Why a Larger Nonpolar Molecule Can Have a Higher Boiling Point

A larger nonpolar molecule can have a higher boiling point than a smaller polar molecule because its London dispersion forces can be stronger overall than the smaller molecule’s dipole-dipole attractions. Being polar gives a molecule an additional type of intermolecular attraction, but it does not automatically give it the higher boiling point.

The comparison depends on the whole molecules: their electron clouds, polarizability, size, shape, and the combined attractions between neighbouring molecules. A large nonpolar molecule can therefore hold together as a liquid more strongly than a smaller polar molecule.

## Why polarity does not decide the boiling point on its own

Imagine comparing two liquids:

- Hexane, \(\ce{C6H14}\), is a relatively large, nonpolar molecule.
- Propanone (acetone), \(\ce{C3H6O}\), is a smaller, polar molecule.

Which should boil at the higher temperature?

You might predict propanone. Its molecules have permanent dipoles, so they experience dipole-dipole attractions as well as London dispersion forces. Hexane has no permanent dipole, so it experiences London dispersion forces only.

That prediction sounds reasonable, but the measured normal boiling points tell a different story:

| Substance | Molecular formula | Molar mass (g mol\(^{-1}\)) | Polarity | Normal boiling point |
|---|---|---:|---|---:|
| Hexane | \(\ce{C6H14}\) | 86.18 | Nonpolar | \(68.7^\circ\text{C}\) |
| Propanone | \(\ce{C3H6O}\) | 58.08 | Polar | \(56.1^\circ\text{C}\) |

Both values are approximate and refer to a pressure of 1 atm.

Hexane boils \(12.6^\circ\text{C}\) higher than propanone. Its stronger overall intermolecular attractions more than compensate for the absence of permanent dipoles.

[IMAGE: Side-by-side structural comparison of a hexane molecule, CH3-CH2-CH2-CH2-CH2-CH3, and a propanone molecule, CH3-C(=O)-CH3. Show hexane as a longer carbon chain with a larger electron cloud and no permanent dipole. Show propanone with its polar carbonyl bond, partial negative charge on oxygen, and partial positive charge towards the carbonyl carbon. Include neighbouring molecules with dotted intermolecular attractions, distinguishing temporary London dispersion attractions around hexane from permanent dipole-dipole attractions between propanone molecules. Do not depict intermolecular attractions as covalent bonds.]

01What makes London dispersion forces stronger?

The key is the size and flexibility of the electron cloud.

Electrons are constantly moving around a molecule. Their distribution is not perfectly even at every instant. A temporary uneven distribution can create a temporary dipole, even in a molecule that has no permanent dipole.

This temporary dipole can distort the electron cloud of a nearby molecule. The neighbouring molecule develops an induced dipole, and the two molecules attract each other. This attraction is called a London dispersion force.

Every molecule experiences London dispersion forces. This includes polar molecules such as propanone and ethanol.

However, larger molecules generally have more electrons and more easily distorted electron clouds. This property is called polarizability.

A more polarizable molecule can produce stronger temporary and induced dipoles. Its London dispersion attractions with neighbouring molecules are therefore generally stronger.

Consider the electron counts of our original pair:

• Hexane, \(\ce{C6H14}\), has 50 electrons.
• Propanone, \(\ce{C3H6O}\), has 32 electrons.

The electron count is calculated by adding the electrons contributed by each neutral atom.

For hexane:

\[ N_e = 6(6)+14(1)=50 \]

For propanone:

\[ N_e = 3(6)+6(1)+8=32 \]

Hexane has a substantially larger electron cloud. It is more polarizable, allowing stronger overall London dispersion attractions to develop between its molecules.

Propanone has an additional attraction that hexane does not have: dipole-dipole attraction.

But the presence of an extra force does not automatically make the total attraction stronger. In this comparison, hexane’s dispersion attractions are sufficient to give it the higher boiling point.

A polar molecule experiences more than one type of attraction

It is important not to think that polar molecules experience dipole-dipole forces instead of London dispersion forces.

The main attractions to consider are:

Propanone is an interesting case. Its carbonyl group, \(\ce{C=O}\), produces a permanent molecular dipole.

However, pure propanone cannot form conventional hydrogen bonds with itself. Its oxygen can accept a hydrogen bond from another suitable molecule, but propanone has no hydrogen attached to oxygen, nitrogen, or fluorine to donate one.

