[AlphaScent]

Attched is a scheme for a reaction that I am going to have to do.  It is a step where I do not really have any idea of what is going on.  In talking with my boss, he says it forms a ketene (R-C=C=O).  I know this can go through a Diels-Alder type cycloaddition.  But where and how is the ketene formed?  My first guess is that it has to do with the olefin acting nucelophilically.  Am I wrong, I do not know.  Any nudging or insight would be greatly appreciated. 

Cheers

[discodermolide]

You make the acid anhydride, then eliminate to form R-C=C=O which does a [2+2] cycloaddition with the C=C.

Usually it is done with the acid chloride. I don't have time to go through in detail ATM, sorry.

[AlphaScent]

Yea I said mixed anhydride to him and he said no.....ok that helps, thanks disco.  Ill post a mechanism when I have one and you can critique it.

[discodermolide]

On iPhone , it probably forms a cyclic one with the tertiary alcohol, which eliminates to give the olefin, ketene and CO2.

[AlphaScent]

No worries.  Not life or death.

That got me thinking.  An acetate is not a bad leaving group do a base, potassium acetate, removes a proton and forms the ketene and the acetate leaves.  To be honest, that way of thinking is suspect to me.  The hydroxyl must be removed, but if it leaves as water it takes the electrons with it.  Attached is my thought process.  Still working.

[AlphaScent]

I understand that 5 bonds are formed, not gonna happen. 

[discodermolide]

I think this forms:


which then looses acetate to give the ketene

which does the cycloaddition


and  the C=C migrates into the ring to give the product. (note I did not add the chiral flag!)

[orgopete]

I think this forms:


which then looses acetate to give the ketene

which does the cycloaddition


and  the C=C migrates into the ring to give the product. (note I did not add the chiral flag!)

I can see the acetate forming and the mixed anhydride, but I'm not seeing the rest. If it does a 2+2 addition, I'm not getting how the product forms.

[discodermolide]

I re-drew it!
Done in refluxing acetic anhydride the double bond will surely migrate?

[orgopete]

Yup, I just wasn't seeing it. No migration needed, just eliminate to double bond of product.

[AlphaScent]

I figured the hydroxyl was going to have to also react with the acetic anhydride.  So the elimination of the acetates has to do with a base (potassium acetate?) removing protons to allow for the elimination of the acetate groups and the formation of the olefins.  That is the only part still cloudy.  The rest, including the bond migration to a more substituted position makes sense to me.

attached is what I am thinking.  Just shaky on the how.

[discodermolide]

What you drew looks ok to me. The K acetate acts as a base as you said. The rest is history:)

[orgopete]

... elimination of the acetate groups and the formation of the olefins.  That is the only part still cloudy.  The rest, including the bond migration to a more substituted position makes sense to me.

Why do you think the elimination gives the least substitution and then migrates to the most substituted with potassium acetate? Why not a simple elimination to the most substituted?

[AlphaScent]

I honestly sir, connot give a good reason why.  I think my drawing is wrong now that you mention it.  It would most eliminate at the 4 position, Giving the tri-substituted olefin. 

If there is a reason for it to go from the least substituted to the most substituted, I am not seeing it.

[zsinger]

Orgopete-
Are you thinking bond angle reasons?
        -Zack