How to Rank Boiling Points When Intermolecular Forces Compete

Which would need the highest temperature to boil: octane, butan-1-ol, or 1-bromobutane?

The first prediction might be butan-1-ol. Its molecules can form hydrogen bonds, while octane molecules cannot. Another student might choose 1-bromobutane because it has the greatest molar mass.

Both predictions sound reasonable. Neither gives the correct ranking.

The key to ranking boiling points when several intermolecular forces compete is to compare the whole molecular structures and the combined attractions between their molecules. You must consider hydrogen bonding, dipole-dipole attractions, London dispersion forces, molecular size, and shape. There is no universal ranking that works for every comparison.

Before working through the method, keep one condition in mind: all boiling-point comparisons below assume pure substances at the same external pressure. Normal boiling points are measured at \(101.325\ \mathrm{kPa}\).

01What actually determines a boiling point?

Think about what happens when a liquid boils.

Its molecules move from being relatively close together in the liquid to being widely separated in the gas. The attractions between neighbouring molecules must be overcome sufficiently for this separation to occur.

The covalent bonds holding the atoms together within each molecule do not normally break.

For example, when octane boils:

\[ \ce{C8H18(l) -> C8H18(g)} \]

The octane molecules remain \(\ce{C8H18}\). They simply move further apart.

This is important because boiling point is determined by intermolecular attractions, not by the strength of the covalent bonds within a molecule.

If you need to refresh this distinction, see Intermolecular vs Intramolecular Forces in HSC Chemistry.

Why do stronger attractions generally mean a higher boiling point?

A liquid has a vapour pressure. This is the pressure exerted by molecules that have escaped from the liquid into the gas phase.

As temperature increases, more molecules have enough kinetic energy to escape, so the vapour pressure increases.

Boiling occurs when:

\[ \text{Vapour pressure}=\text{External pressure} \]

Imagine two liquids at the same temperature.

If the intermolecular attractions in liquid A are stronger overall, its molecules generally have greater difficulty escaping into the gas phase. Its vapour pressure will therefore be lower than that of liquid B.

Liquid A must be heated to a higher temperature before its vapour pressure reaches the external pressure.

Therefore:

Stronger overall intermolecular attractions generally produce a higher boiling point.

However, this is a comparison of the combined behaviour of the liquid. Boiling point is not a direct measurement of one particular intermolecular force.

02How is a boiling point measured?

It helps to understand where the boiling-point data actually comes from.

A common method uses a simple distillation apparatus. A pure liquid is heated in a flask, and its vapour passes through a thermometer and into a water-cooled condenser.

The measurement follows a clear sequence.

Step 1: Heat the pure liquid.

The liquid is placed in the distillation flask and heated. Its molecules gain kinetic energy and some escape into the space above the liquid.

Step 2: Measure the vapour temperature.

The vapour rises through the flask neck and passes the thermometer bulb.

The thermometer should be positioned in the vapour stream, near the entrance to the side arm. Placing it deep inside the liquid would measure the liquid temperature rather than the temperature of the escaping vapour.

Step 3: Condense the vapour.

The vapour travels through the condenser, where cooling water removes energy. The vapour condenses back into a liquid and collects in the receiving flask.

Step 4: Record the temperature plateau.

For a pure substance boiling at constant pressure, the temperature remains approximately constant while the liquid changes into gas.

This stable temperature is the boiling point at the pressure of the experiment.

What does the experiment tell us?

Before taking measurements, we would predict that a liquid with stronger overall intermolecular attractions should require a higher temperature to boil.

The experiment tests that prediction by measuring the temperature at which each pure substance boils.

There are two important controls:

• Pressure: A liquid boils at a different temperature when the external pressure changes. All substances must be compared at the same pressure.
• Thermometer position: The thermometer must measure the vapour temperature, not the temperature of the heating source or the liquid at the bottom of the flask.

A normal boiling point specifically refers to a pressure of \(101.325\ \mathrm{kPa}\).

This gives us a reliable way to compare experimental boiling-point data with predictions based on molecular structure.

