Why Methanol Boils Higher Than Methane: Molecular Attractions

At room temperature, methane is a gas, but methanol is a liquid. Both molecules contain just one carbon atom, and their molar masses aren't dramatically different. So why does methanol have to be heated to about 65 °C before it boils, while methane boils at approximately −162 °C?

The answer is the attractions between whole molecules. Methanol molecules form hydrogen bonds with one another. Methane molecules cannot. This difference makes methanol much harder to separate into individual gas-phase molecules, even though boiling does not break the covalent bonds within either molecule.

## The boiling-point difference: 226 °C

At standard atmospheric pressure (101.325 kPa), the normal boiling points are approximately:

| Substance | Molecular formula | Molar mass (g mol⁻¹) | Normal boiling point |
|---|---|---:|---:|
| Methane | CH₄ | 16.04 | −161.5 °C |
| Methanol | CH₃OH | 32.04 | 64.7 °C |

Methanol boils 226.2 °C higher than methane.

That is a substantial difference for two molecules containing only one carbon atom each. The explanation isn't simply that methanol is heavier. We need to look at the complete molecular structures and identify the intermolecular forces each substance can experience.

[IMAGE: Side-by-side structural comparison of methane, CH4, and methanol, CH3OH. Show methane's symmetrical tetrahedral arrangement of four C-H bonds and methanol's carbon bonded to three hydrogen atoms and an O-H group. Annotate the polar C-O and O-H bonds, partial charges on oxygen and hydrogen, and the overall non-polar character of methane versus the polar character of methanol.]

01Compare the whole molecular structures

Methane: symmetrical and non-polar

Methane, \(\ce{CH4}\), has a tetrahedral molecular shape. Its four C-H bonds are arranged symmetrically around the central carbon atom.

The small bond polarities cancel because of this symmetry. Methane has no overall permanent dipole.

It also has no suitable hydrogen-bond donor or acceptor sites. Its only intermolecular attractions are London dispersion forces.

These attractions occur because electron distributions fluctuate momentarily, creating temporary dipoles that induce dipoles in neighbouring molecules. They are present even in non-polar substances.

Methanol: polar, with hydrogen bonding

Methanol, \(\ce{CH3OH}\), has a different arrangement.

Its oxygen atom is much more electronegative than carbon and hydrogen. It attracts the shared electrons in the C-O and O-H bonds towards itself.

This produces partially charged regions:

• Oxygen has a partial negative charge, \(\delta^-\).
• The hydrogen attached directly to oxygen has a partial positive charge, \(\delta^+\).
• The carbon-containing end of the molecule is relatively less negative than the oxygen end.

The bond dipoles do not cancel. Methanol is a polar molecule, so neighbouring molecules also experience dipole-dipole attractions.

But there is an additional attraction that makes the biggest difference: hydrogen bonding.

The hydrogen attached to oxygen in one methanol molecule is attracted to a lone pair on the oxygen atom of another methanol molecule.

This is an intermolecular hydrogen bond, represented by a dotted line:

\[ \ce{CH3-O-H \bond{...} O(H)-CH3} \]

The solid O-H bonds belong to individual molecules. The dotted interaction connects two separate molecules.

Notice that methanol experiences all three important types of intermolecular attraction:

Methanol also has a larger electron cloud than methane. It contains 18 electrons per molecule, compared with methane’s 10. This generally gives methanol stronger London dispersion forces as well.

So we should avoid an oversimplified explanation such as “methanol boils higher because it has hydrogen bonds”.

A better explanation is that methanol has stronger overall intermolecular attractions because it can hydrogen bond, has permanent dipole-dipole attractions, and experiences stronger dispersion forces.

Why does this raise the boiling point?

Imagine trying to separate two groups of molecules.

In liquid methane, neighbouring molecules attract one another through relatively weak London dispersion forces. Methane molecules can separate and enter the gas phase at a very low temperature.

In liquid methanol, neighbouring molecules experience stronger overall attractions. In particular, the hydrogen bonds between methanol molecules make separation more difficult.

More thermal energy is needed before methanol molecules can escape the liquid in sufficient numbers for boiling to occur.

There is a useful way to describe this precisely.

A liquid has a vapour pressure: the pressure exerted by its vapour when the liquid and vapour are in equilibrium. At a given temperature, a liquid with stronger overall intermolecular attractions generally has a lower vapour pressure because fewer molecules escape into the vapour.

Boiling occurs when:

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

At standard atmospheric pressure, that external pressure is \(101.325\ \mathrm{kPa}\).

Methanol must reach a much higher temperature than methane before its vapour pressure reaches this value.

