Experiment Number 2: Solubility and Intermolecular Forces
Intermolecular forces are important concepts in organic chemistry. We can use it to explain: solubility,
extraction, thin-layer chromatography, paper chromatography, gas chromatography, and column
chromatography, among other topics.
Recall that there are five types of IM forces: London dispersion forces, dipole-dipole, H-bonding, iondipole, and ion-ion. Indicate the strongest type of intermolecular force present in the following
compounds:
Compound
Butane
CH3CH2OH
NaCl
CH3OCH3
IM Force
Compound
CH4
CH3COCH3
NaCH3O
octane
IM Force
Physical Properties and IM Forces:
The strength of intermolecular forces determines the physical properties (i.e. melting point, boiling point,
solubility, etc…) of a compound. Generally speaking a molecule that has stronger intermolecular forces
will have a higher / lower (circle one) melting and boiling points.

Rank the following compounds in decreasing boiling point.
o CH3OCH3
o CH3CH2OH
o CH3CH2NH2

Rank the following compounds in decreasing boiling point
o CH3CH2OH
o CH3(CH2)6CH2OH
Solubility and Intermolecular Forces
In the dissolving process the solute is usually the smaller quantity and it is the substance that is being
dissolved whereas the solvent is the larger quantity and the substance that does the dissolving.
The dissolving process can be broken down into the following steps.
Solute-Solute Interactions
The attractive forces (intermolecular) that hold the solute molecules together must be overcome, breaking
the molecules apart. This step requires energy and the potential energy increases here.
Solvent-Solvent Interactions
The attractive forces (intermolecular) that hold the solvent molecules together must be overcome,
breaking the molecules apart thus making room for the solute molecules. This step requires energy and
the potential energy increases here also.
Solute-Solvent Interactions
During this process, the molecules of the solute interact with the molecules of the solvent. The solvent
particles surround the solute particles and form attractive forces between them. This step decreases the
potential energy.
In order to dissolve, the solvent molecules and the solute molecules must break apart from each other and
this process raises the energy. The new interactions between the solvent and the solute must be strong
enough to lower the energy back down.
We can look at IM forces to compare the interactions before dissolving to the interactions after
dissolving. If the net forces are stronger (or similar with regard to IM forces) after dissolving, then the
solute will dissolve in the solvent.
Let’s examine the following scenarios.
Solvent IM Forces
Solvent IM Forces
Weak
London Dispersion
Strong
(Polar or
H-bonding)
Strong
(Polar or
H-bonding)
Weak
London Dispersion
Strong
(Polar or
ionic)
Weak
London Dispersion
Solvent-Solute
IM Forces
Weak
London Dispersion
Strong
(dipole-dipole or
H-bonding)
Weak
Solubility
Soluble
Soluble
Insoluble
Now, let’s see if we can make a few predictions regarding solubility. Predict if the following solutes
(benzophenone, malonic acid, and biphenyl) will be soluble, partially soluble, or insoluble in the solvents
listed (water, methanol, and hexane).
Compound
Water
H2O
Benzophenone
O
Malonic acid
O
HO
Biphenyl
O
OH
Methanol
CH3OH
Hexane
PART A
To one mL of water and 1 mL of hexane, add dropwise methanol, 1-butanol, and 1-octanol. We can
classify the outcome as soluble, partially soluble, or insoluble. It is important to note what happens upon
the addition of the first drops. Look carefully for the appearance of mixing lines in the solution.
Methanol
1-butanol
1-octanol
1 mL. of Water
1 mL. of Hexane
PART B
Place about 30 mg of benzoic acid into three test tubes and add 1.0 ml of DI water to the first test
tube, 1.0 ml of 1.0 M NaOH to the second test tube, and 1.0 ml of 1.0 M HCl to the third test tube. Shake
the test tube for 20 seconds and note your observations in the table below. Indicate whether the solid is
soluble (dissolves completely) or insoluble (none of it dissolves). Now take the tube containing benzoic
acid and 1.0 M NaOH and add 6.0 M HCl dropwise while stirring until the mixture is acidic. Note the
results of this addition in the table below.
Place about 30 mg of ethyl-4-aminobenzoate into three test tubes and add 1.0 ml of DI
water to the first test tube, 1.0 ml of 1.0 M NaOH to the second test tube, and 1.0 ml of 1.0 M HCl to the
third test tube. Shake the test tube for 20 seconds and note your observations in the table below. Indicate
whether the solid is soluble (dissolves completely) or insoluble (none of it dissolves). Now take the tube
containing ethyl-4-aminobenzoate and 1.0 M HCl and add 6.0 M NaOH dropwise while stirring until the
mixture is basic. Note the results of this addition in the table below.
Compounds
Benzoic Acid
Water
Solvents
1.0 M NaOH
1.0 M HCl
O
OH
Add 6.0 M HCl
Ethyl-4-aminobenzoate
O
H 2N
O
Add 6.0 M NaOH
Part B Chemical Reactions
Complete the reaction below and label the Lewis acid, Lewis base, nucleophile, and electrophile. Is the
product or reactant more water soluble. Explain why.
O
+ OH -
OH
Complete the reaction below and label the Lewis acid, Lewis base, nucleophile, and electrophile. Is the
product or reactant more water soluble. Explain why.
O
H 2N
O
+
H+