Author Topic: Solvent for oxidation to GVL  (Read 1977 times)

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Dextrose

  • Guest
Solvent for oxidation to GVL
« on: August 17, 2003, 11:37:00 PM »
Does anybee know of a solvent for oxidation to GVL?
GVL seems pricey to SWIM, so he figured that there must be some diol or other solvent, usable to the copper chromite method.

According to solventfree.kmno4.oxidation.pdf,
1,4-Pentanediol yields alpha-Methyl-gamma-Butyrolactone - would ingestion of this lactone, give an effect other than poisoning?

It is new to SWIM, that DHV also is active (according to

https://www.thevespiary.org/rhodium/Rhodium/chemistry/ghb.html

.
How about DHB? Has this one been bioassayed?

SWIM is not a skilled chemist, hence the speculation...

Thanks in advance!


Rhodium

  • Guest
You should work on your nomenclature skills...
« Reply #1 on: August 18, 2003, 12:41:00 AM »
It helps a lot to draw each molecule and compare them with each other to be able to answer questions like yours, at least until you are skilled enough to visualize them all in your head by looking at their name. Try drawing the structures next time you wonder about something similar.

Does anybee know of a solvent for oxidation to GVL? GVL seems pricey to SWIM, so he figured that there must be some diol or other solvent, usable to the copper chromite method. According to

https://www.thevespiary.org/rhodium/Rhodium/pdf/solventfree.kmno4.oxidation.pdf

, 1,4-Pentanediol yields alpha-Methyl-gamma-Butyrolactone - would ingestion of this lactone, give an effect other than poisoning?


You have answered your own questions, as "alpha-Methyl-gamma-Butyrolactone" = GVL = gamma-Valerolactone

It is new to SWIM, that DHV also is active (according to

https://www.thevespiary.org/rhodium/Rhodium/chemistry/ghb.html

.
How about DHB? Has this one been bioassayed?


No, and it never will be.

D is short for delta-, and in this context it means that the substituent (the hydroxy group) is situated on the fifth carbon atom down the chain from the other end (DVL is 5-hydroxy-valerate). As the butyric acid molecule (as I assume the B in your DHB abbreviation is short for) doesn't have more than four carbon atoms, there is no way you could place a hydroxy group on a fifth (the delta) carbon. If you add a fifth carbon, it isn't butyric acid anymore, but rather valeric acid (pentanoic acid).

Dextrose

  • Guest
Thanks
« Reply #2 on: August 18, 2003, 01:01:00 AM »
Thanks for the reply, Rhodium.

SWIM can visualize and draw most (if not all) tryptamines and phenethylamines, and most of the PEA-intermediates - but GHB and friends, hasn't been just as easy figuring out, as the others.

SWIM is learning day by day.

Anyway, SWIM was assuming, that 4-Methyl-gamma-Butyrolactone equaled para-Methyl-gamma-Butyrolactone, like pMA = 4-MA...

Thanks for the clarification, although this one still confuses SWIM.

SWIM will read Vogel and other literature, UTFSE, and keep asking newbee questions!  :P