← Module 7: Organic Chemistry

NSWChemistrySyllabus dot point

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 OpusReviewed by Better Tuition Academy9 min answer

Have a quick question? Jump to the Q&A page

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=OC=O. They differ in what else is attached to the carbonyl carbon.

Class Structure Suffix Example
Aldehyde IMATH_5 -al propanal IMATH_6
Ketone IMATH_7 -one propan-2-one IMATH_8
Carboxylic acid IMATH_9 -oic acid propanoic acid IMATH_10

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

Formation: oxidation of alcohols

  • Primary alcohol (Rβˆ’CH2OHR-CH_2OH) β†’[O]\xrightarrow{[O]} aldehyde (Rβˆ’CHOR-CHO) β†’[O]\xrightarrow{[O]} carboxylic acid (Rβˆ’COOHR-COOH).
  • Secondary alcohol (Rβˆ’CH(OH)βˆ’Rβ€²R-CH(OH)-R') β†’[O]\xrightarrow{[O]} ketone (Rβˆ’COβˆ’Rβ€²R-CO-R'). Stops there.
  • Tertiary alcohol: no oxidation.

The oxidant is acidified K2Cr2O7K_2Cr_2O_7 (orange to green) or acidified KMnO4KMnO_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 Cβˆ’CC-C bonds, which does not happen under HSC conditions.

Physical properties

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

  • Aldehydes and ketones have C=OC=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(NH3)2]+[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(NH3)2]++3OHβˆ’β†’RCOOβˆ’+2Ag+4NH3+2H2ORCHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow RCOO^- + 2Ag + 4NH_3 + 2H_2O.
  • Ketone: no reaction.

Fehling's solution or Benedict's solution. Cu2+Cu^{2+} ions in alkaline tartrate or citrate complex, blue.

  • Aldehyde: brick-red precipitate of Cu2OCu_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 (pKapK_a about 4 to 5). They ionise partially in water:

RCOOH(aq)+H2O(l)β‡ŒRCOO(aq)βˆ’+H3O(aq)+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:

2CH3COOH+Mg→(CH3COO)2Mg+H22CH_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):

2CH3COOH+Na2CO3β†’2CH3COONa+H2O+CO22CH_3COOH + Na_2CO_3 \rightarrow 2CH_3COONa + H_2O + CO_2

With bases (neutralisation) to give salt plus water:

CH3COOH+NaOH→CH3COONa+H2OCH_3COOH + NaOH \rightarrow CH_3COONa + H_2O

With alcohols (esterification, H2SO4H_2SO_4 catalyst, reflux) to give an ester plus water. See the esters dot point.

Distinguishing all four classes

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

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

Common traps

Writing a positive Tollens' for a ketone. Ketones do not give a silver mirror. Only aldehydes do.

Forgetting the cyclic dimer. When explaining the high boiling point of carboxylic acids, mention the dimer explicitly; it is what marks the answer.

Confusing the test for "carbonyl group" with the test for "aldehyde". The carbonyl is in both aldehydes and ketones. Tollens and Fehling distinguish aldehyde from ketone, not carbonyl from non-carbonyl.

Oxidising a carboxylic acid further. Under HSC conditions, carboxylic acids are the terminus of the oxidation pathway. They do not go further.

Naming a propanone "propan-2-one" only. Acetone and propan-2-one are both accepted but the IUPAC name is propan-2-one (or simply propanone, since the locant is unambiguous on a 3-carbon chain).

In one sentence

A primary alcohol oxidises to an aldehyde and then to a carboxylic acid, a secondary alcohol oxidises only to a ketone, you tell aldehyde from ketone with the Tollens' silver mirror or Fehling's brick-red Cu2OCu_2O tests, and you identify a carboxylic acid by its effervescence with carbonate or its reaction with metals.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

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 β†’

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

Test 1: Tollens' reagent (ammoniacal AgNO3AgNO_3). Warm gently in a clean test tube.

  • Propanal (aldehyde): forms a silver mirror on the glass. CH3CH2CHO+2[Ag(NH3)2]++3OHβˆ’β†’CH3CH2COOβˆ’+2Ag+4NH3+2H2OCH_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 (Na2CO3Na_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 CO2CO_2 is released; blue litmus turns red. 2CH3CH2COOH+Na2CO3β†’2CH3CH2COONa+H2O+CO22CH_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 β†’

Boiling point is governed by intermolecular forces (IMFs).

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

Alcohols have a polar Oβˆ’HO-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 Oβˆ’HO-H and a polar C=OC=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.

Related dot points