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Inquiry Question 4: How do carbonyl-containing compounds form, behave and how can they be distinguished?

Investigate the structural formulae, properties and reactions of aldehydes, ketones and carboxylic acids, including their formation by oxidation of alcohols and chemical tests that distinguish them

A focused answer to the HSC Chemistry Module 7 dot point on the carbonyl compounds. The oxidation pathway from alcohols, the Tollens and Fehling/Benedict tests that distinguish aldehydes from ketones, acidity of carboxylic acids, and worked HSC past exam questions.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-22

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
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In plain English

Think of an alcohol as a rung on a ladder that oxidation pushes upward, one rung at a time. A primary alcohol (one $CH_2OH$ group with two H's to lose) can be pushed up to an aldehyde, and pushed again to a carboxylic acid, the top rung it can reach. A secondary alcohol only has one rung available: it becomes a ketone and then gets stuck, because a ketone's carbonyl carbon has no H left for the oxidant to remove. A tertiary alcohol cannot even take the first step, since its OH-bearing carbon has no H at all. Once you know which rung a compound sits on, two cheap test-tube reactions tell you exactly which one it is: Tollens' silver-mirror test lights up only for an aldehyde (because only an aldehyde can still be oxidised further, donating electrons to reduce the silver), and fizzing with sodium carbonate lights up only for a carboxylic acid (because only the acid is sour enough to release $CO_2$ from a carbonate). A ketone gives a negative result to both tests, which is itself diagnostic: no mirror and no fizz means "the rung with nowhere further to go and no acidic proton to react with carbonate."

What this dot point is asking

NESA wants you to identify the structural feature of each carbonyl class, predict whether a given alcohol oxidises to an aldehyde, a ketone or a carboxylic acid, describe the chemical tests that distinguish aldehyde from ketone (Tollens, Fehling, Benedict), and explain why carboxylic acids are weak acids that react with metals, carbonates and bases.

The answer

The three functional groups

All three contain a carbonyl group $C=O$ . They differ in what else is attached to the carbonyl carbon.

Class	Structure	Suffix	Example
Aldehyde	$R-CHO$	-al	propanal $CH_3CH_2CHO$
Ketone	$R-CO-R'$	-one	propan-2-one $CH_3COCH_3$
Carboxylic acid	$R-COOH$	-oic acid	propanoic acid $CH_3CH_2COOH$

An aldehyde has at least one $H$ on the carbonyl carbon; a ketone has two carbons on it; a carboxylic acid has an $-OH$ directly on the carbonyl carbon.

Formation: oxidation of alcohols

Primary alcohol ( $R-CH_2OH$ ) $\xrightarrow{[O]}$ aldehyde ( $R-CHO$ ) $\xrightarrow{[O]}$ carboxylic acid ( $R-COOH$ ).
Secondary alcohol ( $R-CH(OH)-R'$ ) $\xrightarrow{[O]}$ ketone ( $R-CO-R'$ ). Stops there.
Tertiary alcohol: no oxidation.

The oxidant is acidified $K_2Cr_2O_7$ (orange to green) or acidified $KMnO_4$ (purple to colourless). To stop a primary alcohol at the aldehyde, distil the aldehyde off as it forms (aldehydes have lower boiling points than alcohols). To go to the acid, reflux with excess oxidant.

Carboxylic acids cannot be made from ketones without breaking $C-C$ bonds, which does not happen under HSC conditions.

The reaction scheme below shows the full oxidation ladder as one diagram, with the reagent and condition labelled on each arrow:

Physical properties

Boiling point trend (same carbon number): alkane < aldehyde/ketone < alcohol < carboxylic acid.

Aldehydes and ketones have $C=O$ dipole-dipole forces but no hydrogen bonding, so they boil above alkanes but below alcohols.
Alcohols hydrogen-bond and boil higher.
Carboxylic acids form cyclic dimers in the liquid phase, with two hydrogen bonds per pair, so they boil highest of all.

