November 22, 2024, 07:37:06 PM
Forum Rules: Read This Before Posting


Topic: 2 problems  (Read 27405 times)

0 Members and 2 Guests are viewing this topic.

Offline Heory

  • Full Member
  • ****
  • Posts: 175
  • Mole Snacks: +17/-2
Re: 2 problems
« Reply #30 on: February 11, 2010, 10:25:26 AM »
Oxy, I like your initial idea better. Base is not present according to the problem. I think the major product depends on the confirmation of the carbamate.

Offline Oxy

  • Regular Member
  • ***
  • Posts: 17
  • Mole Snacks: +2/-0
Re: 2 problems
« Reply #31 on: February 11, 2010, 10:52:40 AM »
After stewie griffin posted the document and due to calculation of Chem3D, I'm not sure about the first idea. I'm try finding a more reasonable way. Base is not required if proton isn't lost. Hmm... Heory, could you get the answer?

Offline Heory

  • Full Member
  • ****
  • Posts: 175
  • Mole Snacks: +17/-2
Re: 2 problems
« Reply #32 on: February 11, 2010, 07:21:32 PM »
No I don't have the answer, nor would I have the chance to get it.

Offline tmartin

  • Full Member
  • ****
  • Posts: 114
  • Mole Snacks: +15/-2
Re: 2 problems
« Reply #33 on: February 12, 2010, 07:36:15 AM »
 :-\   I really got my hopes up waiting for the answer too... now we'll just have to argue these points over and over again like Groundhogs Day!  :P

Offline Oxy

  • Regular Member
  • ***
  • Posts: 17
  • Mole Snacks: +2/-0
Re: 2 problems
« Reply #34 on: February 13, 2010, 10:47:59 AM »
I asked a friend for help :P. Here is his solution.
Due to inductive effect, the C-Br sigma* MO (b) has lower energy than the other one.
The lone pair (1) interact with C-O sigma* MO better than the other one and lower its energy.
So energy between lone pair (1) and C-Br sigma* MO (a) is larger than energy between lone pair (1) and C-Br sigma* MO (b).
Conclusion, path A is preferred.

Sponsored Links