In an undergraduate organic experiment, we are required to identify the functional group present via solubility tests (with H2O, HCl, NaOH and NaHCO3) and Classification Tests (Brady's, Tollen's, Iodoform, FeCl3, Aqueous Bromine, Lucas and Chromic Acid Test).
The results:
The unknown is found to be insoluble in all the solvent used in the solubility test. The results from the classification test have also shown that the unknown contains ketone. The IR spectra contains a strong peak at ~1700cm-1 which indicates the presence of a carbonyl (C=O) bond. *IR attached* The preliminary boiling point for the liquid unknown is ~ 212 degree Celsius.
Since it is most probably a ketone, I proceeded with the second part of the experiment (Crystalline derivatives for identification). Other lab mates whose unknowns have been suspected to have ketone/aldehyde managed to form semicarbazone in minutes/seconds. I am rather worried as my liquid unknown took a very long time (hours) to form semicarbazone. Could it be that the unknown has been falsely analysed?
Could you please describe the specific experimental conditions that you followed with the physical parameters.
You should try to form oxime of the compound using hydroxyl amine hydrochloride and then confirm it by IR or TLC or GC
Hi! I think that you should not be much worried…if you get a solid you may now be able to compare the melting point…
Also, yours colleagues time reaction may not be compared with your time reaction. Different ketones will have different reaction times… also that if your mates’ compound may be an aldehyde (also with a C=O) this class of compounds are much more reactive (faster) than ketones…
By the boiling point you provide your ketone must have a higher molecular weight, i.e., around the C=O may have bulky groups and although the C=O is planar (sp2) bulky groups around would slower the reaction rate. (with the boiling point you provide may be you ketone belongs to the “carvone” family…
Good work and good luck.
NMR Spectrum shows that it is p-methoxyacetophenone.
Thank you for the info Dr Americo.
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