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Intermolecular forces solutions
1. Indicate the type of intra molecular bonding that holds each of these compounds together,
then consider what types of intermolecular forces would exist between molecules of the same
kind, e.g. between CO2 molecules or between H2O molecules.
O
NaCl
CO2
H2O
I2
H3C
C
O
H
H
OH
acetic acid
C
H
C
C
C
C
C
H
C
O- Na+
H
sodium benzoate
2
NaCl: This is an ionic compound. The electronegativity difference between Na and Cl is large,
and we know that Na loses an electron to become nearly as electronically stable as Ne, and
Cl gains one to become more like Ar. The NaCl bond is formed mostly by the electrostatic
force, balanced by e-/e- repulsion. In bulk, like liquid and solid form, the electrostatic interactions still dominate.
CO2: This is a covalently bonded compound, O=C=O, with mildly polar bonds. The electronegativity difference between C and O is small. In bulk CO2, the interactions are governed by
dispersion forces and multipole forces due to the quadrupole moment of CO2, a higher
moment than the dipole moment, and usually much weaker. In the solid, CO2molecules line
up end to end and stack in such lines.
H2O: Water is a covalently bonded molecule with a large dipole moment. In addition the large
electronegativity of oxygen withdraws much of the bonding electron density from the hydrogens, leaving more or less “bare” protons out there. This leads to hydrogen bonding in bulk
water, as hydrogens are attracted to the two lone pairs of electrons on the oxygens. Liquid
water consists of momentary clusters and cyclic rings of weakly H-bonded water molecules.
The important forces, in order from most to least important, are H-bonding, induction and
dispersion.
Acetic acid is a weak acid that dissolves in water to yield a proton and the acetate ion,
H3CCOO-. The acetate ion is covalently bonded and stable. In solution, the double-bonded O
presents a good target for H-bonding. In pure acetic acid, we would expect H-bonding
between the OH hydrogen and the =O.
Sodium benzoate is analogous to acetic acid, except that we would expect it to be a little
more hydrophobic due to the 6-membered hydrocarbon ring. The covalent benzoate group, a
negative ion, binds Na+ through electrostatic attraction. In water, we’d expect the 6-membered rings to be “buried” together, presenting the COO- moeity to the liquid.
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Intermolecular forces solutions
1. Rank these compounds in order of increasing boiling point.
C
O-Na+
sodium acetate
O
B
H3C
C
D
O
H
H
H
H
C
C
C
C
C
C
E
H
H
H
H
H3C
C
OH
cyclohexanone
CH3 H3C
C
HH
HH
acetic acid
CH3
C
C
HH
HH
pentane
C
H
H3C
C
HH
O
A
CH3
isopentane
The larger the intermolecular interaction, the higher the boiling point. We would expect the
ionic compounds, Na-acetate and acetic acid to have the highest boiling points, and we’d
give the edge to Na-acetate because of strong ionic bonding. The other three compounds
have large hydrophobic regions, and therefore would have lower BPs. Of those, cyclohexanone has a double-bonded oxygen, which is a H-bonding site, and that would hold it in liquid
form to a slightly higher temperature. Of pentane and isopentane, both completely hydrophobic, isopentane is more compact and “shows” more hydrogen atoms directly to the
solution, so we’d expect it to have the lowest boiling point. The order is E, D, C, B, A.
3. What causes C28H58 to be a nonpolar solid at room temperature ? This compound is a
pure hydrocarbon (just carbon and hydrogen). It “prefers” to stick together than to dissolve
into solution, and those weak cohesive forces, in a molecule so large, add up to make it a
solid. Think of the progression: CH4, C2H6, C3H8 and C4H10 are gases at room temperature.
Compounds with up to 15 carbons are liquids, and the rest are solids.
4 & 5. Which of the following compounds would you expect to dissolve in water (#3) and
n-hexane (C6H14, #4).
B
A
C
O
H3C
C
D
H3C
O-Na+
sodium acetate
CH3CH2OH
ethanol
Br2
O
E
H3C
H
C
NH2
isopropyl amine
H
H
H
C
C
C
C
H
C H
C H
H
HH
cyclohexanone
Water: Do dissolve in water, a compound has to be at least slightly polar, present some
H-bonding opportunities, or split into ions that aren’t too hydrophobic. That leaves Br2 out.
Cyclohexanone would only likely be slightly soluble in water because of the large hydrophobic ring attached to the =O.
Hexane is a hydrophobic solvent so we’d expect it to dissolve hydrophobic molecules, like
Br2. “Like dissolves like” is a good rule of thumb to remember.
Intermolecular forces solutions
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6. Which compound might you expect to evaporate most rapidly at room temperature:
(a) C8H18 (b) C8H17OH (c) C8H17NH2 (d) C7H15COOH (e) C6H14 ?
The pure hydrocarbons, C6H14 and C8H18 would boil at the lowest temperatures because
they lack any means to H-bond, like the compounds with -OH, -COOH or -NH2 groups. Of
the two, the lightest one should boil/evaporate at a lower temperature, or most rapidly at
room temperature.