Observations:
- Dilute sulphuric acid had a irritating odour.
- Propanone had a pungent odour while ethanal had a pungent and fruity odour.
- Dilute sulphuric acid and sodium hydroxide solution were colourless.
Part 1
Adding sodium hydrogensulphite solution
- For ethanal, the solution turned milky and white spongy precipitate was formed. After pouring away the solution, there were white solid crystal remaining in the test tube.
- For propanone, there wer 2 colourless immisible layers formed.
- Both of the solution became turned warm.
Part 2
Treating with 2,4-dinitrophenylhydrazine
- 2,4-dinitrophenylhydrazine solution was yellow in colour.
- For ethanal, there was yellow precipitate formed.
- For propanone, there was also yellow precipitate formed.
Part 3
- Acidified dichromate
- Potassium dichromate solution was orange in colour.
- For ethanal, the solution turned from orange to green and the test tube turned warm.
- For propanone there was no observable colour and the solution remained orange.
- Fehling’s solution
- Fehling’s solution A was pale blue in colour.
- After adding Fehling’s solution B, there was precipitate formed and the precipitate redissolved with more Fehling’s solution B added.
- The solution was dark blue after the precipitate redissovled.
- For ethanal, the solution turned dark green and after a long period of time, there was brown solid deposited at the bottom.
- For propanone, there was no observable change.
- Tollens’ reagent
- Ammonia solution had a pungent smell.
- The reaction mixture turned dirty green after the reaction.
- For ethanal, there was silver mirror formed.
- For propanone, there was no observation.
Part 4
Iodoform test
- Iodine solution was dark brown in colour.
- For ethanal, there was pale yellow precipitate and there was a clinical smell.
- For propanone, there was also pale yellow precipitate with a clinical smell.
Discussion:
Precautions:
Propanone and ethanal are flammable. So, these liquids must be kept away from direct flame. Flammable
Ketones and aldehydes
Ketones are versatile compounds which can be converted to a number of useful functional groups through reduction, nucleophilic addition or condensation reactions. Ketones and aldehydes are important series in preparation of other compounds and they are commonly prepared by oxidizing alcohol which is done in this experiment. Ketone also plays a very important part in organic synthesis. Ketones and aldehydes can be synthesised into many other chemicals. Reactions involving ketones include nucleophilic addition reactions to the carbon-oxygen double bond to form an –OH group in the compound with the addition of a nucleophilic group.
Reaction of carbonyl compounds
- Using 2,4-dinitrophenylhydrazine
This is used in our experiment. 2,4-DNP reacts with the carbonyl group for a condensation reaction with the elimination of a water molecule.
Take propanone as an example,
The product formed is a yellow colour precipitate, so we can easily distinguish the presence of C=O group.
This can also help us to identify the carbonyl compound as the precipitate collected has a sharp melting point. By using the melting point test, we can find out the melting point of the crystals formed and compare the result with a data book to find out the carbonyl compound.
However, when doing this test, we should only add 1 to 2 drops of ethanal/ propanone into the test tube. If too much carbonyl compound was added into the test tube, it will dissolve the product formed as ethanal and propanone are both good polar and nonpolar solvent.
So, we cannot see any precipitate if too much are added into the test tube.
Yellow precipitate was formed
-
Tollens’ reagent (Distinguish aldehyde from ketone)
The formula of this reagent is Ag(NH3)2+. As this reagent is not very stable, it must be prepared freshly in laboratory.
To prepare the reagent, aqueous ammonia can be added in a continuous fashion directly to silver nitrate solution. At first, silver oxide will be formed and precipitate out, but as more ammonia solution is added the precipitate dissolves and the solution becomes clear as diamminesilver(I) is formed. At this point the addition of the ammonia should be stopped.
This reagent is used in the silver mirror test. In this test, when there is the presence of aldehyde group, there would be formation of silver mirror.
The equation of this reaction is as below
[Ag(NH3)2]+ (aq) + e- → Ag (s) + 2 NH3 (aq)
RCHO (aq) + 3 OH- → RCOO- + 2 H2O + 2 e-
The aldehyde acts as an reducing agent where [Ag(NH3)2]+ was reduced to Ag(S) , the formation of silver mirror. This reaction is very useful to extinguish aldehyde from ketone as ketone does not show this reaction.
Silver mirror formed in a flask
The colour of the product mixture after the reaction.
- Fehling’ reagent
Aldehydes are also oxidized by the Fehling’s solution. This reagent is also prepared freshly in the laboratory.
It is made initially as two separate solutions, known as Fehling's A and Fehling's B. Fehling's A is a blue aqueous solution of , while Fehling's B is a clear solution of aqueous and a strong alkali (commonly ).
Equal volumes of the two mixtures are mixed together to get the final Fehling's solution, which is a deep blue colour.
In this final mixture, aqueous tartrate ions from the dissolved Rochelle salt to Cu2+ (aq) ions from the dissolved copper(II) sulfate, as giving the 4- complex.
The tartarate ions, by complexing copper prevent the formation of Cu(OH)2 from the reaction of CuSO4.2H2O and NaOH present in the solution.
