Physical properties and intermolecular forces (ESCKQ)
Have the learners research the safety data for various compounds, especially those
being used in the experiments in this section, as a way of linking the
properties of organic molecules with their molecular structure.
The types of intermolecular forces that occur in a substance will affect its physical
properties, such as its phase, melting point and
boiling point. You should remember from the kinetic theory of matter
(see Grade
10), that the phase of a
substance is determined by how strong the forces are between its particles. The weaker
the forces, the more likely the substance is to exist as a gas. This is because the
particles are able to move far apart since they are not held together very strongly. If
the forces are very strong, the particles are held closely together in a solid
structure. Remember also that the temperature of a material affects the energy
of its particles. The more energy the particles have, the more likely they are to be
able to overcome the forces that are holding them together. This can
cause a
change in
phase.
Figure 4.52 shows the three
phases of water. Note that we are showing two-dimensional figures when in reality these
are three-dimensional.
The effects of intermolecular forces
The forces between molecules that bind them together are known as
intermolecular forces. Intermolecular forces allow us to
determine which substances are likely to dissolve in other substances, and what
the melting and boiling points of substances are. Without intermolecular forces
holding molecules together we would not exist.
Intermolecular forces
Intermolecular forces are forces that act between molecules.
Remember from Grade
11 that
a dipole molecule
is a molecule that has its charge unevenly distributed. One end of the molecule
is slightly positive and the other is slightly negative. An overview of the
different types of intermolecular forces that are discussed in this chapter are
given below:
Dipole-dipole forces
When one dipole molecule comes into contact with another dipole
molecule, the positive pole of the one molecule will be
attracted to the negative pole of the other, and the molecules
will be held together in this way.
One special case of this is hydrogen bonding:
Do not confuse hydrogen bonds with intramolecular
covalent bonds. Hydrogen bonding is an example of a
scientist naming something, believing it to be one thing
when in fact it was another. In this case the strength
of the hydrogen bonds misled scientists into thinking
this was an intramolecular bond when it is really just a
strong intermolecular force.
Hydrogen bonds
As the name implies, this type of intermolecular bond
involves a hydrogen atom. When a molecule
contains a hydrogen atom covalently bonded to a
small, highly electronegative atom (e.g.
\(\text{O}\), \(\text{N}\) or \(\text{F}\)) this
type of intermolecular force can occur. The
highly electronegative atom on one molecule
attracts the hydrogen atom on a nearby molecule
(see Figure 4.53).
van der Waals forces
Induced-dipole forces
Dipole-induced-dipole intermolecular forces
are also sometimes called
London forces or
dispersion forces.
In non-polar molecules the electronic charge is
usually evenly distributed but it is possible
that at a particular moment in time, the
electrons might not be evenly distributed
(remember that the electrons are always moving
in their orbitals). The molecule will have a
temporary dipole. When this happens,
molecules that are next to each other attract
each other very weakly.
Dipole-induced-dipole forces
These forces exist between dipoles and non-polar
molecules. The dipole induces a dipole in the
non-polar molecule leading to a weak, short
lived force which holds the compounds together.
In this chapter we will focus on the effects of van der Waals forces and hydrogen bonding on
the physical properties of organic molecules.
Viscosity
Viscosity is the resistance to flow of a liquid. Think how easy it is to pour water compared
to syrup or honey. The water flows much faster than the syrup or honey.
Viscosity
Viscosity is a measure of how much a liquid resists flowing. i.e. The higher the
viscosity, the more viscous a substance is.
You can see this if you take a cylinder filled with water and a cylinder filled with glycerol
(propane-1,2,3-triol). Drop a small metal ball into each cylinder and note how easy it
is for the ball to fall to the bottom (Figure 4.55). In the glycerol the ball falls slowly,
while in the water it falls faster.
As implied by the definition, substances with stronger intermolecular forces are more viscous
than substances with weaker intermolecular forces. The stronger the intermolecular
forces the more the substance will resist flowing. The greater the internal friction the
more a substance will slow down an object moving through it.
If the glass sheet is tilted to either side during the resistance to flow activity
the drops will run into each other and the activity will not work. The drops
should reach the finish line in the order of: alcohol (least viscous),
water, oil, syrup (most viscous).
Resistance to flow
Take a sheet of glass at least \(\text{10}\) by \(\text{15}\) \(\text{cm}\) in size.
Using a water-proof marker draw a straight line across the width of the glass
about \(\text{2}\) \(\text{cm}\) from each end.
Place the glass flat on top of two pencils, then carefully put a drop of water on one
end of the line. Leave at least \(\text{2}\) \(\text{cm}\) space next to the
water drop, and carefully place a drop of alcohol. Repeat this with a drop of
oil and a drop of syrup.
Slowly and carefully remove the pencil from the end opposite the drops (make sure you
don't tilt the glass to either side in the process).
Which drop moves fastest and reaches the end line first?
Which drop moves slowest?
The fastest moving substance has the least resistance to flow, and therefore has the
least viscosity (is the least viscous). The slowest moving substance has the
most resistance to flow, and therefore has the most viscosity (is the most
viscous).
Density
Density
Density is a measure of the mass per unit of volume.
The solid phase is often the most dense phase (water is one noteworthy exception to this).
This can be explained by the strong intermolecular forces found in a solid. These forces
pull the molecules together, which results in more molecules in one unit of volume than
in the liquid or gas phases. The more molecules in a unit volume the denser, and heavier
that volume of the substance will be.
Density can be used to separate different liquids, with the more dense liquid settling to the
bottom of the container, while the less dense liquid floats on top. If you throw a leaf
into a river or pond the leaf will float. If you instead throw a rock (with the same
surface area and volume) into the river or pond the rock will sink. This is due to the
different densities of the two substances: rocks are more dense than water while leaves
are less dense than water.