03Identify the intermolecular forces from the whole molecule

Before comparing boiling points, identify which attractions each molecule can experience.

There are three main categories to recognise in HSC Chemistry.

1. London dispersion forces

London dispersion forces occur between all molecules, including non-polar molecules.

Electrons are constantly moving. At any instant, their distribution may be slightly uneven, producing a temporary dipole.

This temporary dipole can distort the electron cloud of a neighbouring molecule, inducing another dipole. The resulting electrostatic attraction is a London dispersion force.

The strength of these forces generally increases when molecules have:

• More electrons.
• Larger, more easily distorted electron clouds.
• Greater effective surface contact with neighbouring molecules.

The ability of an electron cloud to be distorted is called its polarisability.

A larger molecule generally has a more polarisable electron cloud, although molecular shape also matters.

2. Dipole-dipole attractions

Polar molecules have a permanent dipole.

This means one region of the molecule is partially positive, while another is partially negative.

When two polar molecules approach each other, their oppositely charged regions attract.

For example, a carbonyl group contains a polar C=O bond. In propanone, the molecule has a permanent dipole, so propanone molecules experience dipole-dipole attractions.

Remember that a polar bond does not automatically make an entire molecule polar. You must consider the molecular geometry and whether the individual bond dipoles cancel.

3. Hydrogen bonding

Hydrogen bonding is a particularly strong, directional intermolecular attraction.

For the HSC model, look for two features:

• A hydrogen atom directly bonded to nitrogen, oxygen, or fluorine. This provides a hydrogen-bond donor.
• A lone pair on nitrogen, oxygen, or fluorine on a neighbouring molecule. This provides a hydrogen-bond acceptor.

For example, alcohol molecules contain an O-H group. Their oxygen atoms also have lone pairs.

This means one alcohol molecule can donate a hydrogen bond to the oxygen atom of another alcohol molecule.

One detail is particularly important: hydrogen bonding is not an additional force that exists instead of London dispersion and dipole-dipole attractions. Molecules that hydrogen-bond also experience London dispersion forces, and generally have permanent dipoles.

In HSC comparisons, it is useful to identify hydrogen bonding separately because it has specific structural requirements and is generally a particularly strong intermolecular attraction.

Can a molecule hydrogen-bond with itself?

This is where many otherwise good boiling-point predictions go wrong.

Consider two molecules:

• Ethanol, \(\ce{CH3CH2OH}\).
• Dimethyl ether, \(\ce{CH3OCH3}\).

Both contain oxygen, and both have lone pairs on oxygen.

However, only ethanol has a hydrogen directly bonded to oxygen.

Ethanol can donate and accept hydrogen bonds. Dimethyl ether can accept hydrogen bonds, but it cannot donate conventional hydrogen bonds to another dimethyl ether molecule.

This means pure ethanol can form hydrogen bonds between its own molecules, while pure dimethyl ether cannot.

An ether can still hydrogen-bond with water because water molecules can donate hydrogen bonds to the ether oxygen.

The distinction between hydrogen-bond donors and acceptors is essential when comparing the boiling points of alcohols, ethers, and other oxygen-containing molecules.

04A reliable method for ranking competing boiling points

Now combine the three types of attraction with the physical features of the molecules.

Use the following process whenever an HSC question asks you to rank boiling points.

Step 1: Check that you are comparing molecular substances

Make sure the substances consist of discrete covalent molecules.

The method in this article applies to covalent molecular substances. Ionic compounds and giant covalent networks require a different structural explanation.

Step 2: Identify all intermolecular forces

Start with London dispersion forces. Every molecule has them.

Then check whether each molecule is polar and experiences dipole-dipole attractions.

Finally, look for hydrogen-bond donors and acceptors.

Do not assume that every molecule containing oxygen can hydrogen-bond with itself.

Step 3: Compare molecular size and polarisability

Look at the entire molecule, not just its functional group.

Consider:

• Number of electrons.
• Molar mass.
• Presence of large, polarisable atoms.
• Size and extent of the electron cloud.

