Oxidation and Tests for Carbonyl Compounds
A concise revision guide to the oxidation of aldehydes and the tests that tell aldehydes from ketones: Fehling’s or Benedict’s solution, Tollens’ reagent, acidified dichromate(VI), 2,4-dinitrophenylhydrazine, the iodoform test.
- 17B.8i
- 17B.8iv
- 17B.8v
What these spec points say
- 17B.8i understand the reactions of carbonyl compounds with: Fehling’s or Benedict’s solution, Tollens’ reagent and acidified dichromate(VI) ions In equations, the oxidising agent can be represented as [O]
- 17B.8iv understand the reactions of carbonyl compounds with: 2,4-dinitrophenylhydrazine (2,4-DNPH), as a qualitative test for the presence of a carbonyl group and to identify a carbonyl compound given data for the melting temperatures of derivatives The equation for this reaction is not required
- 17B.8v understand the reactions of carbonyl compounds with: iodine in the presence of alkali
Before You Start
This page follows Aldehydes and Ketones: Structure, Naming and Properties and uses the oxidation of alcohols from Topic 6 Organic Chemistry I.
- An aldehyde has a hydrogen atom on its C=O carbon; a ketone does not.
- Acidified potassium dichromate(VI) is an oxidising agent: it turns from orange to green when it is reduced.
- In equations the oxidising agent is written as [O].
Oxidation of Aldehydes
Aldehydes are easily oxidised to carboxylic acids. Ketones are not oxidised by the same reagents.
CH₃CHO + [O] → CH₃COOH
- The reagent is potassium dichromate(VI), K₂Cr₂O₇, acidified with dilute sulfuric acid, and the mixture is heated under reflux.
- The orange dichromate(VI) ion, Cr₂O₇²⁻, is reduced to the green chromium(III) ion, Cr³⁺.
- A ketone leaves the solution orange.
Why ketones resist oxidation
- Oxidising an aldehyde removes the hydrogen on the C=O carbon and puts an –OH in its place. Only a C–H bond is broken.
- A ketone has no hydrogen on its C=O carbon. Oxidation would need a strong C–C bond to break.
- Because they are so easily oxidised, aldehydes are reducing agents. Every test below uses this.
The ionic equation
3CH₃CHO + Cr₂O₇²⁻ + 8H⁺ → 3CH₃COOH + 2Cr³⁺ + 4H₂O
Oxidation routes from primary and secondary alcohols, and the colour change with acidified dichromate(VI): green with an aldehyde, still orange with a ketone.
Remember: Acidified dichromate(VI) also turns green with primary and secondary alcohols. It separates an aldehyde from a ketone, but it does not prove that a compound is an aldehyde.
Tollens’ Reagent: the Silver Mirror Test
Tollens’ reagent is a colourless solution of silver nitrate in aqueous ammonia. It contains the complex ion [Ag(NH₃)₂]⁺ and is a mild oxidising agent.
Method
- Add a few drops of NaOH(aq) to silver nitrate solution: a brown precipitate of silver oxide forms.
- Add dilute ammonia drop by drop until the precipitate just dissolves.
- Add the carbonyl compound and warm the tube in a water bath at about 60 °C.
| Compound | Observation | Explanation |
|---|---|---|
| aldehyde | a silver mirror forms on the inside of the tube | aldehyde oxidised; Ag⁺ reduced to Ag |
| ketone | no change: colourless solution | ketones are not oxidised |
- The aldehyde is oxidised to a carboxylic acid, which forms its carboxylate ion in the alkaline solution.
- The silver ions are reduced to silver: Ag⁺(aq) + e⁻ → Ag(s).
- Overall: CH₃CHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → CH₃COO⁻ + 2Ag + 4NH₃ + 2H₂O.
- A water bath is used because carbonyl compounds are flammable. The reagent is made fresh and never stored.
Making Tollens’ reagent and using it: a silver mirror with an aldehyde, no change with a ketone, and silver ions gaining electrons.
Exam focus: State the reagent, the condition and both observations: Tollens’ reagent, warm; an aldehyde gives a silver mirror; a ketone gives no change.
Quick Check: Reasoning With Oxidation Tests
Pick the one accurate statement in each set, using compounds that are new to this page.
Fehling’s and Benedict’s Solutions
Fehling’s solution is a deep blue alkaline solution containing complexed copper(II) ions, Cu²⁺. Like Tollens’ reagent, it is a mild oxidising agent.
| Compound | Observation on warming | Explanation |
|---|---|---|
| aldehyde | blue solution gives a brick-red precipitate | aldehyde oxidised; Cu²⁺ reduced to copper(I) oxide, Cu₂O |
| ketone | solution stays blue | ketones are not oxidised |
- The aldehyde is oxidised to a carboxylic acid, present as its carboxylate ion in the alkaline solution.
- Copper(II) is reduced to copper(I): 2Cu²⁺ + 2OH⁻ + 2e⁻ → Cu₂O + H₂O.
- Overall: CH₃CHO + 2Cu²⁺ + 5OH⁻ → CH₃COO⁻ + Cu₂O + 3H₂O.
- Benedict’s solution is a similar blue alkaline copper(II) solution and gives the same result.
Fehling’s solution warmed with an aldehyde and with a ketone: a brick-red precipitate of copper(I) oxide, or a solution that stays blue.
Exam focus: Give the colour and the state: a brick-red precipitate, not “the solution turns red”. Name the solid as copper(I) oxide, Cu₂O.
Check: Telling Aldehydes From Ketones
Predict the result of oxidising tests and explain them as redox reactions, with compounds not used on this page.
2,4-Dinitrophenylhydrazine: Detecting and Identifying
2,4-Dinitrophenylhydrazine (2,4-DNPH) reacts with the C=O group of both aldehydes and ketones.
Detecting a carbonyl group
- Add a few drops of the unknown to a solution of 2,4-DNPH.
- A yellow-orange precipitate shows that an aldehyde or a ketone is present.
- Alcohols, carboxylic acids and esters give no precipitate, so the test is specific to aldehydes and ketones.
Identifying the carbonyl compound
The precipitate is a solid derivative (a 2,4-dinitrophenylhydrazone). Each carbonyl compound gives a derivative with its own sharp melting temperature.
- Filter off the precipitate.
- Recrystallise it from the minimum volume of hot solvent to purify it.
- Filter, wash and dry the crystals.
- Measure the melting temperature of the purified derivative.
- Compare the value with a table of known derivatives.
Worked example
The purified derivative of an unknown ketone melts at 116 °C. Data: propanone derivative 126 °C, butanone derivative 116 °C, pentan-3-one derivative 156 °C.
2,4-DNPH in two stages: an orange precipitate detects the carbonyl group, then the purified derivative is identified from its melting temperature.
Exam focus: 2,4-DNPH shows that a carbonyl group is present; it does not separate aldehydes from ketones. Follow it with Tollens’ reagent, and use the melting temperature of the derivative to name the compound.
The Iodoform Test
The iodoform test detects a methyl group next to a carbonyl group. The reagent is iodine in aqueous sodium hydroxide (alkaline I₂), and the mixture is warmed gently.
- A positive result is a pale yellow precipitate of tri-iodomethane (iodoform), CHI₃, with an antiseptic smell.
- It is given by compounds containing the CH₃CO– group: ethanal and methyl ketones such as propanone and butanone.
- Ethanal is the only aldehyde that gives a positive result.
CH₃COCH₃ + 3I₂ + 4OH⁻ → CHI₃ + CH₃COO⁻ + 3I⁻ + 3H₂O
- The other organic product is the salt of a carboxylic acid with one carbon atom fewer than the starting compound.
- Alcohols with the CH₃CH(OH)– group also react: the reagent first oxidises them to CH₃CO–.
- These are ethanol and secondary alcohols with a methyl group on the C–OH carbon, such as propan-2-ol.
| Positive: yellow precipitate | Negative: no precipitate |
|---|---|
| ethanal, CH₃CHO | propanal, CH₃CH₂CHO |
| propanone, CH₃COCH₃ | pentan-3-one, CH₃CH₂COCH₂CH₃ |
| butanone, CH₃COCH₂CH₃ | methanol, CH₃OH |
| ethanol and propan-2-ol | propan-1-ol, CH₃CH₂CH₂OH |
The two groups that give tri-iodomethane, CH₃CO– and CH₃CH(OH)–, shown in the compounds that contain them, beside two that do not.
The iodoform test: alkaline iodine, gentle warming, a pale yellow precipitate of tri-iodomethane, and which compounds give it.
Exam focus: A pale yellow precipitate with iodine in alkali shows a CH₃CO– group, or a CH₃CH(OH)– group in an alcohol. Pentan-2-one is positive; pentan-3-one is negative.
