November 01, 2024, 03:39:19 PM
Forum Rules: Read This Before Posting


Topic: formation of 4-oxo-2-hexenal  (Read 2872 times)

0 Members and 1 Guest are viewing this topic.

Offline AlphaScent

  • Full Member
  • ****
  • Posts: 644
  • Mole Snacks: +24/-7
  • Gender: Male
formation of 4-oxo-2-hexenal
« on: June 02, 2014, 03:13:44 PM »
This should be very simple.  I am having a tough time with part of it.  The formation of the ketone is stumping me.

Is it at all like my attempt or does water attack and then the hemi-acetal opens the ring through a removal of the proton forming the ketone and aldehyde.

So simple and I just cant make it make sense...
If you're not part of the solution, then you're part of the precipitate

Offline Dan

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 4716
  • Mole Snacks: +469/-72
  • Gender: Male
  • Organic Chemist
    • My research
Re: formation of 4-oxo-2-hexenal
« Reply #1 on: June 02, 2014, 03:36:57 PM »
You've skipped too far ahead with your bromination.

What is the full mechanism for the bromination and why have you brominated at that position?
My research: Google Scholar and Researchgate

Offline AlphaScent

  • Full Member
  • ****
  • Posts: 644
  • Mole Snacks: +24/-7
  • Gender: Male
Re: formation of 4-oxo-2-hexenal
« Reply #2 on: June 02, 2014, 04:34:42 PM »
I am writing the mechanism out and it hits me.  I do not exactly know which position to brominate. What makes sense is that when the double bond is formed thourgh E2 elimination at the 2 position, the proton closest to the carbonyl (aldehyde) would be the most labile. 

Also, this is enol ether chemistry.  Which I think there is the hydrolysis of enol ether going on.  I have to look some more.  Ill post what I have and we can go from there.
If you're not part of the solution, then you're part of the precipitate

Offline AlphaScent

  • Full Member
  • ****
  • Posts: 644
  • Mole Snacks: +24/-7
  • Gender: Male
Re: formation of 4-oxo-2-hexenal
« Reply #3 on: June 02, 2014, 04:51:13 PM »
The pyridine is there to scavenge the HBr produced from the formation the radical.

THF and acetone are solvents. 

The HBr and water can also complete the acid catalyzed hydrolysis of an enol ether.  This would help with the formation of the compound.

Slowly but surely....
If you're not part of the solution, then you're part of the precipitate

Offline Dan

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 4716
  • Mole Snacks: +469/-72
  • Gender: Male
  • Organic Chemist
    • My research
Re: formation of 4-oxo-2-hexenal
« Reply #4 on: June 02, 2014, 06:36:24 PM »
Since light or a radical initiator are not given in the question, I would not assume a radical pathway. Google furan bromination.
My research: Google Scholar and Researchgate

Offline AlphaScent

  • Full Member
  • ****
  • Posts: 644
  • Mole Snacks: +24/-7
  • Gender: Male
Re: formation of 4-oxo-2-hexenal
« Reply #5 on: June 03, 2014, 09:49:05 AM »
So I think where I went wrong is immediately assuming since it was NBS that it had to be allylic bromination through a radical mechanism.  But I started thinking (with a nudge from Dan) that NBS is just a constant, low concentration source of bromine.  So I thought about it and after some google help, a article said the bromination of furan happens through a 1,4-addition type mechanism.  Attached is the second half of the mechanis, then the bromination mechanism.
Does this seem more logical? Does to me.

Thanks Dan
If you're not part of the solution, then you're part of the precipitate

Offline discodermolide

  • Chemist
  • Sr. Member
  • *
  • Posts: 5038
  • Mole Snacks: +405/-70
  • Gender: Male
    • My research history
Re: formation of 4-oxo-2-hexenal
« Reply #6 on: June 03, 2014, 11:38:59 AM »
Initially it might be better to invoke the oxygen lone pair to move the electrons around.
Development Chemists do it on Scale, Research Chemists just do it!
My Research History

Offline AlphaScent

  • Full Member
  • ****
  • Posts: 644
  • Mole Snacks: +24/-7
  • Gender: Male
Re: formation of 4-oxo-2-hexenal
« Reply #7 on: June 03, 2014, 12:20:17 PM »
I think that is even better.  Thanks Disco!!
If you're not part of the solution, then you're part of the precipitate

Sponsored Links