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The Double Solubility Rule An Experimental Test. Meyerhoffer’s double solubility rule: The solubility of a mixture consisting of a dl pair of enantiomers that crystallize in the form of a conglomerate will be twice the solubility of chiral enantiomer.
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The Double Solubility Rule An Experimental Test
Meyerhoffer’s double solubility rule: The solubility of a mixture consisting of a dl pair of enantiomers that crystallize in the form of a conglomerate will be twice the solubility of chiral enantiomer. Can the double solubility rule be verified?
What’s the total pressure in each case? PB PA Stopcock closed Solid A Solid B Stopcock opened PA+PB PA+PB Solid A Solid B
Typical Phase Diagrams in Chiral Systems A. Compound formation
2. Solid solution with a maximun or minimum in melting ln x = (fusHd /R)[1/Tfus(d)-1/Tfus] 3. Conglomerate When x = 0.5 then Tfus = Tfusdl
Consider pure crystalline d: Pd = Ade-subH/RT Simlarly for the l form: Pl = Ale-subH/RT Ad =Al ; subHd = subHl Daltons Law of Partial Pressures: PT = Pd + Pl Pdl = 2Ade-subH/RT Therefore Pdl/ Pl = 2
In a real dl mixture Pdl =Adle-subHdl/RT Pdl/ Pd = 2? 2 = Adle-subHdl/RT/ Ade-subHd/RT Furthermore subHdl = fusHdl + vapHdl ; subHd = fusHd + vapHd and vapHd vapHdl What aboutfusHdl and fusHd ? 2 = Adle-fusHdl/RT/ Ade-fusHd/RT 2 = Adl/ Ade- fusH/RT if Adl/ Ad = 1 then fusH = RTln(2) fusH = 1.73kJ mol-1 at T = 300 K
Conglomerates ln x = (fusHd /R)[1/TfusA-1/Tfus] When x = 0.5 then Tfus = Tfusdl
Conglomerates: T /K, fusHA /kJ mol-1 TfusA fusHA fusHdl Tfus dl Tfus fusH (calcd) o-chloromandelic acid 392.5 24.7 20.1 358.5 359.5 3.6 hydrobenzoin 420.5 34.3 31.4 393 393 2.9 1,2-dichloroacenaphthene 375 21.3 20.5 339 340 0.8 phenylhydracrylic acid 391 32.6 29.7 366 366 2.9 ln x = (fusHd /R)[1/TfusA-1/Tfus] fusH = 1.73kJ mol-1 at T = 300 K
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