A larger electron cloud is generally more polarisable, producing stronger London dispersion forces.

Molar mass is useful evidence, but it is not a complete measure of intermolecular attraction.

Step 4: Consider molecular shape

Two molecules can have the same molecular formula and molar mass but different boiling points.

An extended molecule can often make more effective contact with neighbouring molecules than a compact, highly branched molecule.

This can increase London dispersion attractions.

For a deeper explanation, see How Molecular Shape Explains Boiling-Point Differences.

Step 5: Compare the combined attractions

This is the most important step.

Do not automatically write:

\[ \text{Hydrogen bonding} > \text{Dipole-dipole} > \text{London dispersion} \]

as a universal ranking of boiling points.

For molecules of similar size, hydrogen bonding often produces a substantial increase in boiling point.

However, a large non-polar molecule can have stronger overall attractions than a much smaller polar molecule. A molecule containing a highly polarisable atom can also have substantial London dispersion forces.

You must decide which effects dominate in the particular comparison.

Step 6: State the ranking and justify it

A strong HSC explanation should identify:

1. The intermolecular forces present in each substance.
2. The relevant differences in molecular size, polarisability, and shape.
3. Which substance has stronger overall intermolecular attractions.
4. Why this produces a higher or lower boiling point.

Avoid vague statements such as “it has stronger forces because it is bigger”.

Explain what being bigger changes.

07Common mistakes when ranking boiling points

08Questions and solutions

Try each question before reading its solution. The questions progress from identifying familiar forces to comparing structures where several effects compete.

Question 1

Rank the following substances in order of increasing normal boiling point:

• Ethane: \(\ce{CH3CH3}\)
• Ethanal: \(\ce{CH3CHO}\)
• Ethanol: \(\ce{CH3CH2OH}\)

Explain your ranking using intermolecular forces.

Solution 1

Increasing boiling-point order: ethane < ethanal < ethanol.

Ethane is non-polar, so its molecules experience London dispersion forces only.

Ethanal contains a polar carbonyl group, \(\ce{C=O}\). Its molecules experience London dispersion forces and dipole-dipole attractions. However, ethanal does not have a hydrogen directly bonded to oxygen, so it cannot form conventional hydrogen bonds with other ethanal molecules.

Ethanol contains an O-H group. Its molecules experience London dispersion forces, dipole-dipole attractions, and hydrogen bonding.

The additional hydrogen bonding in ethanol produces stronger overall intermolecular attractions than in ethanal.

Therefore, ethanol has the highest boiling point, while ethane has the lowest.

Question 2

Propan-1-ol and propan-2-ol are structural isomers with the molecular formula \(\ce{C3H8O}\).

Their approximate normal boiling points are:

Both molecules contain an O-H group and can form hydrogen bonds.

Explain why propan-1-ol has the higher boiling point, even though both substances experience the same categories of intermolecular forces.

Solution 2

Propan-1-ol has the higher boiling point because its more extended molecular structure allows more effective contact between neighbouring molecules, strengthening their overall intermolecular attractions.

Both substances have the same molecular formula and molar mass.

Both contain an O-H group, meaning both can donate and accept hydrogen bonds. They also experience dipole-dipole attractions and London dispersion forces.

The difference is their molecular structure.

Propan-1-ol has a more extended arrangement, while propan-2-ol has a more compact arrangement because the OH-bearing carbon is in the middle of the carbon skeleton.

The more compact structure reduces the effectiveness of intermolecular contact and changes the overall attractions between molecules.

Therefore, even though both alcohols can hydrogen-bond, their boiling points are not identical.

The important lesson is that identifying the types of intermolecular forces is only the first part of the comparison. Molecular shape still matters when the molecules have similar functional groups and molar masses.

Question 3

Three pure substances have the following structures and normal boiling points:

(a) Rank the substances in order of increasing boiling point.

(b) Explain why 1-chlorobutane has a higher boiling point than hexane, despite hexane having a longer carbon chain.

