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

Part 5. The chemistry of ALDEHYDES and KETONES

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

Part 5.7 The oxidation of aldehydes and ketones and chemical tests & iodoform reaction


INDEX of ALDEHYDES and KETONES revision notes

All Advanced A Level Organic Chemistry Notes

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Oxidation of aldehydes and ketones

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

(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

 

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

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).


Qualitative ORGANIC functional group tests for aldehydes and ketones

CHEMICAL TEST FOR

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

Note

(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. 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

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 hydrogen 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 CI3CO 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

-

 


Doc Brown's Advanced Level Chemistry Revision Notes

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INDEX of ALDEHYDE and KETONE revision notes

 All Advanced Organic Chemistry Notes

 
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