Octanol Water Partition

Transcription

1 Another phase change H 3 C H 2 H 2 H 2 C C C OH C C C H2 H2 H2 A ater A octanol K o [ Aoctanol ] [ A ] ater C H2 octanol aqueous solution

2 Why octanol? drug uptake animal testing mortality related to K o oral absorption (to needs) drug must first pass through lipid bilayers in the intestinal epithelium drug must be hydrophobic enough to partition into the lipid bilayer drug must be hydrophilic enough to avoid retention, non selective effects C mortality K o

3 Why octanol? environmental transport sorption to organic matter H 3 C C uptake by organisms H2 organic matter and organisms are octanol like H 2 H 2 H 2 C C C OH C C C H2 H2 H2

4 Wide range of K o O H 3 C CH compound K o benzene phenol trichloroethene phenanthrene ,2,5,5 -tetrachlorobiphenyl

5 i = 3.7x10 3 1/ i = x i = 2.7x ater for ~4 octanol (Octanol) 8 octanol for ~100000aters (Water) K io C io C i V V o i io pure n-octanol = 0.16L/mol (at 25 o C) Use approximation or calculate Vmix =(0.75)(0.16) + (0.25)(0.018) = 0.12 L mol-1

6 At equilibrium: K o [A ] oct [A ] K o x oct V oct x V x oct V x V oct 1 V oct 1 V oct H 2 O octanol aqueous solution K o V oct V oct

7 At equilibrium: molar volume of octanol pure octanol, V oct = 0.16 L mol 1 ater in octanol: 1 :4 V oct = 0.12 L mol 1 molar volume of ater H 3 C 1 octanol per 12,500 H 2 Os V ~= L mol 1 K C H2 o H 2 H 2 H 2 C C C OH C C C H2 H2 H2 [ A ] oct [ A ] H 2 O oct V V oct octanol aqueous solution

8 Assumptions sat even at saturation, solute molecules ill not be near each other octanol present in ater does not affect oct 1 to 10 for most compounds [ A ] K oct o [ A ] oct V V oct

9 Is K o related to aqueous solubility? log K o vs. log C sat (L) K K o [ A ] oct [ A ] sat oct o sat sat C ( L) 1 1 C ( L) C ( L) V oct oct sat o oct log K log C ( L) log logv oct

10

11 Experimental determination of K o shake flask measure solute distribution in octanol and ater phases (should be) limited to K o < 10 5 poor reproducibility among researchers e.g., DDT, log K o range of 4.89 to 6.91; over 60 different papers (Pontolillo and Eganhouse, 2001) need for reliable estimation method

12 K o Pontolillo and Eganhouse (2001)

13 Pontolillo and Eganhouse (2001)

14 original data correctly cited erroneous data (incorrectly cited) Pontolillo and Eganhouse (2001)

15 Example: lindane 10 6 mole of lindane is added to 100 ml separatory funnel containing 10 ml of octanol and 90 ml of ater. -hexachlorocyclohexane At equilibrium and 25C, hat concentration (M) of lindane ill be found in the ater?

16

17 Lindane K o = hexachlorocyclohexane f f total moles in ater total moles in octanol and ater [ lin] V [ lin] V [ lin] V [ lin] o K [ lin] K [ lin] [ lin] f f o o o o o [ lin] V K [ lin] V [ lin] V o o V K V V o o

18 Lindane K o = f f f n V K V 3.78 (10 10 ml) 90 ml V o o lin, lin, T 90 ml -hexachlorocyclohexane 6 9 (0.0015)(10 mol) mol n mol [ lin] V L f n 9 lin, 8 M

19 Estimation of K o related to partitioning in other solvents butanol, hexane not much data related to aqueous solubility and activity coefficients (Table 7.3, book) log K o alogc sat b ' log K o alog b related to retention time in chromatography (Figure 7.7 next slide)

20 C-18 column, MeOH: Water system, use reference compounds

21 Example estimate the K o of lindane using its aqueous solubility and T m = 112 C -hexachlorocyclohexane no Table 7.3, SGI (2003)

22 Lindane alkane? PCB? sat log K 0.85log C ( L) 0.78 C sat (L) o solubility of liquid or subcooled liquid no melting for octanol ater partition p C L C s p * sat sat L ( ) ( ) * s

23 Lindane C sat (s) = M p s */p L *? =? =? p ln log Tm 1 p T * s * L -hexachlorocyclohexane

24 Lindane What is the torsional bond number for lindane? A. 0 B. 5 C hexachlorocyclohexane SP30.5SP20.5RING

25 Lindane What is the rotational symmetry number for lindane? A. 1 B. 2 C. 4 D. 12 -hexachlorocyclohexane

26 Lindane C sat (s) = M p s */p L * = 0; = 1 p ln (0) 2.3log(1) 1 p * s * L p ln 1.98 p p p * s * L p * * s L * * L ps

27 Lindane p C L C p C * sat sat L 4.60 ( ) * s sat 3.74 ( L) 10 M sat log K 0.85log C ( L) 0.78 o log K 0.85( 3.74) 0.78 o log 3.96 o K measured log K o = 3.78