2. Test for phenol
PROCEDURE: Add several drops of bromine water into a test tube which if filled with 1cm³ of the unknown.
OUTCOME: A white precipitate formed indicates a phenol present; no change shows a carboxylic acid.
EXPLANATION: Aqueous phenol decolourises bromine water to form a white precipitate of 2,4,6-tribromophenol
The presence in phenol of the –OH group increases the susceptibility of the benzene ring to electrophilic attack. The oxygen in the –OH group has two lone pairs of electrons, these can overlap with the delocalised π electrons. Overall, the π electron charge density is increased (especially at the 2, 4 and 6 positions). This is why this reaction happens only with phenol and not with the carboxylic acid.
SAFETY: Follow general safety procedures and bromine water can be toxic and corrosive so always make reference to hazcard 15.
EQUIPMENT: Test tube
Pipette
Bromine water
3. Test for carbonyl group
PROCEDURES: To 1cm³ of 2,4-dinitrophenylhydrazine solution and 1cm³ of ethanol, add several drops of the test substance using a pipette. Heat with a water bath if necessary.
OUTCOMES: A colour change to orange/yellow indicates the presence of a carbonyl group, which will be an aldehyde or a ketone. No colour change indicates that it is an alcohol, carboxylic acid, ester or phenol.
EXPLANATION:
The equation for the reaction of ethanal with 2,4-dinitrophenylhydrazine is:
SAFETY: Follow general safety procedures and 2,4-dinitrophenylhydrazine is explosive and toxic so always make reference to hazcard 30 also be aware that ethanol is highly flammable so make reference to hazcard 40.
EQUIPMENT: Test tube
Pipettes
2,4 DNPH solution
1cm³ of Ethanol
Water Bath (if necessary)
4. Test for an aldehyde
PROCEDURES: Put about 1cm³ of 0.05M silver nitrate solution into a very clean test tube and add 3 or 4 drops of sodium hydroxide solution. Drop by drop; add ammonia solution until the precipitate of silver oxide nearly dissolves. Add a few drops of unknown to the tube and shake gently. Place tube in a beaker of warm water, note observations and immediately rinse out test tube.
OUTCOMES: A change from clear to a silver mirror precipitate on the side of the test tube indicates the presence of an aldehyde. No change indicates a ketone.
EXPLANATION: Aldehydes are easily oxidised to acids. Ketones cannot be oxidised, as there is no place for the oxygen from the oxidising agent.
When the silver nitrate is mixed with ammonia to form Tollen's reagent, the complex ion [Ag (NH3) 2]+ is formed.
This is reduced to silver during the process; Tollen's reagent is the oxidising agent.
SAFETY: Follow general safety procedures and silver nitrate solution, sodium hydroxide solution and ammonia solution are all corrosive so always make reference to hazcards 87, 91 and 6.
EQUIPMENT: Test tube
Pipette
Silver nitrate solution
Sodium hydroxide solution
Ammonia solution
Water Bath
5. Test to find an alcohol
PROCEDURES: Add 1cm³ of potassium dichromate into a test tube also add 1cm³ of sulphuric acid to the test tube and add a few drops of the unknown and warm using a water bath.
OUTCOMES: A colour change to green shows an alcohol is present. No colour change shows an ester.
EXPLANATION:
Primary alcohols are oxidised by acidified dichromate to aldehydes and then to acids.
C2H5OH + [O] --> CH3CHO + H2O [O] is the oxygen from the oxidising agent then CH3CHO + [O] --> CH3COOH
If this is done with a secondary alcohol, only a ketone forms.
In each oxidation, the dichromate is reduced to Cr3+ (green).
No reaction occurs with a tertiary alcohol.
SAFETY: Follow general safety procedures and potassium dichromate is toxic so always make reference to hazcard 78 also sulphuric acid is extremely corrosive so make reference to hazcard 98.
EQUIPMENT: Test tube
Pipette
Potassium dichromate solution
Sulphuric acid
Water bath
ANALYSIS OF RESULTS
Now that I have identified the functional groups of the unknowns using the chemical techniques, I will now use the spectra provided to confirm these results.
Unknown A, I have identified to be an alcohol. Using firstly the mass spectrum the relative atomic mass (RAM) of the compound is approximately 46. I can now make an estimation of the molecular formula. C2H6O is my estimation.
Using information from the infrared spectrum, there is a strong, broad absorbance at wavelength 3200-3600. This confirms that an OH bond is present showing an alcohol.
Finally using the NMR spectrum I can identify the number of protons and the environments and positions they are in. For example a chemical shift at 3.5-5.5 shows a R-OH proton present. From all this information I conclude Unknown A to be ethanol C2H6O.
Unknown B, I have identified to be an ester. Using the mass spectrum I can see that the RAM of the compound is 88. My estimation of the molecular formula is C4H8O2.
From the infra red spectrum I can see that there is a strong, sharp absorbance at 1680-1750 showing the presence of a C=O bond also there is a strong absorbance at 100-1300 showing the presence of a C-O bond. This shows that there is an ester present.
Using the NMR spectrum I can find the exact positioning of the protons present. There are 8 protons present in 3 different environments. A chemical shift at 3.3-4.3 shows -O-CH2-R is present. From all the information present I have deduced Unknown B to be ethyl ethanoate C4H8O2.
Unknown C I have identified to be an aldehyde. From the mass spectrum the RAM is 44 so my estimation of the molecular formula is C2H4O.
From the infrared spectrum there is a strong, sharp absorbance at 1680-1750 showing a C=O bond (present in an aldehyde). O
From the NMR the offset at 9.8 shows a chemical shift of R-C-H. From this information I conclude the aldehyde I identified is ethanal C2H4O.
There are no spectra for the Unknown D compound as it is a phenol. Although a mass spectrum would be useful to confirm the RAM of phenol.
EVALUATION
In this experiment I am happy with the results I received, as they all appear to be correct.
The chemical techniques I used only gave the functional groups of the unknowns and I cannot find out what exactly the unknown is by just using these techniques. I had to use the physical techniques as well.
With all the chemical techniques there was always the possibility that an error could occur through the contamination of equipment, chemicals or the unknowns themselves. This would result in an incorrect result so therefore I paid particular care when taking part in this experiment.
The mass spectrum gave me the relative atomic mass of the unknown this can be used to find a molecular formula however this could be appropriate to more than one functional group so I would not be able to use this technique on its own to find the identity of an unknown.
The infrared spectrum doesn’t simply make a distinction between an aldehyde and ketone so other tests would be required if this situation occurred.
The NMR spectrum was very useful as it gave most information needed to identify an unknown.
Another technique I could use to identify an unknown which I didn’t use would be finding the melting and boiling points. There would however be difficulties with impurities so I would have to make sure the unknown was pure.