Solubility in water decreases with chain length. Short-chain carbonyls (acetone, propanal, ethanoic acid) are fully miscible because the polar functional group hydrogen bonds with water. Beyond about C5, the alkyl chain dominates and solubility falls.

Tests that distinguish aldehyde from ketone

Tollens' reagent (silver mirror test). $[Ag(NH_3)_2]^+$ in alkaline solution. Warm gently in a clean glass test tube.

Aldehyde: silver metal deposits on the glass as a silver mirror. $RCHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow RCOO^- + 2Ag + 4NH_3 + 2H_2O$ .
Ketone: no reaction.

Fehling's solution or Benedict's solution. $Cu^{2+}$ ions in alkaline tartrate or citrate complex, blue.

Aldehyde: brick-red precipitate of $Cu_2O$ forms on warming.
Ketone: no reaction.

Both tests work because aldehydes are easily oxidised to carboxylates; ketones are not. Only aliphatic aldehydes give a positive Fehling's; aromatic aldehydes are negative. Tollens works for both.

A third option is to oxidise with acidified dichromate. Both aldehydes and primary alcohols decolourise (orange to green); ketones do not. If you suspect aldehyde, the silver mirror confirms it.

Reactions of carboxylic acids

Carboxylic acids are weak acids ( $pK_a$ about 4 to 5). They ionise partially in water:

RCOOH_{(aq)} + H_2O_{(l)} \rightleftharpoons RCOO^-_{(aq)} + H_3O^+_{(aq)}

They undergo all the standard acid reactions.

With reactive metals (Mg, Zn, Fe) to give salt plus hydrogen:

2CH_3COOH + Mg \rightarrow (CH_3COO)_2Mg + H_2

With carbonates and hydrogencarbonates to give salt plus water plus carbon dioxide (this is the diagnostic test, since aldehydes, ketones and alcohols do not react):

2CH_3COOH + Na_2CO_3 \rightarrow 2CH_3COONa + H_2O + CO_2

With bases (neutralisation) to give salt plus water:

CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O

With alcohols (esterification, $H_2SO_4$ catalyst, reflux) to give an ester plus water. See the esters dot point.

Because a carboxylic acid is a weak acid, its titration against a strong base such as NaOH has an equivalence point ABOVE pH 7, not at pH 7:

Worked example: mass of silver deposited by Tollens' reagent

Question. A student adds excess Tollens' reagent to 0.444 g of propanal ( $CH_3CH_2CHO$ ) and warms gently until reaction is complete. Calculate the theoretical mass of silver metal deposited, to 3 significant figures. ( $M(\text{propanal}) = 58.08\ \text{g mol}^{-1}$ , $M(\text{Ag}) = 107.87\ \text{g mol}^{-1}$ .)

Step 1: write the balanced equation.

CH_3CH_2CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow CH_3CH_2COO^- + 2Ag + 4NH_3 + 2H_2O

The mole ratio of propanal to silver is 1:2.

Step 2: moles of propanal.

n(\text{propanal}) = \frac{m}{M} = \frac{0.444\ \text{g}}{58.08\ \text{g mol}^{-1}} = 0.0076446\ \text{mol}

Step 3: moles of silver (1:2 ratio).

n(\text{Ag}) = 2 \times 0.0076446\ \text{mol} = 0.0152892\ \text{mol}

Step 4: mass of silver.

m(\text{Ag}) = n \times M = 0.0152892\ \text{mol} \times 107.87\ \text{g mol}^{-1} = 1.6494\ \text{g}

Step 5: round to 3 significant figures (matching 0.444 g, which has 3 s.f.).

m(\text{Ag}) = 1.65\ \text{g}

Marker's note: the most common error here is using a 1:1 ratio instead of the correct 1:2 ratio from the balanced half-equation; always check the coefficient of $Ag$ in the equation before scaling moles. A ketone in the same mass would deposit ZERO silver, since it gives no reaction with Tollens' reagent.

