Why Like Dissolves Like Is Useful, but Not the Whole Story

A bottle of cooking oil and a glass of water make a stubborn pair. Shake them together and the oil breaks into tiny droplets, but leave the mixture alone and two layers form again. Yet ethanol and water mix completely. Both ethanol and oil contain carbon and hydrogen, so why does water treat them so differently?

The useful answer is **like dissolves like because molecules dissolve most readily when the interactions they can make with the solvent compensate for the interactions that must be disrupted**. Molecular polarity, hydrogen bonding, and the balance between polar and non-polar regions all matter.

But there is an important catch: you must compare the *whole molecular structure*, not just spot one functional group. And dissolving a substance does not normally mean breaking its covalent bonds. Those are two separate ideas that are easy to confuse in HSC Chemistry.

## Why molecular structure matters more than one functional group

Imagine comparing ethanol and a long-chain alcohol. Both contain an -OH group, so both can form hydrogen bonds with water. You might predict that they should have similar solubilities.

That prediction misses something important. Ethanol has a short non-polar carbon chain. A long-chain alcohol has a much larger non-polar region. As the carbon chain grows, the influence of the polar -OH group becomes less significant relative to the rest of the molecule.

A molecule is not just its most noticeable functional group. Its overall structure determines how it interacts with its surroundings.

[IMAGE: Three clearly drawn molecular structures side by side: ethanol, butan-1-ol and octan-1-ol. Show every carbon and oxygen atom, the O-H bond, and the increasing size of each non-polar hydrocarbon region. Label polar and predominantly non-polar regions without suggesting that the entire molecule has a sharp physical boundary between them.]

01Look at the entire structure, not just the polar group

Consider ethanol, with the structural formula \(\ce{CH3CH2OH}\).

It has two different regions:

• The hydroxyl group (\(\ce{-OH}\)) is polar. It can form hydrogen bonds with water.
• The ethyl group (\(\ce{CH3CH2-}\)) is predominantly non-polar. It interacts mainly through dispersion forces.

Ethanol can therefore interact with both polar and non-polar substances. Its polar region is sufficiently important, relative to its short carbon chain, for ethanol to mix completely with water.

Now compare it with octan-1-ol, \(\ce{CH3(CH2)7OH}\). It still has exactly one hydroxyl group, but its non-polar carbon chain is much longer.

The hydroxyl group can still hydrogen bond with water. However, the much larger hydrocarbon region makes the overall molecule less compatible with water. Consequently, octan-1-ol has very limited solubility in water.

This gives us an important distinction:

Having a polar functional group does not automatically make an entire molecule highly polar or water-soluble.

Three structural features are particularly useful when making predictions.

1. Overall molecular polarity

A molecule’s polarity depends on both the polarity of its bonds and its molecular shape.

For example, carbon dioxide contains two polar carbon-oxygen bonds. However, its linear structure means the two bond dipoles point in opposite directions and cancel.

The molecule is therefore non-polar overall.

This is why simply counting polar bonds is not enough. You need to consider their directions and the shape of the complete molecule.

2. Hydrogen-bonding ability

Hydrogen bonding can make a substantial difference to how well a molecule interacts with water.

A hydrogen-bond donor has a hydrogen atom covalently bonded to a highly electronegative atom, such as oxygen. A hydrogen-bond acceptor has an available lone pair that can participate in hydrogen bonding.

Ethanol can both donate and accept hydrogen bonds.

Propanone, \(\ce{CH3COCH3}\), provides a useful contrast. It has a polar carbonyl group, \(\ce{C=O}\), whose oxygen can accept hydrogen bonds from water. However, propanone cannot donate hydrogen bonds because it has no hydrogen bonded directly to oxygen, nitrogen, or fluorine.

Despite this difference, propanone is completely miscible with water.

That is a useful warning against memorising rules such as “molecules need an -OH group to dissolve in water”.

3. The size of the non-polar region

The size and shape of the hydrocarbon region also matter.

Consider these four substances:

The table shows why functional groups must be considered alongside the rest of the structure.

Butan-1-ol can hydrogen bond with water, but its four-carbon chain limits its compatibility with the solvent. Hexane has no comparable polar group to provide favourable interactions with water.

