Transcript lecture five

Why mixtures mix
 Consider a glass of wine. Why do alcohol, water, &
pigment mix together?
 There must be attractive forces.
Intramolecular forces is the force of
attraction between atoms in a molecule
Intermolecular forces is the attraction between
molecules
Intermolecular forces are responsible for many properties of molecular
compounds, including crystal structures (e. g. the shapes of snowflakes), melting
points, boiling points,
Types of Intermolecular Forces of
attraction
1) Ionic (Ionic bonds form if the EN is 1.7 or greater)
2) dipole – dipole (EN is around 0.5-1.7)
3) H-bonding( is a type of dipole-dipole)
4) London forces or Van der Waals Forces (exist in all
molecules, but are especially important in nonpolar covalent molecules, where EN is less than
0.5).
 Ionic intermolecular forces of attraction are
strongest
 Dipole-dipole are not as strong as ionic
 Hydrogen bonds are about five times
stronger than regular dipole-dipole bonds.
 London forces are weakest force of
attraction.
London forces
 Non-polar molecules do not have dipoles like polar





molecules. How, then, can non-polar compounds form
solids or liquids?
London forces are named after Fritz London (also
called van der Waal forces)
London forces are due to small dipoles that exist in
non-polar molecules.
These forces exist between all molecules
Because electrons are moving around in atoms there
will be instants when the charge around an atom is not
symmetrical
The resulting tiny dipoles cause attractions between
atoms/molecules
Dipole-dipole interaction
 This happens in all molecules having polar bonds.
 But Polar molecules are held to each other more
strongly than the non-polar molecules of comparable
molecular weight.
 occur when polar molecules are attracted to one
another.
 Electrostatic attraction between (+) and (-) ends of
molecules
 This accounts for the attraction one water molecule
has for another water molecule.
H-bonding
H
O
H
H
O
H
 H-bonding is a special type of dipole - dipole attraction that is very strong
 It occurs when hydrogen is covalently bonded to a very electronegative
atom like N, O, or F
 because of the small size of hydrogen, it’s positive charge can get very
close to the negative dipole of another molecule.
 H is so strongly positive that it will sometimes exert a pull on a “lone pair”
in a non-polar compound
 The high EN (electronegativity difference) between two atoms like of
NH, OH, and HF bonds cause these to be strong forces (about 5x stronger
than normal dipole-dipole forces).
Calculate the EN for HCl and H2O
Soln: HCl = 2.9-2.1 = 0.8
H2O = 3.5-2.1 = 1.4
 They are given a special name (H-bonding) because compounds
containing these bonds are important in biological systems
Molecules with 3 Atoms
Even though the C-O bond is polar, the bonds
cancel each other out because the molecule is linear
the dipole moments are equal and in opposite
directions. Therefore CO2 is non-polar.
CO2
HCN
SO2
The dipole moment between H-C points in
the direction of C. The dipole moment points
between C-N points in the direction of the N.
Therefore the dipole vectors are additive and
HCN is polar
SO2 is a polar molecule because the S-O dipole
Moments don’t cancel each other out due to
the angle
Which of the following molecules are polar (have a
dipole moment)?H2O, CO2, SO2, and CH4
O
S
dipole moment
polar molecule
dipole moment
polar molecule
H
O
C
O
no dipole moment
nonpolar molecule
H
C
H
H
no dipole moment
nonpolar molecule
10.2
Molecules with 4 Atoms
CCl4 is non-polar
CHCl3 is polar
Formation of Intermolecular H- bonds
1.Between non-identical molecules
a. alcohol and water
H
R
O
H
O
H
R
O
H
H
O
H
O
R
H
b. chloroform and pyridine
Cl
Cl
C
Cl
H
N
c. acid and water
O
O
R
H
H
C
H
O
O
H
H
2. Between identical molecules
a. NH3 molecule
H
H
H
N
H
N H
H
b. hydrocyanic acid
H
C
H
N
c. alcohols
O
O
R
H
H
R
C
N
d. acids
H
O
R
O
C
O
O
H
e. amines
H
R
H
N
C
H
N
R
H
H
R
N
R
R
Intramolecular H- bond results to ring formation or
chelation. Ortho substituted are usually involved in
chelation, because meta and para substituents are already
distal group as far as effective H- bonding is concerned.
