Distinguishing between aldehydes
and ketones
Adehydes and ketones can be structural isomers of each
other. Aldehydes are produced by the oxidation of a primary
alcohol and have the C=O on end carbon.
Ketones are produced by the oxidation of a secondary alcohol
and have the C=O on a carbon atom in the middle of the carbon
chain.
Aldehydes can be further oxidised to carboxylic acids, while
ketones are not oxidised further.
The tests we use to distinguish between aldehydes and
ketones all involve oxidising the aldehyde but not the ketone.
While acidified dichromate or permanganate will distinguish
between aldehydes and ketones, they are strong oxidising
agents which will also change colour in the presence of alcohol
or other reagents.
Oxidising agents which oxidise aldehydes are:
• Tollen’s reagent
• Benedict solution
• Fehling’s solution
The ‘silver mirror’.
You’re more likely to
get a mirror with a very
clean test tube.
Tollen’s reagent
Tollen’s reagent is [Ag(NH3)2]+ which, when reduced, forms
Ag(s). It must be freshly prepared.
Silver nitrate
solution
A few drops of
NaOH to form a
precipitate.
Add ammonia solution
till the precipitate
dissolves.
Add a few drops of the
aldehyde or ketone, shake, and
warm gently.
The ketone remains colourless,
the aldehyde will react.
If you are lucky you will get a
‘silver mirror’ as elemental
silver forms on the inside of
the test tube.
Less spectacular, but just as
valid is the formation of a grey
or black precipitate, also of
elemental silver.
A grey precipitate
of silver.
Tollens’ Test
Tollens reagent is a complex of Ag(NH3)2+ .
When heated with an aldehyde a redox reaction occurs
producing a silver mirror on the inner surface of the
test tube.
The aldehyde is oxidised to a carboxylic acid.
The reduction half-equation is
• Ag+(aq) +
e
 Ag(s)
•If Tollens’ reagent is heated with a ketone or an
alcohol no reaction occurs.
Tollens’ Test
The overall reaction is :
RCHO + 2Ag(NH3)2OH
RCOONH4 + 2Ag + H2O + 3NH3
Silver mirror
Task – write the balanced redox reactions for the
oxidation of propan-1-ol to propanal using Cr2072- /H+
and give all colour changes.
(X3) 3 CH3CH2CH2OH
3 CH3CH2COH + 2H
6H++ + 2e6e-
Cr2072- + 14H+ + 6e-
2Cr3+ + 7H20
Full balanced redox equation
3CH3CH2CH2OH + Cr2072-+ 8H+
3CH3CH2COH + 2Cr3++ 7H20
Benedict solution
Benedict solution is an alkaline solution of Cu2+,
complexed with citrate ions to keep it in solution.
It is a mild oxidising agent which is reduced to Cu+.
In the alkaline solution the Cu+ is in the form of
Cu2O which is a brick-red precipitate which is a
positive test for an aldehyde.
Take about 2 mL of
Benedict solution in each of
two test tubes.
Add a few drops of
aldehyde to one tube,
and ketone to the other
tube, and shake to mix.
Heat the mixture by putting the
tubes in hot water. Shake several
times to mix.
A reaction has occurred in the left hand
(aldehyde) tube, but not in the right hand
(ketone) tube.
If you wait long
enough you will see
the red-brown
precipitate of Cu2O
form.
Benedicts and Fehlings
Aldehydes reduce the copper (II) ions in both Fehlings and
Benedicts solution to form a reddish brown copper (I) oxide
which is precipitated
The reaction is :
RCHO + 2Cu
2+
+ 2OH-
RCOO- + Cu2O + 3H+
Brick red
precipitate
Fehling’s solution
Like Benedict solution, Fehling’s contains alkaline Cu2+, but
Fehling’s uses potassium tartrate to complex the copper.
The mixture is freshly prepared:
Pour a little Fehling’s A
solution into each test
tube.
Add the ‘B’ solution until
a precipitate forms.
Keep adding ‘B’ solution
until the precipitate has
redissolved and the
solution is a clear, dark
blue.
A reaction occurs in the aldehyde tube as Cu2O forms. No
reaction occurs in the ketone tube.
Add a few drops of
aldehyde and ketone to
separate tubes, shake,
and heat in a beaker of
hot water.
Glucose is an aldehyde and will form a mirror
with Tollens and will also give a positive test
with Benedicts and Fehlings solutions
Do you remember doing this test for sugars in
yr10 ?
