Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: mir on May 21, 2006, 08:49:30 AM
-
I am studying radical cyclization reaction of anilidyl radicals. The term "oxidative entry" is often showing up in the litterature, but I cant find a definition of it. What is it?
-
Strange, since there is no oxidation in that step. I would just call it a radical-olefin cyclization. Do you have any other examples where they call it oxidative?
-
The carbon on the right hand side of the double bond did start at an oxidation state of 0 then stayed 0 in the radical intermediate and finally became +1 in the final product. So, it was oxidized. Its a pain to view organic mechanisms that way, but it can sometimes help.
-
Didnt thought of that. Thank you :-)
Radical reactions is pretty interesting. maybe I take 10sp in radical chemistry next year.
-
The carbon on the right hand side of the double bond did start at an oxidation state of 0 then stayed 0 in the radical intermediate and finally became +1 in the final product. So, it was oxidized. Its a pain to view organic mechanisms that way, but it can sometimes help.
On an olefin you count one of the carbons as in the alcohol oxidation state and one in the alkane oxidation state, right? I guess I don't see the oxidation of the carbon on the right of the alkene. If you count the one on the left as in the alcohol oxidation state, then it stays the same and the one on the right stays in the alkane oxidation state.
Maybe it's just the way crazy organic chemists glaze over formal oxidation states....
To me it seems that the oxidation really happens when the N-S bond breaks since the S takes an electron from N. The hydride abstraction at the end is a reduction, if anything, right?