Introduction
In this post, I will present the preparation of 5-iodovanillin by electrophilic aromatic substitution of vanillin. The electrophile in the reaction is iodine which is generated in-situ by the oxidation of potassium iodide with chlorine bleach. The overall reaction can be summarized as follows:
The mechanism of the reaction is simple: iodine attacks the ring onto the 5 position generating a positively charged intermediate stabilized by resonance with the hydroxy group on the ring. The proton in the 5 position is then removed by a base to restore aromaticity and yield the final product. Please notice how, thanks to the highly electron-rich vanillin ring, the reaction happens without requiring the use of any catalyst to further activate the halogen by polarization.
Experimental part
In a 100mL flask, 0.87g of vanillin (5.7mmol, 1eq.) and 1.26g of potassium iodide (7.6mmol, 1.3eq) are dissolved in 25mL of 95% ethanol. The mixture is chilled in an ice bath and 14.5mL of a 3.5% sodium hypochlorite solution is added, under constant stirring, over 20 minutes.
After the addition, the mixture is removed from the ice bath and let stirring at room temperature for another 20 minutes. The excess iodine and sodium hypochlorite are neutralized with 0.98g of sodium thiosulphate. The solution is acidified with hydrochloric acid until complete precipitation of the product, cooled, and filtered on a Buchner funnel. The solid on the filter is rinsed with cold water a couple of times. The solid is transferred to a beaker and recrystallized from an ethanol-water mixture. To do so the product is dissolved in a minimal quantity of boiling ethanol and water is added until cloudiness is observed. The solution is left to cool to room temperature and the crystalline product is recovered by vacuum filtration. The obtained solid is dried over sodium hydroxide. A total amount of 1g of 5-iodovanillin is obtained corresponding to a 63% yield.
