# LogP of halogen derivatives of aromatics and aliphatics explanation

Benzene logP = 2.13 (logP=log([benzene in octan-1-ol]/[benzene in water])) Fluorobenzene logP = 2.27 (slight increase, slightly more hydrophobic) Chlorobenzene logP = 2.84 Bromobenzene logP = 2.99

So as we go down group 7, logP or hydrophobicity increases. What can explain these trends? Furthermore on aliphatic compounds, we observe this trend which confuses matters more.

-Fluorine DECREASES logP slightly (by -0.17) on an aliphatic compound. The book doesn't give a reference as to which alpiphatic compound this is and which hydrogen is substituted, but oh well. -Chlorine increases logP (by 0.39) on the same aliphatic compound. -Bromine increases logP even more (by 0.60) on the same aliphatic compound.

So a few questions: 1. Why does a fluorine substitution decrease logP slightly on the aliphatic but increase slightly on the aromatic? 2. Why do chlorine and bromine substitutions increase logP in aliphatics and slightly more in aromatics.

Answers to these trends would be greatly appreciated, thanks.

Here is a chart of the solubility of the species in water and the log of the partition coefficient you noted.

                           solubility       solubility
MW       g/100ml water    moles/liter   log(P)
Benzene         78.11    0.179 (1)        0.0229         2.13
Flororbenzene   96.103   0.155 (2)        0.0161         2.27
Cholorbenzene  112.56    0.050 (2)        0.0044         2.84
Bromobenzene   157.01    0.041 (2)        0.0026         2.99


(1) Wikipedia

(2) Cation Complexes of Compounds Containing Carbon-Carbon Double Bonds. VI. The Argentation of Substituted Benzenes
L. J. Andrews, R. M. Keefer
J. Am. Chem. Soc., 1950, 72 (7), pp 3113–3116
DOI: 10.1021/ja01163a085

There is a Wikipedia chart in article on Ionic radii.

 Ion     radii (picometers)
F-      119
Cl-     167
Br-     182


The halide on a benzene ring is going to be relatively covalently bound. It definitely won't dissociate to form ions and only very moderately participate in hydrogen bonding.

So as the size of the halide increases the "volume" of the molecule increases. Thus the molecule would preferentially like to be in the non-polar phase.

Here is a scaled image of fluorobenzene molecule:

and a scaled image of the bromobenzene molecule.

Haloalkanes

The "anomalous" fluoroalkane is explained in a different manner. To get an n-alkane that has a boiling point above room temperature we need to go to n-pentane.

                   BP                 solubility      solubility
(deg C)   MW       g/100ml water   millimoles/liter
n-Pentane         36        72.15    0.040 (1)         0.55
1-Fluoropentane   63        90.14    1.270 (2)         14.1
1-Chloropentane  108       106.6     0.197 (3)         1.85
1-Bromopentane   130       151.05    0.127 (4)         0.838


(1) Wikipedia
(2) NIH
(3) Wikipedia
(4) Handbook of Aqueous Solubility Data, Second Edition By Samuel H. Yalkowsky, Yan He, Parijat Jain

There are two competing effects.

(1) The size of atom bound to the alkane.

Br > Cl > F > H

Scaled model of Fluoroethane

(2) The second effect is the electronegativity of the X group in the $\text{Alkane}\text{X}$ bond.

 F    >   Cl   >   Br   >   H
4.0      3.2      3.0      2.2


The fluorine atom has a slightly larger radius in a fluoroalkane than the corresponding C-H group. But the halide groups allow more dipole forces. So by "volume" the overall effect is:

 fluoroalkane
alkane
choloroalkane
bromoalkane

• Why does increased volume lead to the molecule preferring the non polar phase? I don't understand how increased dipole forces reduce the volume of the fluoroalkane. Furthermore why wouldn't the volume of chloro and bromoalkanes be reduced in this case? Also, why is it that fluoroalkanes actually prefer water MORE than alkanes? – Laksh Nov 6 '15 at 1:44
• See this problem for an answer to the volume effect. chemistry.stackexchange.com/questions/40108/… – MaxW Nov 6 '15 at 1:53