Handbook of Solvents - George Wypych - ChemTech - Ventech!

5.3 Methods of calculation of solubility parameters 263
Group
Small 2
Van Krevelen 5
F, (cal cm 3 ) 1/2 mol -1
Hoy 7
-CN 410 480 354.6
-CHCN- (438) 548.5 (440.6)
-F (122) 80 41.3
-Cl 270 230 205.1
-Br 340 300 257.9
-I 425 - -
The Scatchard equation is correct only for nonpolar substances because they have only
dispersive interactions between their molecules. Small eliminated from his consideration
the substances containing hydroxyl, carboxyl and other groups able to form hydrogen
bonds.
This method has received its further development due to Fedors’ work, 4 who extended
the method to polar substances and proposed to represent as an additive sum not only the attraction
energy but also the molar volumes. The lists of such constants were published in
several works. 5-8 The most comprehensive set of contributions to cohesion energy can be
found elsewhere. 9
Askadskii has shown 10 that Fedors’ supposition concerning the additivity of contributions
of volume of atoms or groups of atoms is not quite correct because the same atom in an
environment of different atoms occupies different volume. In addition, atoms can interact
with other atoms in different ways depending on their disposition and this should be taken
into account for computation of cohesion energy. Therefore, a new scheme of the solubility
parameters calculation was proposed that takes into account the nature of an environment of
each atom in a molecule and the type of intermolecular interactions. This approach is similar
to that described in the work of Rheineck and Lin. 6
∑
⎛ ΔΕ
⎞ i ⎜ ⎟
i
δ= ⎜ ⎟
⎜NA∑ΔVi
⎟
⎝ i ⎠
12 /
[5.3.3]
where:
NA Avogadro number
ΔEi increment (contribution) to cohesion energy of atom or group of atoms
ΔVi increment to the van der Waals volume of atom
The volume increment ΔVi of an atom under consideration is calculated as volume of
sphere of the atom minus volumes of spherical segments, which are cut off on this sphere by
the adjacent covalently-bound atoms:
where:
4 3 1 3
ΔV1 = πR −∑ π hi ( 3R
−hi)
[5.3.4]
3 3
i