(c) Explain why butan-1-ol has a higher boiling point than both substances.

Solution 3

(a) The increasing boiling-point order is hexane < 1-chlorobutane < butan-1-ol.

(b) 1-Chlorobutane has a higher boiling point than hexane because it is polar and experiences dipole-dipole attractions in addition to London dispersion forces.

Hexane is non-polar and experiences London dispersion forces only.

1-Chlorobutane contains a polar C-Cl bond and has a permanent molecular dipole. It therefore experiences dipole-dipole attractions as well as London dispersion forces.

Although hexane has a longer carbon chain, the additional dipole-dipole attractions in 1-chlorobutane contribute to its greater overall intermolecular attractions in this comparison.

This produces the measured difference:

\[ 78.4^\circ\mathrm{C}>68.7^\circ\mathrm{C} \]

(c) Butan-1-ol has the highest boiling point because its molecules form hydrogen bonds with neighbouring molecules.

Butan-1-ol also experiences London dispersion forces and dipole-dipole attractions.

The hydrogen bonding provides an additional strong intermolecular attraction that is absent in both 1-chlorobutane and hexane.

This gives butan-1-ol the highest boiling point, despite its lower molar mass.

The question demonstrates why you must consider the type of intermolecular attraction as well as molecular size.

Question 4

Two substances have similar molar masses:

• n-Pentane: \(\ce{CH3CH2CH2CH2CH3}\), molar mass \(72.15\ \mathrm{g\,mol^{-1}}\).
• Ethoxyethane (diethyl ether): \(\ce{CH3CH2OCH2CH3}\), molar mass \(74.12\ \mathrm{g\,mol^{-1}}\).

Their normal boiling points are:

(a) Which substance has the higher boiling point?

(b) Explain why ethoxyethane does not have the higher boiling point, despite experiencing dipole-dipole attractions as well as London dispersion forces.

(c) Does the result mean London dispersion forces are always stronger than dipole-dipole attractions? Explain.

Solution 4

(a) n-Pentane has the slightly higher boiling point, at \(36.1^\circ\mathrm{C}\), compared with \(34.6^\circ\mathrm{C}\) for ethoxyethane.

(b) Ethoxyethane experiences dipole-dipole attractions and London dispersion forces, but n-pentane has sufficiently strong London dispersion attractions to produce a slightly higher boiling point.

Both molecules have similar molar masses and therefore broadly similar overall sizes.

Ethoxyethane is polar because of its C-O bonds and molecular geometry. It experiences both London dispersion and dipole-dipole attractions.

However, it cannot form conventional hydrogen bonds with other ethoxyethane molecules because it has no hydrogen directly bonded to oxygen.

n-Pentane is non-polar and experiences London dispersion forces only.

Its extended carbon chain allows effective contact between neighbouring molecules, producing substantial dispersion attractions.

For this particular pair, those attractions are sufficient to give n-pentane the slightly higher boiling point.

(c) No. This result does not establish that London dispersion forces are always stronger than dipole-dipole attractions.

It shows that intermolecular force categories cannot be ranked using one universal hierarchy.

The overall result depends on molecular size, polarisability, shape, and the combined attractions between neighbouring molecules.

In this case, the dispersion attractions in n-pentane are sufficient to outweigh the additional dipole-dipole attractions in ethoxyethane.

The small difference of \(1.5^\circ\mathrm{C}\) also reminds us that qualitative predictions become less certain when two substances have very similar boiling points.

Question 5

Three pure molecular substances have the same molecular formula, \(\ce{C4H10O}\), and the same molar mass, approximately \(74.12\ \mathrm{g\,mol^{-1}}\).

Their structures and normal boiling points are shown below.

A student makes the following argument:

“All three substances have the same molecular formula, so their intermolecular attractions should be similar. Ethoxyethane contains an oxygen atom with lone pairs, so it can hydrogen-bond with itself. Also, A and B both contain an O-H group, so they should have almost identical boiling points.”

Evaluate this argument.