A.
O H
F
B.
H3C
C CH
O
C.
O
C
O H
O
CH
H
CH
O
CH2CH3
STRUCTURE AND PHYSICAL
PROPERTIES
1. MELTING POINT
2. BOILING POINT
3. SOLUBILITY
BOILING POINT
 One of the most revealing of all physical properties for
a chemical substance is its boiling point. Boiling point
reflects the strength of the intermolecular attractive
forces that hold the molecules of a substance together
in a condensed phase, and as such, it is useful to
compare the boiling points for related compounds to
see how structural differences account for the
differences in intermolecular attractions. The trends in
boiling points for various groups of compounds helps
in understanding how size, shape, and functional
group polarity affect boiling point.
The
is the temperature
which a liquid
Thenormal
boiling boiling
point ispoint
the temperature
at whichatthe
boils
when the vapor
external
pressure
1 atm.is equal to the external
(equilibrium)
pressure
of aisliquid
pressure.
The
boiling point is the temperature at which the
(equilibrium) vapor pressure of a liquid is equal to the external
pressure.
11.8
FACTORS THAT AFFECTS BOILING
POINT/ MELTING POINT
1. Molecular Weight/Size
The bigger the molecule, the higher its boiling point.
Thus, for a series of related compounds, the higher the
molecular weight, the higher the boiling point. Note
the trend for the first five straight-chain alkanes:
Van der Waals attractive forces increases as the hydrocarbon
chain increases so that the boiling point is high.
Compound Formula (mol. wt.)
B.P.
methane
CH4(16)
-164°C
ethane
C2H6 (30)
-88°C
propane
C3 H8 (44)
-42°C
n-butane
C4 H10 (58)
0°C
n-pentane
C5 H12 (72)
36°C
2. Branching in the hydrocarbon
lowers the boiling point
 In the straight chain hydrocarbon molecules can
approach each other so stronger Van der Waals
attractive force.
 In branched chain hydrocarbon, there will be no close
approach of molecules so that there is weak van der
waals attractive force.
Example:
CH3
CH3
C
CH3
<
CH3
CH2
CH2
CH3
neopentene
n-pentene
CH2
CH3
3. Dipole moment – measured in
Debye Units (D)
μ= e X d ( magnitude of charge X distance)


differences in electronegativity
The more polar the higher is the boiling point
 Molecular nitrogen (N2) and carbon monoxide (CO)
have identical molecular weights: 28g/mol. Which has
the higher boiling point?
4. Association
Associated liquids - liquids whose molecules are held
together by H- bonds. Boiling points are abnormally high.
The more associated the liquid, the higher the boiling point.
Example: methanol
CH3OH can form 1 H- bond
CH3COOH can form 2 H- bonds
5.
CH3
CH
CH3
CH3
CH2
CH3
CH3
C
CH
In propene, diffused pi electron cloud would not allow
the molecules to be closer as there will be repulsion.
Propane molecule interact with each other more
strongly than propene molecule.
In propyne, the pi electron cloud is tighter so that
molecule can approach each other closer, so stronger
and therefore higher boiling point than propane.
)
The b.following
are set of organic compounds. Determine which
one has a higher boiling point or melting point. Why?
a.)
OH
benzene
van der waals
b.)
O H
phenol
van der waals & H-bonding
O
H
O
H
NO2
NO2
The polar nitro group provides additional attractive forces
between molecule
O
H
O
H
NO2
NO2
 intramolecular H-bonding is invoked in the ortho
position
 When OH and NO2 groups are tied up internally by a
hydrogen bond, their
effectivity in forming
intermolecular hydrogen bonding is lessened.
O
O H
O
H
H
O
N
O
O
N
O
H
intramolecular H-bond
O
intermolecular H-bonding
c.)
O H
NH2
Lone pair from N can easily be delocalized to the ring, whereas,
lone pair in phenol is not as easily delocalized because of the strong
attracting power of this atom from the lone pair.
d.)
H3C
CH2
CH2
CH2
NH2
H3C
CH2
CH2
CH2
OH
Butanol forms stronger H-bonding because O is more
electronegative than N.
e.)
H3C
CH2
O
CH2
CH3
very volatile (low boilingpoint), molecules cannot form H-bonding,
only van der waals
f.)
