Exercise 1
LIQUID-LIQUID EXTRACTION
ØOrganic compounds may be obtained from
their natural sources through extraction using
a solvent
ØThe desired organic compound from the
leaves can be pulled out from the leaf matrix
due to its solubility in the extracting solvent
Extraction Methods:
1. Solid-liquid extraction
2. Liquid-liquid extraction or Liquid/Solvent
Partitioning
1. Solid-liquid extraction
2.1 Liquid-liquid extraction: Single extraction
The desired organic compound distribute between the two liquids according
to its solubility in each of the liquids (solvent)
At equilibrium, the quantitative measure of how an organic compound
(solute) distributes between the two liquids (layers) is the distribution or
partition coefficient, KD.
Distribution or Partition coefficient, KD
KD = SO/SA = CO/CA
SO = solubility of solute in organic layer (g/mL)
SA = solubility of solute in aqueous layer (g/mL)
CO = concentration of solute organic layer (g/mL)
CA = concentration of solute aqueous layer (g/mL)
Example:
KD = 8/2 = 4
Sample Problem 1
A nearly saturated solution of 0.50g hyoscyamine in 150mL
water is to be extracted into 150mL diethyl ether. How much
hyoscyamine would be extracted into the diethyl ether layer
in this process?
KD =
𝑜𝑟𝑔𝑎𝑛𝑖𝑐 𝑠𝑜𝑙𝑢𝑏𝑖𝑙𝑡𝑦
𝑤𝑎𝑡𝑒𝑟 𝑠𝑜𝑙𝑢𝑏𝑖𝑙𝑖𝑡𝑦
1.44𝑔 ℎ𝑦𝑜𝑠𝑐𝑦𝑎𝑚𝑖𝑛𝑒
100𝑚𝐿 𝑑𝑖𝑒𝑡ℎ𝑦𝑙 𝑒𝑡ℎ𝑒𝑟
KD =
0.354𝑔 ℎ𝑦𝑜𝑠𝑐𝑦𝑎𝑚𝑖𝑛𝑒
100𝑚𝐿 𝑤𝑎𝑡𝑒𝑟
𝐊𝐃 = 𝟒. 𝟎𝟕
Given: 0.50g hyoscyamine in the aqueous solution
If " 𝒙 " is the gram quantity of hyoscyamine extracted into the diethyl ether layer,
then "0.50g − 𝒙 " would remain in the aqueous layer after equilibrium is established.
𝑥
150𝑚𝐿 𝑑𝑖𝑒𝑡ℎ𝑦𝑙 𝑒𝑡ℎ𝑒𝑟
𝟒. 𝟎𝟕 =
0.50𝑔 − 𝑥
150𝑚𝐿 𝑤𝑎𝑡𝑒𝑟
𝒙 = 𝟎. 𝟒𝟎𝒈
% 𝐑𝐞𝐜𝐨𝐯𝐞𝐫𝒚 =
#.%#& '()*+,)'- ./01,/+234'
x100
#.5#& ./01,/+234' 34 0*3&34+6 1+276'
% 𝐑𝐞𝐜𝐨𝐯𝐞𝐫𝒚 = 80%
𝒙 = 𝟎. 𝟒𝟎𝒈
= 𝟎. 𝟏𝟎𝒈
remained in
the aqueous
layer
KD=
𝟎.𝟒𝒈
𝟎.𝟏𝟎𝒈
KD = 𝟒. 𝟎
2.2 Liquid-liquid extraction: Multiple extraction
Sample Problem 2
A solution of 0.50g hyoscyamine in 150mL water
is to be extracted into diethyl ether. Instead of
using one 150mL portion, the solvent was split
into three 50mL portions. How much hyoscyamine
would be extracted with this method?
𝑥
50𝑚𝐿 𝑑𝑖𝑒𝑡ℎ𝑦𝑙 𝑒𝑡ℎ𝑒𝑟
𝟒. 𝟎𝟕 =
0.50𝑔 − 𝑥
150𝑚𝐿 𝑤𝑎𝑡𝑒𝑟
𝑥
50𝑚𝐿 𝑑𝑖𝑒𝑡ℎ𝑦𝑙 𝑒𝑡ℎ𝑒𝑟
𝟒. 𝟎𝟕 =
0.21𝑔 − 𝑥
150𝑚𝐿 𝑤𝑎𝑡𝑒𝑟
𝑥
50𝑚𝐿 𝑑𝑖𝑒𝑡ℎ𝑦𝑙 𝑒𝑡ℎ𝑒𝑟
𝟒. 𝟎𝟕 =
0.09𝑔 − 𝑥
150𝑚𝐿 𝑤𝑎𝑡𝑒𝑟
1.44𝑔 ℎ𝑦𝑜𝑠𝑐𝑦𝑎𝑚𝑖𝑛𝑒
100𝑚𝐿 𝑑𝑖𝑒𝑡ℎ𝑦𝑙 𝑒𝑡ℎ𝑒𝑟
KD =
0.354𝑔 ℎ𝑦𝑜𝑠𝑐𝑦𝑎𝑚𝑖𝑛𝑒
100𝑚𝐿 𝑤𝑎𝑡𝑒𝑟
𝐊𝐃 = 𝟒. 𝟎𝟕
𝒙 = 𝟎. 𝟐𝟗𝒈
first extraction; 0.21g (0.50g-29g)
remained in the aqueous layer
𝒙 = 𝟎. 𝟏𝟐𝒈
second extraction; 0.09g (0.21g-0.12g)
remained in the aqueous layer
𝒙 = 𝟎. 𝟎𝟓𝒈
third extraction; 0.04g (0.09g-0.05g)
remained in the aqueous layer
Summary
% 𝐑𝐞𝐜𝐨𝐯𝐞𝐫𝒚 =
#.%8& '()*+,)'- 9/01,/+234'
x100
#.5#& 9/01,/+234' 34 0*3&34+6 1+276'
% 𝐑𝐞𝐜𝐨𝐯𝐞𝐫𝒚 = 92%
Greater than single extraction method (80%)
Main reference
https://chem.libretexts.org/Bookshelves/Organi
c_Chemistry/Book%3A_Organic_Chemistry_Lab
_Techniques_(Nichols)/4%3A_Extraction/4.4%3
A_Extraction_Theory