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Offline Heory

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2 problems
« on: February 08, 2010, 06:19:32 AM »
I have no idea about these problems. Could anybody do me a favor? Thanks.

1. The molecule illustrated below can react through either Path A or Path B to form salt 1 or salt 2 . In both instances the carbonyl oxygen functions as the nucleophile in an intramolecular alkylation. What is the preferred reaction path for the transformation in question?

2. Explain the "abnormal" diastereoselection of the reaction.


Offline cpncoop

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Re: 2 problems
« Reply #1 on: February 08, 2010, 09:47:15 AM »
I think that the answer to question 1 can be solved by reviewing Baldwin's rules for ring closure:

www.indiana.edu/~msvlab/.../Baldwins%20Rules.pdf

Offline Heory

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Re: 2 problems
« Reply #2 on: February 08, 2010, 10:06:23 AM »
I think that the answer to question 1 can be solved by reviewing Baldwin's rules for ring closure:

www.indiana.edu/~msvlab/.../Baldwins%20Rules.pdf

I don't think it can be solved by Baldwin's rules.  :)

Offline azmanam

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Re: 2 problems
« Reply #3 on: February 08, 2010, 11:29:54 AM »
on first glance, I was going to say part a was neighboring group participation, but now I'm not so sure...
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Offline stewie griffin

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Re: 2 problems
« Reply #4 on: February 08, 2010, 01:39:17 PM »
azmanam I don't think neighboring group participation is too far off. Part one seems eerily like a nitrogen mustard (See http://en.wikipedia.org/wiki/Nitrogen_mustard and http://en.wikipedia.org/wiki/Sulfur_mustard). Since N is more nucleophilic than O, it would make sense to form the three membered ring with N, then have the carbonyl oxygen reopen the three membered N-containing ring.

Offline Heory

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Re: 2 problems
« Reply #5 on: February 08, 2010, 09:19:29 PM »
azmanam I don't think neighboring group participation is too far off. Part one seems eerily like a nitrogen mustard (See http://en.wikipedia.org/wiki/Nitrogen_mustard and http://en.wikipedia.org/wiki/Sulfur_mustard). Since N is more nucleophilic than O, it would make sense to form the three membered ring with N, then have the carbonyl oxygen reopen the three membered N-containing ring.

It does look like a nitrogen mustard but after all it's an amide, so the nucleophilicity of the nitrogen atom is much poorer than that of the one of a nitrogen mustard. I think the oxygen atom of the amide would act as a nucleophilic reagent directly without formation of an azirdine or an oxazolidone (another Br kicked out). I want to find out in which path in the TS the lone pair of the amide oxgen overlaps the anti C-Br sigma bond better, which would solved this problem, but I don't know how, since I don't know the bond lengths and the bond angles.


Offline tmartin

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Re: 2 problems
« Reply #6 on: February 08, 2010, 09:49:27 PM »
For #1 Heory's last structure made me think of the preferred confirmation of esters due to the lone pair.  Even if neighboring group participation does not play a huge role, I would think it could still offer a slight rate increase to favor path B

#2 is interesting, is there a given Lewis acid or is it general?

Offline Heory

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Re: 2 problems
« Reply #7 on: February 09, 2010, 07:55:41 AM »
LA is general

Offline stewie griffin

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Re: 2 problems
« Reply #8 on: February 09, 2010, 07:56:20 AM »
Two things:
1) I don't think the answer comes from bond lengths and angle strain. Both the nitrogen and oxygen are sp2 so they have approximately the same angles. And I just don't feel that the small difference in bond lengths is going to have any significant influence here.
2) I think the N is more nucleophilic than you would expect. Sure an amide is less nucleophilic than an amine due to resonance. Here we have a carbamate though... so both the N and O are donating electron density into the carbonyl. I would argue therefore that the N is doing less resonance in a carbamate than it does in an amide, and thus the N is a fine nucleophile here. See Tetrahedron Letters 2001, 42, 1799  for an example of a mild N-alkylation of carbamates with alkyl halides.

Offline tmartin

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Re: 2 problems
« Reply #9 on: February 09, 2010, 08:08:07 AM »
For #2 my initial thinking would be the Lewis acid is blocking the top face of the enone, so the nucleophile comes in on the same side as the other substituent.

Also, my earlier comment about ester geometry probably isn't going to be a big factor in this problem, and stewie has won me over to the anchimeric assistance argument.

Offline azmanam

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Re: 2 problems
« Reply #10 on: February 09, 2010, 08:12:17 AM »
what class is this for and what chapter are you studying (and from what book)?

I know I've seen problem 1 before... but I just can't place it.  It's driving me crazy.  Maybe with some context, I'll remember and be able to help better.
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Offline Heory

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Re: 2 problems
« Reply #11 on: February 09, 2010, 08:16:46 AM »
On thinking furthur I agree with you more than before. I've no access to the data base so could you show me a picture of the reactions in the literature?

Offline Heory

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Re: 2 problems
« Reply #12 on: February 09, 2010, 08:20:17 AM »
These problems are from lectures of Evans' group.
See the attachment.

Offline azmanam

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Re: 2 problems
« Reply #13 on: February 09, 2010, 08:47:53 AM »
yes!  that's where i'd seen it.  Of course Evans doesn't give the answer.  Whew.  You have no idea how crazy it was making me.  I had all my lecture notes from every grad class I'd ever taken, plus both volumes of Carey & Sundberg, plus March.  I'll have to think about it more, now.
Knowing why you got a question wrong is better than knowing that you got a question right.

Offline Oxy

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Re: 2 problems
« Reply #14 on: February 09, 2010, 10:53:39 AM »
I think that:

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