Ethanol and propan-1-ol both contain an \(\ce{O-H}\) group, so their molecules can form hydrogen bonds with one another.

This distinction becomes useful when you compare different functional groups rather than simply counting polar bonds.

02Boiling point measures the combined attractions

Why do stronger intermolecular attractions lead to a higher boiling point?

Picture molecules moving around in a liquid. Some have enough energy to escape into the space above the liquid. These escaping molecules create vapour pressure.

As the liquid is heated, its vapour pressure increases. Boiling occurs when the vapour pressure equals the external pressure.

For a pure liquid at a fixed external pressure:

• Stronger overall intermolecular attractions generally produce a lower vapour pressure at a given temperature.
• The liquid must reach a higher temperature before its vapour pressure matches the external pressure.
• Its boiling point is therefore higher.

The important word is overall. Boiling point reflects the combined effect of the attractions throughout the liquid. It does not measure the strength of just one type of intermolecular force.

There is another distinction worth keeping clear.

When hexane boils, its molecules separate from one another. The covalent bonds holding the carbon and hydrogen atoms together remain intact.

The physical change is:

\[ \ce{C6H14(l) -> C6H14(g)} \]

The molecular formula does not change. Only the physical state changes.

The same applies to propanone:

\[ \ce{C3H6O(l) -> C3H6O(g)} \]

Boiling overcomes intermolecular attractions sufficiently for molecules to escape the liquid. It does not normally break the covalent bonds within those molecules.

For a more detailed explanation of this distinction, see Why Boiling Water Does Not Break Its Covalent Bonds.

03A useful experimental comparison

Suppose you wanted to investigate whether a large nonpolar molecule can boil above a smaller polar molecule that forms hydrogen bonds.

A particularly useful comparison is:

• n-Octane, \(\ce{C8H18}\), a large, nonpolar alkane.
• Propan-1-ol, \(\ce{C3H8O}\), a smaller, polar alcohol that forms hydrogen bonds.

Before looking at the results, make a prediction.

Propan-1-ol has London dispersion forces, dipole-dipole attractions, and hydrogen bonding. Octane has London dispersion forces only.

If the type of intermolecular force were the only factor, you might predict that propan-1-ol would boil higher.

The measured normal boiling points disagree.

Octane boils approximately \(28.5^\circ\text{C}\) higher.

Its much larger, more polarizable electron cloud produces substantial London dispersion attractions throughout the liquid. These attractions give octane greater overall cohesion than propan-1-ol, despite the latter’s hydrogen bonding.

How would a chemist measure these boiling points?

A fair experimental comparison requires controlled conditions. Simply heating two liquids in separate beakers and recording when they appear to boil is not particularly reliable.

A suitable method uses a distillation apparatus.

The apparatus works as follows:

1. Place a pure liquid sample and anti-bumping granules in the distillation flask.
2. Heat the flask gently until the liquid boils.
3. Vapour travels upwards through the flask and past the thermometer bulb.
4. The vapour enters the condenser, where cooling water removes energy and the vapour returns to liquid.
5. The thermometer records the temperature of the vapour as the liquid boils. Record the stable temperature.

The thermometer bulb belongs in the vapour path, near the side-arm connection, rather than deep inside the liquid.

Pressure must also be recorded. A liquid boils at a lower temperature when the external pressure is reduced. Therefore, two boiling-point measurements must be compared at the same pressure.

The normal boiling point is specifically defined at 101.325 kPa, approximately 1 atm.

The octane and propan-1-ol values above are reference data compiled by the National Institute of Standards and Technology (NIST). They are not results from an experiment performed for this article.

The experimental comparison tests the original prediction: the polar alcohol does not necessarily have the higher boiling point. Molecular size and polarizability can outweigh the additional attractions associated with polarity and hydrogen bonding.

04The contrasting case: when a smaller polar molecule boils higher

Now consider ethanol and hexane.

Ethanol is much smaller than hexane, yet ethanol boils at a higher temperature.

Why?

Ethanol contains an \(\ce{O-H}\) group. Its molecules can form hydrogen bonds with neighbouring ethanol molecules.

These hydrogen bonds provide substantial additional attraction in the liquid. In this particular comparison, they are enough to give ethanol a higher boiling point than hexane.

