Reactions of Carbonyls (Edexcel International A Level Chemistry): Revision Note
Carbonyls - Reactions
Carbonyl compounds can undergo a number of reactions, some of which allow you to distinguish between an aldehyde and a ketone
Oxidation - with acidified potassium dichromate(VI) solution, Tollens' reagent or Fehling's / Benedict's solution
Reduction - with lithium tetrahydridoaluminate / lithium aluminium hydride in dry ether
Nucleophilic addition - with HCN or acidified KCN
Non-specific carbonyl testing
2,4-dinitrophenylhydrazine (2,4-DNPH) - to identify a carbonyl compound
Iodoform test - iodine in the presence of alkali
Acidified potassium dichromate(VI)
Aldehydes can be oxidised to form carboxylic acids using acidified potassium dichromate(VI) solution
K2Cr2O7 (aq) with sulfuric acid, H2SO4
General reaction of an aldehyde with an oxidising agent
Ketones are very resistant to being oxidised
This is because ketones do not have a readily available hydrogen atom, like aldehydes do
An extremely strong oxidising agent would be needed for oxidation of a ketone to take place
Such oxidation will likely oxidise a ketone in a destructive way, breaking a C-C bond
Heating with acidified potassium dichromate can distinguish between an aldehyde and a ketone
The aldehyde would be oxidised, and you would see an orange to green colour change
The ketone would not be oxidised, so you would see no colour change
Tollens' reagent
Tollens' reagent contains the silver(I) complex ion [Ag(NH3)2]+
This is formed when aqueous ammonia is added to a solution of silver nitrate
Tollens' reagent is also known as ammoniacal silver nitrate
If gently warmed with Tollens' reagent, an aldehyde will become oxidised
The silver(I) complex ion solution, [Ag(NH3)2]+, is colourless
As the aldehyde is oxidised, it causes the [Ag(NH3)2]+ ions to become reduced to solid metallic silver, Ag
This is why a positive test result is called a "silver mirror"
Positive Test Result:
When Tollens' reagent is gently warmed with an aldehyde, the silver mirror is formed
This is the positive test result
When Tollens' reagent is gently warmed with a ketone, no silver mirror will be seen, as the ketone cannot be oxidised by Tollens' reagent, so no reaction takes place
This is a negative test result
The Ag+ ions in Tollens’ reagent are oxidising agents, oxidising the aldehyde to a carboxylic acid and getting reduced themselves to silver atoms
Fehling's Solution
Fehling’s solution is a solution containing copper(II) ions dissolved in sodium hydroxide, which act as the oxidising agent
Benedict's solution is exactly the same as Fehling's solution but the copper(II) ions are dissolved in sodium carbonate
If an aldehyde is warmed with Fehling's solution, the aldehyde will be oxidised and a colour change will take place
Fehling's solution is blue, because of the copper(II) complex ions present
During the reaction, as the aldehyde is oxidised to a carboxylic acid, the blue Cu2+ ions are reduced to Cu+ ions and a brick red precipitate is formed
The brick red precipitate is copper(I) oxide
If a ketone is warmed with Fehling's solution, no reaction takes place as the ketone will not be oxidised, so the solution will remain blue
The copper(II) ions in Fehling’s solution are oxidising agents, oxidising the aldehyde to a carboxylic acid and getting reduced themselves to copper(I) ions in the Cu2O precipitate
Examiner Tips and Tricks
You are expected to know all of the above methods which can be used to distinguish between an aldehyde and a ketone! However, Tollens' reagent is the most commonly used method, if trying to identify an unknown sample for example.
