You would need to reduce 1-Bromo-3-Chloro-5,5-dimethylhydantoin with KBr, NaBr or other bromides in the presence of an acid in order to efficiently
use it as a bromine source:
1-Bromo-3-Chloro-5,5-dimethylhydantoin(s) + 3KBr(aq) + H2SO4(aq) => 2Br2(l) + K2SO4(aq) + KCl(aq) + 5,5-dimethylhydantoin(s)
Using this exact stoichiometry is very important else you get other pathways reduce the yield of Br2.
If you want only the Br in 1-Bromo-3-Chloro-5,5-dimethylhydantoin you could reduce it with other reductants instead of bromides, but then you would
obtain only half mol Br2 per one mol 1-Bromo-3-Chloro-5,5-dimethylhydantoin. For example, like you did with NaNO2, forming NaCl, NaNO3 and Br2.
Sulfites and similar reducing agents would also work. However, like I said, it would be a waste of the oxidative power to do so while
the reagent can be used to oxidize bromides instead.
[Edited on 7-2-2010 by entropy51] |