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.
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
particularly useful for identifying aldehydes or ketones, although the
observations can distinguish aldehydes from ketones.
|
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.
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. |
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 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