This gives us three useful comparisons:

• Hexane boils above propanone: the larger nonpolar molecule has stronger overall attractions.
• Ethanol boils above hexane: hydrogen bonding allows the smaller polar molecule to win.
• Octane boils above propan-1-ol: sufficiently strong overall dispersion attractions can outweigh the additional attractions in a smaller hydrogen-bonding molecule.

None of these results contradicts the others.

They show why you must consider the whole molecular structure rather than memorising a fixed ranking of intermolecular forces.

05A reliable method for comparing boiling points

When an HSC question asks you to explain an unexpected boiling-point trend, work through these decisions.

1. Check that you are comparing like conditions.

Are the substances pure? Are the boiling points measured at the same external pressure?

If the pressure differs, the boiling-point comparison may not tell you what you think it does.

2. Identify all intermolecular attractions.

Remember that every molecule experiences London dispersion forces.

Then check for:

• Permanent molecular dipoles, which produce dipole-dipole attractions.
• Suitable hydrogen-bond donors and acceptors, which allow hydrogen bonding.

3. Compare molecular size and polarizability.

Look at the number of electrons, molar mass, and overall size.

A larger electron cloud is generally more polarizable, producing stronger overall dispersion attractions. Remember that molar mass is a useful clue, not a direct measurement of intermolecular force strength.

4. Consider molecular shape as well.

Shape influences how effectively neighbouring molecules can approach and interact. This is particularly useful when comparing molecules with the same molecular formula.

See Why Branched and Straight-Chain Isomers Boil Differently for that separate comparison.

5. Compare the total attractions and explain the boiling-point trend.

Do not simply write that one molecule is polar and the other is nonpolar.

Explain why the combined attractions in one liquid are greater, and connect that difference to the energy needed for molecules to escape into the gas phase.

A strong HSC explanation might read:

Although propanone is polar and experiences dipole-dipole attractions, hexane has a larger, more polarizable electron cloud. Its London dispersion attractions are sufficiently strong to produce greater overall intermolecular cohesion in the liquid. More energy is therefore required for hexane molecules to escape into the gas phase, giving hexane the higher boiling point.

That explanation compares the whole molecules, identifies the relevant forces, and connects them to the observed physical property.

06Worked examples

07Questions and solutions

Question 1

At 1 atm, n-pentane, \(\ce{C5H12}\), has a normal boiling point of \(36.1^\circ\text{C}\). Propanone, \(\ce{C3H6O}\), has a normal boiling point of \(56.1^\circ\text{C}\).

(a) Calculate the difference between their boiling points.

(b) Explain why propanone has the higher boiling point, despite having a lower molar mass.

Solution 1

Propanone has the higher boiling point, by \(20.0^\circ\text{C}\).

Step 1: Identify the intermolecular forces.

Both substances experience London dispersion forces.

Propanone is polar because of its carbonyl group, so it also experiences dipole-dipole attractions. Pentane is nonpolar and does not experience permanent dipole-dipole attractions.

Step 2: Compare molecular size.

Pentane has a molar mass of approximately \(72.15\ \mathrm{g\\,mol^{-1}}\), compared with \(58.08\ \mathrm{g\\,mol^{-1}}\) for propanone.

Pentane has a larger electron cloud, so its London dispersion attractions are generally stronger. However, propanone’s permanent dipole produces additional attractions between its molecules.

For this particular pair, the combined attractions in propanone are greater.

Step 3: Calculate the difference.

\[ \begin{aligned} \Delta T_b &=56.1-36.1\\\ &=20.0^\circ\mathrm{C} \end{aligned} \]

This demonstrates that a smaller polar molecule can boil higher than a larger nonpolar molecule.

It also shows why molecular size alone cannot predict every boiling-point trend.

Question 2

Ethanol and propanone are both polar molecules containing oxygen.

Their normal boiling points are:

• Ethanol: \(78.4^\circ\text{C}\).
• Propanone: \(56.1^\circ\text{C}\).

A student argues that both substances should experience hydrogen bonding because they contain oxygen.

(a) Explain why this statement is incorrect.

(b) Identify the intermolecular attractions experienced by each substance.

(c) Explain why ethanol has the higher boiling point, even though it has a lower molar mass.

(d) Calculate the boiling-point difference.

Solution 2

Ethanol has the higher boiling point because its molecules can form hydrogen bonds with one another, whereas pure propanone cannot form conventional hydrogen bonds with itself.