Melting and boiling points
Intermolecular forces affect the boiling and melting points of substances. Substances with
weak intermolecular forces will have low melting and boiling points as less energy
(heat) is needed to overcome these forces. Those with strong intermolecular forces will
have high melting and boiling points as more energy (heat) is required to overcome these
forces. When the temperature of a substance is raised beyond it's melting or boiling
point the intermolecular forces are not weakened. Rather, the molecules have enough
energy to overcome those forces.
An ether is a compound that contains two alkyl chains (e.g. methyl, ethyl) joined at
an angle by an oxygen atom (\(-\)O\(-\)).
Name
Main
intermolecular
forces
Molecular
Mass (\(\text{g.mol$^{-1}$}\))
Melting point (℃)
Boiling point (℃)
Phase
(at 25℃)
ethane
induced-dipole
\(\text{30,06}\)
\(-\text{183}\)
\(-\text{89}\)
gas
dimethyl ether
dipole-dipole
\(\text{46,06}\)
\(-\text{141}\)
\(-\text{24}\)
gas
chloroethane
dipole-dipole
\(\text{64,5}\)
\(-\text{139}\)
\(\text{12,3}\)
gas
pentane
induced-dipole
\(\text{72,12}\)
\(-\text{130}\)
\(\text{36}\)
liquid
propan-1-ol
hydrogen bonds
\(\text{60,08}\)
\(-\text{126}\)
\(\text{97}\)
liquid
ethanol
hydrogen bonds
\(\text{46,06}\)
\(-\text{114}\)
\(\text{78,4}\)
liquid
butan-1-ol
hydrogen bonds
\(\text{74,1}\)
\(-\text{90}\)
\(\text{118}\)
liquid
ethanoic acid
hydrogen bonds
\(\text{60,04}\)
\(\text{16,5}\)
\(\text{118,5}\)
liquid
Table 4.9: Relationship between intermolecular forces and melting point,
boiling point and physical state.
As the intermolecular forces increase (from top to bottom in Table 4.9) the
melting and boiling points increase. The stronger the intermolecular forces the more
likely a substance is to be a liquid or a solid at room temperature.
Worked example 29: Comparing physical properties
Given:
Melting point (℃)
Boiling point (℃)
propane
\(-\text{188}\)
\(-\text{42}\)
butanoic acid
\(-\text{7,9}\)
\(\text{163,5}\)
bromoethane
\(-\text{118}\)
\(\text{38,5}\)
diethyl ether
\(-\text{116,3}\)
\(\text{34,6}\)
Fill in the table below:
Name
Main intermolecular
forces
Molecular
formula
Molecular
mass (\(\text{g.mol$^{-1}$}\))
Phase
(at 25 ℃)
propane
butanoic acid
bromoethane
diethyl ether
Draw the structural representation of each molecule
What are the intermolecular forces that each molecule will
experience?
propane has only single carbon-carbon bonds
and no other functional group. It will therefore have
induced-dipole forces only.
butanoic acid has the carboxylic acid
functional group. It can therefore form hydrogen bonds.
bromoethane has the highly electronegative
bromine atom. This means that it will form dipole-dipole
interactions with neighbouring molecules.
diethyl ether will have induced-dipole
forces, however due to the flexible nature of the
molecule is can also have dipole-induced-dipole
(stronger van der Waals interactions) and dipole-dipole
forces.
Do these forces make sense with the melting and boiling points
provided?
Propane has the lowest melting and boiling points and the weakest
interactions. The next lowest melting and boiling points are for
bromoethane and diethyl ether, which both have dipole-dipole
interactions, the next strongest intermolecular forces. The highest
melting and boiling points are for butanoic acid which has strong
hydrogen bonds. Therefore these forces do make sense.
Calculate the molecular mass of these molecules
propane - \(\text{C}_{3}\text{H}_{8}\), molecular mass =
\(\text{44,08}\) \(\text{g.mol$^{-1}$}\)
butanoic acid - \(\text{C}_{4}\text{H}_{8}\text{O}_{2}\),
molecular mass = \(\text{88,08}\)
\(\text{g.mol$^{-1}$}\)
bromoethane - \(\text{C}_{2}\text{H}_{5}\text{Br}\),
molecular mass = \(\text{108,95}\)
\(\text{g.mol$^{-1}$}\)
diethyl ether - \(\text{C}_{4}\text{H}_{10}\text{O}\),
molecular mass = \(\text{74,10}\)
\(\text{g.mol$^{-1}$}\)
What will the phase of each compound be at \(\text{25}\)
\(\text{℃}\)?
To determine the phase of a molecule at \(\text{25}\) \(\text{℃}\) look
at the melting and boiling points:
The melting and boiling points of propane are both below
\(\text{25}\) \(\text{℃}\), therefore the molecule
will be a gas at \(\text{25}\) \(\text{℃}\).
The melting points of butanoic acid, bromoethane and dimethyl
ether are below \(\text{25}\) \(\text{℃}\),
however the boiling points of all three molecules are
above \(\text{25}\) \(\text{℃}\). Therefore these
molecules will be liquids at \(\text{25}\)
\(\text{℃}\).
Fill in the table
Name
Main intermolecular
forces
Molecular
formula
Molecular
mass (\(\text{g.mol$^{-1}$}\))
Phase
(at 25 ℃)
propane
induced-dipole
\(\text{C}_{3}\text{H}_{8}\)
\(\text{44,08}\)
gas
butanoic acid
hydrogen bonds
\(\text{C}_{4}\text{H}_{8}\text{O}_{2}\)
\(\text{88,08}\)
liquid
bromoethane
dipole-dipole
\(\text{C}_{2}\text{H}_{5}\text{Br}\)
\(\text{108,95}\)
liquid
diethyl ether
dipole-dipole
\(\text{C}_{4}\text{H}_{10}\text{O}\)
\(\text{74,10}\)
liquid
In the investigation of boiling and melting points experiment learners are required
to work with ethyl methanoate. This can irritate your eyes, skin, nose and
lungs. Please remind the learners to be careful and wear the appropriate safety
equipment when handling all chemicals. The safety equipment includes gloves,
safety glasses and protective clothing. Learners should also keep all open
flames away from their experiment as they are handling flammable substances.
