TECH 0705 Post-Lab Questions: Seperating Acids and Neutral Compunds by Solvent Extraction

1. Calculate percent recovery... not hard
2. When the first washing was done with NaOH, the strong attraction of the OH- pulled both compounds into the aqueous solution, leaving nothing to react with the acidification of the NaHCO3 layer.
OR
Appartently, you had benzoic acid, naphthalene, and 2-naphthol dissolved in ether. If you mistakenly extracted that solution with aqueous NaOH, then:

C6H5COOH + NaOH ===> C6H5COO-Na+ + H2O

C10H7OH + NaOH ===> C10H7O-Na+ + H2O

C10H8 + NaOH ===> No reraction

So both the benzoate and the 2-naphtholate would have gone into the water solution and the naphthalene remained in the ether solution. If you had extracted the ether solution with NaHCO3 solution next, only the naphthalene would be left in the ether solution, nothing would go into the NaHCO3 solution, and no crystals would be formed on acidification of the NaHCO3 solution.


3. p-tert-butylphenoxide anion (The sodium salt of p-tert-butylphenol)

4.Dichloromethane would sepaprate into the lower layer of the separatory funnel instead of the top, making extraction more difficult.  Instead of using the funnel's stopcock, one would need to extract the top layer from above, leaving the bottom layer intact for further washes. (Check density)
 The only change you would make is you would save the opposite layer. For example, if you were extracting an organic into the organic layer (tbme), you would save the top layer and discard the bottom (aqueous). If you used dichloromethane instead, you would save the bottom layer (MeCl2) and discard the top (aqueous).


5. Make a flowchart (http://i.imgur.com/gji6F.gif) also refer to (http://www.sas.upenn.edu/~zhall/flowchart.html) for details and cool stuff and unicorns and cute puppies




6. Lower/wider melting point range than expected implies less purtiy, nuff said