Edexcel A-Level Chemistry 9CH0 · Year 13 · Topic 17 (17.1–17.16)

Part 2: Carbonyl properties, redox reactions and identification

Reviewed 9 October 2026.

Connect carbonyl structure to intermolecular forces, use Edexcel reduction conditions and interpret a sequence of selective chemical tests.

Identify what is attached to C=O

An aldehyde contains –CHO: the carbonyl carbon is bonded to at least one hydrogen. A ketone contains –CO– between two carbon groups. Methanal, HCHO, has two hydrogens attached to its carbonyl carbon; propanone, CH₃COCH₃, has two methyl groups. The carbonyl carbon belongs in the parent-chain carbon count.

Oxygen attracts the shared electrons more strongly, making C=O polar, Cδ+–Oδ−. This local polarity explains attack by nucleophiles. Do not write full formal charges on the starting neutral C=O group. Carboxylic acids, esters and amides also contain C=O, but they are different functional groups and do not normally give the aldehyde/ketone 2,4-DNPH test.

Why a carbonyl can hydrogen-bond with water but not with itself

Simple aldehydes and ketones have London forces and permanent dipole–dipole attractions between their molecules. They have no O–H or N–H hydrogen to donate a conventional hydrogen bond, so they do not hydrogen-bond to one another. For comparable small molecules their boiling temperatures are usually above corresponding non-polar hydrocarbons but below hydrogen-bonded alcohols.

The carbonyl oxygen has lone pairs and can accept a hydrogen bond from the Hδ+ of a water O–H bond. Thus small aldehydes and ketones are appreciably soluble in water. As the non-polar hydrocarbon portion grows, it disrupts a greater proportion of water’s hydrogen-bond network, and solubility generally decreases.

Show the interaction as C=O···H–O–H, with the dotted hydrogen bond between carbonyl O and water H and the O···H–O arrangement approximately linear. A hydrogen on the aldehyde carbon is not a hydrogen-bond donor in this model. Boiling concerns forces between carbonyl molecules; dissolving concerns interactions between carbonyl and water molecules.

Oxidise aldehydes; reduce both classes

Warm an aldehyde with acidified potassium dichromate(VI): the orange solution turns green as Cr(VI) is reduced to Cr(III), while the aldehyde becomes a carboxylic acid. Ketones resist oxidation by the usual aldehyde test reagents because comparable oxidation would require breaking a C–C bond. “Ketones can never be oxidised” is too broad.

Lithium tetrahydridoaluminate, LiAlH₄, in dry ether reduces an aldehyde to a primary alcohol and a ketone to a secondary alcohol. Follow with controlled aqueous/acid work-up to protonate the oxygen-containing product. Dry conditions are essential during reduction because LiAlH₄ reacts with water, consuming reagent and releasing heat and hydrogen. This is supervised laboratory chemistry, with no naked flame around ether.

Worked transformation: CH₃COCH₂CH₃ becomes CH₃CH(OH)CH₂CH₃. The C=O carbon receives H; oxygen becomes OH. The product is butan-2-ol, not butan-1-ol, because reduction retains the carbon skeleton and oxygen’s position. Edexcel explicitly specifies LiAlH₄ in dry ether here; do not replace its conditions with aqueous sodium borohydride conditions.

RCHO + [O] → RCOOH
RCHO + 2[H] → RCH₂OH
RCOR′ + 2[H] → RCH(OH)R′

Use observations to answer a defined question

Add 2,4-dinitrophenylhydrazine (2,4-DNPH, Brady’s reagent): aldehydes and ketones form a yellow/orange precipitate. The useful observation is a precipitate, not merely an orange colour. It supports a reactive aldehyde/ketone carbonyl but does not distinguish the two classes. The derivative equation is not required by Topic 17.8(iv).

