July 14, 2020, 03:59:43 PM
Forum Rules: Read This Before Posting


Topic: Problem of the Week - 6/8/09  (Read 18095 times)

0 Members and 1 Guest are viewing this topic.

Offline azmanam

  • Chemist
  • Sr. Member
  • *
  • Posts: 1417
  • Mole Snacks: +160/-24
  • Mediocrity is a handrail -Charles Louis d'Secondat
Problem of the Week - 6/8/09
« on: June 07, 2009, 08:09:51 PM »
I came across this tasty mechanism while putting together a literature group meeting on non-enzymatic desymmetrization I gave last week.  This is a three part-er.  We'll start with part 1.  After we get the mechanism, I have 2 follow up questions.

QUESTION:  Provide the mechanism for the following transformation

(there are a lot of cations in the pot... It may be easier to stick with aluminum throughout)
Knowing why you got a question wrong is better than knowing that you got a question right.

Offline nj_bartel

  • Sr. Member
  • *****
  • Posts: 1487
  • Mole Snacks: +76/-42
Re: Problem of the Week - 6/8/09
« Reply #1 on: June 09, 2009, 04:30:23 AM »
That's after work up I'm assuming?

Offline azmanam

  • Chemist
  • Sr. Member
  • *
  • Posts: 1417
  • Mole Snacks: +160/-24
  • Mediocrity is a handrail -Charles Louis d'Secondat
Re: Problem of the Week - 6/8/09
« Reply #2 on: June 09, 2009, 04:51:39 AM »
Quote
That's after work up I'm assuming?

the LAH?  no.  add LAH before workup.
Knowing why you got a question wrong is better than knowing that you got a question right.

Offline nj_bartel

  • Sr. Member
  • *****
  • Posts: 1487
  • Mole Snacks: +76/-42
Re: Problem of the Week - 6/8/09
« Reply #3 on: June 09, 2009, 05:03:09 AM »
I meant the product.  At any rate, this is way beyond my level, but this is what I could come up with.

Hydride irreversibly deprotonates the alpha proton and exits as hydrogen gas.  Reaction stalls here until addition of LAH.  LAH reduces both carbonyls.  This tri-anion molecule is stabilized due to the fact that aluminum complexes the two oxygen anions to form a 6-membered ring.  The carbanion then kicks off an ethoxide and forms an enolate.  The second ethoxide is kicked off by the carbonyl reforming.  The reformed carbonyl is then reduced by a hydride, and you're back to the 6 membered ring, now a dianion.  I'm stuck on how to get the enolate to an olefin.

This is probably all wrong though, so hey  :P  Sorry for typing it up and not drawing it out.  Time consuming and it's late.  Let me know if something doesn't make sense, and I'll draw that part out.

Offline azmanam

  • Chemist
  • Sr. Member
  • *
  • Posts: 1417
  • Mole Snacks: +160/-24
  • Mediocrity is a handrail -Charles Louis d'Secondat
Re: Problem of the Week - 6/8/09
« Reply #4 on: June 09, 2009, 07:05:14 AM »
Yeah, It'd be helpful to see it drawn out with arrow pushing and all.

I'll tell you first, though, that dianions are involved... but not trianions.

edit: and yes, the product is isolated after standard protic workup (prolly Rochelle's salt in this case to break up the aluminum emulsion... but that's not important)
« Last Edit: June 09, 2009, 07:15:22 AM by azmanam »
Knowing why you got a question wrong is better than knowing that you got a question right.

Offline spirochete

  • Chemist
  • Full Member
  • *
  • Posts: 516
  • Mole Snacks: +49/-9
  • Gender: Male
Re: Problem of the Week - 6/8/09
« Reply #5 on: June 09, 2009, 01:53:32 PM »
Is it simply deprotonation of the the diester, followed by exhaustive reduction of both esters to alcohols using LAH, followed by E1CB making a pi bond using the electron pair created in the first step?

Offline azmanam

  • Chemist
  • Sr. Member
  • *
  • Posts: 1417
  • Mole Snacks: +160/-24
  • Mediocrity is a handrail -Charles Louis d'Secondat
Re: Problem of the Week - 6/8/09
« Reply #6 on: June 09, 2009, 03:05:49 PM »
good guess, but no.  complete reduction of the esters doesn't give a diol until workup... until then they're both deprotonated alkoxides.  E1CB won't eliminate out an alkoxide (to give O2-).
Knowing why you got a question wrong is better than knowing that you got a question right.

Offline spirochete

  • Chemist
  • Full Member
  • *
  • Posts: 516
  • Mole Snacks: +49/-9
  • Gender: Male
Re: Problem of the Week - 6/8/09
« Reply #7 on: June 09, 2009, 03:11:27 PM »
Darn it.  I was thinking maybe Some of the spent aluminum could complex with the alkoxide and allow it to leave.

