conditions)
Secondary alcohol → ketone (dead end)
Primary alcohol → aldehyde → carboxylic acid (must use distillation to stop at aldehyde)
Activities
Activity A
Test Result Prediction — Complete the Table
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.
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
Check Your Understanding
Question 1. A student adds Fehling's solution to pentan-3-one and warms it in a water bath. What does the student observe?
ABrick-red precipitate forms — pentan-3-one is oxidised by Cu²⁺
BSolution turns green — Cr³⁺ is produced as the ketone is reduced
CNo colour change — pentan-3-one is a ketone and does not react with Fehling's solution
DSilver mirror forms on the test tube — Ag⁺ is reduced by the ketone
Question 2. Which correctly explains why propanal has a lower boiling point than propan-1-ol, despite both being C3 compounds containing oxygen?
APropanal is more polar than propan-1-ol because the C=O bond is stronger than C–O
BPropan-1-ol can both donate and accept hydrogen bonds via its O–H group; propanal can only accept H-bonds (via C=O lone pairs) but cannot donate — resulting in weaker intermolecular forces for propanal
CPropanal has lower molecular mass than propan-1-ol and therefore weaker dispersion forces
DPropan-1-ol has more hydrogen atoms than propanal, giving it more intermolecular contact points
Question 3. A student treats an unknown compound with Tollens' reagent and observes a silver mirror. What can the student conclude?
AThe compound is a ketone — ketones reduce Ag⁺ to Ag⁰ when heated
BThe compound is an aldehyde — only aldehydes (which have H on the carbonyl carbon) reduce Ag⁺ to Ag⁰ under Tollens' conditions
CThe compound is a primary alcohol — primary alcohols give a silver mirror with Tollens' reagent
DThe compound could be either an aldehyde or a ketone — both reduce Ag⁺ to Ag⁰
Question 4. Which statement correctly explains why ketones cannot be further oxidised to carboxylic acids under normal laboratory conditions?
AThe carbonyl carbon in a ketone has no H atom — oxidation requires removing H from the carbonyl carbon, which is impossible for a ketone
BKetones have a higher molar mass than carboxylic acids, making the reaction endothermic and thermodynamically unfavourable
CThe C=O bond in ketones is stronger than the C=O bond in carboxylic acids, preventing the reaction
DKetones are non-polar and do not dissolve in the aqueous K₂Cr₂O₇/H⁺ solution
Question 5. Propanone (acetone) is described as a "universal solvent." Which combination of properties best explains this?
APropanone has a high boiling point, making it stable enough to dissolve most solutes
BPropanone can donate H-bonds because it contains an oxygen atom near H atoms
CPropanone is a ketone and all ketones are miscible with water and organic solvents
DPropanone has a polar C=O (lone pairs accept H-bonds from water, compatible with polar solvents) and two non-polar CH₃ groups (compatible with organic solvents), giving it dual polarity
Revisit Your Initial Thinking
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Short Answer Questions
Question 64 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 75 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 86 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.
Multiple Choice Answers
Q1 — C: No colour change — ketone doesn't react with Fehling's.
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, no Cu₂O precipitate forms. A and D describe aldehyde results. B describes the K₂Cr₂O₇ colour change (irrelevant here — and K₂Cr₂O₇ wouldn't change either for a ketone).
Q2 — B: Propan-1-ol donates AND accepts H-bonds; propanal only accepts.
Propan-1-ol (BP 97°C) has an O–H group — it both donates H-bonds (O–H to neighbouring O lone pair) 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 (weaker than H-bonding) → lower BP. A is wrong — polarity of the C=O bond does not directly compare to C–O in this way. C is partially true (MW 58 vs 60) but the MW difference is negligible — IMF type is the dominant factor.
Q3 — B: Only aldehydes give a positive Tollens' test.
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 under Tollens' conditions. Ketones (no H on carbonyl C) give no silver mirror. Primary alcohols do NOT give a positive Tollens' test under standard conditions — Tollens' is specific to aldehydes. A and D are wrong — ketones give NEGATIVE Tollens'.
Q4 — A: No H on carbonyl C in ketones.