Tollens’ reagent contains ammoniacal silver(I). Gentle warming with an aldehyde can produce a silver mirror or grey silver deposit as Ag(I) is reduced to Ag. Fehling’s or Benedict’s solution contains complexed Cu(II); a typical aliphatic aldehyde gives a brick-red precipitate of Cu₂O on warming. Ketones do not give these positive aldehyde results under normal test conditions. Aromatic aldehydes may not reduce Fehling’s/Benedict’s reliably; Tollens’ is the broader comparison.

These reagents are alkaline, so the immediate organic oxidation product is a carboxylate ion. Acidification would give the carboxylic acid. Use fresh, separate portions for each test. Tollens’ reagent must be prepared and disposed of promptly under the laboratory’s instructions; never store it or allow residues to dry.

RCHO + 2Ag⁺ + 3OH⁻ → RCOO⁻ + 2Ag + 2H₂O
RCHO + 2Cu²⁺ + 5OH⁻ → RCOO⁻ + Cu₂O + 3H₂O
Carbonyl tests: observation and justified inference
Reagent and conditionsPositive observationInference and limitation
2,4-DNPHYellow/orange precipitateAldehyde or ketone supported; not a class distinction
Tollens’, warm gentlySilver mirror/depositA reducing aldehyde supported within the proposed candidates
Fehling’s/Benedict’s, warmBrick-red Cu₂O precipitateTypical aliphatic aldehyde; aromatic examples need care
Acidified dichromate(VI), warmOrange to greenOxidisable group; primary/secondary alcohols also react

Identify a carbonyl using a purified derivative

After producing a 2,4-DNPH derivative, collect the crystals, recrystallise them in a suitable solvent, wash sparingly with cold solvent and dry. Measure a melting range slowly near the expected transition and compare it with supplied derivative data. Recrystallisation separates the solid derivative from soluble impurities; drying removes solvent that could depress or broaden its melting range.

Original illustrative data: candidate derivatives A, B and C melt at 118–120°C, 145–147°C and 162–164°C. A dry sample melting sharply at 145–147°C supports B. A crude sample melting at 136–144°C is not sound evidence for a fourth substance: contamination commonly depresses and broadens the range. These are teaching data, not reference melting points.

A melting match supports an identity within the candidate list; combine it with formula, tests and spectra if a unique structure is needed. Use a desiccator or other suitable solid-drying method, not loose anhydrous salt mixed into the crystals. Keep commercial DNPH reagent appropriately wetted and follow its safety instructions.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.

Q1. Explain why propanone can dissolve in water although pure propanone does not form intermolecular hydrogen bonds.Show answer

The carbonyl O accepts hydrogen bonds from water’s O–H hydrogens. Pure propanone has no O–H or N–H hydrogen to donate, so its own molecules interact by London and permanent dipole forces instead.

Q2. An unknown gives a DNPH precipitate but no Tollens’ silver mirror. Which class is supported among simple aldehydes and ketones?Show answer

A ketone. DNPH establishes the aldehyde/ketone carbonyl, while the negative Tollens’ result argues against an aldehyde under suitable working conditions. DNPH alone would not distinguish them.

Q3. Give the reagent, essential solvent condition and organic product for reduction of butanal.Show answer

LiAlH₄ in dry ether, followed by suitable aqueous work-up. CH₃CH₂CH₂CHO + 2[H] → CH₃CH₂CH₂CH₂OH, butan-1-ol. The carbon count stays at four.

Q4. Why is an orange-to-green dichromate result alone insufficient to identify an aldehyde?Show answer

Primary and secondary alcohols can also reduce dichromate(VI). Establish the carbonyl with DNPH and combine with a more selective aldehyde test; observations must be interpreted within the candidate set.

Q5. A derivative melts broadly below the expected range. Give a likely cause and a targeted improvement.Show answer

Residual solvent or other impurities can lower and broaden the range. Recrystallise appropriately and dry the crystals before measuring again slowly. Simply quoting a database value does not establish that the sample is pure.

Sources

Sources and examiner guidance (reviewed 9 October 2026)

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