The graph represents the important relationship: stronger intermolecular attractions generally mean a lower vapour pressure at the same temperature and a higher normal boiling point.

02How was the boiling-point difference measured?

The explanation comes from combining molecular structure with experimental measurements.

The experimental question is straightforward:

At the same external pressure, what temperature does each pure substance reach when it boils steadily?

There is an important problem with the simplest possible experiment.

Methanol is a liquid at room temperature, so it can be heated in a suitable boiling-point apparatus. Methane, however, is already a gas at room temperature and atmospheric pressure.

Putting methane in an open flask and heating it won’t produce a useful boiling-point measurement. The methane simply escapes as a gas.

The experiment therefore needs two different arrangements.

Measuring methanol

A suitable arrangement uses:

1. A boiling flask containing pure liquid methanol.
2. A controlled heat source beneath the flask.
3. A thermometer positioned in the vapour path.
4. A condenser connected to the flask.
5. A receiving vessel for the condensed vapour.

As the methanol is heated, vapour rises past the thermometer. When steady boiling occurs, the temperature is recorded.

At approximately \(101.325\ \mathrm{kPa}\), the measured temperature is about \(64.7^\circ\mathrm{C}\).

Measuring methane

Methane needs a different arrangement because it must first be cooled sufficiently to exist as a liquid.

A cryogenic, pressure-controlled equilibrium cell can hold liquid methane and its vapour while the temperature and pressure are monitored.

The pressure is maintained at \(101.325\ \mathrm{kPa}\). As the liquid methane is warmed, the temperature at which steady boiling occurs is recorded.

The normal boiling point is approximately \(111.7\ \mathrm{K}\), equivalent to \(-161.5^\circ\mathrm{C}\).

The important experimental controls are:

• The same external pressure for both measurements.
• Pure samples, without contaminants.
• Calibrated temperature sensors.
• Correct thermometer placement.
• A stable temperature reading during boiling.

The observations establish the boiling-point difference. They do not directly measure hydrogen bonds or show individual molecules attracting one another. Those interactions provide the molecular explanation for the measured result.

03A contrasting case: Why oxygen alone isn’t enough

Consider ethanol and dimethyl ether.

Both have the molecular formula \(\ce{C2H6O}\), and both have a molar mass of approximately \(46.07\ \mathrm{g\\,mol^{-1}}\).

However, their normal boiling points are very different.

Ethanol contains an O-H bond. It can donate a hydrogen bond through its partially positive hydrogen and accept a hydrogen bond through a lone pair on its oxygen.

Dimethyl ether contains an oxygen atom with lone pairs, so it can accept hydrogen bonds from suitable molecules such as water. However, it has no hydrogen directly bonded to oxygen, nitrogen, or fluorine. It cannot donate conventional hydrogen bonds to neighbouring dimethyl ether molecules.

Consequently, pure ethanol can form hydrogen bonds between its own molecules, whereas pure dimethyl ether cannot form the same type of self-association.

This comparison exposes a common misconception: simply finding an oxygen atom in a molecular formula does not establish that a pure substance can hydrogen bond with itself.

The arrangement of the atoms matters.

The same reasoning applies to methanol. Its O-H group allows its molecules to form hydrogen bonds with one another. Methane has no equivalent functional group.

For another important comparison, see Why Branched and Straight-Chain Isomers Boil Differently. Molecular shape can influence boiling points too, but the methanol-methane comparison is primarily about the different types of intermolecular attraction.

04What actually happens when methanol boils?

There is one final distinction worth getting right: boiling does not break the covalent bonds inside methanol molecules.

Consider what happens before and after boiling.

In liquid methanol, neighbouring molecules are relatively close together. Hydrogen bonds form between the O-H group of one molecule and the oxygen atom of another.

When methanol boils, molecules gain enough energy to overcome the intermolecular attractions holding them together in the liquid.

The molecules separate and enter the gas phase:

\[ \ce{CH3OH(l) -> CH3OH(g)} \]

Each molecule remains \(\ce{CH3OH}\).

Its carbon-hydrogen, carbon-oxygen, and oxygen-hydrogen covalent bonds remain intact. It is the attractions between separate molecules that are overcome sufficiently for the phase change to occur.

This distinction is important when explaining why methanol has a high boiling point. Its O-H covalent bond is not responsible for the high boiling point. The hydrogen bonding between separate methanol molecules is.