In all of these reactions (Tollen’s, Benedict and
Fehling’s), the aldehyde is oxidised to the
carboxylic acid while no reaction occurs to the
ketone.
Task – write the balanced redox reactions for the
oxidation of the oxidation of propan-2-ol to propanone
using Mn04- /H+ and give all colour changes.
(X5) 5 CH3CHOHCH3
5 CH3COCH3 + +10H
2H++++ 10e2e-
(X2) 2MnO
Mn044-- ++16H
8H++ ++ 10e5e-
2Mn
Mn2+ + 8H
4H20
Full balanced redox equation
5CH3CHOHCH3 + 2Mn04-+ 6H+
5CH3COCH3 + 2Mn2++8H20
Producing an Aldehyde from a Primary Alcohol
When forming the aldehyde (ethanal) from ethanol the
alcohol and dichromate must be added to the hot
concentrated H2SO4
We use a distillation technique to collect a volatile
product
Producing an Aldehyde from a Primary Alcohol
using distillation
Dropping funnel with K2Cr2O7
and ethanol
Distillation
flask with hot
H2SO4
The alcohol/dichromate mixture is added to the acid and
only the first 2-3 mls of distillate is collected – why?
The alcohol/dichromate mixture is added to the acid and
only the first 2-3 mls of distillate is collected – why?
To make sure that the immediate production of ethanal
with a lower boiling point of 21 deg C was vapourised and
collected quickly and was not allowed to be converted to
ethanoic acid.
To make sure a reaction goes to completion one such as:
•Primary alcohol is completely oxidised to form a
carboxylic acid
•A secondary alcohol is oxidised to form a ketone
You would use a reflux arrangement
Making Aspirin – using methylsalicylate
methylsalicylate
Acetic
anhydride
Aspirin
Carboxylic
acid
Carboxylic Acids
O
R
C
Named with -oic on the end
They are all organic acids which are
weak acids
OH
Reactions of Carboxylic Acids
Carboxylic acids undergo 4 types of
substitution reactions (of the –OH)
Reactions of Carboxylic Acids
1. Forming acid chlorides from carboxylic
acids – reagents are PCl5,PCl3 or SOCl2
acid chlorides are named at the end
by the –oyl group
Functional group of the acid chloride
O
R
C
Cl
Reactions of Acid Chlorides
2. Forming esters from acid chloride – reagents are
a primary alcohol
Ester
From acyl chloride
O
R
C
CH3COCl + CH3OH
ethanoyl chloride + methanol
O
+
HCl
R’ From alcohol
CH3COOCH3 + HCl
methyl ethanoate + hydrogen
chloride
Reactions of Acyl Chlorides
3. acyl chlorides form amides – reagent
ammonia and heat
Functional
group of the
amide
CH3COCl + 2NH3
ethanoyl chloride + ammonia
O
R
C
NH2
CH3CONH2 + NH4Cl
ethanamide
+
ammonium
chloride
Reactions of Acyl Chlorides
4. Acyl chlorides forming
N - substituted amides – reagent amine
O
N substituted amide
CH3
O
C
Cl
+ NH2 CH3
R
CH3
Ethanoyl chloride + aminomethane
C
NH
R’
O
C
+ HCl
NH CH3
N – methyl ethanamide
+ hydrochloric acid
Reactions of Acyl Chlorides
Acyl chlorides react with water to form
acidic solutions
CH3
O
C
Cl H
O
H
ethanoyl chloride + water
CH3
O
C
O
ethanoic acid
H
+ HCl
+ hydrogen
chloride
Making Acyl Chlorides
Carboxylic acids react with PCl3, PCl5 or
SOCl2 (not HCl) to form Acyl chlorides
by substituting the –OH for a Cl.
CH3
O
C
OH
Ethanoic acid + PCl5
PCl5
CH3
O
C
Cl
ethanoyl chloride
Lysergic acid diethylamide (LSD)
ESTERS
Carboxylic acids react with alcohols, in the presence of conc
sulfuric acid as a catalyst, to form esters.
The reagents are heated together to bring about a reaction.
Any excess acid is neutralised by the addition of sodium
carbonate.
If ethanoic acid is reacted with methanol the ester,
methyl ethanoate is formed.
O
H3C
O
H3C
O
+ CH3OH
C
OH
C
H3O+
O
CH3
heat
H3C
+ H2O
C
O
CH3
Hydrolysis of esters
The hydrolysis of an ester in aqueous solution results
in the break up of the ester and the formation of an
alcohol and the carboxylic acid or carboxylate ion
*(depending on the pH of the solution).