02What actually happens when a substance dissolves?

Before mixing two substances, their molecules interact with other molecules of the same substance.

For example:

• Water molecules form an extensive network of hydrogen bonds.
• Ethanol molecules interact through hydrogen bonding and dispersion forces.
• Hexane molecules interact predominantly through dispersion forces.

When a solute dissolves, some existing intermolecular attractions must be disrupted so that the molecules can mix.

New attractions form between the solute and solvent molecules.

The important question is whether those new interactions make mixing favourable overall.

Think of it as rearranging a group of people seated at tables. To introduce someone new, existing groups may need to separate and reorganise. Whether the new arrangement is favourable depends on the interactions in both arrangements.

The analogy is limited, of course: molecules do not make conscious decisions, and solubility is governed by molecular interactions and thermodynamics.

For water and ethanol, hydrogen bonds can form between the two types of molecules. These interactions help compensate for the water-water and ethanol-ethanol attractions that are disrupted.

For water and hexane, the new interactions are much less favourable. Hexane cannot replace the hydrogen-bonding interactions between water molecules with interactions of comparable strength.

The result is that the liquids separate into layers.

There is another important qualification. Solubility is not determined by intermolecular attraction alone. The overall thermodynamic balance also includes the increase in entropy associated with mixing. Temperature and the relative amounts of solute and solvent can therefore affect the final result.

This is why like dissolves like is a useful prediction, rather than an exact law.

For a broader explanation of when this prediction works and when it needs qualification, see When Does Like Dissolves Like Actually Predict Solubility?.

03Testing the rule: a simple solubility investigation

How could you test whether molecular structure actually influences solubility?

A useful investigation compares several liquids with water while keeping the experimental conditions consistent.

Imagine four dry, labelled test tubes arranged upright in a test-tube rack.

1. Add 5.0 mL of distilled water to each tube.
2. Add 5.0 mL of a different liquid to each tube: ethanol, propanone, butan-1-ol, and hexane.
3. Seal each tube with a suitable stopper.
4. Invert each tube ten times to mix its contents.
5. Return the tubes to the rack and allow them to stand for 10 minutes at 25°C.

Use the same volumes, temperature, mixing procedure, and settling time for every sample.

Before looking at the expected results, make a prediction.

If the presence of an -OH group were the only important factor, would butan-1-ol behave just like ethanol?

No. The longer carbon chain changes the overall molecular structure, so we should expect different behaviour.

The following table gives the expected qualitative outcomes. These are illustrative predictions for the investigation, not measurements from an experiment.

The hexane forms the upper layer because it is less dense than water.

But notice something important about butan-1-ol. Two layers do not mean that absolutely no butan-1-ol has dissolved. Some butan-1-ol is present in the water-rich layer, and some water is present in the butanol-rich layer.

This distinction matters when interpreting experimental observations.

What the first design cannot tell us

Suppose a student shakes the water and hexane mixture and immediately records that it has become cloudy.

The student might conclude that the liquids have mixed.

That conclusion is premature. Shaking can disperse one liquid into tiny droplets throughout the other. This creates an emulsion, which can look cloudy even though the liquids have not formed a single homogeneous solution.

Allowing the mixture to settle helps distinguish a temporary dispersion from a stable single phase.

There is another limitation. Observing one clear phase at a single mixing ratio does not necessarily prove that two liquids are miscible in all proportions.

A revised investigation should test several different ratios. If quantitative solubility is required, the equilibrium composition of each phase must also be measured using a suitable analytical method.

04Worked examples: applying the rule to real molecular structures

05Boiling and dissolving do not normally break covalent bonds

There is a second misconception worth separating from solubility.

A student might hear that boiling separates molecules, then assume that the covalent bonds inside those molecules must also break.

That is not what happens.

Consider a beaker of water being heated to its boiling point.

What happens during boiling?

Water molecules attract one another through intermolecular hydrogen bonding.

As the water is heated, the molecules gain kinetic energy. At the boiling point, bubbles of water vapour form throughout the liquid.

Energy is required to separate the water molecules from one another.