Offline azmanam

  • Chemist
  • Sr. Member
  • *
  • Posts: 1417
  • Mole Snacks: +160/-24
  • Mediocrity is a handrail -Charles Louis d'Secondat
Re: Problem of the Week - 6/8/09
« Reply #8 on: June 09, 2009, 03:20:09 PM »
the key (obviously) is getting rid of that one oxygen atom.  Although it seems arbitrary to choose aluminum as the dominant cation, its role will end up being important.  I think if part of nj_bartel's guess and part of spirochete's guess can work together, we might get there.  (but which part...)
Knowing why you got a question wrong is better than knowing that you got a question right.

Offline spirochete

  • Chemist
  • Full Member
  • *
  • Posts: 516
  • Mole Snacks: +49/-9
  • Gender: Male
Re: Problem of the Week - 6/8/09
« Reply #9 on: June 09, 2009, 08:25:25 PM »
Can you really make a 6 membered ring using AlH3 as a lewis acid?  I would think it could only bond to one oxygen, otherwise it would have a minus 2 charge.

Offline Dan

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 4716
  • Mole Snacks: +467/-72
  • Gender: Male
  • Organic Chemist
    • My research
Re: Problem of the Week - 6/8/09
« Reply #10 on: June 10, 2009, 05:18:46 AM »
Ok, this might be a bit silly but hey, if nobody shared their silly ideas where would we be?
My research: Google Scholar and Researchgate

Offline azmanam

  • Chemist
  • Sr. Member
  • *
  • Posts: 1417
  • Mole Snacks: +160/-24
  • Mediocrity is a handrail -Charles Louis d'Secondat
Re: Problem of the Week - 6/8/09
« Reply #11 on: June 10, 2009, 10:13:32 AM »
nope, not silly at all.  That's essentially it, in fact.  The key step is the expulsion of the aluminum alkoxide to rid the organic piece of one of the oxygen atoms.  I'll draw it below the way they drew it in the paper - it's essentially the same, just arrows in a little different order.  Aluminum was important, imho, because it can chelate well to give the 6-membered ring, and because it's fairly oxophillic allowing for the key oxygen extrusion step.

So that's part one.  Here's part two:

QUESTION (2): While the allylic alcohol is indeed the major (and desired) product, a non-trivial amount of the saturated alcohol (shown below) is formed as a minor (and undesired) byproduct.  The authors contend the mechanism which accounts for the formation of the major product cannot readily account for the formation of the byproduct.  Rather, they propose that the mechanism forks to provide an alternative mechanism for the formation of the saturated byproduct: that is, one of the intermediates in the mechanism we already have can proceed in one of two directions - the direction we have drawn to give the major product, and a second direction which leads to the byproduct. 

Provide a mechanism for the formation of the undesired saturated byproduct.


As said above, the mechanism will start the same, then at some point diverge to give the undesired byproduct.
Knowing why you got a question wrong is better than knowing that you got a question right.

Offline azmanam

  • Chemist
  • Sr. Member
  • *
  • Posts: 1417
  • Mole Snacks: +160/-24
  • Mediocrity is a handrail -Charles Louis d'Secondat
Re: Problem of the Week - 6/8/09
« Reply #12 on: June 10, 2009, 02:49:18 PM »
Quote
Can you really make a 6 membered ring using AlH3 as a lewis acid?  I would think it could only bond to one oxygen, otherwise it would have a minus 2 charge.

No, AlH3 will not chelate 2 oxygen atoms as a lewis acid.  Remember, there are 3 cations in solution (Al, Na, Li).  They are likely transmetalating throughout the mechanism.  AlH3 can coordinate to one oxygen atom.  After it loses a second hydride and becomes AlH2OR, it can then coordinate to a second oxygen atom as AlH2(OR)2-Li+.  By the time the reaction's done and worked up, it will end up as Al(OR)3.

This is a quote from the methodology paper:

Quote
For the sake of clarity, some of the species are depicted as cyclic aluminum structures although two aluminum (and/or lithium) atoms could actually be involved in a noncyclic intermediate
Knowing why you got a question wrong is better than knowing that you got a question right.

Offline azmanam

  • Chemist
  • Sr. Member
  • *
  • Posts: 1417
  • Mole Snacks: +160/-24
  • Mediocrity is a handrail -Charles Louis d'Secondat
Re: Problem of the Week - 6/8/09
« Reply #13 on: June 15, 2009, 05:07:10 AM »
I'll leave this one up this week.  Anyone up for a take at the formation of the saturated byproduct in part 2 of the question?

http://www.chemicalforums.com/index.php?topic=33985.msg130418#msg130418
Knowing why you got a question wrong is better than knowing that you got a question right.

Offline Squirmy

  • Full Member
  • ****
  • Posts: 199
  • Mole Snacks: +24/-7
Re: Problem of the Week - 6/8/09
« Reply #14 on: June 15, 2009, 12:20:56 PM »
The most logical would be the alpha-beta unsaturated aldehyde undergoing 1,4-reduction, the resulting enolate picking up a proton, and reduction of the aldehyde. Over-reduction is common for alpha,beta unsaturated aldehydes.

The question is where is the enolate picking up a proton under those conditions? Workup, I guess.

Sponsored Links