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

05A reliable way to answer boiling-point comparisons

When comparing two molecular substances, use this sequence:

1. Compare the complete molecular structures. Identify the atoms, functional groups, and molecular geometry.
2. Determine molecular polarity. Decide whether the bond dipoles cancel or produce a permanent molecular dipole.
3. Identify the intermolecular forces. Consider London dispersion, dipole-dipole attractions, and hydrogen bonding between molecules of the pure substance.
4. Compare the overall attractions. Consider molecular size and electron count as well as the types of intermolecular forces.
5. Connect the attractions to boiling point. Stronger overall intermolecular attractions generally mean lower vapour pressure at a given temperature and a higher boiling point at the same external pressure.

For methane and methanol, the comparison is:

The key is to explain how the molecular structure creates the intermolecular attractions, rather than simply listing which forces are present.

06Questions and solutions

Question 1

Methane has a normal boiling point of −161.5 °C, while methanol has a normal boiling point of 64.7 °C.

Explain why methanol has a significantly higher boiling point.

Solution 1

Methanol has a higher boiling point because its molecules experience stronger overall intermolecular attractions than methane molecules.

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

Methanol is polar and experiences London dispersion forces, dipole-dipole attractions, and hydrogen bonding. Its O-H group allows the hydrogen atom of one molecule to attract the lone pair on the oxygen atom of another.

More energy is therefore needed to separate methanol molecules during boiling. Its vapour pressure is lower at a given temperature, so it must reach a higher temperature before its vapour pressure equals atmospheric pressure.

Question 2

A student makes the following statement:

“Methanol has a higher boiling point than methane because its O-H covalent bonds are stronger. When methanol boils, these bonds must be broken to allow the molecules to separate.”

Identify the errors in this explanation. Write the correct equation for methanol boiling and explain what happens to the intermolecular and covalent bonds.

Solution 2

The student’s explanation is incorrect because boiling overcomes intermolecular attractions, not the covalent bonds within methanol molecules.

The correct equation is:

\[ \ce{CH3OH(l) -> CH3OH(g)} \]

During boiling, methanol molecules gain sufficient energy to separate from their neighbouring molecules.

The hydrogen bonds, dipole-dipole attractions, and London dispersion forces between molecules are overcome sufficiently for the molecules to enter the gas phase.

However, the covalent O-H and C-O bonds remain intact. Each molecule entering the gas phase is still \(\ce{CH3OH}\).

The high boiling point is therefore caused by strong intermolecular attractions, particularly hydrogen bonding, rather than the strength of the O-H covalent bond.

Question 3

A student wants to experimentally compare the normal boiling points of methane and methanol.

They place each substance in a separate open vessel at room temperature and atmospheric pressure. They heat both vessels and observe that methanol boils, while methane remains a gas.

The student concludes that methane does not have a measurable boiling point.

(a) Explain why the student’s conclusion is incorrect.

(b) Describe a suitable experimental arrangement for measuring methane’s normal boiling point.

(c) State two variables that must be controlled to make the comparison valid.

Solution 3

(a) Methane does have a normal boiling point, but it is already a gas at room temperature and atmospheric pressure.

Its normal boiling point is approximately \(-161.5^\circ\mathrm{C}\).

The student cannot observe methane boiling in an open vessel at room temperature because there is no liquid methane present to undergo the liquid-to-gas phase change.

The absence of visible boiling does not mean the substance has no boiling point.

(b) A cryogenic, pressure-controlled equilibrium cell should be used.

The methane must first be cooled sufficiently to form a liquid.

The cell should maintain an external pressure of \(101.325\ \mathrm{kPa}\) while the liquid methane is warmed. A calibrated temperature probe records the temperature when steady boiling occurs.

The measured normal boiling point should be approximately:

\[ T_b=111.7\ \mathrm{K} \]

Converting to Celsius:

\[ T_b=111.7-273.15 \]

\[ \boxed{T_b=-161.5^\circ\mathrm{C}} \]

(c) Two important controlled variables are external pressure and sample purity.

The external pressure must be the same because boiling point depends on pressure.

Both samples must also be sufficiently pure because impurities can change the measured boiling behaviour.

Thermometer calibration and positioning should also be consistent with the measurement method.

Question 4

Pure methanol normally boils at approximately \(64.7^\circ\mathrm{C}\) when the external pressure is \(101.325\ \mathrm{kPa}\).

A student places methanol in an apparatus operating at a much lower external pressure. The methanol now boils at \(50^\circ\mathrm{C}\).

The student concludes:

“The hydrogen bonds between methanol molecules must have become weaker because methanol is boiling at a lower temperature.”

Evaluate this conclusion. Explain why the boiling temperature changes even though the chemical identity and molecular structure of methanol remain unchanged.

Solution 4

The student’s conclusion is incorrect. The lower boiling temperature results from the reduced external pressure, not the disappearance or removal of hydrogen bonding.