Step 1: Identify the hydrogen-bonding requirements.

Hydrogen bonding requires a hydrogen atom covalently bonded to a highly electronegative atom such as oxygen, nitrogen, or fluorine, together with a suitable electron-rich acceptor.

Ethanol contains an \(\ce{O-H}\) bond. Its hydrogen can participate in hydrogen bonding with the oxygen of a neighbouring ethanol molecule.

Propanone contains a carbonyl oxygen that can accept hydrogen bonds, but it has no suitable hydrogen-bond donor group.

Therefore, pure propanone molecules do not form conventional hydrogen bonds with one another.

Step 2: Identify all attractions.

Step 3: Explain the boiling-point difference.

Ethanol has a lower molar mass and fewer electrons than propanone, so its London dispersion attractions are generally weaker.

However, hydrogen bonding provides additional attraction between ethanol molecules. The combined attractions in liquid ethanol are sufficient to produce the higher boiling point.

Step 4: Calculate the difference.

\[ \begin{aligned} \Delta T_b &=78.4-56.1\\\ &=22.3^\circ\mathrm{C} \end{aligned} \]

Ethanol boils \(22.3^\circ\text{C}\) higher than propanone.

The important distinction is that having an oxygen atom does not automatically mean a molecule can hydrogen-bond with identical molecules.

Question 3

A student heats pure liquid hexane until it boils.

They write the following explanation:

The energy supplied breaks the carbon-carbon and carbon-hydrogen covalent bonds, allowing the individual atoms to separate and form a gas.

Evaluate this explanation. Describe what actually happens to the hexane molecules during boiling, and write an appropriate chemical equation for the change of state.

Solution 3

The explanation is incorrect. Boiling hexane separates intact hexane molecules rather than breaking their internal covalent bonds.

Step 1: Distinguish the two types of attraction.

Within each hexane molecule, carbon and hydrogen atoms are held together by covalent bonds.

Between separate hexane molecules, there are London dispersion forces.

These attractions have different roles:

• Covalent bonds hold the atoms together within each molecule.
• London dispersion forces attract neighbouring molecules in the liquid.

Step 2: Explain what happens during boiling.

As hexane is heated, its molecules gain kinetic energy and move more rapidly.

At its boiling point, the molecules can overcome enough of the intermolecular attractions holding them in the liquid to escape into the gas phase.

The carbon-carbon and carbon-hydrogen bonds remain intact.

Step 3: Write the correct equation.

\[ \ce{C6H14(l) -> C6H14(g)} \]

The chemical identity of hexane does not change. It remains \(\ce{C6H14}\), but its physical state changes from liquid to gas.

The energy supplied during boiling is associated with separating molecules against their intermolecular attractions, not breaking the covalent bonds within them.

Question 4

A student wants to compare the boiling points of n-octane and propan-1-ol.

They obtain the following results:

The second reading is illustrative experimental data, not a quoted reference value.

The student concludes that octane must have stronger overall intermolecular attractions because its measured boiling point is higher.

(a) Explain the flaw in this conclusion.

(b) Describe how the experiment should be improved to allow a fair comparison.

(c) State the normal boiling points of the two substances and explain what the correctly controlled comparison shows.

Solution 4

The student’s conclusion is not justified because the two substances were measured at different external pressures.

Step 1: Explain the effect of pressure.

Boiling occurs when the vapour pressure of a liquid equals the external pressure.

Reducing the external pressure means the liquid needs a lower vapour pressure to begin boiling. It therefore boils at a lower temperature.

Propan-1-ol was measured at 80.0 kPa, while octane was measured at 101.3 kPa.

Its recorded boiling point is therefore not directly comparable with octane’s normal boiling point.

Step 2: Improve the experiment.

The student should:

• Use pure samples of both substances.
• Use the same calibrated distillation apparatus for both measurements.
• Position the thermometer bulb in the vapour path.
• Maintain the same external pressure for both tests.
• Record the stable temperature during boiling.

For a comparison of normal boiling points, both substances must be measured at 101.325 kPa.

Step 3: Compare the reference values.

At approximately 101.3 kPa:

• n-Octane boils at approximately \(125.7^\circ\text{C}\).
• Propan-1-ol boils at approximately \(97.2^\circ\text{C}\).

The correctly controlled comparison shows that octane has the higher normal boiling point.