The Copper (II) ion is reduced to copper (I) oxide which is a red ppt, and in some cases, to copper metal (copper mirror). This is also useful to distinguish aldehyde from ketone and aromatic aldehyde as both ketone and aromatic aldehyde does not show any reaction.
[O]
2Cu2+(aq) Cu2O(S)
Left: Fehling’s reagent Right: Product mixture
- Nucleophilic addition reaction
In the experiment, we used hydrogensulphite as a nucleophile to attack the electron deficient carbon in the C=O. The reaction mechanism of this reaction is as follow:
(Where Nuc represents the nucleophilic species which, in this case, is hydrogensulphite ion)
The final product was an alcohol.
For the 2nd step of the reaction, water is already sufficient to provide the H+ ion as methoxide ion is more basic than hydroxide ion, so the methoxide ion will take a proton from water forming an alcohol and leaving a hydroxide ion.
However, depending on the reactivity of the nucleophile, there are two possible general scenarios:
Strong nucleophiles (anionic) add directly to the C=O to form the intermediate alkoxide. The alkoxides are then protonated on work-up with dilute acid.
Examples of such nucleophiles are: LiAlH4, NaBH4 (H-)
Weaker nucleophiles (neutral) require that the C=O be activated prior to attack of the Nu.
This can be done using an acid catalyst which protonates on the Lewis basic O and makes the system more electrophilic.
Examples of such nucleophiles are : H2O, ROH, R-NH2
- Oxidation with acidified dichromate (VI) solution
Aldehyde can be further oxidized to carboxylic acid.
The product formed will be a carboxylic acid.
In the reduction of dichromate ion, the dichromate ion (orange) is turned to chromium ion (green).
Ketone may undergo further oxidation to form carboxylic acid with hot acidified potassium permanganate and under reflux. This reaction has very high activation energy because this requires breaking the strong C-C bond. So, ketone usually does not react with acidified dichromate solution in normal conditions.
Therefore, acidified dichromate solution can be used to distinguish aldehyde from ketone.
Cr2O72- Cr3+
Besides acidified dichromate solution, acidified potassium permanganate solution can also be used as an oxidizing agent.
The equation of its reduction is:
MnO4- + 8H+ +5e- Mn2+ + 4H2O
However, this oxidation test cannot be used to confirm that the sample is a aldehyde as there are also many other compounds that can be oxidized by these 2 strong oxidizing agent such as ethene which turned to ethen-1,2-diol.
- Iodoform test
Iodoform test can be used to test for methyl ketone group ( -COCH3) or methyl alcohol group ( -CH2OHCH3 group)
The equation of this test is
The final products contain CHI3 which is insoluble in water and has a clinical smell. It exists in yellow precipitate in the mixture.
If a secondary alcohol is present, it is oxidized to a ketone by the hypohalite (hydroxide depicted below should be hypohalite and the water≡H2O should be hypohalic acid≡HXO):
If a methyl ketone is present, it reacts with the hypohalite in a three-step process:
(1) R-CO-CH3 + 3 OX- → R-CO-CX3 + 3
(2) R-CO-CX3 + OH− → + −CX3
(3) + −CX3 → + CHX3
When a - methyl carbonyl compounds react with iodine in the presence of a base, the hydrogen atoms on the carbon adjacent to the carbonyl group (a hydrogens) are subsituted by iodine to form tri iodo methyl carbonyl compounds which react with OH - to produce iodoform and carboxylic acid (2):
Reaction mechanism
The hydrogen atoms on the methyl group are slightly acidic and can be removed with hydroxide. The carbanion formed then react with iodine molecules to give a iodide ion and a monoiodonated methyl carbonyl derivate.
Introduction of the first iodine atom (owing to its electronegativity) makes the remaining hydrogens of the methyl group more acidic. Hence a base-catalized iodination of a monohalogenated methyl carbonyl derivate occurs at the carbon that is already substituted. Finally a tri iodo methyl carbonyl derivate is formed.
The next step is a nucleophilic attack by hydroxide on the carbonyl carbon atom. A carbon - carbon bond cleavage occurs and a triiodomethanide ion departs. The triiodomethanide ion is unusually stable. Its negative charge is dispersed by the three negative iodine atoms. In the last step a proton transfer takes place between carboxylic acid and triiodomethanide ion to form ultimately carboxylate ion and iodoform.
Other method of distinguishing carbonyl compound
Using IR spectrum
Using the IR spectrum, if the compound contains a carbonyl (C=O) group, there will be a stretch in the wave number 1670-1820 with high intensity.
Below is the IR spectrum of propanone
In the test with Tollens’ reagent, we need to use a very clean test tube. Why?
This is because if the test tube contains tap water or other impurities, there may be precipitate formed when silver ion is added into the test tube as tap water contain a small amount of Cl-.
Moreover, if the test tube is not clean, the surface of the inner wall of the test tube may not be smooth, so the silver may not be evenly distributed on the wall of the test tube which makes silver mirror not able to be formed.
Reference: