Pre-university Advanced Level Organic Chemistry: The oxidation of aldehydes & ketones, tests

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Part 5. The chemistry of ALDEHYDES and KETONES - OXIDATION

Doc Brown's Chemistry Advanced Level Pre-University Chemistry Revision Study Notes for UK KS5 A/AS GCE IB advanced level organic chemistry students US K12 grade 11 grade 12 organic chemistry tests for aldehydes and ketones reagents observations deductions

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Part 5.7 The oxidation of aldehydes and ketones and chemical tests with Tollen's reagent and Fehlings and Benedict's solutions plus the iodoform reaction for identification of functional groups

Oxidation of aldehydes and ketones

Readily oxidisable aldehydes gives a relatively stable carboxylic acid e.g.

using a mixture of dil. sulfuric acid and potassium dichromate(VI), even at room temperature you see an orange to green colour change in the aqueous reagent.

Ketones are not usually readily oxidised by this reagent, so it will often distinguish an aldehyde from a ketone, BUT, lots of other organic compounds are readily oxidised by acidified dichromate solution e.g. primary and secondary alcohols, so its not a useful test for aldehydes.

(i) ethanal ==> ethanoic acid

aldehydes and ketones nomenclature (c) doc b +  [O] ===>aldehydes and ketones nomenclature (c) doc b

(ii) 2-methylpropanal ==> 2-methylpropanoic acid

aldehydes and ketones nomenclature (c) doc b+  [O] ===>(c) doc b

(iii) butanal ===> butanoic acid

aldehydes and ketones nomenclature (c) doc b+  [O] ===>aldehydes and ketones nomenclature (c) doc b

(iv) pentanal ==> pentanoic acid

aldehydes and ketones nomenclature (c) doc b+  [O]  ===>aldehydes and ketones nomenclature (c) doc b

The redox half-equation for oxidising an aldehyde to a carboxylic acid is:

R-CHO  +  H2O  ===> R-COOH  +  2H+  + 2e-

The redox half-equation for oxidising a primary alcohol to an aldehyde is:

R-CH2OH  ===>  R-CHO  +  2H+  + 2e-

Ketones are NOT readily oxidised because it would involve breaking strong bonds in the carbon chain.


Aldehydes are readily oxidised by:

Potassium dichromate(VI)/sulfuric acid solution.

This gives the free carboxylic acid:

RCHO + [O]  ===> RCOOH

The colour changes from the orange of the dichromate(VI) ion to the green chromium(III) ion.

Cr2O72-(aq)  == reduced ==> Cr3+(aq)

(i) reduction half-equation: Cr2O72–(aq) + 14H+(aq) + 6e ===> 2Cr3+(aq) + 7H2O(l)

(ii) oxidation half-equation for ethanal: CH3CHO(aq)  +  H2O(l)  ===> CH3COOH(aq)  +  2H+(aq)  +  2e-

To obtain the fully balanced redox equation you add (i) + 3 x (iii)

(think 6 electron change to balance 1 : 3 ratio of equations (i) and (ii)

3CH3CHO(aq)   +  Cr2O72–(aq) +  8H+(aq)  ===>  3CH3COOH(aq)   +  2Cr3+(aq) +  4H2O(l)

More details of this reaction can be found on ...

Part 4.5 Controlled oxidation of alcohols/aldehydes with selected oxidising agents

Potassium manganate(VII) in acidic, neutral or alkaline solution.

In acid/neutral conditions you get the free carboxylic acid:

RCHO + [O]  ===> RCOOH

In acid conditions the purple manganate(VII) ion MnO4-  is reduced to the almost colourless Mn2+ ion.

Under neutral and alkaline conditions the purple manganate(VII) ion is reduced to the insoluble brown precipitate of

In alkaline conditions you get the carboxylate anion - as the salt of the carboxylic acid:

RCHO + [O] + OH-  ===> RCOO-  +  H2O


Unlike ketones, aldehydes are oxidised to the carboxylate ion when testing with Fehling's solution or Tollen's reagent  (see below).

Non of the above oxidations and observations are directly particularly useful for identifying aldehydes or ketones, although the observations can distinguish aldehydes from ketones.


Qualitative ORGANIC functional group tests for aldehydes and ketones


Aldehydes chemical test (R–CHO, R = H, alkyl or aryl) to distinguish from ketones (R2C=O, R = alkyl or aryl) and also reducing sugars.


(1) Test (b)(i) and (ii) can be used to distinguish aldehydes (reaction) and ketones (no reaction).

(2) Aromatic aldehydes do NOT give a positive result with (b)(ii) Benedict's or Fehling's reagent).

(3) Reducing sugars may also give a positive test with (b)(i) and (ii) reagents e.g. glucose (aldohexose) but not fructose? (ketohexose)?

(a) Add a few drops of the suspected carbonyl compound to Brady's reagent (2,4–dinitrophenylhydrazine solution)

Often quoted as just 24DNPH !

(a) A yellow–orange precipitate forms with both types of carbonyl compound.

Specific identification

The precipitate may be collected, recrystallised and from it's unique melting point of the crystals, the aldehyde or ketone can be identified from data tables.

The aldehyde or ketone 2,4–dinitrophenylhydrazone is formed

R2C=O + (NO2)2C6H3NHNH2 ==>

(NO2)2C6H3NHN=CR2 + H2O

(R = H, alkyl or aryl)

This tells you it's an aldehyde or ketone, but you can't distinguish them, read on below!

