Author Topic: What about amount per extraction?  (Read 1849 times)

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FreeBee

  • Guest
What about amount per extraction?
« on: October 21, 2003, 08:51:00 AM »
So its obvious from the fine theoretical work posted here by our fearless leader that three EQUAL extractions is the way to go.  But is there a rule for calculating how much solvent to use in each extraction as it relates to the mass of the material that is to be extracted and as well the total mass of the solution being extracted?  

To make it a more practical example, could someone illustrate the required amount of DCM to use to recover the finished honey from say a 2.5 KG NaBH4 post reduction solution (flooded workup) ?

Rhodium

  • Guest
calculating the solubility product
« Reply #1 on: October 22, 2003, 12:03:00 PM »
could someone illustrate the required amount of DCM to use to recover the finished honey from say a 2.5 KG NaBH4 post reduction solution (flooded workup)?

You cannot determine it beforehand without knowing the partition coefficient for MDMA between DCM and aqueous alcohol (and I doubt anyone has tabulated that anywhere). It will also differ if you have more or less salts dissolved in the solution, as well as if you have used methanol, ethanol or isopropanol as the alcohol, and how much water you flooded it with.

But - if you adhere to a standard protocol, and measure how much product you get in the first, second and third extraction, you can calculate the solubility product as in

Post 464877

(Chromic: "Okay, more guessing", Chemistry Discourse)
and then from that you can work out the optimal number of extractions, and how large each of them should be to retrieve 99.5% of all your product.

java

  • Guest
Re: Properties of solvents
« Reply #2 on: November 23, 2003, 06:24:00 PM »

Critical compilation of scales of solvent  parameters.

 
           Part I. Pure, non-hydrogen bond donor solvents
      
Pure Appl. Chem.,          Vol. 71, No. 4, pp. 645-718, 1999
      

http://www.iupac.org/publications/pac/1999/71_04_pdf/abboud.pdf


     
Excerp......
       
           It has long been known that solvents often affect chemical reactivity, this involving, e.g., the shift of the position of chemical equilibria (thermodynamic aspect) as well as significant changes in reaction rate constants (kinetic aspect). Physical properties, particularly the frequencies and intensities of transitions in IR, UV-visible, fluorescence, NMR and ESR spectroscopies are also known to be affected by solvents.

 These phenomena are consequences of differences in the solvation of reagents and products (thermodynamic effects) or reagents activated complexes (kinetic effects). Differential solvation of species in the ground and excited states accounts for the spectral  phenomenology indicated above. Differences in solvation of a given solute  in two different solvents determine the size of the corresponding partition coefficient.



wylderness

  • Guest
regarding miscable solvents, and A/B
« Reply #3 on: July 10, 2004, 10:30:00 PM »
First, I want to say thank you to all of you guy's here. I thought when SWIM started his present endeavour, he could just go to Rhodium's (MOST EXCELLENT) site, pick a synth, use clean glass, and get the job done..... HAHAHAH.. was SWIM wronng! And he would not have been anywhere as far along as je is at solving most of SWIMS problems if i hadnt been using this excellent hive, and availing myself of your freely shared knowledge. You have my gratitude, and my appreciation..
Now, for my very first question,ever posted at the HIVE, I would put this forward:
SWIM was cleaning up his lab, and had a whole bunch of solvent filled mason jars, that he is SURE have some honey in them (MDMA). Problem is...some of the so,lvents are mixed... The one that most concerns him is the one with water and tone.... Now this is my question... If SWIM wanted to extract whatever honey was there...(from a water/acetone mix) would it work to use a solvent that would dissolve acetone and honey, but not water..(Xylene??) and tehn basify the mix, pushing the amine in to the Xylene and out of the water, or should i add xylene, then acidify it, and extract via the water?
Thanx for reading this!
WYLD at heart in the WYLDERNESS


r2e3

  • Guest
just evap
« Reply #4 on: July 11, 2004, 09:49:00 AM »
acetone bp 56C - remove with distillation
drive off remaining water to yield resultant product.
if it is ungassed then use dcm to extract - evap it and start again with clean dry acetone.
i think you need to be a little more systematic in your approach dude

java

  • Guest
Rubber products chemical resistance
« Reply #5 on: July 29, 2004, 12:19:00 PM »
What rubber hose, gasket material to choose with different solvents see the chemical resistance chart....

http://www.akrongasket.com/charts.html




magstirrer

  • Guest
Personal experience
« Reply #6 on: September 19, 2004, 02:56:00 PM »
On the three layer theme, from personal experience:

Suppose you have benzaldehyde salted out of water, ie two layers: bottom benzaldehyde (denser than water, practically insoluble in water), top water with NaCl and other inorganic salts - in my case, mostly acetates and (hydro)sulfites. Very carefully adding xylene (lighter than water) as not to disturb the liquids indeed causes three layers to form. But a slight shake of the flask and the benzaldehyde layer "connects" to the top xylene and starts to slowly migrate. If you add enough xylene (or toluene, ether or what have you) you might suck the bottom benzaldehyde in a minute. Btw this is a hell of a funny experiment to watch  ;D

Cheers and good luck with your extractions