When first three chemicals (K2MnO4+HNO3+NaOH) were mixed with each other (200 µL H2O, 5 µL 10 M NaOH, 5 µL 0.01 N HNO3, 5µL 1 M K2MnO4) there was no color change. But when I added sucralose (200 µL H2O, 5 µL 10 M NaOH, 5 µL 0.01 N HNO3, 5µL 1 M K2MnO4, 5 µL 1M Sucralose) color of K2MnO4 was changed. Do you have any idea what kinds of reaction product can I get from the reaction?
Given the structure of sucralose, you would most likely get competitive oxidation of the 5 free hydroxy groups to give you a complex mixture of products, many of which may further eliminate chlorine via an E1cb mechanism. This may be further complicated by SN2 displacement of chlorine / michael addition reactions or loss of water.
Sucralose is a chlorinated sugar analog containing 2 diol groups and CH2OH group. They can be easily oxidized under you conditions. For example, CH2OH COOH, diol dialdehyde which can be further oxidized. Manganate will be reduced, perhaps to MnO2 or even to Mn(+2).
Thanks!
@ Dubey & Eric, There was precipitation when reaction was completed. What could be the precipitation? Was that potassium nitrate salt? or any idea?
Goal was to remove chloride from Sucralose by a easy chemical reaction?
Under basic conditions K2MnO4 is likely reduced to MnO2, which would be a brown precipitate.
Boiling it in a large excess of KOH should get rid of chlorine. Either by the formation of cyclic ethers or Sn2 displacement.
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