The boiling point of propanoic acid is approximately \(\text{141}\)
\(\text{℃}\), which means that the cooking oil will be very hot by the
time propanoic acid starts to boil. The learners should be reminded not to place
their hands anywhere near the oil, and to allow it to cool before cleaning the
apparatus.
This is a good opportunity to get the learners to research the hazard data sheets for
butan-1-ol, propanoic acid and ethyl methanoate.
Investigation of boiling and melting points
Aim
To investigate the relationship between boiling points and intermolecular
forces
three test tubes, a beaker, a thermometer, a hot plate
Method
Ethyl methanoate can irritate your eyes, skin, nose and lungs. Keep
open flames away from your experiment and make sure you work in
a well ventilated area.
Label the test tubes 1, 2
and 3. Place \(\text{20}\)
\(\text{ml}\) of butan-1-ol into test tube 1,
\(\text{20}\) \(\text{ml}\) of propanoic acid into test
tube 2, and \(\text{20}\) \(\text{ml}\) of ethyl
methanoate into test tube 3.
Half-fill the beaker with cooking oil and place it on the hot
plate.
Place the thermometer and the three test tubes in the beaker.
Make a note of the temperature when each substance starts to
boil.
Results
Fill in the gaps in the table below. Do the values you obtained match those
reported in literature?
Compound
ethyl methanoate
butan-1-ol
propanoic acid
Molecular
formula
Molecular
mass
Main intermolecular
forces
Literature
boiling point (℃)
\(\text{54}\)
\(\text{118}\)
\(\text{141}\)
Experimental
boiling point (℃)
Draw the structural representation of ethyl methanoate, butan-1-ol and
propanoic acid.
Discussion and conclusion
You should have found that the ethyl methanoate boiled first, then the
butan-1-ol and then the propanoic acid. Ethyl methanoate has some
dipole-dipole interactions, but cannot form a hydrogen bond. The alcohol
(butan-1-ol) can form hydrogen bonds and so has a higher boiling point.
This strong intermolecular force needs more energy to break and so the
boiling point is higher. For propanoic acid hydrogen bonds form between
the carbonyl group on one acid and the hydroxyl group on another. This
means that each molecule of propanoic acid can be part of two hydrogen
bonds (this is called dimerisation, see Figure
4.57) and so the boiling point is even higher for propanoic
acid than for butan-1-ol.
Flammability and vapour pressure
Flammability is a measure of how easy it would be for a substance to catch alight and burn.
The flash point of a substance is the lowest temperature that is likely to form
a gaseous mixture you could set alight. If a liquid has a low enough flash point it is
considered flammable (able to be ignited easily) while those with higher flash points
are considered nonflammable. A substance that is classified as nonflammable can still be
forced to burn, but it will not ignite easily.
When a substance is in the liquid or solid state there will be some molecules in the gas
state. These molecules have enough energy to overcome the intermolecular forces holding
the majority of the substance in the liquid or solid phase. These gas molecules exert a
pressure on the liquid or solid (and the container) and that pressure is the vapour
pressure of that compound. The weaker the intermolecular forces within a substance the
higher the vapour pressure will be. Compounds with higher vapour pressures have lower
flash points and are therefore more flammable.
Vapour pressure
The pressure exerted (at a specific temperature) on a solid or liquid
compound by molecules of that compound that are in the gas phase.
Name
Main intermolecular forces
Vapour pressure (kPa at 20℃)
Flash point (℃)
Flammability
ethane
induced-dipole
\(\text{3 750}\)
\(-\text{135}\)
very high
propane
induced-dipole
\(\text{843}\)
\(-\text{104}\)
very high
dimethyl ether
dipole-dipole
\(\text{510}\)
\(\text{41}\)
very high
butane
induced-dipole
\(\text{204}\)
\(-\text{60}\)
very high
chloroethane
dipole-dipole
\(\text{132,4}\)
\(-\text{50}\)
very high
pentane
induced-dipole
\(\text{57,9}\)
\(-\text{49}\)
very high
propanone
dipole-dipole
\(\text{24,6}\)
\(-\text{17}\)
high
ethanol
hydrogen bonds
\(\text{5,8}\)
\(\text{17}\)
high
water
hydrogen bonds
\(\text{2,3}\)
-
very low
propan-1-ol
hydrogen bonds
\(\text{2}\)
\(\text{22}\)
high
ethanoic acid
hydrogen bonds
\(\text{1,6}\)
\(\text{40}\)
moderate
butan-1-ol
hydrogen bonds
\(\text{0,6}\)
\(\text{35}\)
high
Table 4.10: Relationship between intermolecular forces and the flammability
of a substance.
As the intermolecular forces increase (from top to bottom in Table 4.10) you can
see a decrease in the vapour pressure. This corresponds with an increase in the flash
point temperature and a decrease in the flammability of the substance. Figure 4.58 shows a
few examples.
Types of intermolecular forces
Textbook
Exercise 4.19
Use your knowledge of different types of intermolecular
forces to explain the following statements:
The boiling point of hex-1-ene is much lower than the
boiling point of propanoic acid.
Propanoic acid has hydrogen bonds which are much
stronger than the induced-dipole forces in
hex-1-ene. In order for a liquid to boil the
intermolecular forces must be broken. The
stronger the intermolecular forces, the more
energy it will take to overcome these forces. As
a result the boiling point will be higher for
propanoic acid than for hex-1-ene.
Water evaporates more slowly than propanone.
Water has strong intermolecular forces (hydrogen
bonds) while propanone only has weaker
dipole-dipole forces. Substances with stronger
intermolecular forces take longer to evaporate
than substances with weaker intermolecular
forces.
IUPAC name
Boiling point
(℃)
A
ethane
\(-\text{89,0}\)
B
ethanol
\(\text{78,4}\)
C
ethanoic acid
\(\text{118,5}\)
Which of the compounds listed in the table are gases
at room temperature?
ethane
Name the main type of intermolecular forces for A, B
and C.