O
H3C
CH2
CH2
C
H
butanal
doest not form H-bonding
H3C
CH2
CH2
CH2
butanol
forms H-bonding
OH
g.)
H3C
CH2
CH2
CH2
H3C
OH
MW= 74
Forms only one H-bond
CH2
COOH
MW= 74
forms 2 H-bond between
molecules
h.)
O
R
O
H
O
H
Forms only one H-bond
R
R
H
O
C
C
O H
R
O
forms 2 H-bond between molecules
MELTING POINT
Melting – a change from a highly ordered arrangement of the particles to the
more random arrangement that characterizes liquid.
Melting occurs when the temperature is reached at which the thermal energy is
great enough to overcome the intracrystalline forces that hold particles in
position.
To melt ionic solid enough energy is needed to break the ionic bond.
Example: NaCl ( mpt = 801oC)
To melt molecular solid – enough energy is needed to break the intermolecular
forces
Example: CH4 ( mpt= -183oC)
Electrostatic
Example: CH3OH hydrogen bonding and dipole-dipole
mpt less than 801oC but higher than -183oC
The melting point of a solid
or the freezing point of a
liquid is the temperature at
which the solid and liquid
phases coexist in equilibrium
Freezing
H2O (l)
Melting
H2O (s)
11.8
11.8
Which is of higher melting point?
a.
C
O
H
O
H
C
O
H
OH
OH
o-hydroxybenzalhehyde
- intramolecular H-bond
- less associated with each other
b.
C
O
H
O
C
H
p-hydroxybenzaldehyde
- intermolecular H-bond
- more associated with each other
C
O
H
NH2
NH2
o-aminobenzaldehyde
p-aminobenzaldehyde
C.
C
O
H
C
O
O
H
CH3
O
CH3
o-methoxybenzaldehyde
p-methoxybenzaldehyde
more defined negativity
less defined negativity
stronger interaction
weaker interaction
Both will not form H- bonding
d.) carboxylic acids – with increasing MW exhibit increasing
melting point.
SOLUBILITY
- is dissolution of the structural units ( ions or molecules)
-are separated using up the energy coming from the attraction
between the solute and the solvent molecules
-the energy required to break the bonds between solute
particles is supplied by the formation of bonds between the
solute particles and the solvent molecules
- The solubility of a solute is the maximum quantity of solute
that can dissolve in a certain quantity of solvent or quantity of
solution at a specified temperature.
- The factors that determine solubility are the strength of
IMFs and speed of molecules.
The main factors that have an effect on solubility are:
1.the nature of the solute and solvent -- While only 1 gram of lead (II)
chloride can be dissolved in 100 grams of water at room temperature, 200 grams
of zinc chloride can be dissolved. The great difference in the solubilities of the of
these two substances is the result of differences in their natures.
2.temperature -- Generally, an increase in the temperature of the solution
increases the solubility of a solid solute. A few solid solutes, however, are less
soluble in warmer solutions.
For all gases, solubility decreases as the
temperature of the solution rises.
3. pressure -- For solids and liquid solutes, changes in pressure have practically
no effect on solubility. For gaseous solutes, an increase in pressure increases
solubility and a decrease in pressure decreases solubility. (When the cap on a
bottle of soda pop is removed, pressure is released, and the gaseous solute
bubbles out of solution. This escape of a gas from solution is called
effervescence.)
The rate of solution is a measure of how fast a substance dissolves. Some of the
factors determining the rate of solution are:
1.size of the particles -- When a solute dissolves, the action takes place only at the
surface of each particle. When the total surface area of the solute particles is
increased, the solute dissolves more rapidly. Breaking a solute into smaller pieces
increases its surface area and hence its rate of solution. (Sample problem: a cube with
sides 1.0 cm long is cut in half, producing two pieces with dimensions of 1.0 cm x 1.0
cm x 0.50 cm. How much greater than the surface area of the original cube is the
combined surface areas of the two pieces? 2.0 cm2
2. stirring -- With liquid and solid solutes, stirring brings fresh portions of the
solvent in contact with the solute, thereby increasing the rate of solution.
3. amount of solute already dissolved -- When there is little solute already in
solution, dissolving takes place relatively rapidly. As the solution approaches the
point where no solute can be dissolved, dissolving takes place more slowly.