Oxidation of a carbonyl compound requires removing the H from the carbonyl carbon. In a ketone (R–CO–R'), the carbonyl C is bonded to two other C atoms and has NO H atom available. Without that H, the oxidising agent has nothing to remove at that position — the reaction cannot proceed under normal conditions. B is wrong — this has no chemical basis. C is wrong — bond strength of C=O is similar in both. D is wrong — many ketones are partially water-soluble (propanone is fully miscible with water).
Q5 — D: Polar C=O + non-polar CH₃ gives dual polarity.
Propanone's versatility as a solvent comes from its dual nature: the polar C=O oxygen (lone pairs accept H-bonds from water → miscible with polar/aqueous solvents) + two non-polar CH₃ groups (compatible with non-polar organic solvents like hexane and diethyl ether). A is wrong — propanone has a LOW BP (56°C). B is wrong — propanone has NO O–H bond and cannot donate H-bonds. C is wrong — longer-chain ketones are not miscible with water.
Short Answer Sample Answers
Q6 — Propanal vs propanone (4 marks):
(a) Structural difference: both have molecular 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) Tollens' results: Propanal → silver mirror (positive). The H on the carbonyl carbon of propanal is oxidised; Ag⁺ in Tollens' reagent is reduced to Ag⁰ (silver metal), which deposits as a mirror on the glass. Propanone → no reaction (negative). Propanone has no H on its carbonyl carbon for Tollens' to oxidise; Ag⁺ is not reduced. [1 mark each = 2 marks]
(c) Propanal can be oxidised because the carbonyl carbon has one H atom that can be removed by an oxidising agent, forming propanoic acid (the C=O is converted to –COOH by adding [O]). Propanone cannot be oxidised because the carbonyl carbon has no H atom — the oxidising agent has nothing to remove at that position. Breaking a C–C bond would be required, which does not occur under normal conditions. [1 mark]
Q7 — BP ranking and IMF (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 (electron-donating) is slightly more polar than terminal C=O, giving marginally stronger dipole-dipole forces [1 mark]
Butan-1-ol: O–H H-bonds — can both donate and accept → much stronger IMF than dipole-dipole → large jump in BP [½ mark]
Butanoic acid: O–H H-bonds PLUS dimerisation (carboxylic acids form pairs with two simultaneous H-bonds per pair) → strongest IMF → highest BP [1 mark]
Q8 — Identifying unknowns X, Y, Z (6 marks):
(a) All are C₃H₆O (one degree of unsaturation = C=O). X: positive Tollens' and positive Fehling's → aldehyde. Y: negative Tollens' and negative Fehling's → ketone. Z: positive Tollens' → aldehyde (isomer of X). [1 mark]
(b) C₃H₆O aldehydes: only one possible → propanal (CH₃CH₂CHO). C₃H₆O ketone: only one possible → propanone (CH₃COCH₃). Both X and Z must be propanal as it is the only C₃ aldehyde — they could be different samples of the same compound (or experiment has duplicate). X = Z = propanal; Y = propanone. [2 marks — 1 for structures, 1 for correct identification]
(c) Molecular basis: X is propanal — the carbonyl carbon has one H atom. Tollens' reagent ([Ag(NH₃)₂]⁺) is a mild oxidising agent that removes this H, oxidising the aldehyde to a carboxylate ion; Ag⁺ is simultaneously reduced to Ag⁰ (silver metal), depositing as a silver mirror on the glass. Y is propanone — the carbonyl carbon has NO H atom (bonded to two CH₃ groups). Tollens' cannot remove H from the carbonyl carbon because none exists; Ag⁺ is not reduced, no silver mirror forms. [2 marks]
(d) Propanal oxidation: CH₃CH₂CHO + [O] → CH₃CH₂COOH. Product: propanoic acid. Conditions: excess K₂Cr₂O₇/H₂SO₄, reflux. Colour change: orange → green. [1 mark]
Revisit — Think First
Vanilla and Raspberry — The Same Group, Differently Placed
Return to your structural prediction. The key answer:
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 on its aromatic ring. 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.
Both vanillin and raspberry ketone smell the way they do partly because of this carbonyl group's polarity — C=O is highly polar and volatile at room temperature, which helps these molecules reach your olfactory receptors. The structural difference determines chemistry; the C=O polarity determines the scent experience.
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