A2 Chemistry Unit 4
What functional group do all
carbonyl compounds contain?
δ+
δ-
What is the aldehyde functional
group?
What is the ketone functional
group?
Name the following molecules:
Propanal
Propanone
Methylpropanal
Pentan-3-one
Methanal
Methylbutanone
Describe and explain the boiling point of carbonyl
compounds, referring to the type of bonding.
Methanal is a gas and the other carbonyl compounds are liquid at
room temperature.
The molecule are polar, so dipole-dipole forces as well as van der
Waals forces exist between them. Intermolecular hydrogen bonding
is not possible because neither aldehydes nor ketones have a
hydrogen atom that is sufficiently δ+.
This causes them to:
Have higher boiling points than alkanes and alkenes
Have lower boiling points than alcohols
Describe and explain the solubility of
carbonyl compounds in water.
The lower members of both aldehydes and ketones are soluble in water.
This solubility is due to the hydrogen bonding between the lone pair of
electron in the δ- oxygen in the carbonyl compound and the δ+
hydrogen in a water molecule. Solubility decreases as the number of
carbons increases because the hydrocarbon tail is hydrophobic.
O
δ+
δ+
δ-
H
H
H
δ+
O
H
Describe the smell of carbonyl
compounds.
The lower members of the
homologous series of aldehydes
have pungent odours.
Ketones have much sweeter smells than
aldehydes.
Describe the preparation of aldehydes and give
the equation for the preparation of ethanal.
Aldehydes are prepared by the partial oxidation of a primary alcohol
Reagents:
Primary alcohol
Potassium dichromate (VI) in dilute sulfuric acid
Conditions:
Temperature of about 60°C
Distil of product as it forms
Observation:
Colour change from orange to green as dichromate (VI)
is reduced to chromium (III)
Describe the preparation of ketones and give
the equation for the preparation of propanone.
Aldehydes are prepared by the oxidation of a secondary alcohol
Reagents:
Secondary alcohol
Potassium dichromate (VI) in dilute sulfuric acid
Conditions:
Heat under reflux
When reaction has finished, distil off product
Observation:
Colour change from orange to green as dichromate (VI)
is reduced to chromium (III)
Describe and explain the differences between
aldehydes and ketones with respect to
oxidation.
Aldehydes are readily oxidised whereas ketones are not.
This is due to the presence of the hydrogen atom on the C=O in
aldehydes which is very easily oxidised
When an aldehyde is oxidised:
Describe the reaction of aldehydes with
Fehling’s/Benedict’s solution and give an
equation involving ethanal.
Both Fehling’s and Benedict’s contain copper (II) complexes.
When an aldehyde is warmed with Fehling’s/Benedict’s solution it is
oxidised. The solution is alkaline so the salt of the carboxylic acid is
produced.
The blue solution is reduced to a
red precipitate of copper oxide.
N.B. With ketones there is no reaction and the solution stays blue.
Describe the reaction of aldehydes with
Tollens’ reagent and give an equation
involving ethanal.
Tollens’ reagent is made by adding a few drops of sodium hydroxide to silver nitrate
solution and then dissolving the precipitate in dilute ammonia.
When a few drops of Tollens’ are added to the aldehyde and the
mixture is warmed gently, the aldehyde is oxidised. Because the
solution is alkaline the salt of the carboxylic acid is formed.
The colourless Tollen’s reagent is
reduced to give a silver mirror.
N.B. With ketones there is no reaction and the solution stays colourless.
Describe the reaction of aldehydes with
acidified potassium dichromate (VI) and
give an equation involving ethanal.
When heated with the acidified potassium dichromate the
aldehyde is oxidised to a carboxylic acid.
The orange potassium
dichromate (VI) is reduced
to green Cr (III)
N.B. With ketones there is no reaction and the solution stays orange.
Describe the reduction of an aldehyde,
including the equation for the reduction of
ethanal.
Reducing agent:
Lithium tetrahydridoaluminate (III), LiAlH4
The LiAlH4 acts as a source of H- ions which add on to the δ+
carbon atom – at this point the reactants must be kept dry so it is
carried out in ether solution.
