Both aliphatic and aromatic aldehyde (not ketone) react with Tollen’s reagent. In the reaction, aldehyde is oxidized to form carboxylate and the Ag(NH3)2 positive ion which appear as silver mirror on the wall of the test tube.
An aldehyde will react positively with schiff’s reagent giving pinkish purple colour solution. However it is not a good confirmatory test as benzaldehyde gives negative result and propanone gives positive result.
Similar to alcohol, carbonyl compounds can also react with iodine in warm alkaline solution, NaOH to form yellow precipitate of triiodomethane, CHI3.
Apparatus
Stopper, test tube, water bath, thermometer
Chemical reagents
Iodine solution or potassium iodide, Tollens’ reagent, Schiffs’ reagent, Fehling’s solution, 2.5 M NaOH solution, Dioxane or its equivalent, 2,4-Dinitrophenylhydrazine, ethanol, benzaldehyde, acetone, an unknown carbonyl compound ( X compound)
RESULT
A) Brady’s Test
B) Fehling’s Test
C) Tollen’s Test
D) Schiff’s Test
E) Iodoform Test
DISCUSSION
- An aldehyde or ketone will undergo a general reaction with the Brady reagent, 2, 4-dinitrophenylhydrazine, to produce 2, 4-dinitrophenyl-hydrazone which will appear as orange or yellow precipitates. This reaction is commonly used to ascertain the presence of a carbonyl group in a compound.
- Aldehydes can be distinguished from ketones through several tests. One test involves the use of a Schiff's reagent which will produce a violet-pink solution with aldehydes but not ketones. Some aromatic aldehydes such as vanillin also give a negative result with Schiff‟s test.
- Another test that can distinguish aldehydes from ketones is through weak oxidizing agents such as the Tollen's reagent (ammonium nitrate complex in ammonia solution). A positive reaction is indicated by the formation of a silvery mirror on the side of the tube.
RCHO + 2Ag(NH3)2OH 2Ag (s) + RCO2NH4 + H2O + NH3
- Iodoform test is a useful test for the identification of methyl ketones and secondary methyl carbinols. This test involves a reaction in which the methyl group of the ketone is removed from the molecule and produces iodoform (CHI3). A positive test is indicated by the formation of yellow precipitates or suspension of iodoform.
- Secondary alcohols with a methyl group adjacent to the carbon bearing the hydroxyl group such as ethanol can be oxidized to methyl ketones by “iodine bleach” or hypoiodide. Hence, alcohols such as ethanol will also produce yellow iodoform precipitates as methyl ketones in an iodoform test.
- There is a lot of error happened in this experiment. This can be caused from many aspect such as the environment condition and the human error. The environment condition include the conditions of the reagant in this experiment. Some of the reagant might be expired and this affect some of the final result in this experiment . There are some student that did not handle the experiment properly seriously such as they did not read the instruction correctly and played during this experiment. All of this things affect the final result of the experiment.
CONCLUSION
The objective was achieved as we can identify the carbonyl compounds. Based on the experiment, we can conclude that the unknown carbonyl is an aromatic carbonyl. We can say this because, unknown carbonyl does not react with the Fehling’s reagent. Only aromatic carbonyl such as Phenylpropane does not react with Fehling’s test to form Cu2O which appear as brick-red precipitate.
REFERENCE
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3. http://www.wikipedia.org
