Virginia Wesleyan College
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Norfolk , VA 23502
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Student Research Projects
The Conversion of Vanillin to Isovanillin: An Industrial By-product to a Pharmaceutical Feedstock
Dr. Joyce Easter|
|Course||CHEM 489: Research in the Natural Sciences|
Currently, green chemistry is of growing interest to chemical manufacturers. Green chemistry focuses on products and processes that reduce or eliminate the use and synthesis of hazardous substances to the environment. The application of the twelve principles of green chemistry can improve many processes and make them more environmentally friendly. It is this same application that can be used in processes of drug synthesis which will reduce the necessity for crude oil in developing chemical feedstocks and products. To have a greater impact, this can be applied to morphine, a drug that has widespread usage. As a pain reducer for many different types of ailments, morphine is synthesized initiating with the pharmaceutical feedstock, isovanillin. Isovanillin has been shown to be a building block and a feedstock for many drug syntheses to include PDE4 inhibitors, Ariflo, and (R)-Rolipram. The isovanillin for this synthesis can be obtained via a conversion of vanillin to protocatechualdehyde. The vanillin, obtained from paper waste, can be demethylated to a protocatechualdehyde intermediate which can then be methylated specifically to form isovanillin, the pharmaceutical feedstock.
In order to demethylate the vanillin to form the protocatechualdehyde, a SN2 nucleophilic reaction was attempted using a strong acid, HBr and HCl (Scheme 1). In this reaction, a strong nucleophile Br- and Cl- was used to replace the oxygen on the methyl group. For this particular reaction, conditions such as heat were altered in order to determine their effect on the reaction. After thin-layer chromatography analysis on these separate reactions, it was difficult to determine if protocatechualdehyde had formed so another SN2 reaction was performed using Na-I salt (Scheme 2). The iodine is a strong nucleophile; however, the oxygen on the deprotonated intermediate would then possess a negative charge. The addition of a strong acid would protonate the oxygen to form the protocatechualdehyde.