Distinguishing all four classes

A flowchart that handles alcohol, aldehyde, ketone, carboxylic acid:

Sodium carbonate or blue litmus. Effervescence/red colour identifies the carboxylic acid. Remove it from consideration.
Tollens' reagent on the remaining three. Silver mirror identifies the aldehyde.
Acidified dichromate on the last two. Orange to green identifies the alcohol; orange remains for the ketone.

Examples in context

Example 1. Industrial production of ethanoic acid at Penrice Soda. Although Penrice's main plant in South Australia made carbonate, the NSW import pipeline relied on truck-tankered ethanoic acid produced by catalytic oxidation of ethanol. Air-oxidised ethanol over a copper catalyst gives ethanal as a removable middle product, then ethanoic acid: $CH_3CH_2OH \rightarrow CH_3CHO \rightarrow CH_3COOH$ . Commercial vinegar packed at the Cornwell vinegar plant near Lidcombe is the dilute aqueous version, around 4 to 8 percent acid. The HSC oxidation tree predicts every transformation in the plant flowsheet. Quality control labs verify the product with sodium hydrogen carbonate effervescence and with phenolphthalein titration against NaOH.

Example 2. Distinguishing propanal from propanone in a Stage 6 lab. A NSW HSC depth study common across schools gives students unlabelled propanal and propanone and asks them to identify each chemically. A few drops of Tollens' reagent (silver-ammonia) added and warmed in a water bath produces a silver mirror with propanal (the aldehyde) but no change with propanone (the ketone). Repeating with Fehling's solution gives brick-red $Cu_2O$ with propanal and a clear blue solution with propanone. Students must write the half-equations and explain why the aldehyde, but not the ketone, can lose another H to become a carboxylic acid. This is exactly the test NESA examines in Section II tasks.

Try this

Q1. Write structural formulae for butanal, butan-2-one and butanoic acid, and identify which can be distinguished by Tollens' reagent. [3 marks]

Cue. Butanal is an aldehyde (gives silver mirror); butan-2-one is a ketone (negative); butanoic acid is an acid (negative with Tollens', positive with $Na_2CO_3$ ).

Q2. A 1.50 g sample of butanoic acid is titrated with 0.250 mol L $^{-1}$ NaOH. Calculate the volume of NaOH required for complete neutralisation. [3 marks]

Cue. $n(\text{acid}) = 1.50 / 88.11 = 0.0170$ mol; 1:1 stoichiometry; $V = 0.0170 / 0.250 = 0.0681$ L $= 68.1$ mL.

Q3. A student is given three unknowns A, B and C in unlabelled bottles. A gives a silver mirror with Tollens', B effervesces with sodium hydrogen carbonate, C gives neither reaction. (a) Classify each compound. (b) Write a balanced ionic equation for the carbonate reaction. (c) Predict the oxidation product of A. [2+2+1 marks]

Cue. (a) A aldehyde, B carboxylic acid, C ketone or alcohol. (b) $2RCOOH + CO_3^{2-} \rightarrow 2RCOO^- + H_2O + CO_2$ . (c) Carboxylic acid.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2021 HSC4 marksA student has three unlabelled flasks containing propanal, propan-2-one and propanoic acid. Describe two chemical tests, including expected observations, that would identify each compound.

Show worked answer →

A 4 mark answer needs two tests, the reagents, and the observations for each of the three compounds.

Test 1: Tollens' reagent (ammoniacal $AgNO_3$ ). Warm gently in a clean test tube.

Propanal (aldehyde): forms a silver mirror on the glass. $CH_3CH_2CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow CH_3CH_2COO^- + 2Ag + 4NH_3 + 2H_2O$ .
Propan-2-one (ketone): no reaction, mixture stays colourless.
Propanoic acid: no reaction with Tollens' (carboxylic acid is already at the highest oxidation level reachable here).

Tollens' identifies propanal.

Test 2: Sodium carbonate ( $Na_2CO_3$ solution) or a few drops of universal indicator/blue litmus.