However, the covalent O-H bonds inside each water molecule remain intact.

A water molecule in the vapour phase is still \(\ce{H2O}\), not separate hydrogen and oxygen atoms.

The process is a physical change: liquid water becomes gaseous water.

What happens during dissolving?

Now consider ethanol being mixed with water.

Some water-water and ethanol-ethanol intermolecular attractions are disrupted. New ethanol-water interactions form, including hydrogen bonds.

The individual ethanol molecules remain \(\ce{CH3CH2OH}\), and the individual water molecules remain \(\ce{H2O}\).

Their covalent bonds are not broken simply because the substances have mixed.

The distinction is between separating molecules and separating the atoms within those molecules.

There is one qualification: not every substance dissolves without reacting. Some molecular substances react with water or form ions. You must distinguish an ordinary physical dissolution from a chemical reaction.

For the neutral molecular substances discussed in this article, assume they do not react with the solvent.

For a more detailed explanation of the phase-change distinction, see Why Boiling Water Does Not Break Its Covalent Bonds.

06A practical decision process for HSC Chemistry

When an exam question asks you to predict or explain solubility, avoid jumping straight to the phrase “like dissolves like”.

Work through the molecular structure instead.

Step 1: Draw or inspect the complete molecular structure.

Identify the functional groups, carbon chain length, and molecular shape.

Step 2: Determine overall molecular polarity.

Consider the polarity of individual bonds and whether their dipoles cancel because of the molecular geometry.

Step 3: Identify possible intermolecular interactions.

Look for:

• Hydrogen-bond donors and acceptors.
• Permanent dipoles that can participate in dipole-dipole interactions.
• Dispersion forces, which occur between all molecules.

Step 4: Compare the polar and non-polar regions.

A molecule with a small non-polar region may interact well with water despite having a hydrocarbon section.

A molecule with a much larger non-polar region may have limited water solubility, even if it contains a polar functional group.

Step 5: Consider the interactions before and after mixing.

Explain which solute-solute and solvent-solvent interactions must be disrupted, and which new solute-solvent interactions form.

Step 6: Make a qualified prediction.

State whether the substance is likely to be highly soluble, partially soluble, or poorly soluble. Remember that temperature, composition, and chemical reactions can affect the outcome.

A strong HSC explanation connects the molecular structure to the intermolecular forces and then connects those forces to the observed solubility.

Just writing “both substances are polar” is not enough.

07Questions and solutions

Question 1

A student adds 5.0 mL of hexane to 5.0 mL of water in a stoppered test tube. The tube is inverted several times and left undisturbed for 10 minutes at 25°C.

Describe the final appearance of the mixture. Explain why the liquids behave this way, identifying which liquid forms the upper layer.

Solution 1

The mixture separates into two liquid layers, with hexane forming the upper layer.

Water molecules form hydrogen bonds with neighbouring water molecules. Hexane molecules interact predominantly through dispersion forces.

When the liquids are mixed, the water-water attractions must be disrupted. However, hexane cannot form hydrogen bonds with water, so the new water-hexane interactions are relatively weak.

Mixing is therefore unfavourable, and the liquids separate into two phases.

Hexane is less dense than water, so it floats above the water layer.

Importantly, the water and hexane molecules remain intact. Their covalent bonds are not broken during mixing.

Question 2

A student investigates the solubility of butan-1-ol in water at 25°C.

The investigation provides the following illustrative equilibrium data:

• Maximum solubility: 0.80 g of butan-1-ol per 10.0 g of water.
• Mass of water used: 15.0 g.
• Mass of butan-1-ol added: 2.50 g.

Calculate:

a) The maximum mass of butan-1-ol that can dissolve.

b) The mass of butan-1-ol remaining undissolved.

c) Explain why the presence of an -OH group does not make butan-1-ol completely miscible with water.

Solution 2

A maximum of 1.20 g dissolves, leaving 1.30 g undissolved.

Step 1: Calculate the maximum amount that can dissolve.

The solubility is 0.80 g per 10.0 g of water.