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

At atmospheric pressure, methanol must reach approximately \(64.7^\circ\mathrm{C}\) for its vapour pressure to reach \(101.325\ \mathrm{kPa}\).

At a lower external pressure, methanol reaches the required vapour pressure at a lower temperature. It can therefore boil at \(50^\circ\mathrm{C}\) without changing its molecular structure.

The same types of intermolecular attractions are still present.

This demonstrates why boiling-point comparisons must specify the external pressure. The normal boiling point is defined at \(101.325\ \mathrm{kPa}\), allowing different substances to be compared under the same conditions.

Question 5

A student wants to determine whether hydrogen bonding alone explains the \(226.2^\circ\mathrm{C}\) difference between the normal boiling points of methane and methanol.

They collect the following information:

The student then finds another comparison:

They argue:

“Because ethanol and dimethyl ether have the same molecular formula and molar mass, their boiling-point difference must be entirely due to hydrogen bonding. Therefore, the entire boiling-point difference between methanol and methane must also be caused by hydrogen bonding.”

Evaluate this argument.

Your response should:

• Explain the intermolecular attractions present in methane and methanol.
• Explain why ethanol and dimethyl ether behave differently.
• Assess whether the second comparison proves that hydrogen bonding accounts for the entire methane-methanol boiling-point difference.
• Explain why a difference in boiling temperatures cannot be used directly as a measurement of individual hydrogen-bond strength.

Solution 5

The student’s argument is partly correct, but the final conclusion is not justified. Hydrogen bonding is an important reason methanol boils higher than methane, but the entire \(226.2^\circ\mathrm{C}\) difference cannot be attributed to hydrogen bonding alone.

First, consider methane and methanol.

Methane has 10 electrons and a molar mass of \(16.04\ \mathrm{g\\,mol^{-1}}\). It is non-polar and experiences only London dispersion forces.

Methanol has 18 electrons and a molar mass of \(32.04\ \mathrm{g\\,mol^{-1}}\). Its larger electron cloud produces stronger London dispersion attractions. It is also polar, so it experiences dipole-dipole attractions.

Most importantly, methanol has an O-H group that allows hydrogen bonding between neighbouring molecules.

Therefore, several factors contribute to the stronger overall intermolecular attractions in methanol.

Next, consider ethanol and dimethyl ether.

Both substances have the same molecular formula and molar mass, so they have the same number of electrons.

However, their atoms are connected differently.

Ethanol has the structure:

\[ \ce{CH3CH2OH} \]

Its O-H bond allows it to donate hydrogen bonds, while its oxygen lone pairs allow it to accept hydrogen bonds.

Dimethyl ether has the structure:

\[ \ce{CH3OCH3} \]

Its oxygen has lone pairs, so it can accept hydrogen bonds from substances such as water. However, it has no hydrogen directly bonded to oxygen, so it cannot donate conventional hydrogen bonds to neighbouring dimethyl ether molecules.

Ethanol can therefore form hydrogen bonds with other ethanol molecules, while dimethyl ether cannot form the same self-associated hydrogen-bond network.

This helps explain why ethanol boils at a much higher temperature.

Finally, consider what the second comparison actually proves.

The tempting argument is that ethanol and dimethyl ether have the same formula and molar mass, so hydrogen bonding must be the only difference.

However, equal molecular formula and molar mass do not guarantee identical molecular geometry or polarity. Their different atomic arrangements also affect their permanent dipoles and intermolecular interactions.

The comparison provides strong evidence that the ability to form hydrogen bonds between molecules can substantially increase boiling point. It does not isolate hydrogen bonding as the only possible contributing factor.

The same limitation applies to methane and methanol. Their molecular masses, electron counts, polarities, and intermolecular forces all differ.

Nor can we interpret the \(226.2^\circ\mathrm{C}\) difference as the energy of a hydrogen bond. Boiling point is a temperature at which vapour pressure equals external pressure. It is not a direct measurement of an individual intermolecular attraction.

To calculate the energy required to vaporise a substance, we would need thermochemical information such as its enthalpy of vaporisation. Even that describes the overall liquid-to-gas transition, rather than the strength of one individual hydrogen bond.

The strongest conclusion is that hydrogen bonding is a major contributor to methanol’s higher boiling point, alongside its greater dispersion forces and permanent dipole-dipole attractions.

The next time you encounter an unexpected boiling-point comparison, don’t jump straight to molecular mass or the presence of an O-H group. Compare the complete structures, identify the attractions between molecules of each pure substance, and then connect those attractions to vapour pressure.

That approach will also help when you encounter boiling-point differences caused by molecular branching, shape, or changes in functional groups.