[MalchiorX]

I need some help on a mechanism.
I'm pretty sure it's either E2 or E1cb. I can get the alkene, but I'm missing a hydrogen and i need to break the furanone ring somehow. Help is appreciated.

[reflux]

Is the iodine supposed to be axial in that structure?  If so it can't align itself antiperiplaner with the adjacent H required to form the product shown via E2.

[MalchiorX]

The professor is out of town, and the other organic professor isn't sure. I'm pretty sure the size of the base, and the fact it is a polar, aprotic solvent is driving the reaction to be more E1cb like. I get the same result using either one, what I don't get is the ring opening. Or maybe I'm approaching the whole problem wrong.

[MalchiorX]

Also I can basically say that since it's hard to tell if its axial or not, I just picked it to be in a one position or the other and committed to it.

[azmanam]

I was looking at this yesterday.  I was immediately reminded of a total synthesis example from my 1st year advanced organic class - Nicolaou's synthesis of endiendric acid b.  He used a formal halohydrin formation to 'protect' one alcohol based on proximity to the intermediate iodonium ion.  The other alcohol could be selectively protected as a silyl ether, then the halohydrin was reverted to the alkene and alcohol such that the now free alcohol could be manipulated selectively (see top below).

Maybe something like that is going on here?  The iodide could kick down to form the iodonium ion - and expel the carboxylate leaving group?  Coupla things based on previous responses -  I don't think this is a straight forward E2, thus I don't think the I and H need to be antiperiplanar.  If they were, and it did go through an E2, the lactone would remain intact.  The fact that the lactone breaks into the carboxylate is telling.  If the stereochemistry is as you've drawn, this would support the iodide kicking down, forming the iodonium ion, and releasing the carboxylate (see bottom below).  I just can't get from the iodonium ion to the alkene.  Are there any other reagents or steps besides pyridine?  what exactly is the solvent?

[macman104]

It's not an E2, because you can't be achieve the anti-coplanar transition state.

E1 seems likely because the electron-withdrawing oxygen will stabilize the carbocation, and it's also 3°.

You are correct about the second part though.  You can protonate your C=O bond, and put a positive charge on the carbon, but then I'm not sure where you would go from there.  No matter how I draw it, you are short a hydrogen from somewhere.

Are you sure there aren't any other reagents in the mechanism, that it is run in neat pyridine?  Or even more, maybe it's pyridine hydrochloride, so it's actually an acid-catalyzed mechanism?

Azamanam, really interesting thought, I like that synthesis technique as well.  I also cannot get past the iodonium ion.  There has to be a missing piece of information.

[MalchiorX]

Unfortunately I have included all the information I have. I like the idea of that iodonium ion. My answer involved using E1cb to get the alkene and then having a pyridinium ion attack the keto oxygen. Electrons would shift to fill octets and that would break the C-O bond. Then I had pyridine attack the carbocation left on the alkene forming a positive complex. Then the oxygen with negative charge attacks the pyridine complex leaving another carbanion at the alkene. Then the carbonanion sucks up the hydrogen. Then iodide ion attacks the pyridine complex leaving product and an iodopyridinium complex. I'll see if I can draw out my answer.

[MalchiorX]

Here is the mechanism i proposed

[reflux]

Yeah I initially thought of forming an iodinium intermediate as well... but then you would likely form an allyliodide and thats not mentioning the problems with regiochemistry.  (This kind of transormation can be down, however, with a little Zn/AcOH.)

Can thermolysis of the C-I bond lead to a carbocation (let's lable that C1)?  A hydride shift could lead to the + on the carbon attached to the oxygen (an oxonium ion, the type of which I've never encountered).  Then you could deprotonate at C1 (where the C-I bond was) to kick out the carboxylate and form the double bond.

[MalchiorX]

I think that will still result in missing electrons.

[reflux]

Oops.. realized my mechanism wouldn't work as a drew it out on paper!  I guess one should begin drawing before typing!