Summary of the Tests
| Reagent | Aldehyde | Ketone | Distinguishes |
|---|---|---|---|
| Acidified K₂Cr₂O₇, warm | orange solution turns green | stays orange | aldehyde from ketone |
| Tollens’ reagent, warm | silver mirror | no change | aldehyde from ketone |
| Fehling’s or Benedict’s solution, warm | blue solution gives a brick-red precipitate | stays blue | aldehyde from ketone |
| 2,4-DNPH | yellow-orange precipitate | yellow-orange precipitate | carbonyl from other compounds |
| I₂ in NaOH(aq), warm | pale yellow precipitate with ethanal only | pale yellow precipitate with methyl ketones only | the CH₃CO– group |
- First show that the compound is a carbonyl compound: 2,4-DNPH gives a yellow-orange precipitate.
- Then decide between aldehyde and ketone with Tollens’ reagent or Fehling’s or Benedict’s solution.
- Use iodine in alkali to check for a CH₃CO– group.
- Finally, identify the compound from the melting temperature of its purified 2,4-DNPH derivative.
Exam focus: A full answer to “describe a test” gives the reagent, the conditions, the observation with each compound and what the observation shows.
Check: Choosing and Reading the Tests
Match reagents to observations and decide what each result shows, using compounds not used on this page.
Worked Example: Identifying an Unknown
Compound X has the molecular formula C₄H₈O. It gives a yellow-orange precipitate with 2,4-DNPH. It gives a silver mirror with Tollens’ reagent. It gives no precipitate with iodine in alkali.
Step 3. The negative iodoform test agrees: neither aldehyde has a CH₃CO– group, whereas butanone would give a yellow precipitate.
The oxidation product of butanal would be butanoic acid: CH₃CH₂CH₂CHO + [O] → CH₃CH₂CH₂COOH.
Common Exam Points
Say
- “The aldehyde is oxidised to a carboxylic acid.”
- “Silver ions are reduced to silver, forming a silver mirror.”
- “A brick-red precipitate of copper(I) oxide forms.”
- “Recrystallise the derivative, measure its melting temperature and compare with known values.”
- “A pale yellow precipitate of CHI₃ shows a CH₃CO– group.”
Do not say
- “Goes clear” (say colourless) or “silver precipitate” (say silver mirror).
- “The ketone is oxidised slowly” (it is not oxidised by these reagents).
- “2,4-DNPH is a test for aldehydes” (it detects aldehydes and ketones).
- “Measure the boiling temperature of the precipitate” (it is the melting temperature of the purified derivative).
Watch for
- Acidified dichromate(VI) reacting with alcohols as well as aldehydes.
- Equations with [O]: one [O] per aldehyde, and no water formed.
- Alkaline reagents: the acid is formed as its carboxylate ion, RCOO⁻.
- Ethanal as the only aldehyde, and ethanol as the only primary alcohol, to give tri-iodomethane.
Check: Identifying Unknown Carbonyl Compounds
Combine the results of several tests to work out structures, with unknowns not used on this page.
FAQs
Use these quick answers to check the oxidation of aldehydes and the tests for carbonyl compounds.
Why can aldehydes be oxidised but ketones cannot?
An aldehyde has a hydrogen atom on its carbonyl carbon, which is replaced as the CHO group becomes COOH. A ketone has no such hydrogen, so oxidation would mean breaking a carbon-carbon bond, and mild oxidising agents cannot do that.
What happens in the silver mirror test?
An aldehyde, warmed gently with Tollens’ reagent, coats the tube with a silver mirror. The aldehyde is oxidised to a carboxylic acid, present as its salt, while [Ag(NH₃)₂]⁺ ions are reduced to silver. A ketone leaves the solution colourless.
What does Fehling’s solution show?
On warming, an aldehyde reduces the blue copper(II) complex to a brick-red precipitate of copper(I) oxide, Cu₂O, and is itself oxidised to a carboxylic acid. A ketone is not oxidised, so the solution stays blue.
What does 2,4-DNPH tell me?
2,4-Dinitrophenylhydrazine gives a yellow-orange precipitate with any aldehyde or ketone, so it detects the C=O group but cannot tell them apart. Recrystallising the precipitate and measuring its melting temperature identifies the original compound from data tables.
Which compounds give a positive iodoform test?
Warming with iodine in NaOH(aq) gives a pale yellow precipitate of CHI₃ when the compound contains a CH₃CO group or a CH₃CH(OH) group. That means ethanal, methyl ketones, ethanol and secondary alcohols with the OH on carbon 2.
Copyright notice: This OLS revision content, including the explanations, layout, diagrams, tables and embedded learning structure, is authored for Online Learning System by Dr. Mohammed Al-Fatah. It may not be copied, reproduced, redistributed or adapted without written permission.
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