For details of the 24DNPH reaction Addition - elimination reactions of aldehydes & ketones

(b)(i) warm a few drops of the compound with Tollens' reagent [ammoniacal silver nitrate]

(b)(ii) simmer with Fehling's or Benedict's solution [a blue tartrate complex of Cu2+(aq)]

Tollen's reagent is made by adding a few drops of sodium hydroxide to silver nitrate solution. Dilute ammonia is added until the precipitate just dissolve.

(b) Only the aldehyde produces ...

(i) A silver mirror on the side of the test tube.

(ii) The deep blue solution produces a brown or brick red  precipitate of copper(I) oxide if its an aldehyde


Ketones do not give a response to tests with Tollens' reagent, Fehling's solution or Benedict's solution - no colour changes and no precipitates.

Aldehydes are stronger reducing agents than ketones and reduce the metal ion and are oxidised in the process to a carboxylic acid

i.e. RCHO + [O] ==> RCOOH

(i) reduction of silver(I) ion to silver metal

RCHO + 2Ag+ + H2O ==> RCOOH + 2Ag + 2H+

(ii) reduction of copper(II) to copper(I) i.e. the blue solution of the Cu2+ complex changes to the brown/brick red colour of insoluble copper(I) oxide Cu2O.

RCHO + 2Cu2+ + 2H2O ==> RCOOH + Cu2O + 4H+

For (b)(i)/(ii) technically the oxidation is

RCHO  +  [O]  +  OH-  ===> RCOO-  +  H2O

because the reagents are alkaline.

With (b)(i)/(ii) no reactions with ketones.

Iodoform test

The formation of CHI3, triiodomethane (or old name 'iodoform'.

NaOH(aq) is added to a solution of iodine in potassium iodide solution until most of the colour has gone. The organic compound is warmed with this solution. A yellow solid is formed with the smell of an antiseptic, CHI3, tri–iodomethane, melting point 119oC. This reaction is given by the aldehyde ethanal CH3CHO and all ketones with the '2–one' structure R–CO–CH3  ('methyl ketones'). More details below.

Its a combination of halogenation and oxidation and is not a definitive test for any functional group, it just indicates a possible part of a molecule's structure.

Extra notes on the iodoform reaction

To make the reagent, sodium hydroxide solution is added to a solution of iodine in potassium iodide solution until most of the colour has gone.

You can also use a mixture of potassium iodide dissolved in sodium chlorate(I) solution.

The organic compound is warmed with this solution.

Only the aldehyde ethanal (CH3CHO) and all the ketones with the CH3CO- end grouping give the iodoform reaction.

It is a complicated reaction, in which the hydrogen atoms of the methyl group of the CH3CO grouping are replaced by iodine atoms.

The sodium hydroxide/iodine reagent causes the C-C bond of the CH3CO grouping to break in the process of forming the pale yellow precipitate of CHI3 (triiodomethane, 'iodoform'), that smells like an antiseptic.

The following equilibrium is found in the NaOH/I2 reagent:

I2  +  2OH-    IO-  +  I-  H2O

The iodate(I) ion, IO-, substitutes I for H forming the I3CO grouping.

The electron withdrawing effect of the three iodine atoms weakens the C-C sigma bond allowing for the formation of CHI3.

For R = H or alkyl e.g. CH3, CH2CH3 etc., the overall equation is:

CH3COR  +  3I2  +  4NaOH  ===>  CHI3  +  RCOONa  +  3NaI  +  3H2O

e.g. for:

ethanal: CH3CHO  +  3I2  +  4NaOH  ===>  CHI3  +  HCOONa  +  3NaI  +  3H2O

propanone: CH3COCH3  +  3I2  +  4NaOH  ===>  CHI3  +  CH3COONa  +  3NaI  +  3H2O

butanone: CH3COCH2CH3  +  3I2  +  4NaOH  ===>  CHI3  +  CH3CH2COONa  +  3NaI  +  3H2O

Extra notes on alcohols that give the same iodoform reaction

The iodoform reaction is given by the primary alcohol ethanol CH3CH2OH and all secondary alcohols with the 2–ol structure R–CH(OH)–CH3 and in the test the primary alcohol is first is oxidised to an aldehyde and secondary alcohols to a ketone e.g.

ethanol is first oxidised to ethanal:  CH3CH2OH  +  [O]  ===>  CH3CHO  +  H2O

propan-2-ol is first oxidised to propanone: CH3CH(OH)CH3  +  [O]  ===>  CH3COCH3  +  H2O

butan-2-ol is first oxidised to butanone: CH3CH(OH)CH2CH3  +  [O]  ===>  CH3COCH2CH3  +  H2O

and then the equations to form iodoform itself are as above for ethanal and 'methyl ketones'

Other primary/secondary alcohols may be oxidised to an aldehyde/ketone, but will not be further oxidised to CHI3 and if they are oxidised it will be to a carboxylic acid.

e.g. pentan-1-ol  CH3CH2CH2CH3CH2OH   or  pentan-3-ol  CH3CH2CH(OH)CH2CH3CH3CH2COCH2CH3

All other aldehydes may be oxidised to a carboxylic acid, but will not give an iodoform reaction.

e.g. methanal HCHO  or  propanal CH3CH2CHO

See also Tests for alcohols

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