Its larger, more polarizable electron cloud produces sufficient overall London dispersion attractions to outweigh the additional intermolecular attractions in propan-1-ol.

The original measurements were not a fair comparison, even though the student’s final ranking happens to agree with the reference data.

Question 5

A student is investigating whether hydrogen bonding or London dispersion forces have a greater effect on boiling points.

They collect the following normal boiling-point data at 1 atm:

Two students interpret the results differently.

Student A argues:

“Hydrogen bonding is stronger than London dispersion forces, so every hydrogen-bonding liquid must boil above every nonpolar liquid.”

Student B argues:

“Octane has the highest boiling point, so London dispersion forces must be stronger than hydrogen bonds.”

Evaluate both arguments.

In your response:

(a) Use the data to identify one comparison that challenges each student’s claim.

(b) Explain why neither conclusion establishes a universal ranking of intermolecular force strengths.

(c) Explain what boiling-point data can establish about the overall attractions in a liquid, and what it cannot establish about an individual intermolecular interaction.

Solution 5

Neither student’s conclusion is valid. Both have confused the strength of an individual type of intermolecular interaction with the combined attractions experienced by an entire liquid.

Step 1: Challenge Student A’s argument.

Compare ethanol and hexane.

Ethanol has hydrogen bonding, while hexane has London dispersion forces only.

Ethanol boils at \(78.4^\circ\text{C}\), compared with hexane at \(68.8^\circ\text{C}\).

This is consistent with hydrogen bonding providing substantial additional attraction in ethanol.

However, it does not prove that every hydrogen-bonding molecule must boil above every nonpolar molecule.

Step 2: Challenge Student B’s argument.

Now compare octane and propan-1-ol.

Octane is nonpolar and experiences London dispersion forces only. Propan-1-ol is polar and can form hydrogen bonds.

Despite this, octane boils at \(125.7^\circ\text{C}\), compared with \(97.2^\circ\text{C}\) for propan-1-ol.

Octane has a much larger electron cloud and substantially greater polarizability. Its overall dispersion attractions are sufficient to give it a higher boiling point.

Step 3: Identify what both students have overlooked.

The molecules differ in more than the type of intermolecular force they experience.

They also differ in:

• Number of electrons.
• Molecular size.
• Electron-cloud polarizability.
• Molecular shape.
• The number and combined effect of intermolecular contacts in the liquid.

It is therefore misleading to compare two liquids by looking at only one type of attraction.

Step 4: Distinguish overall cohesion from individual force strength.

A boiling point reflects the collective behaviour of a liquid.

It tells us how the combined intermolecular attractions influence the liquid’s vapour pressure and the temperature required for boiling at a given external pressure.

It does not directly measure the energy of one individual London dispersion interaction or one individual hydrogen bond.

To compare individual interaction energies, additional evidence and appropriate molecular models would be needed.

Final conclusion: Neither London dispersion forces nor hydrogen bonding can be declared universally dominant in every molecular comparison. The relative boiling points depend on the complete molecular structures and the combined intermolecular attractions in each liquid.

This is the key idea behind the original question: a large nonpolar molecule can boil higher than a smaller polar molecule because the total London dispersion attractions can outweigh the additional dipole-dipole attractions or even hydrogen bonding in the smaller molecule.

08Where to take this next

Once you can compare molecular size, polarity, and intermolecular forces, the next useful step is to compare molecules with the same molecular formula.

In that situation, electron count and molar mass are held constant, so molecular shape becomes particularly important. Branching can change how effectively neighbouring molecules interact, producing different boiling points.

Continue with Why Branched and Straight-Chain Isomers Boil Differently.

For the related distinction between covalent bonds within molecules and intermolecular forces between molecules, see Why Boiling Water Does Not Break Its Covalent Bonds.

Reference data and syllabus scope

Normal boiling-point data are rounded from the NIST Chemistry WebBook, including its entries for n-hexane, acetone, ethanol, n-octane, 1-propanol, and n-pentane. Small differences in the final decimal place can occur between reference compilations.

The relevant current NSW course is the Chemistry Stage 6 (2017) syllabus. NESA schedules implementation of the Chemistry 11-12 Syllabus (2025) from 2028, with the first HSC examination under the new syllabus in 2029. This article develops molecular-structure and intermolecular-force reasoning without claiming that this particular comparison is a named requirement of the 2025 syllabus. NESA syllabus information.