Part 5. The chemistry of ALDEHYDES and KETONES - OXIDATION

Part 5.7 The oxidation of aldehydes and ketones including chemical tests using Tollen's reagent, Fehlings and Benedict's solutions plus the iodoform reaction for identification of e.g. 2-one ketone functional group

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

+  [O] ===>

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

+  [O] ===>

(iii) butanal ===> butanoic acid

+  [O] ===>

(iv) pentanal ==> pentanoic acid

+  [O]  ===>

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.

Cr₂O₇^2-(aq)  == reduced ==> Cr³⁺(aq)

(i) reduction half-equation:

Cr₂O₇^2–_(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)   +  Cr₂O₇^2–_(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.

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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  oxidises to an aldehyde and carboxylic acid

or  pentan-3-ol  CH₃CH₂CH(OH)CH₂CH₃  oxidises to a ketone

or CH₃CH₂COCH₂CH₃ cannot be oxidised with breaking carbon chain

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

e.g. methanal HCHO  or  propanal CH₃CH₂CHO do not give the iodoform reaction

See also Tests for alcohols