A - induced-dipole forces
B - hydrogen bonding
C - hydrogen bonding
Account for the difference in boiling point between A
and B.
A has only induced-dipole forces (weak van der Waals
forces). B has hydrogen bonding. Since hydrogen
bonding is a stronger intermolecular force than
van der Waals forces, more energy is required to
separate the molecules of ethanol than the
molecules of ethane. Thus ethanol has a higher
boiling point than ethane.
Account for the difference in boiling point between B
and C.
Both B and C undergo hydrogen bonding. However the
hydrogen bonding in C (carbonyl and hydroxyl
group) is stronger than that of B (hydroxyl
group). This is because C forms a hydrogen
bonding dimer (see Figure 4.57),
while B forms only single hydrogen bonds. Thus C
has a higher boiling point than B.
Draw the structural representations of A, B and C.
Which container (A or B) has the compound with higher
vapour pressure in it? Explain your answer.
B. There are more molecules of B in the vapour phase
than there are of A. More molecules in the
vapour phase means a higher vapour pressure.
Draw the condensed structural formula for each of
these compounds:
Propanoic acid has a vapour pressure of
\(\text{0,32}\) \(\text{kPa}\) at \(\text{20}\)
\(\text{℃}\)
Butan-1-ol has a vapour pressure of \(\text{0,64}\)
\(\text{kPa}\) at \(\text{20}\)
\(\text{℃}\)
Explain the difference in vapour pressure.
Both compounds have the same molecular mass, but
propanoic acid has two sites for hydrogen
bonding (can form a hydrogen bonding dimer, see
Figure 4.57)
while butan-1-ol has only one. So butan-1-ol has
weaker intermolecular forces and will go into
the vapour phase easier than propanoic acid.
Butan-1-ol therefore has a higher vapour
pressure than propanoic acid.
Physical properties and functional groups (ESCKR)
Compounds that contain very similar atoms can have very different properties depending on how
those atoms are arranged. This is especially true when they have different functional
groups. Table 4.11 shows
some properties of different functional groups.
Vanilla has a sweet smell. The smell of ethene and dimethyl ether cling in
your nose though, and the smell of ether can even stay on your skin up to
\(\text{24}\) hours.
Functional
group
Typical
Smell
Example
Formula
Melting point (℃)
Boiling point (℃)
Phase
(at 25℃)
alkane
odourless
ethane
\(\text{C}_{2}\text{H}_{6}\)
\(-\text{183}\)
\(-\text{89}\)
gas
alkene
sweet/musky
ethene
\(\text{C}_{2}\text{H}_{4}\)
\(-\text{169,2}\)
\(-\text{103,7}\)
gas
ether (-O-)
sweet
dimethyl ether
\(\text{C}_{2}\text{H}_{6}\text{O}\)
\(-\text{141}\)
\(-\text{24}\)
gas
haloalkane
almost
odourless
chloro
ethane
\(\text{C}_{2}\text{H}_{5}\text{Cl}\)
\(-\text{139}\)
\(\text{12,3}\)
gas
aldehyde
pungent
fruity
ethanal
\(\text{C}_{2}\text{H}_{4}\text{O}\)
\(-\text{123,4}\)
\(\text{20,2}\)
gas
alcohol
sharp
ethanol
\(\text{C}_{2}\text{H}_{6}\text{O}\)
\(-\text{114}\)
\(\text{78,4}\)
liquid
ester
often fruity
methyl methanoate
\(\text{C}_{2}\text{H}_{4}\text{O}_{2}\)
\(-\text{100}\)
\(\text{32}\)
liquid
alkyne
odourless
ethyne
\(\text{C}_{2}\text{H}_{2}\)
\(-\text{80,8}\)
\(-\text{84}\)
gas
carboxylic acid
vinegar
rancid butter
ethanoic acid
\(\text{C}_{2}\text{H}_{4}\text{O}_{2}\)
\(\text{16,5}\)
\(\text{118,5}\)
liquid
Table 4.11: Some properties of compounds with different functional groups.
Listed in the table are the common smells and other physical properties found for common
functional groups. Only one representative example from each homologous series is
provided. This does not mean that all compounds in that series have exactly the same
properties. For example, short-chain and long-chain alkanes are generally odourless,
while those with moderate chain length (approximately \(\text{6}\) - \(\text{12}\)
carbon atoms) smell like petrol (Table 4.15).
Some specific properties of a few functional groups will now be discussed in more detail.
Physical properties of the alcohols
In humans, ethanol reduces the secretion of a hormone called antidiuretic
hormone (ADH). The role of ADH is to control the amount of water that
the body retains. When this hormone is not secreted in the right
quantities, it can cause dehydration because too much water is lost from
the body in the urine (ethanol is a diuretic). This is why people who
drink too much alcohol can become dehydrated, and experience symptoms
such as headaches, dry mouth, and lethargy. Part of the reason for the
headaches is that dehydration causes the brain to shrink away from the
skull slightly.
Ethane has a relatively low solubility in water.
Because hydroxyl (\(-\text{OH}\)) groups can hydrogen bond, all three
pentanol molecules have a greater solubility in water than ethane.
Ethanol is completely soluble in water in any amount. This is because it
contains a hydroxyl group and has a much shorter non-polar chain length
than pentanol.
pentanol
solubility
(\(\text{g.dm$^{-3}$}\))
ethane
\(\text{0,057}\)
1-pentanol
\(\text{22}\)
2-pentanol
\(\text{45}\)
3-pentanol
\(\text{59}\)
ethanol
soluble
The hydroxyl group (\(-\text{OH}\)) affects the solubility of the
alcohols.
Solubility
Solubility is a measure of the ability of a substance (solid, liquid or gas)
to dissolve in another substance. The amount of the substance than can
dissolve is the measure of its solubility.