Propanal: pH about 7, no effervescence.
Propan-2-one: pH about 7, no effervescence.
Propanoic acid: vigorous effervescence as $CO_2$ is released; blue litmus turns red. $2CH_3CH_2COOH + Na_2CO_3 \rightarrow 2CH_3CH_2COONa + H_2O + CO_2$ .

Sodium carbonate identifies propanoic acid; by elimination, the remaining flask is propan-2-one.

Markers reward (1) the two reagents, (2) the specific observation for each compound including the silver mirror and effervescence, (3) correct elimination logic.

2018 HSC3 marksExplain why the boiling points of carboxylic acids are higher than those of aldehydes, ketones and alcohols of similar molar mass.

Show worked answer →

Boiling point is governed by intermolecular forces (IMFs).

Aldehydes and ketones have a polar $C=O$ but no $O-H$ . Their main IMFs are dipole-dipole plus dispersion. No hydrogen bonding.

Alcohols have a polar $O-H$ , so they hydrogen bond. Each alcohol can donate one H and accept up to two via the O lone pairs.

Carboxylic acids have both a polar $O-H$ and a polar $C=O$ . In the liquid phase, two acid molecules associate as a cyclic dimer, held together by two hydrogen bonds. Effectively the boiling species is twice the molar mass, and breaking the dimer requires breaking two hydrogen bonds.

Hence the order: alkane < aldehyde/ketone < alcohol < carboxylic acid.

Markers reward (1) identifying H-bonding in alcohols and acids only, (2) noting the dimer for acids, (3) ranking the series.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marksClassify each compound as aldehyde, ketone or carboxylic acid, and give its IUPAC name: (a)

CH_3CH_2CH_2CHO

; (b)

CH_3COCH_2CH_3

; (c)

CH_3CH_2CH_2COOH

Show worked solution →

A 3-mark identify needs the correct class AND the correct IUPAC name for each.

(a) $CH_3CH_2CH_2CHO$: The carbonyl carbon carries one H, so this is an aldehyde. Four carbons in the chain gives the stem "but-", plus the aldehyde suffix "-al": butanal.
(b) $CH_3COCH_2CH_3$: The carbonyl carbon carries two carbon substituents, so this is a ketone. Four carbons total, carbonyl on C2: butan-2-one.
(c) $CH_3CH_2CH_2COOH$: The carbonyl carbon carries an -OH directly, so this is a carboxylic acid. Four carbons: butanoic acid.

Marking criteria: 1 mark per compound for the correct class and matching IUPAC name (both required for the mark).

foundation4 marksState the reagent(s) and condition(s), and the observed colour or visual change, for each conversion: (a) propan-1-ol to propanal only; (b) propan-1-ol to propanoic acid; (c) propan-2-ol to propan-2-one.

Show worked solution →

A 4-mark identify needs the reagent, the condition, and the observation for each of the three conversions.

(a) Propan-1-ol to propanal only: Acidified $K_2Cr_2O_7$ , gentle heat, distilling the aldehyde off as it forms (not refluxed). Colour change: orange to green as the dichromate is reduced.
(b) Propan-1-ol to propanoic acid: Acidified $K_2Cr_2O_7$ in excess, under reflux. Colour change: orange to green.
(c) Propan-2-ol to propan-2-one: Acidified $K_2Cr_2O_7$ , under reflux (a secondary alcohol oxidises fully to the ketone in one step, since only one H is available on the C-OH carbon). Colour change: orange to green.

Marking criteria: 1 mark for (a)'s distillation condition specifically (not reflux), 1 mark for (b)'s excess/reflux condition, 1 mark for (c)'s reagent and condition, 1 mark for correctly stating the orange-to-green colour change applies to all three.

core5 marksA 2.20 g sample of pure ethanoic acid is dissolved in water and titrated to the equivalence point with 0.400 mol L

^{-1}

NaOH solution. Calculate the volume of NaOH solution required, to 3 significant figures. (

M(\text{CH}_3\text{COOH}) = 60.05\ \text{g mol}^{-1}

Show worked solution →

Step 1: write the neutralisation equation (1:1 stoichiometry).

CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O

One mole of acid reacts with exactly one mole of NaOH.

Step 2: moles of ethanoic acid.

n(\text{CH}_3\text{COOH}) = \frac{m}{M} = \frac{2.20\ \text{g}}{60.05\ \text{g mol}^{-1}} = 0.036636\ \text{mol}

Step 3: moles of NaOH required (1:1 ratio).

n(\text{NaOH}) = 0.036636\ \text{mol}

Step 4: volume of NaOH solution.

V = \frac{n}{c} = \frac{0.036636\ \text{mol}}{0.400\ \text{mol L}^{-1}} = 0.091590\ \text{L}

Step 5: convert to mL and round to 3 significant figures (matching 2.20 g, which has 3 s.f.).

V = 91.6\ \text{mL}

Marking criteria: 1 mark for the correct balanced 1:1 equation, 1 mark for correct moles of acid, 1 mark for correctly carrying the 1:1 mole ratio, 1 mark for the volume calculation, 1 mark for the correct final answer to 3 significant figures with correct units (mL or L).

core5 marksThe titration curve below shows pH against volume of 0.400 mol L

^{-1}

NaOH added to 25.0 mL of the ethanoic acid solution from the previous question. (a) Identify the approximate equivalence point volume and explain why the pH at equivalence is above 7, not exactly 7. (b) Name a suitable indicator for this titration and justify your choice using the graph.

Show worked solution →

Reading the curve. The pH starts low (around 3, typical of a weak acid), rises gradually through a buffer region, then shows a steep vertical jump centred near 91.6 mL, before levelling off at high pH (around 12) with excess NaOH.

(a) Equivalence point and pH above 7. The equivalence point is at approximately 91.6 mL, matching the stoichiometric volume calculated from the mole ratio. At equivalence, the solution contains only sodium ethanoate, $CH_3COONa$ , dissolved in water. The ethanoate ion $CH_3COO^-$ is the conjugate base of a weak acid, so it hydrolyses water:

CH_3COO^-_{(aq)} + H_2O_{(l)} \rightleftharpoons CH_3COOH_{(aq)} + OH^-_{(aq)}

This produces excess $OH^-$ , making the equivalence-point pH greater than 7 (basic), unlike a strong acid/strong base titration where the equivalence point is exactly pH 7.

(b) Indicator choice. Phenolphthalein (colour-change range approximately pH 8.2 to 10.0) is suitable, because its colour-change range lies entirely within the steep vertical section of the curve, close to the true equivalence point above pH 7. Methyl orange (range approximately pH 3.2 to 4.4) would change colour far too early, well before the equivalence point, and would give an inaccurate (too low) titre.

Marking criteria: (a) 1 mark for the correct approximate equivalence volume read from the graph, 1 mark for identifying the solution at equivalence as sodium ethanoate, 1 mark for the hydrolysis equation or equivalent reasoning explaining pH above 7. (b) 1 mark for naming phenolphthalein specifically, 1 mark for justifying using the steep region of the graph coinciding with its colour-change range (not just "it works for weak acids").

core6 marksA student is given four unlabelled solutions: ethanol, ethanal, propan-2-one and ethanoic acid. Design a test sequence using no more than two reagents that identifies all four, stating reagents, conditions, and expected observations at each stage.

Show worked solution →

Reagent 1: solid $NaHCO_3$ (or aqueous $Na_2CO_3$ ), added to a small sample of each.

Ethanoic acid: vigorous effervescence ( $CO_2$ gas released), confirming the carboxylic acid. Remove from further testing.
Ethanol, ethanal, propan-2-one: no observable reaction (no gas). These three remain to be distinguished.

Reagent 2: Tollens' reagent (ammoniacal $AgNO_3$ ), warmed gently in a water bath, applied to the remaining three.

