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Second intermediate reaction
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Topic: Second intermediate reaction (Read 7382 times)
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Lithium
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Second intermediate reaction
«
on:
July 30, 2008, 07:04:13 PM »
I wanted to know if someone could show me the mechanism for this problem as my txt doesnt clearly define what the first second, and third intermediates are(question 14) and finally for question 16 im torn between d and b. I was informed it was either d or e can someone provide me some good feed back on this? Thanks
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azmanam
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Re: Second intermediate reaction
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Reply #1 on:
July 31, 2008, 11:17:09 AM »
According to this page (
http://www.cem.msu.edu/~reusch/VirtTxtJml/special1.htm#birch
), it seems the second intermediate is the neutral radical after the radical anion has been protonated. The site explains that electron donating groups (like amines) cause initial protonation at the meta position to the EDG, and second protonation is always para to the first protonation.
Thus, A will end up being the product, but I say the answer is E. The first intermediate is the radical anion with the anion meta to the amine, and the radical para to the anion (and ortho to the amine). The second intermediate should be the protonated, neutral radical (see below) with the radical ortho to the amine. The third intermediate should be the anion of that structure, and the product will be the reduced diene.
edit: wow that picture is huge. sorry for the size...
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azmanam
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Re: Second intermediate reaction
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Reply #2 on:
July 31, 2008, 11:25:24 AM »
(i think 16 will be b). I don't see an answer e.
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Lithium
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Re: Second intermediate reaction
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Reply #3 on:
July 31, 2008, 11:35:51 AM »
Thanks for the help as far as 16 e is none of the above. I was informed that it could be D because it gets flattenout can that be true?
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azmanam
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Re: Second intermediate reaction
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Reply #4 on:
July 31, 2008, 11:37:43 AM »
Quote
it could be D because it gets flattenout
what do you mean by that?
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Lithium
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Re: Second intermediate reaction
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Reply #5 on:
August 01, 2008, 11:24:42 AM »
Can I get more feed back im getting told by some C is the second step and D is the second step for ? 14 and now e. How sure are you of that answer as far as accuracy?
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azmanam
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Re: Second intermediate reaction
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Reply #6 on:
August 01, 2008, 11:27:45 AM »
Not 100%. I'm extrapolating based on the information at the linked site. If anyone else has insight, I'm sure it would be helpful. Or ask your instructor.
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macman104
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Re: Second intermediate reaction
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Reply #7 on:
August 01, 2008, 11:44:33 AM »
It's a stupid question. Depends on how far you break up the intermediates. For example the site azmanam used shows 2 intermediates (radical anion, anion), but organic-chemistry.org has 3 (radical anion, neutral radical, anion)
http://www.organic-chemistry.org/namedreactions/birch-reduction.shtm
I do not believe it's C, that should not be a valid intermediate in the process.
So I guess I'd go with E as well. And I agree the final product would be A.
Do you have notes that you took when you were going over the birch reduction? How did your professor show the process?
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Lithium
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Re: Second intermediate reaction
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Reply #8 on:
August 01, 2008, 12:57:06 PM »
No my professor never went through the mechanism!!! we use the following as our txt
http://www.ochem4free.info/node/143
its a free organic txt online. I think it talks about it on chapter 20 or 19. I just cant make sense of the mechanism.
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macman104
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Re: Second intermediate reaction
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Reply #9 on:
August 01, 2008, 03:15:08 PM »
I can't access the textbook. It isn't free, but you must have a login that your school provides you, which I'm sure they pay for (I wouldn't advise giving us the login either so we can see the book, just in case you thought of doing that).
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