The hydroxyl group generally makes the alcohol molecule polar and therefore
more likely to be soluble in water. However, the carbon chain
resists solubility, so there are two opposing trends in the alcohols. Alcohols
with shorter carbon chains are usually more soluble in water than those with
longer carbon chains.
Alcohols tend to have higher boiling points than the hydrocarbons
because of the strong hydrogen bond between hydrogen atoms of one hydroxyl group
and the oxygen atom of another hydroxyl group.
Physical properties of haloalkanes
There can be more than one halogen substituted for a hydrogen atom on one haloalkane.
The more halogens are substituted the less volatile the
haloalkane becomes. Take for example the haloalkane series shown in Figure
4.59.
Volatility
The tendency of molecules at the surface of a compound to enter the gas
phase. The more volatile a compound the more likely this is.
For every extra chlorine atom on the original methane molecule the
volatility of the compound decreases. This can be seen by the increase in both
the melting and boiling point (Table 4.12) as one goes from chloromethane
through to tetrachloromethane. The more halogen atoms in the compound the
stronger the intermolecular forces are, which leads to higher melting and
boiling points.
Common Name
Number Cl atoms
Melting point (℃)
Boiling point (℃)
chloromethane
\(\text{1}\)
\(-\text{97,4}\)
\(-\text{24,2}\)
dichloromethane
\(\text{2}\)
\(-\text{96,7}\)
\(\text{39,6}\)
trichloromethane
\(\text{3}\)
\(-\text{63,5}\)
\(\text{61,2}\)
tetrachloromethane
\(\text{4}\)
\(-\text{22,9}\)
\(\text{76,7}\)
Table 4.12: Melting and boiling points of haloalkanes with increasing
numbers of chlorine atoms.
Properties of carbonyl compounds
Carboxylic acids are weak acids, in other words they only dissociate
partially. Why does the carboxyl group have acidic properties? In the carboxyl
group, the hydrogen tends to separate itself (dissociate) from the oxygen atom.
In other words, the carboxyl group becomes a source of positively-charged
hydrogen ions (\(\text{H}^{+}\)). This is shown in Figure 4.60.
The carboxylic acid functional group is soluble in water. However, as the number of
carbon atoms in the attached carbon chain increases the solubility decreases.
This is discussed in greater detail in the next section.
Remember that carboxylic acids form hydrogen bonding dimers (the formation is called
dimerisation) as shown in Figure 4.61.
The ability of a molecule to hydrogen bond leads to increased melting and boiling
points when compared to a similar molecule that is unable to hydrogen bond.
Similarly, the ability of a molecule to form a hydrogen bonding dimer leads to
increased melting and boiling points when compared to similar molecules that can
only form one hydrogen bond (Table 4.13).
Molecule
Hydrogen bonds per molecule
Melting point (℃)
Boiling point (℃)
ethane
0
-183
-89
ethanol
1
-114
\(\text{78,4}\)
ethanoic acid
2
\(\text{16,5}\)
\(\text{118,5}\)
Table 4.13: The melting and boiling points of similar organic
compounds that can form different numbers of hydrogen bonds.
Physical properties of ketones
Hydrogen bonds are stronger than the van der Waals forces found in ketones. Therefore
compounds with functional groups that can form hydrogen bonds are more likely to
be soluble in water. This applies to aldehydes as well as to ketones.
Physical properties and functional groups
Textbook Exercise 4.20
Refer to the data table below which shows the melting point
and boiling point for a number of organic compounds with
different functional groups.
Formula
Name
Melting point
(℃)
Boiling point
(℃)
\(\text{C}_{4}\text{H}_{10}\)
butane
\(-\text{137}\)
\(\text{0}\)
\(\text{C}_{4}\text{H}_{8}\)
but-1-ene
\(-\text{185}\)
\(-\text{6,5}\)
\(\text{C}_{4}\text{H}_{10}\text{O}\)
butan-1-ol
\(-\text{90}\)
\(\text{118}\)
\(\text{C}_{4}\text{H}_{8}\text{O}_{2}\)
butanoic acid
\(-\text{7,9}\)
\(\text{163,5}\)
\(\text{C}_{5}\text{H}_{12}\)
pentane
\(-\text{130}\)
\(\text{36}\)
\(\text{C}_{5}\text{H}_{10}\)
pent-1-ene
\(-\text{165,2}\)
\(\text{30}\)
\(\text{C}_{5}\text{H}_{12}\text{O}\)
pentan-1-ol
\(-\text{78}\)
\(\text{138}\)
\(\text{C}_{5}\text{H}_{10}\text{O}_{2}\)
pentanoic acid
\(-\text{34,5}\)
\(\text{186,5}\)
At room temperature (approx.
\(\text{25}\)\(\text{°C}\)), which of the
organic compounds in the table are:
gases
butane, but-1-ene
liquids
All except butane and but-1-ene
Look at an alkane, alkene, alcohol and carboxylic
acid with the same number of carbon atoms:
How do their melting and boiling points
compare?
The carboxylic acids have the highest melting
and boiling points, then the alcohols,
then the alkanes and the alkenes have
the lowest melting and boiling points.
Looking at those compounds with four carbon
atoms:
Butanoic acid has the highest boiling (and
melting) point, the next highest boiling
(and melting) point is butan-1-ol, then
butane (the alkane) followed by
but-1-ene (the alkene).
Looking at those compounds with five carbon
atoms:
Pentanoic acid has the highest boiling (and
melting) point, the next highest boiling
(and melting) point is pentan-1-ol, then
pentane (the alkane) followed by
pent-1-ene (the alkene).
Explain why their melting points and boiling
points are different?
The carboxylic acids have the strongest
intermolecular forces due to their
ability to form two hydrogen bonds
(dimerisation). The alcohols can only
form one hydrogen bond leading to lower
melting and boiling points. The alkenes
and alkanes cannot form hydrogen bonds.
