Author Topic: questions about the Acid synt from Elymoclavine  (Read 2605 times)

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flipper

  • Guest
questions about the Acid synt from Elymoclavine
« on: May 11, 2002, 03:04:00 PM »
I wanna asked some thing about this new LSD synt on Rhodiums page.

LSD synt from Elymoclavine

(http://www.public.asu.edu/~junkie/lsd.htm)

Question 1:

What is "Elymoclavine"? Is it easy too aquire?

Question 2:

I'm really bad in redox reactions. I just don't understand how to see which electrons from which atom is moved too whatever other atom. In my eyes the Elymoclavine is reduced on the bond between the the O and the H and the bond between the C and the H so the loose electrons can form a new bond and the two H atoms can form H2 gas. I must be wrong because the text says that the Elymoclavine is oxidized. Maybe it is, otherwise there are two electrons missing. As you can see I don't understand shit of it So I wanna asked somebody if he can show me the Half-reactions from step 1 till the last step. ( I also don't understand the reduction with Zink)

Question 3:

Are the tradional solvent's used in this reaction? (Benzene,Toluene,Xylene an Methanol((step 1)) ).

Thanks  ;D

You've gotta love me.

lugh

  • Guest
Clavine Alkaloids
« Reply #1 on: May 11, 2002, 06:49:00 PM »
It's been found as a minor constituent in Rivea corymbosa and Ipomoea tricolor; various companies sell it  :)

Lilienthal

  • Guest
That synthesis scheme is a highly speculative ...
« Reply #2 on: May 11, 2002, 07:10:00 PM »
That synthesis scheme is a highly speculative draft. I'm pretty sure the first and the last step don't work.
And as you can see the end product in that scheme is not LSD, the double bond is in the wrong position.

Lilienthal

  • Guest
In organic chemistry you are not counting ...
« Reply #3 on: May 12, 2002, 08:10:00 AM »
In organic chemistry you are not counting electrons or write down half reactions...  :) . You only have to remember the following two rules. From the left to the right it's a reduction, from the right to the left it's an oxidation:

triple bond __> double bond __> single bond
acid / ester / amide __> aldehyde / ketone / imine __> alcohol / amine __> alkane

That's all (at least for 99% of all organic reactions  :) ).

flipper

  • Guest
saying thanks
« Reply #4 on: May 12, 2002, 01:09:00 PM »
Thanks.  :)   ;D   ;)
I asked the question about the electrons to find out what the ratio should be, if that's possible. I don't know because I always fucked it up.
Is it possible?  ::)  If not how do I find out what the ratio should be?
 

You've gotta love me.

Rhodium

  • Guest
If you look carefully at the bottom of the ...
« Reply #5 on: May 12, 2002, 09:23:00 PM »
If you look carefully at the bottom of the Elymoclavine document, you will see that a lot of REFERENCES is provided, where all of the things you are asking about can be read.

Lilienthal

  • Guest
Ok, I looked up the reference today (Tet. Lett.
« Reply #6 on: May 14, 2002, 07:10:00 AM »
Ok, I looked up the reference today (Tet. Lett. 19, 1627 (1977) T.-C. Choong, H. R. Shough). And that's indeed what they did!

I just wonder why this article is only partially indexed in Beilstein, while the next article in that volume is completely in there  :o .

You can't calculate the amount of activated manganese dioxide in this kind of oxidations. Remember that you are using a suspended solid MnO2. A rule of thumb is to use 10 to 20 equivalents by weight.

BTW, the second reaction is cyanide-catalyzed  :) .