Remember that HI will reduce iodides and alcohols
ROH + HI <=> RI + H2O
RI + HI => RH + I2 (the reaction that might interfere for you)
I suspect that you will have better yields by making the chloro or bromo acetone, then treating it with a solution of NaI in acetone
RCl + NaI => RI + NaCl (ppt)
Note that the iodide should be anhydrous, as water will allow the NaCl to stay in solution; its precipitation is what drives the reaction.
Sodium bromide is slightly soluble in acetone, using the organic bromide may not give quite as good yields, or require an excess of sodium iodide.
You might be able to one-pot this by chloronating an excess of acetone will marble chips to scavange the HCl formed, then add finely powdered NaI or
CaI2 and stir for a time. |