5
$\begingroup$

In 1979, Karel Frantisek Wiesner performed a tremondous total synthesis in $59$ steps of the 13-desoxydelphonine. In one of those steps there is a very tricky Wagner-Meervien rearrangement. I tried to find something by myself, I also look on the net, I found this and you can see page $8$ they have the same problem than me.

What is the mechanism of this rearrangement ?

Wagner-Meervien Rearrangemeent

Just to be a bit more clear, the acetate which looks to appear magically obviously don't, it comes from the previous step because of the solvant, they just heat several days after so the solvant is still the same as you can see in the picture below. enter image description here

If someone has an idea, I would be glad! Thank you in advance :)

Improve this question
$\endgroup$
4
  • 4
    $\begingroup$ Something is missing here. There's an extra acetate in the product. Is glacial acetic acid the solvent? $\endgroup$
    – Zhe
    Commented Nov 20, 2016 at 15:11
  • 1
    $\begingroup$ Also, if you look at the reaction on page of your link. The structure after this rearrangement is the same as the one after reacting the intermediate with CAN and acetic acid. That's probably where the acetate comes from, but it also means that the structure of the final product is wrong. I recommend looking at the original reference for what the actual product is. $\endgroup$
    – Zhe
    Commented Nov 20, 2016 at 15:15
  • $\begingroup$ @Zhe if you look carefully the structures are not the same, a methyle is transform as aldehyde with CAN and AcOH. $\endgroup$
    – ParaH2
    Commented Nov 20, 2016 at 15:30
  • $\begingroup$ Oh drat... But we're still missing an acetate. I guess the previous reaction is in acetic acid/acetic anhydride mixture. If they're just taking this pot forward, then I guess that's where the acetate comes from... $\endgroup$
    – Zhe
    Commented Nov 20, 2016 at 15:31

2 Answers 2

Reset to default
3
$\begingroup$

My literal back of the envelope solution. Notable points:

  1. I invoked a nonclassical carbocation in the first step of the rearrangement. This was the easiest way to explain the stereochemistry of the new acetate group.
  2. This is a reaction from 1959-1960. It's almost certainly racemic. I think the product as drawn is actually the enantiomer of the one that you would expect from the reactant.

enter image description here

Improve this answer
$\endgroup$
0
$\begingroup$

My bet would be the following hydride shift, followed by nucleophilic attack of the acetate: enter image description here

Improve this answer
$\endgroup$
2
  • $\begingroup$ Though I must admit that Zhe's answer is probably correct. Still weird that nonclassical carbocation. Probably possible because of the +M effect of the OMe group. $\endgroup$
    – logical x 2
    Commented Nov 20, 2016 at 16:08
  • $\begingroup$ I actually thought of something like this to start. But I couldn't get the correct stereochemistry on the acetate group, so it was back to the drawing board... $\endgroup$
    – Zhe
    Commented Nov 20, 2016 at 16:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.