Using a direct proportion:

\[ m_{\text{dissolved}}= \frac{0.80}{10.0}\times15.0 \]

\[ =1.20\text{ g} \]

Step 2: Calculate the excess butan-1-ol.

\[ m_{\text{excess}}= m_{\text{added}}-m_{\text{dissolved}} \]

\[ =2.50-1.20 \]

\[ =1.30\text{ g} \]

Step 3: Explain the molecular behaviour.

Butan-1-ol contains a hydroxyl group that can form hydrogen bonds with water.

However, it also contains a four-carbon hydrocarbon chain, which is predominantly non-polar. This region reduces the molecule’s overall compatibility with water.

Consequently, butan-1-ol has limited solubility rather than complete miscibility.

The calculation demonstrates that a substance can dissolve to a certain extent before reaching its saturation limit.

Question 3

An investigation produces the following observations at 25°C:

• Water and propanone form one clear phase.
• Water and butan-1-ol form two liquid phases after mixing and settling.

A student makes two conclusions:

1. Butan-1-ol does not dissolve in water because two layers are visible.
2. Propanone must form stronger attractions with water than water molecules form with each other because the mixture forms one phase.

Evaluate both conclusions.

Propose a revised investigation that would distinguish complete miscibility, partial miscibility, and a temporary emulsion.

Solution 3

Both conclusions are incorrect or too strong. Two layers do not mean zero solubility, and complete miscibility does not prove that every solute-solvent attraction is stronger than every solvent-solvent attraction.

First conclusion: Butan-1-ol does not dissolve in water.

This is incorrect because butan-1-ol has limited water solubility.

Its hydroxyl group can form hydrogen bonds with water, allowing some butan-1-ol molecules to enter the water-rich phase.

However, its larger non-polar region limits the amount that can dissolve. Once the mixture exceeds the relevant solubility limit, two phases form.

Second conclusion: Propanone-water attractions must be stronger than water-water attractions.

This is not necessarily true.

Propanone can accept hydrogen bonds from water through its carbonyl oxygen. These interactions contribute to favourable mixing.

However, solubility depends on the overall balance between:

• Attractions within the original substances.
• New attractions between solute and solvent.
• The effects of mixing and entropy.

A single clear phase does not establish that every new intermolecular interaction is stronger than every original interaction.

A revised investigation:

The student should:

1. Prepare several mixtures using different water-to-solute ratios.
2. Keep the temperature and mixing procedure consistent.
3. Allow each mixture to settle for a standard period.
4. Observe whether one phase or two phases remain.
5. Use an appropriate analytical method to measure the composition of each phase if quantitative solubility is required.

A temporary cloudy emulsion should not be mistaken for a homogeneous solution. Allowing sufficient time for droplets to separate helps identify this problem.

Testing several ratios also avoids assuming that a single clear mixture proves complete miscibility.

Question 4

Water is heated to its boiling point at atmospheric pressure. Separately, ethanol is mixed with water at room temperature.

A student claims that both processes require breaking covalent bonds because molecules are being separated.

Explain why this statement is incorrect.

In your response:

a) Identify the interactions disrupted during boiling.

b) Explain what happens to the covalent bonds within water molecules.

c) Describe the intermolecular changes when ethanol dissolves in water.

d) Suggest one observation that could help distinguish vaporisation from chemical decomposition.

Solution 4

Neither ordinary boiling nor the dissolution of ethanol in water requires breaking the covalent bonds within the molecules. Both processes involve changes in intermolecular interactions.

a) Interactions disrupted during boiling

During boiling, water molecules separate from neighbouring water molecules.

Energy is required to overcome the intermolecular hydrogen-bonding attractions that hold the liquid together.

b) Covalent bonds within water

The covalent O-H bonds remain intact.

Each molecule entering the gas phase is still \(\ce{H2O}\). Boiling changes the physical state, not the chemical identity of water.

c) Interactions during ethanol dissolution

Ethanol molecules separate from other ethanol molecules, and some water-water attractions are disrupted.

New hydrogen bonds form between ethanol and water molecules.

The covalent bonds within both ethanol and water remain intact.

d) Distinguishing vaporisation from decomposition

Collecting the water vapour and condensing it back into a liquid provides useful evidence.