However, the alkenes are more reactive
than the alkanes due to the double bond
(unsaturated) and so the melting and
boiling points of an alkene are lower
than those of an alkane with the same
number of carbon atoms.
Physical properties and chain length (ESCKS)
Remember that the alkanes are a group of organic compounds that contain carbon and hydrogen
atoms bonded together. The carbon atoms link together to form chains of varying lengths.
We have already mentioned that the alkanes are relatively unreactive because of their
stable C-C and C-H bonds. The boiling points and melting points of these molecules are
determined by their molecular structure and their surface area.
The more carbon atoms there are in an alkane, the greater the surface area available for
intermolecular interactions.
This increase in intermolecular attractions leads to higher melting and boiling points. This
is shown in Table 4.14.
Formula
Name
Molecular mass
(\(\text{g.mol$^{-1}$}\))
Melting point (℃)
Boiling point (℃)
Phase
(at 25℃)
\(\text{CH}_{4}\)
methane
\(\text{16,04}\)
\(-\text{182}\)
\(-\text{162}\)
gas
\(\text{C}_{2}\text{H}_{6}\)
ethane
\(\text{30,06}\)
\(-\text{183}\)
\(-\text{89}\)
gas
\(\text{C}_{3}\text{H}_{8}\)
propane
\(\text{44,08}\)
\(-\text{188}\)
\(-\text{42}\)
gas
\(\text{C}_{4}\text{H}_{10}\)
butane
\(\text{58,1}\)
\(-\text{137}\)
\(\text{0}\)
gas
\(\text{C}_{6}\text{H}_{14}\)
hexane
\(\text{86,14}\)
\(-\text{95}\)
\(\text{68,5}\)
liquid
\(\text{C}_{8}\text{H}_{18}\)
octane
\(\text{114,18}\)
\(-\text{57}\)
\(\text{125,5}\)
liquid
\(\text{C}_{20}\text{H}_{42}\)
icosane
\(\text{282,42}\)
\(\text{37}\)
\(\text{343}\)
solid
Table 4.14: The physical properties of some alkanes.
Notice that when the molecular mass of the alkanes is low (i.e. there are few carbon atoms),
the organic compounds are gases because the intermolecular forces are weak. As
the number of carbon atoms and the molecular mass increases, the compounds are more
likely to be liquids or solids because the intermolecular forces are stronger.
The larger a molecule is, the stronger the intermolecular forces are between the molecules.
This is one of the reasons why methane (\(\text{CH}_{4}\)) is a gas at room temperature
while hexane (\(\text{C}_{6}\text{H}_{14}\)) is a liquid and icosane
(\(\text{C}_{20}\text{H}_{42}\)) is a solid.
Be careful when comparing molecules with different types of
intermolecular forces. For example, small molecule with hydrogen bonding can
have stronger intermolecular forces than a large molecule with only van der
Waals forces.
Table 4.15 shows some other
properties of the alkanes that also vary with chain length.
Name
Density
(\(\text{g.dm$^{-3}$}\))
Flash
point (℃)
Smell
Phase
(at 25℃)
methane
\(\text{0,66}\)
\(-\text{188}\)
odourless
gas
ethane
\(\text{1,28}\)
\(-\text{135}\)
odourless
gas
propane
\(\text{2,01}\)
\(-\text{104}\)
odourless
gas
butane
\(\text{2,48}\)
\(-\text{60}\)
like petrol
gas
hexane
\(\text{650}\)
\(-\text{26}\)
like petrol
liquid
octane
\(\text{700}\)
\(\text{13}\)
like petrol
liquid
icosane
\(\text{790}\)
\(>\text{113}\)
odourless
solid
Table 4.15: Properties of some of the alkanes.
Density increases with increasing molecular size. Note that compounds that are gaseous are
much less dense than compounds that are liquid or solid.
Remember that the flash point of a volatile molecule is the lowest temperature at which that
molecule can form a vapour mixture with air and be ignited. The flash point increases
with increasing chain length meaning that the longer chains are less flammable (Table 4.10) although
they can still be ignited.
It is partly the stronger intermolecular forces that explain why petrol (mainly
octane (\(\text{C}_{8}\text{H}_{18}\))) is a liquid, while candle wax
(\(\text{C}_{23}\text{H}_{48}\)) is a solid. If these intermolecular forces did
not increase with increasing molecular size we would not be able to put liquid
fuel into our cars or use solid candles.
The change in physical properties due to chain length does not only apply to the
hydrocarbons. The solubility of ketones in water decreases as the chain length
increases. The longer the chain length, the stronger the intermolecular interactions
between the ketone molecules. This means that more energy is required to overcome those
interactions and the molecule is therefore less soluble in water. Long chains can also
fold around the polar carbonyl group and stop water molecules from bonding with it.
This same idea can be applied to all the compounds with water-soluble functional groups. The
longer the chain becomes, the less soluble the compound is in water.
Physical properties and chain length
Textbook Exercise 4.21
Refer to the table below which gives information about a
number of carboxylic acids, and then answer the
questions that follow.
Condensed
formula
Common name
Source
IUPAC name
Melting point
(℃)
Boiling point
(℃)
formic acid
ants
methanoic acid
\(\text{8,4}\)
\(\text{101}\)
\(\text{CH}_{3}\text{COOH}\)
vinegar
ethanoic acid
\(\text{16,5}\)
\(\text{118,5}\)
propionic acid
milk
propanoic acid
\(-\text{20,8}\)
\(\text{141}\)
\(\text{CH}_{3}(\text{CH}_{2})_{2}\text{COOH}\)
butyric acid
butter
\(-\text{7,9}\)
\(\text{163,5}\)
valeric acid
valerian root
pentanoic acid
\(-\text{34,5}\)
\(\text{186,5}\)
\(\text{CH}_{3}(\text{CH}_{2})_{4}\text{COOH}\)
caproic acid
goat
skin
\(-\text{3,4}\)
\(\text{205,8}\)
enanthic acid
vines
heptanoic acid
\(-\text{7,5}\)
\(\text{223}\)
\(\text{CH}_{3}(\text{CH}_{2})_{6}\text{COOH}\)
caprylic acid
goat
milk
\(\text{16,7}\)
\(\text{239,7}\)
Fill in the missing spaces in the table by writing
the formula, common name or IUPAC name.