Next, a solution of aqueous acid is added which protonates the Oformed in the first step.
A primary alcohol is formed.
Describe the reduction of an aldehyde,
including the equation for the reduction of
propanone.
Reducing agent:
Lithium tetrahydridoaluminate (III), LiAlH4
The LiAlH4 acts as a source of H- ions which add on to the δ+
carbon atom – at this point the reactants must be kept dry so it is
carried out in ether solution.
Next, a solution of aqueous acid is added which protonates the Oformed in the first step.
A secondary alcohol is formed.
Why does LiAlH4 reduce carbonyl
compounds but not alkenes?
LiAlH4 is a reducing agent that reacts
specifically with polar π-bonds.
State and explain the conditions needed
for the reaction between hydrogen
cyanide and a carbonyl compound.
pH 8 – this provides the CN- ion catalyst
Draw the mechanism for the reaction
of hydrogen cyanide and ethanal.
2-hydroxy-propanenitrile
Draw the mechanism for the reaction
of hydrogen cyanide and propanone.
2-hydroxy-2-methylpropanenitrile
What is the difference between
the two reactions?
The ketone reaction happens at a
slower rate.
Describe the mechanism for the reaction of
hydrogen cyanide and a carbonyl compound.
The lone pair of electrons on the carbon atom of the CN- ion form a
bond with the δ+ carbon atom in the carbonyl compound.
At the same time, the π-electrons in the C=O group move to the
oxygen atom.
The anion formed in the first step removes a proton from an
HCN molecule to form the organic product and regenerate the
CN- catalyst.
Why is it important that the pH is
neither too high or too low?
If it is too low, there are
not enough CN- ions for
the first step.
If it is too high there are not
enough HCN molecules for the
second step.
Describe and explain the optical activity of
the product when hydrogen cyanide adds
on to a carbonyl compound.
The product is a racemic mixture of both enatiomers so it does
not rotate the plane of plane-polarised light.
This is because the carbonyl compound is planar around the C=O
group so the cyanide ion can attack from above or below the
plane.
Describe how carbonyl compounds react
with compounds containing an H2Ngroup.
The lone pair of electrons on the nitrogen atom acts a
nucleophile and forms a bond with the δ+ carbon atom in the
C=O group.
A water molecule is then lost and a C=N bond is formed.
Describe how 2,4-dinitrophenylhydrazine
can be used to test for a carbonyl group.
Add a few drops of 2,4-dinitrophenylhydrazine to a
solution of the compound.
Simple aldehydes and
ketones give a yellow
precipitate.
Aromatic aldehydes and
ketones give an orange
precipitate.
Give the overall equation for the reaction
between a carbonyl compound and 2,4dinitrophenylhydrazine.
What type of reaction occurs and
what type of mechanism is it?
Condensation reaction –
with the loss of a water
molecule
Addition-elimination
Describe how you can make and purify a derivative of a
carbonyl compound using 2,4-dinitrophenylhydrazine.
Explain how this removes
impurities.
In step 2 the derivative will recrystallise but the impurities
will remain in solution.
The third step, washing with cold solvent, will then rinse of
any impurities that did crystallise.
How can the purified product be used to
identify the original carbonyl compound?
Find the melting temperature and compare to
known data.
What is the reaction of carbonyl
compounds with iodine in alkali called?
The iodoform reaction
What type of carbonyl compounds
undergo the iodoform reaction?
Ethanal and methyl ketones
Describe how to carry out the iodoform
reaction, including observations.
Warm the organic substance with either:
A mixture of iodine and NaOH solution
 A solution of potassium iodide in sodium chlorate (I)
A pale yellow
precipitate forms
Medical smell
Describe how the iodoform reaction
occurs, including equations.
Sodium hydroxide solution is added to iodine solution to form iodate (I)
ions:
These substitute into the –CH3 group next to the C=O group, forming a
CI3C=O group.
The three halogen atoms and the oxygen atom have an electronwithdrawing effect which weakens and breaks the σ-bond between the
two carbon atoms, forming iodoform:
Write the overall equation for the
iodoform reaction.