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zbde00
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[*] posted on 19-2-2005 at 01:24
how to make this ester


My substrate is
[img]http://ccebbs.com/forum/showimg.asp?id=9323[/img]
I want to make a phosphoric ester at only the 3-OH.
How can i do it?
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zbde00
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[*] posted on 19-2-2005 at 01:27


the picture cann't be seen correctly here.
you can see it in this website
http://ccebbs.com/forum/dispq.asp?lid=9323
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[*] posted on 19-2-2005 at 06:58


Oh no I can't.
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chloric1
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[*] posted on 19-2-2005 at 08:45


That kind of claritiy can only compare to the clarity one experiences after a long night of binge drinking. LMAO!



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BromicAcid
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[*] posted on 19-2-2005 at 09:37


Since I can't read Chinese I can only assume you mean gallic acid or methane/carbon tetrachloride which appears in the banner. Since neither are esters you probably mean something in the text of the page, a text which I cannot read, I'm afraid that not many people here speak Chinese, if you want a better response please provide a formula.

[Edited on 2/19/2005 by BromicAcid]




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runlabrun
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[*] posted on 20-2-2005 at 01:45


i think the refernce to the page is for the molecule picture....
Its 3,4,5-trihydroxycyclohex-1-enecarboxylic acid.
So he is refering to the OH group on position 3... making the phosphoric ester at this position...

Ok, now thats sorted i must say i dont know how you would do this.... but now that the whole substrate and chinese text is sorted someone else may be able to help with the ester formation.

-rlr
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[*] posted on 20-2-2005 at 06:34


"That kind of claritiy can only compare to the clarity one experiences after a long night of binge drinking. LMAO! "

At 3 in the afternoon? OK, it was Saturday so I had been out drinking the night before, but I can usually manage to muster that degree of clarity anyway.
(For what it's worth, when I looked at that page it was incomprehensible, and not even recognisable as Chinese, nor did the structures show up.)
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zbde00
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[*] posted on 21-2-2005 at 05:27


thank everyone.
the reference is only the picture.


I'm a chinese. I'm very glad to meet with everyone in the forum.
Welcome to China.
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zbde00
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[*] posted on 21-2-2005 at 05:33


runlabrun is correct.
I will try my best to draw a new picture in a new post.
thank everyone
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zbde00
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smile.gif posted on 21-2-2005 at 05:52
how to make this ester in moderate yield


how to make this ester in moderate?
see the picture in the attachment

untitled.gif - 2kB
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sparkgap
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[*] posted on 21-2-2005 at 06:11
Proposed synthesis


I'm not sure about the correctness of this, but it looks good to me.

Substrate --(CH<sub>3</sub>;)<sub>2</sub>CO + H<sup>+</sup>--> acetal

acetal --P<sub>2</sub>O<sub>5</sub>--> phosphate

phosphate --H<sub>3</sub>O<sup>+</sup>--> desired product

I used non-OTC methods since I don't think the substrate would be OTC anyway. :D

(edit: the only side product I can see forming is the isomer phosphorylated at the -OH opposite the double bond.)

sparky (^_^)

[Edited on 21-2-2005 by sparkgap]




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zbde00
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[*] posted on 21-2-2005 at 06:32


Thank sparkgap.
The method you proposed is good.
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alchemie
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[*] posted on 22-2-2005 at 16:10


Quote:
Originally posted by sparkgap
Substrate --(CH<sub>3</sub>;)<sub>2</sub>CO + H<sup>+</sup>--> acetal


Why wouldn´t the 3-OH react with the acetone to form mixed acetals?


Another approach would be to oxidize selectively the 3-OH.

Several reagents accomplish this task, I'll give a couple of examples:

-K2MnO4 under fase-transfer conditions oxidizes allylic but doesn't touch primary or secondary alcohols. Ref: Tetrahedron Letters, 30, 19 , 1989.

-NaNO2 and Ac2O oxidizes primary, allylic and benzylic but not secondary alcohols. http://pubs.rsc.org/ej/P1/2000/B001215G.pdf

Then react with (CH<sub>3</sub>;)<sub>2</sub>C(OCH<sub>3</sub>;)<sub>2</sub> to form the acetal, reduce to the alcohol and phosphorylate.

[Edited on 23-2-2005 by alchemie]

[Edited on 23-2-2005 by alchemie]

[Edited on 23-2-2005 by alchemie]

[Edited on 23-2-2005 by alchemie]
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alchemie
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[*] posted on 22-2-2005 at 17:04


see also scheme 2 in this page:

http://www.albmolecular.com/features/tekreps/vol04/no48/

They selectively protect a diol as an acetonide in the presence of an acrolein (wich you would obtain from oxidation of an allylic alcohol).
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sparkgap
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[*] posted on 23-2-2005 at 00:18


My reasoning in assuming no (or more precisely, very little) mixed acetals would form was because of entropy considerations.

But I must admit your approach is more elegant. :D

sparky (^_^)




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alchemie
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[*] posted on 23-2-2005 at 11:53


sparkgap:

The way I suggested is pretty much more complicated than your's.

If the two secondary alcohols form the acetal preferentialy then your's is much better. I tried to look up something about alcohol reactivity towards acetal formation but didn´t find anything.

Another problem in both approaches is the carboxylic acid. Maybe it stays there and just watches;), or maybe it has to be protected as well.
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[*] posted on 24-2-2005 at 06:25


thank alchemie,sparkgap
you shed light on my question...
I learn much from your posts.
Do you have some writings about learning organic chemistry?
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sparkgap
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[*] posted on 24-2-2005 at 06:45


zbde00:

Very much welcome. :)

Best way to learn organic chemistry? The only advice I can give is to read books on it. McMurry and Carey are pretty good in explaining the basics. If you deem yourself ready to tackle advanced organic chemistry, you might want to check out March.

About tackling organic synthesis, Warren writes good. I can't for the life of me comprehend Corey. :(

Oh, and yes, practice makes near-perfect. :D

alchemie:

I gave my superior this question the other day. She told me there is no use hoping for chemoselectivity since they most likely have equal reactivities. Thus, forming the acetal results in two possible products, one to be phosphorylated subsequently, and the other to be separated out.

She also said the carboxylic acid is not very likely to interfere. :D

sparky (^_^)




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