Condensed
formula
Common name
Source
IUPAC name
Melting
point
(℃)
Boiling
point
(℃)
\(\text{HCOOH}\)
formic acid
ants
methanoic acid
\(\text{8,4}\)
\(\text{101}\)
\(\text{CH}_{3}\text{COOH}\)
acetic acid
vinegar
ethanoic acid
\(\text{16,5}\)
\(\text{118,5}\)
\(\text{CH}_{3}\text{CH}_{2}\text{COOH}\)
propionic acid
milk
propanoic acid
\(-\text{20,8}\)
\(\text{141}\)
\(\text{CH}_{3}(\text{CH}_{2})_{2}\text{COOH}\)
butyric acid
butter
butanoic acid
\(-\text{7,9}\)
\(\text{163,5}\)
\(\text{CH}_{3}(\text{CH}_{2})_{3}\text{COOH}\)
valeric acid
valerian root
pentanoic acid
\(-\text{34,5}\)
\(\text{186,5}\)
\(\text{CH}_{3}(\text{CH}_{2})_{4}\text{COOH}\)
caproic acid
goat
skin
hexanoic acid
\(-\text{3,4}\)
\(\text{205,8}\)
\(\text{CH}_{3}(\text{CH}_{2})_{5}\text{COOH}\)
enanthic acid
vines
heptanoic acid
\(-\text{7,5}\)
\(\text{223}\)
\(\text{CH}_{3}(\text{CH}_{2})_{6}\text{COOH}\)
caprylic acid
goat
milk
octanoic acid
\(\text{16,7}\)
\(\text{239,7}\)
Draw the structural representation of butyric acid.
Give the molecular formula for caprylic acid.
\(\text{C}_{8}\text{H}_{16}\text{O}_{2}\)
Draw a graph to show the relationship between
molecular mass (on the x-axis) and
boiling point (on the y-axis)
Describe the trend you see.
As the molecular mass increases, so does the
boiling point OR boiling point increases
with an increase in molecular mass
(something along these lines).
Suggest a reason for this trend.
Longer chains allow for greater van der Waals
interactions leading to higher boiling
points OR an increase in the surface
area of the molecules leads to greater
intermolecular forces which leads to
increasing boiling points OR the forces
holding the molecules in the solid phase
are increasing and so the boiling point
increases, this is because the compounds
with larger molecular mass have longer
chains (something along these lines).
Refer to the data table below which shows the melting point
and boiling point for a number of organic compounds with
different functional groups.
Formula
Name
Melting point
(℃)
Boiling point
(℃)
\(\text{C}_{4}\text{H}_{10}\)
Butane
\(-\text{137}\)
\(\text{0}\)
\(\text{C}_{5}\text{H}_{12}\)
Pentane
\(-\text{130}\)
\(\text{36}\)
\(\text{C}_{6}\text{H}_{14}\)
Hexane
\(-\text{95}\)
\(\text{68,5}\)
\(\text{C}_{4}\text{H}_{8}\)
But-1-ene
\(-\text{185}\)
\(-\text{6,5}\)
\(\text{C}_{5}\text{H}_{10}\)
Pent-1-ene
\(-\text{165,2}\)
\(\text{30}\)
\(\text{C}_{6}\text{H}_{12}\)
Hex-1-ene
\(-\text{140}\)
\(\text{63}\)
At room temperature (approx. 25 °C), which of
the organic compounds in the table are:
gases
butane, butene
liquids
pentane, pentene, hexane, hexene
In the alkanes:
Describe what happens to the melting point
and boiling point as the number of
carbon atoms in the compound increases.
The melting and boiling points increase as
the number of carbon atoms increase.
Explain why this is the case.
As the number of carbon atoms increases, so
does the surface area of the molecule.
This leads to greater intermolecular
forces, which are responsible for the
increase in melting and boiling points.
Fill in the table below. Under boiling point put 1 for the
compound with the lowest boiling point, 2 for the next
lowest boiling point until you get to 6 for the compound
with the highest boiling point. Do not use specific
boiling point values, but rather use your knowledge of
intermolecular forces.
Draw the structural representations for each of the following
compounds and answer the question that follows.
but-2-yne
The prefix but- tells us that there are four carbon
atoms in the longest chain. The suffix -2-yne
tells us that there is a triple bond between the
second and third carbon atoms.
hex-2-yne
The prefix hex- tells us that there are six carbon
atoms in the longest chain. The suffix -2-yne
tells us that there is a triple bond between the
second and third carbon atoms.
pent-2-yne
The prefix pent- tells us that there are five carbon
atoms in the longest chain. The suffix -2-yne
tells us that there is triple bond between the
second and third carbon atoms.
Which of these compounds will have the highest
viscosity?
hex-2-yne - All three compounds are alkynes. There is
an increase in chain length with but-2-yne
having the shortest chain and hex-2-yne the
longest. The van der Waals forces will be
stronger in the longest chain compound,
hex-2-yne, because it has more sites for the van
der Waals forces to act (the surface area is
larger). Hex-2-yne will therefore have the
highest viscosity as the molecules will have the
greatest resistance to flow.
Physical properties and branched groups (ESCKT)
In a straight chain molecule, the carbon atoms are connected to at most two other
carbon atoms. However, in a branched molecule some carbon atoms are connected to
three or four other carbon atoms. This is the same principle that was discussed
with primary, secondary and tertiary alcohols.
Straight chains always have higher boiling points than the equivalent molecule with branched
chains. This is because the molecules with straight chains have a larger surface area
that allows close contact. Branched chains have a lower boiling point due to a smaller
area of contact. The molecules are more compact and cannot get too close together,
resulting in fewer places for the van der Waals forces to act.