4. temperature -- For liquids and solid solutes, increasing the temperature not only
increases the amount of solute that will dissolve but also increases the rate at which
the solute will dissolve. For gases, the reverse is true. An increase in temperature
decreases both solubility and rate of solution.
Solvation can be done in 2 ways:
1.Use of lonepairs -------aprotic solvents
2. Use of H-bonding ------protic solvents
Protic Solvents - solvent containing H that is attracted to O and
N and appreciably acidic.
Example: CH3OH and H2O ( can solvate both + and – ion)
Solvation of NaCl using H2O
Aprotic solvent – polar solvent of moderately high dielectric constants, which do
not contain acidic hydrogen.
Example:
Dimethylsulfoxide
CH3
S CH3
O
O
N,N-dimethylformamide(DMF)
CH3
H
C
O
N
CH3
Sulfolane
O
O
Solubility of Acids and Bases
General Rule:
•
Strong acids are soluble in strong
bases and vise versa.
•
Weak bases are soluble in strong
acids and vise versa.
 Solubility of amines in dilute HCl is associated
with the tendency of the lone pair of electron of N
to bond with the proton.
H2N
+ HCl
H3N+
salt formation
-
Cl
 In general aliphatic amines are soluble in dilute HCl.
When alkyl groups are bulky in secondary and tertiary
amines, solubility in dilute HCl decreases. This is a
consequence of steric inhibition of the approach of
the acid to bond with the lone pair or the instability
of the salt formed as a result of steric overcrowding.
CH3
H3C
H3C
NH2
CH3
CH3
CH3
 Triphenylamine is not soluble in dilute HCl because
of unavailability of the lone pair for coordination
with the acid as result of resonance effects or effects
due to electron delocalization.
N
 Amides are not soluble in dilute HCl as simple
amines because of lesser availability of lone pair
of electron of the N of amides compared with
simple amines. The lone pair of amides is
delocalized towards the carbonyl carbon.
O
R C
Amide
NH2
R
NH2
amine
 Disubstituted amides, however, in contrast to the
simple amides are more soluble in dilute HCl. In
these disubstituted amides the alkyl groups
increases the availability of the lone pair of
electrons on the nitrogen for the acid to
coordinate with.
O
R
C
N
R
R
Solubility of unsaturated noncyclic hydrocarbons and
some aromatic hydrocarbons in cold concentrated
H2SO4 is a consequence of the availability of pi electron
for coordination with proton.
CH3CH=CH2 + H2SO4
CH3CH2-CH3
H
+
H2SO4
+
H
H
ROR
+
H2SO4
R-O-R
+
H
ROH
+
H2SO4
R-O-H
+
O
R-C-OR
OH
+
H2SO4
R-C
OR
Solubility of organic compounds in dilute sodium hydroxide is
a consequence of the presence of acidic hydrogen.
OH
O Na +
S-H
+
CH3 - N
+
aqNaOH
Na
N
+
+
HOH
O
+
O
+
aq NaOH
O
O
OH
HOH
O
O
O
+
S Na
aqNaOH
O
N H
HOH
CH2
N
CH2
O
O
N
O
O
+
aq NaOH
O
Na
+ +
HOH
In chelated phenols, acidic H is tied
up as a hydrogen bond, thus
insoluble in dilute NaOH.
H
C
O
O H
 Ketones and aldehydes are insoluble in dilute
NaOH even though they posses acidic H, because
the acidity of H is too weak to allow dissolution in
dilute NaOH.
 Solubility of organic compounds in dilute sodium
bicarbonate(NaHCO3) can reflect strength in
acidity of these systems. Only carboxylic acid,
sulfonic acids, and sulfinic acids are soluble in
dilute sodium bicarbonate solution(NaHCO3).
 Phenols and aliphatic alcohols which are regarded as
weak acids do not dissolved in NaHCO3, but 2, 4, 6trinitrophenol is an exception, because the 3 NO2
increases the acidity of phenol.
 Solubility of chloroform in organic bases such as
pyridine and trimethylamine is a consequence of Hbonding.
 Solubility of unsaturated noncyclic hydrocarbons and
some aromatic hydrocarbons in cold concentrated
H2SO4 is a consequence of the availability of pi electron
for coordination with proton.