JefferyH - 17-9-2014 at 11:46
I have an aromatic aldehyde contaminated with maybe 1-2% its weight in halogen.
Can I add just enough bisulfite, with some forming the aldehyde adduct, which may react with the halogen forming sodium hydrogen sulfate and forming
the hydrohalic acid which I can then neutralize with a base?
Or does the bisulfite adduct not react with the halogens, meaning I will need to add a stochiometric ammount of bisulfite and then neutralize it all
with a base to reobtain my aldehyde?
[Edited on 17-9-2014 by JefferyH]
bbartlog - 17-9-2014 at 12:43
I would expect the oxidization of the bisulfite by the halogen to be the primary reaction, and the bisulfite adduct to be formed only by whatever was
left. And yes you would presumably want to neutralize the acid, maybe with bicarbonate (no sense using strong base).
However, I question your description of the situation: wouldn't an aldehyde be oxidized fairly rapidly if contaminated with unreacted halogen? How do
you know this is what you have?
JefferyH - 17-9-2014 at 14:05
I performed the oxidation with the classical Oxone + NaBr. The original write up calls for a stochiometric amount of bromide, but it seems a catalytic
amount does fine as well. It operates in a cycle. Oxone oxidizes NaBr to form Hypobromous Acid, some of this reacts with NaBr to form bromine, of
which is in equilibrium with the hypobromous. There is a red color that forms, and the red color disappears as the alcohol is oxidized. I actually
remedied the situation by adding in just enough alcohol to make the rest of the halogen color disappear, so I need not to worry about the bisulfite
addition for now. The yields seem to be fantastic from what I've noticed.... 95%+ at least.