Shown in Figure 4.68 are three
isomers that all have the molecular formula \(\text{C}_{5}\text{H}_{12}\). They are all
alkanes with the only difference being branched groups leading to longer or shorter main
chains.
As shown in Table 4.16 the
properties of these three compounds are significantly different. The melting points have
significant differences and the boiling points steadily decrease with an increasing
number of branched groups.
Name
Melting point (℃)
Boiling point (℃)
Density
(\(\text{g.dm$^{-3}$}\))
Vapour pressure
(kPa at 20℃)
Flash point (℃)
pentane
\(-\text{130}\)
\(\text{36}\)
\(\text{621}\)
\(\text{57,9}\)
\(-\text{49}\)
2-methyl
butane
\(-\text{160}\)
\(\text{27,7}\)
\(\text{616}\)
\(\text{77}\)
\(-\text{51}\)
2,2-dimethyl
propane
\(-\text{16,6}\)
\(\text{9,5}\)
\(\text{586}\)
\(\text{147}\)
\(<-\text{7}\)
Table 4.16: Properties of compounds with different numbers of branched
groups.
Although the densities of all three compounds are very similar, there is a decrease in
density with the increase of branched groups. The vapour pressure increases with
increasing branched groups and the flash point of 2,2-dimethylpropane is significantly
higher than those of pentane and 2-methylbutane.
It is interesting that the common trend of melting and boiling points is not followed here -
2,2-dimethylpropane has the highest melting point, lowest boiling point and is the least
flammable although all three compounds are very flammable. Symmetrical
molecules (such as 2,2-dimethylpropane) tend to have higher melting points than similar
molecules with less symmetry due to their packing in the solid state.
Once in the liquid state they follow the normal trends.
Physical properties and branched chains
Textbook Exercise 4.22
Structural Formula
Boiling point
(℃)
\(\text{78}\)
\(\text{51}\)
Give the IUPAC name for each compound
1-chlorobutane and 2-chloro-2-methylpropane
Explain the difference in the melting points
The branched haloalkane (2-chloro-2-methylpropane)
will have less surface area and thus weaker
intermolecular forces than the straight chain
haloalkane (1-chlorobutane). Less energy will be
needed to break the intermolecular forces of
2-chloro-2-methylpropane, and therefore a lower
boiling point is observed.
There are five isomers with the molecular formula
\(\text{C}_{6}\text{H}_{14}\).
Draw the structural representations of:
2-methylpentane and 2,2-dimethylbutane (two of the
isomers)
What are the names of the other three isomers?
hexane, 3-methylpentane and 2,3-dimethylbutane
Draw the semi-structural representations of these
three molecules.
The melting points of these three isomers are:
\(-\text{118}\), \(-\text{95}\) and
\(-\text{130}\) \(\text{℃}\). Assign the
correct melting points to the correct isomer.
Give a reason for your answers.
The lowest melting point will be with the molecule
with the weakest intermolecular forces. This
molecule is 2,3-dimethylbutane as it is the
isomer with the most branched chains. The
highest melting point will be with the molecule
with the strongest intermolecular forces. This
molecule is hexane as it is the isomer with no
branched groups.
Therefore the melting points will be:
2,3-dimethylbutane: \(-\text{130}\)
\(\text{℃}\)
3-methylpentane: \(-\text{118}\)
\(\text{℃}\)
hexane: \(-\text{95}\)
\(\text{℃}\)
Physical properties of organic compounds
Textbook Exercise 4.23
The table shows data collected for four organic compounds
(A - C) during a practical
investigation.
Compound
Molecular Mass
(\(\text{g.mol$^{-1}$}\))
Boiling point (℃)
A
\(\text{CH}_{3}\text{CH}_{2}\text{CH}_{3}\)
\(\text{44,08}\)
\(-\text{42}\)
B
\(\text{CH}_{3}\text{CHO}\)
\(\text{44,04}\)
\(\text{20}\)
C
\(\text{CH}_{3}\text{CH}_{2}\text{OH}\)
\(\text{46,06}\)
\(\text{78}\)
Is compound A saturated or unsaturated? Give
a reason for your answer.
Saturated. It contains only single carbon-carbon bonds.
To which homologous series does compound B
belong?
aldehydes
Write down the IUPAC name for each of the following
compounds:
B
ethanal
C
ethanol
Refer to intermolecular forces to explain the difference in
boiling points between compounds A and
C.
Between alkane molecules/molecules of compound A/propane
molecules there are only weak van der Waals
forces/intermolecular forces.
Between alcohol molecules/molecules of compound D/ethanol
molecules there are strong hydrogen bonds (as well as
weak van der Waals forces).
More energy is needed to overcome the intermolecular
forces/hydrogen bonds between alcohol molecules/ethanol
molecules/molecules of compound C
OR
Less energy is needed to overcome the intermolecular
forces/van der Waals forces between alkane
molecules/propane molecules/molecules of compound A
Which one of compounds B or
C will have the highest vapour pressure
at a specific temperature? Give a reason for your
answer.
B. A lower boiling point (or weaker
intermolecular forces) means that more molecules will go
into the vapour phase, resulting in a higher vapour
pressure.
Give the IUPAC names for the following compounds and answer the
questions that follow.
Boiling point = \(\text{118}\) \(\text{℃}\)
butan-1-ol
Boiling point = \(\text{99}\) \(\text{℃}\)
butan-2-ol
Boiling point = \(\text{82,5}\) \(\text{℃}\)
2-methylpropan-2-ol
Explain the difference in boiling points.
The hydrogen bonding hydroxyl group is located on an end
carbon atom in butan-1-ol. This allows it to form
hydrogen bonds more easily than butan-2-ol. The branched
group in 2-methylpropan-2-ol means that it is even more
difficult for hydrogen bonds to form between these
molecules (weak intermolecular forces). The greater the
intermolecules forces, the more energy that is required
to overcome them, and the higher the boiling point will
be.