If the condensed substance is still water, with the same characteristic properties, this supports a physical phase change rather than chemical decomposition.

Chemical decomposition would produce different chemical substances.

The central distinction is that intermolecular attractions act between molecules, whereas covalent bonds hold atoms together within molecules.

Question 5

Two molecular substances, A and B, have the same molecular formula, \(\ce{C4H10O}\), and each contains one hydroxyl group.

Their structural formulae are:

• A: \(\ce{CH3CH2CH2CH2OH}\) (butan-1-ol).
• B: \(\ce{(CH3)3COH}\) (2-methylpropan-2-ol).

The following illustrative comparison data are supplied:

The boiling points are approximate, rounded values supplied for this question.

A student makes three predictions:

1. A and B should have identical water solubilities because they have the same molecular formula and functional group.
2. B should be less soluble in water because its lower boiling point indicates weaker intermolecular forces.
3. When B boils, its O-H and C-O covalent bonds must break because its molecules are separating.

Evaluate all three predictions.

Use molecular shape, intermolecular interactions, and the difference between boiling and dissolving to construct your explanation.

Solution 5

All three predictions are incorrect. Substance B is more water-compatible despite having a lower boiling point, and neither boiling process requires breaking the covalent bonds within the alcohol molecules.

This problem is difficult because it combines three properties that are related to intermolecular forces but do not measure the same thing.

Prediction 1: Same molecular formula means identical solubility.

This is incorrect.

Both substances contain four carbon atoms, ten hydrogen atoms, and one oxygen atom. They also contain the same hydroxyl functional group.

However, their molecular structures are different.

Substance A has a straight carbon chain. Substance B has a more compact, branched structure, with three methyl groups attached to the carbon bearing the hydroxyl group.

Branching changes the shape and the effective non-polar surface exposed to the surrounding water.

The branched structure of B makes its hydrocarbon region less disruptive to water’s intermolecular interactions. Its hydroxyl group can still form hydrogen bonds with water.

Consequently, B is much more compatible with water and is completely miscible under the stated conditions.

The key lesson is that molecular formula and functional group alone do not determine solubility. Molecular shape also matters.

Prediction 2: Lower boiling point means lower water solubility.

This is the most tempting mistake.

Substance B has a lower boiling point than substance A. This indicates that less energy is required to separate the molecules in the pure liquid at atmospheric pressure.

The more compact branched structure of B reduces the effectiveness of intermolecular contact between its hydrocarbon regions, reducing the contribution of dispersion forces to the cohesion of the pure liquid.

However, boiling point measures the cohesion between molecules in the pure substance.

Water solubility measures how favourably the substance interacts with water compared with remaining in its own phase.

These are different comparisons.

B can therefore have weaker overall cohesion in its pure liquid while having greater compatibility with water.

Prediction 3: Boiling breaks the O-H and C-O covalent bonds.

This is incorrect.

When either alcohol boils, the molecules separate from one another and enter the gas phase.

The intermolecular attractions between alcohol molecules are overcome, but the covalent bonds within each molecule remain intact.

A molecule of butan-1-ol in the gas phase is still butan-1-ol. A molecule of 2-methylpropan-2-ol in the gas phase is still 2-methylpropan-2-ol.

No decomposition is implied by boiling.

The deeper conclusion

This comparison exposes an important limitation of using one physical property to predict another.

• Molecular shape influences both intermolecular interactions and solubility.
• Boiling point depends on the attractions between molecules in a pure liquid.
• Solubility depends on the balance of interactions when a solute enters a particular solvent.
• Neither process normally requires breaking the covalent bonds within the molecules.

A lower boiling point does not automatically mean lower solubility, just as having an -OH group does not automatically mean complete miscibility.

08Where to take this next

The next useful step is learning to distinguish two different questions:

1. How strongly do molecules attract one another in a pure liquid? This helps explain boiling-point trends.
2. How favourably do those molecules interact with a different solvent? This helps explain solubility.

Molecular size, shape, polarity, and intermolecular forces influence both properties, but they do not influence them in identical ways.

Once you keep those two comparisons separate, you can explain why molecules with the same functional group, or even the same molecular formula, can behave very differently.