Ethanal: a silver mirror forms on the inside of the test tube (aldehyde is oxidised, reducing $Ag^+$ to $Ag$ ).
Ethanol: no reaction (an alcohol is not oxidised by Tollens' reagent under these mild conditions).
Propan-2-one: no reaction (a ketone has no H on the carbonyl carbon to allow further oxidation).

Distinguishing ethanol from propan-2-one. Since Tollens' gives a negative result for both, a third observation is needed: acidified $K_2Cr_2O_7$ turns from orange to green with ethanol (oxidisable primary alcohol) but stays orange with propan-2-one (already fully oxidised, cannot be oxidised further under these conditions). This is a NECESSARY third reagent because ethanol and propan-2-one cannot be told apart by carbonate or Tollens' alone.

Marking criteria: 1 mark for the carbonate test isolating the acid with correct observation, 1 mark for correct elimination logic, 1 mark for the Tollens' test isolating the aldehyde with correct observation (silver mirror), 1 mark for correctly noting propan-2-one and ethanol both give no reaction with Tollens', 1 mark for proposing acidified dichromate as the necessary distinguishing step for the last two, 1 mark for the correct colour-change observation for that final step.

exam7 marksAn unknown organic liquid Y has molecular formula

C_4H_8O

. Warmed gently with Tollens' reagent, Y gives no silver mirror. Warmed with acidified

K_2Cr_2O_7

, the solution stays orange (no colour change). Evaluate what these two results allow you to conclude about the structure of Y, propose a structure consistent with both observations and the molecular formula, and explain why a single test would have been insufficient.

Show worked solution →

This is a 7-mark EVALUATE: markers reward reasoned elimination across both pieces of evidence plus an explicit judgement on why one test alone would not suffice.

Step 1: interpret the molecular formula: $C_4H_8O$ has one degree of unsaturation (compare to saturated $C_4H_{10}O$ , which has two more H atoms), consistent with a single $C=O$ or $C=C$ or a ring, but not both. A carbonyl compound with this formula could be an aldehyde (e.g. butanal) or a ketone (e.g. butan-2-one); both fit $C_4H_8O$ .
Step 2: interpret the negative Tollens' result: No silver mirror rules out an aldehyde, since any aldehyde would reduce the ammoniacal silver complex to metallic silver. This is strong evidence AGAINST butanal (or any $C_4H_8O$ aldehyde isomer).
Step 3: interpret the negative dichromate result: No colour change with acidified $K_2Cr_2O_7$ rules out an easily oxidised functional group: it rules out a primary or secondary alcohol (which would turn the solution green) and is also consistent with "no aldehyde present" (since dichromate also oxidises aldehydes, giving the same colour change).
Step 4: combine the evidence: Together, both negative results eliminate alcohol AND aldehyde as candidates. The only carbonyl-containing structural class left that fits $C_4H_8O$ , contains no reactive $O-H$ , and gives no reaction with either oxidant is a ketone: the carbonyl carbon has no H atom, so it cannot be oxidised by either reagent.
Step 5: propose a structure: Butan-2-one, $CH_3COCH_2CH_3$ , fits $C_4H_8O$ exactly and is unreactive to both tests.
Step 6: why one test alone is insufficient: A negative Tollens' result alone is also consistent with a plain alcohol, an ether, or a ketone, since none of these give a silver mirror; it does not on its own rule out an alcohol. A negative dichromate result alone is consistent with a ketone or a tertiary alcohol (also unreactive), and would not distinguish those from each other. Only by combining BOTH negative results, alcohol is ruled out by the dichromate test and aldehyde is ruled out by the Tollens' test, does the evidence converge uniquely on a ketone.

Marker's note: full marks require (1) correctly reading the degree of unsaturation from the molecular formula, (2) using each negative result to eliminate a specific functional class rather than just restating "no reaction", (3) a structure that is fully consistent with $C_4H_8O$ , and (4) an explicit statement of why neither test alone is conclusive, this last point is what separates a band-6 evaluate from a band-4 description.