1.0 g of vanillin was dissolved in 20 mL of ethanol. To this solution, 1.17 g of
ID: 517440 • Letter: 1
Question
1.0 g of vanillin was dissolved in 20 mL of ethanol. To this solution, 1.17 g of sodium iodide was added, then cooled to 0 °C with an ice/water bath. 11.0 mL of aqueous sodium hypochlorite solution (5.25% w/w) was added dropwise to the stirred reaction mixture at a rate of 1 drop per second. Once the addition was complete, the mixture was warmed to room temperature and stirred for 10 minutes.Then, 10 mL of sodium thiosulfate solution (10% w/v) was added, then 6 mL of 3M hydrochloric acid was added. The aryl iodide should precipitate at this point.The ethanol was removed using a rotary evaporator. The flask was cooled for 10 minutes, then the precipitate was collected through vacuum filtration. Crude product was weighed to be 1.92 grams. The product was recrystallized from ethanol and the recrystallized product mass was recorded to be 0.83 grams.There are three possible monoiodo products of this reaction. Draw their structures. Explain why the other two were not formed in this reaction.
Explanation / Answer
This green chemistry reaction yields 4-hydroxy 3-iodo 5-methoxy benzaldehyde as the product by the electrophilic aromatic substitution reaction. This is due to the stability of the aromatic ring due to the resonance. Two more mono iodo products are possible as given above, may be due the electrophilic substitutions at 2 and 6 positions. These are not formed due to the positioning of hydroxy group which is o- directing and the aldehyde group which is m- directing. The affinity for the electrophilic substitution is guided to the 3 position, hence the single product. The products not formed are 4-hydroxy 2-iodo 5-methoxy benzaldehyde and 4-hydroxy 6-iodo 5-methoxy benzaldehyde.
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