Reduction of Carbonyls
Aldehydes are reduced to primary alcohols and ketones are reduced to secondary alcohols
This is done with lithium tetrahydridoaluminate / lithium aluminium hydride, LiAlH4, in dry ether
The LiAlH4 generates a hydride ion nucleophile, :H-
The hydride ion reduces the carbonyl group in an aldehyde or a ketone
When this reaction takes place, it is an example of a nucleophilic addition reaction
Reduction equation for an aldehyde
Reduction equation for a ketone
Nucleophilic addition
Many of the reactions which carbonyl compounds undergo are nucleophilic addition reactions
The carbonyl group -C=O, in aldehydes and ketones is polarised
The oxygen atom is more electronegative than carbon drawing electron density towards itself
This leaves the carbon atom slightly positively charged and the oxygen atom slightly negatively charged
The carbonyl carbon is therefore susceptible to attack by a nucleophile, such as the cyanide ion
General Mechanism with an aldehyde
General Mechanism with a ketone
In both reactions, the nucleophile (Nu) attacks the carbonyl carbon to form a negatively charged intermediate which quickly reacts with a proton
Addition of HCN to carbonyl compounds
The nucleophilic addition of hydrogen cyanide to carbonyl compounds is a two-step process, as shown below
In step 1, the cyanide ion attacks the carbonyl carbon to form a negatively charged intermediate
In step 2, the negatively charged oxygen atom in the reactive intermediate quickly reacts with aqueous H+ (either from HCN, water or dilute acid) to form 2-hydroxynitrile compounds,
e.g. 2-hydroxypropanenitrile
This reaction is important in organic synthesis, because it adds a carbon atom to the chain, increasing the chain length
The products of the reaction are hydroxynitriles
The nitrile group is the priority functional group so it is attached to carbon 1 and results in the suffix -nitrile
The hydroxyl group is not the priority functional group so the hydroxyl group is named using the hydroxy- prefix, rather than the -ol suffix
2,4-dinitrophenylhydrazine (2,4-DNPH)
2,4-dinitrophenylhydrazine (also known as 2,4-DNPH) is a reagent which detects the presence of carbonyl compounds (compounds with -C=O group)
The carbonyl group of aldehydes and ketones undergoes a condensation reaction with 2,4-dinitrophenylhydrazine
A condensation reaction is a reaction in which two molecules join together and a small molecule (such as H2O or HCl) is eliminated
The product formed when 2,4-DNPH is added to a solution that contains an aldehyde or ketone is a deep-orange precipitate which can be purified by recrystallisation
The melting point of the formed precipitate can then be measured and compared to literature values to find out which specific aldehyde or ketone had reacted with 2,4-DNPH
Ketones and aldehydes react with 2,4-DNPH in a condensation reaction
The test tube on the left shows a negative 2,4-DNPH test and the tube on the right shows a positive test
Examiner Tips and Tricks
The 2,4-DNPH test is especially useful as other carbonyl compounds such as carboxylic acids and esters do not give a positive result
Iodoform test
Tri-iodomethane (also called iodoform) forms a yellow precipitate with methyl ketones
Methyl ketones are compounds that have a CH3CO-group
Ethanal also contains a CH3CO- group and therefore also forms a yellow precipitate with iodoform
The reagent is heated with an alkaline solution of iodine
This reaction involves a halogenation and hydrolysis step
In the halogenation step, all three H-atoms in the -CH3 (methyl) group are replaced with iodine atoms, forming a -CI3 group
The intermediate compound is hydrolysed by an alkaline solution to form a sodium salt (RCO2– Na+) and a yellow precipitate of CHI3
The reaction of ethanal with iodoform results in the formation of a yellow CHI3 precipitate
The overall reaction equation is
CH3COCH3 + 3I2 + 4OH– → CH3COO– + CHI3 + 3I– + 3H2O
Alcohols and the iodoform test
If an alcohol contains a R-CH(OH)-CH3 group it will also produce a yellow precipitate in this test
These include all methyl secondary alcohols (such as hexan-2-ol) and the primary alcohol ethanol
Iodine in sodium hydroxide reacts to form iodate(I) ions which will oxidise the secondary alcohols to ketones
Iodination of carbonyls under acidic conditions
If propanone is treated with iodine solution under acid conditions a different reaction takes place
A hydrogen atom in propanone is replaced by an iodine atom, producing iodopropanone
CH3COCH3 + I2 → CH3COCH2I + H+ + I–
The iodopropanone is colourless and forms slowly
This makes this a very useful reaction to study kinetics as the yellow-brown colour of the iodine fades so it is an ideal reaction to monitor using colorimetry
By varying the concentrations of propanone, iodine and hydrogen ions you can easily determine the reaction orders and overall rate equation
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