Hydrolysis in acid produces the alcohol + carboxylic acid
CH3CH2COOCH3 + H2O / H+
methyl propanoate

CH3CH2COOH
propanoic acid
+ CH3OH
methanol
Hydrolysis of esters
Hydrolysis in NaOH soln gives alcohol + the sodium salt of
the carboxylic acid.
CH3CH2COOCH3 + NaOH
methyl propanoate
 CH CH COO Na
3
2
sodium
propanoate

+
+ CH3OH
methanol
hydrolysing an
ester (methyl salicylate)
OH
C O
CH3
Methyl salicylate
O
Write the products if we hydrolyse
it in acid ie H2O/H+
hydrolysing an ester (methyl salicylate)
if we hydrolyse it in acid ie H2O/H+
the products are
OH
C O
H
+
CH3OH
O
Salicylic acid
methanol
Write the products if we hydrolyse
it in alkaline conditions ie
H2O/NaOH
OH
C O – Na
+
+
CH3OH
O
sodium salicylate
methanol
Turn to the last page in the booklet
Ester Hydrolysis read the method
(the ester is methyl salicylate)
Change the method as follows:
Weigh out 4.8 grams of NaOH place in boiling
flask then add 20mls of water (careful! it may
get very hot)
Measure out and add 5mls of oil of
wintergreen (methyl salycilate) to flask
Place boiling chips in boiling flask and reflux
carefully for 30 mins
Reflux – heating mixture without
losing volatile substances
Distillation – using the deferring boiling
points of substances to separate them
Write equations using structural formulae for the formation
of the following esters
a) ethyl methanoate from ethanol and methanoic acid
b) butyl propanoate from butan-1-ol and propanoic acid
complete the following scheme giving all
Formula
H+
Ethanoyl chloride
CH
Formula?
3COCl
C
Reagents?
2H5OH
NH3 / ethanol
Ethyl ethanoate OH-
C2H5OH
CH3COOH + C2H5OH
CH3COOC2H5
Formula?
ethanamide
CH3CONH2
salt
+ ethanol
CH3COO- + C2H5OH
Reagents?
+ Conc H2SO4
Ethanoic acid
CH3COOH
C2H5NH2
/ ethanol
N-ethylethanamide
CH3CONH2C2H5
Fats and oils
Fats and oils (lipids) are all triesters made from glycerol
(propane-1,2,3-triol) and three long chain carboxylic acids
(fatty acids) as shown below.
Glycerol is an example of a”triol” which has three -OH groups
present. Each of these can form an ester link with a different
carboxylic acid, for example the fat called stearin.
O
H
H C
OH
+
R1COOH
H C
OH
+
R2COOH
H C
OH
H
Glycerol
+

R3COOH
3 fatty acids
H2C
O
C
O
R1
HC
O
C
O
R2
H2C
O
C
R3
Fat or oil
+ 3H2O
Soap
The three ester links present in these molecules can be broken
(or hydrolysed) by heating with sodium hydroxide solution.
This releases the original glycerol molecule plus the sodium
salts of the long chain fatty acids which are soaps. This
“saponification” process is shown in the diagram below.
O
H
H
O
HH 2CC O
C
O
O
C
O
HC O
C
2
HC
H2C
H2C
OR
1
R2
C
O
OR C
O
3
O
C
R1
H C
OH
H C
OH
H C
OH
H
R2+ 3NaOH
R3
Heat
H C
OH + R1COONa+
H C
OH + R COONa+
2
H C
OH + R 3COONa+
H
3 soap molecules
Soaps work because the tail of the molecule is a long non-polar
hydrocarbon chain (from the fatty acid) which readily
dissolves grease and dirt (as “like dissolves like”). Then the
ionic carboxylate ion readily dissolves in water (which is also
polar) and is able to carry away the grease with it in the rinse
water.
HO
CH2
C
CH2
CH2
CH
CH2
CH2
CH
CH
CH
CH2
CH2
CH2
CH2
O
polar carboxylate
head (dissolves in
water)
non-polar hydrocarbon
tail (dissolves grease)
CH3
CH2
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This is your homework Tim!
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is an excellent revision website run by the Auckland
UniversityTo get on to this site you must use the following details –
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ID: 919
Password: compound
Log on using your own username and the supplied
password and ID
Go on and try the organic section
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