In your response:

(a) Rank A, B, and C in order of decreasing boiling point.

(b) Explain why A has a higher boiling point than B, even though both can form hydrogen bonds.

(c) Explain why C cannot form conventional hydrogen bonds with itself but can form hydrogen bonds with water.

(d) Explain what happens to the covalent bonds within each molecule when these substances boil.

Solution 5

(a) The decreasing boiling-point order is A > B > C.

\[ \boxed{ \text{Butan-1-ol} > \text{2-methylpropan-1-ol} > \text{ethoxyethane} } \]

(b) A has a higher boiling point than B because both substances can hydrogen-bond, but A has a less compact molecular structure and experiences stronger overall intermolecular attractions.

Both A and B are alcohols.

Each molecule contains an O-H group that can donate hydrogen bonds and an oxygen atom with lone pairs that can accept hydrogen bonds.

Both therefore experience:

• London dispersion forces.
• Dipole-dipole attractions.
• Hydrogen bonding.

Their molecular formula and molar mass are also identical.

The important difference is their shape.

Butan-1-ol has an extended, unbranched carbon chain. In contrast, 2-methylpropan-1-ol has a branched, more compact carbon skeleton.

The extended structure of A allows more effective contact between neighbouring molecules, increasing the overall London dispersion attractions.

Both substances can hydrogen-bond, but their remaining intermolecular attractions are not identical.

Consequently, butan-1-ol has the higher boiling point.

(c) C cannot form conventional hydrogen bonds with itself because it has no hydrogen directly bonded to oxygen, but it can accept hydrogen bonds from water.

Ethoxyethane contains an oxygen atom with two lone pairs.

This means its oxygen can accept a hydrogen bond from another molecule.

However, both hydrogen atoms and carbon atoms in ethoxyethane are connected through C-H bonds. There is no O-H, N-H, or F-H bond that allows ethoxyethane to donate a conventional hydrogen bond.

Therefore, two ethoxyethane molecules cannot form conventional hydrogen bonds with each other.

Now consider water.

Water contains O-H bonds, so its molecules can donate hydrogen bonds to the oxygen atom in ethoxyethane.

The interaction can be represented as:

\[ \ce{H-O-H\bond{...}O(CH2CH3)2} \]

The dotted interaction represents a hydrogen bond between water’s partially positive hydrogen and ethoxyethane’s partially negative oxygen.

This reveals an important distinction:

A molecule does not need an O-H group to accept a hydrogen bond from another substance. However, it needs an appropriate donor and acceptor interaction to form conventional hydrogen bonds with identical molecules in its pure liquid.

The ability to hydrogen-bond with water therefore does not automatically mean a substance can hydrogen-bond with itself.

(d) The covalent bonds within all three molecules remain intact during ordinary boiling.

Boiling separates whole molecules by overcoming intermolecular attractions.

For example, butan-1-ol changes from:

\[ \ce{CH3CH2CH2CH2OH(l)} \]

to:

\[ \ce{CH3CH2CH2CH2OH(g)} \]

The molecules remain butan-1-ol.

The C-C, C-H, C-O, and O-H covalent bonds within each molecule are not broken by the ordinary boiling process.

The same principle applies to 2-methylpropan-1-ol and ethoxyethane.

The student’s argument therefore contains two major errors:

• Having the same molecular formula does not guarantee the same molecular shape or intermolecular attractions.
• Having an oxygen atom with lone pairs does not automatically allow a molecule to hydrogen-bond with identical molecules.

This is why ranking boiling points requires more than identifying a functional group. You must examine the complete molecular structure and consider how its molecules interact in the liquid.

09Where to take this next

Once you can rank boiling points by considering all competing intermolecular forces, the next useful skill is explaining why boiling-point trends change across homologous series and between structural isomers.

In particular, learn to connect increasing carbon-chain length, molecular branching, and electron-cloud polarisability to London dispersion forces.

These ideas allow you to move beyond simply identifying intermolecular forces and make much stronger predictions about the physical properties of unfamiliar molecular substances.