Year 12 Chemistry Module 7 · Organic Chemistry ⏱ ~45 min 5 MC · 3 Short Answer Lesson 13 of 23 Core

Aldehydes & Ketones, Structure, Properties & Tests

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.

Today's hook: Vanilla and raspberry, both get their signature flavours from compounds with a C=O group. So why does one form a silver mirror in a test tube while the other does nothing?

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Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

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Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

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Connect: meet the carbonyl family

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.

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Key Structures & Test Equations

Aldehyde

R–CHO  (terminal C=O, H on carbonyl C)

suffix –al · IUPAC locant not needed (always C1) · CₙH₂ₙO

Ketone

R–CO–R'  (internal C=O, NO H on carbonyl C)

suffix –one · locant needed (C5+) · CₙH₂ₙO (same formula, isomers)

Tollens' Test (aldehyde only)

R–CHO + 2[Ag(NH₃)₂]⁺ → R–COO⁻ + 2Ag⁰ + …

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

+XP

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

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

Explain: locate and name the carbonyl

📚 Core Content

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.

Migrated comparison of propanal vs propanone carbonyl chemistry

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

IUPAC Naming Summary

ALDEHYDES, suffix –al

            HCHO, methanal

            CH₃CHO, ethanal

            CH₃CH₂CHO, propanal

            CH₃CH₂CH₂CHO, butanal

            CH₃CH₂CH₂CH₂CHO, pentanal

Locant not needed, always C1

KETONES, suffix –one

            CH₃COCH₃, propanone (C3, no locant)

            CH₃COCH₂CH₃, butan-2-one

            CH₃COCH₂CH₂CH₃, pentan-2-one

            CH₃CH₂COCH₂CH₃, pentan-3-one

Locant required for C5+ to show C=O position

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. 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. The C–H bond of –CHO cannot donate H-bonds.

Quick check: Which of the following is an aldehyde?

Represent: connect structure to properties

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)

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 a strong answer from a top-band 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.

True or False: Propanone can donate hydrogen bonds to other propanone molecules because it contains oxygen.

Respond: interpret distinguishing tests

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.

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.

Four labelled test tubes compare aldehyde positive and ketone negative results. Tollens' reagent deposits reflective silver on the inside glass wall for the aldehyde while the ketone control remains clear. Fehling's solution forms a particulate brick-red copper(I) oxide precipitate for the aldehyde while the ketone control remains uniformly blue. — A positive Tollens' test deposits Ag(s) on the glass wall; a positive Fehling's test forms brick-red Cu₂O(s) in the liquid. Ketone controls do neither.

Complete Test Results Summary

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

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."

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.

A student adds Fehling's solution to butan-2-one and heats it. What does the student observe?

Feedback: explain the oxidation boundary

Beyond the syllabus. Structures, naming, properties and the link to alcohol oxidation are core. Tollens', Fehling's and Benedict's tests are identification enrichment — they are not prescribed in Module 7, so treat the detailed reagent chemistry as extension rather than exam-required recall.

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

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.

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

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. The pathway "secondary alcohol → ketone" is a dead end for further oxidation.

Complete: Ketones resist oxidation because the carbonyl carbon has no _____ atom. In contrast, an aldehyde has a _____ on the carbonyl carbon which can be removed by a mild oxidising agent like Tollens' reagent. The product when an aldehyde is oxidised is a _____ acid.

Apply: solve carbonyl unknowns

🔬 Worked Examples

Example 1 Straightforward

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

(a) CH₃CH₂CH₂CHO: carbonyl at chain end with H on same carbon (–CHO). Aldehyde. 4 carbons → butanal. Tollens': positive (silver mirror).

(b) CH₃COCH₂CH₂CH₃: C=O between C2 and C3. Internal carbonyl, no H on carbonyl C. Ketone. 5 carbons, C=O at C2 → pentan-2-one. Tollens': negative.

(c) HCHO: H₂C=O, two H atoms on carbonyl C. Aldehyde. → methanal (formaldehyde). Tollens': positive.

(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.

Answer:
(a) Butanal, aldehyde, Tollens' positive (silver mirror)
(b) Pentan-2-one, ketone, Tollens' negative
(c) Methanal, aldehyde, Tollens' positive
(d) Pentan-3-one, ketone, Tollens' negative

Example 2 Intermediate

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.

Identify IMF type for each:
Butane: non-polar alkane → dispersion forces only
Butanal: aldehyde → C=O dipole-dipole + dispersion; cannot H-bond between molecules
Butan-2-one: ketone → C=O dipole-dipole + dispersion; cannot H-bond between molecules
Butan-1-ol: alcohol → O–H H-bonding (donor + acceptor) + dispersion
Butanoic acid: carboxylic acid → O–H H-bonding + dimerisation (2 simultaneous H-bonds per pair)

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.

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 3 Extended 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⁺.

(a) Classify by test results:
A (C₄H₈O): Tollens' + and Fehling's + → aldehyde
B (C₄H₈O): Tollens' −, Fehling's −, K₂Cr₂O₇ stays orange → no oxidation → ketone
C (C₄H₁₀O): Tollens' −, Fehling's −, but K₂Cr₂O₇ → green → oxidisable but not aldehyde → secondary alcohol
D (C₄H₈O): Tollens' + and Fehling's + → aldehyde (isomer of A)

(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)

(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.

(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.

🔬 Predict, Then Reveal +8 XP

Two compounds: propanal (CH₃CH₂CHO) and propanone (CH₃COCH₃). Before reading, predict which gives a positive result with Tollens' reagent (silver mirror test) and explain why the other does not.

Your predictionExpert answerCompare

Complete the Learn phase to unlock Practice.

Practice: classify, predict and explain

✏️ Activities

Activity A, Test Result Prediction

For each compound below, predict the result of each test (positive/negative and describe the observation). Justify each prediction.

Propanal (CH₃CH₂CHO), Tollens' · Fehling's · K₂Cr₂O₇/H⁺
Pentan-3-one (CH₃CH₂COCH₂CH₃), Tollens' · Fehling's · K₂Cr₂O₇/H⁺
Butan-1-ol (primary alcohol), Tollens' · Fehling's · K₂Cr₂O₇/H⁺
2-methylpropan-2-ol (tertiary alcohol), Tollens' · Fehling's · K₂Cr₂O₇/H⁺

Activity B, BP Ranking Practice

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

📝 Check Your Understanding

Short Answer Questions

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.

Show All Answers

Multiple Choice Answers

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).

Short Answer Sample Answers

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]

Vanilla and Raspberry, The Same Group, Differently Placed

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.

Review: retrieve the carbonyl evidence

Check what actually stuck

🏆 Review

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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