Why Carboxylic Acids Form Cyclic Hydrogen-Bonded Dimers
Discover how carboxylic acids form cyclic dimers through two reciprocal hydrogen bonds. Learn how molecular structure enables this arrangement, how IR spectroscopy provides evidence, and why boiling disrupts intermolecular attractions without breaking covalent bonds.
Take two molecules of ethanoic acid, \(\ce{CH3COOH}\). Each molecule has one O-H hydrogen, so you might predict that the pair can form just one hydrogen bond. But there’s another important feature in the structure: a carbonyl oxygen.
When two carboxylic acid molecules approach each other, each O-H group can form a hydrogen bond with the carbonyl oxygen of the other molecule. The result is a pair held together by two reciprocal hydrogen bonds, forming a structure called a cyclic dimer.
The reason this arrangement is possible comes down to the way the two oxygen-containing parts of the carboxyl group work together.
01Why two carboxylic acid molecules form a dimer
Start with the carboxyl functional group:
\[ \ce{-COOH} \]
It contains two different structural features:
- A carbonyl group, \(\ce{C=O}\).
- A hydroxyl group, \(\ce{O-H}\).
These features give each carboxylic acid molecule two important abilities. It can donate a hydrogen bond, and it can accept a hydrogen bond.
One molecule can both donate and accept
Consider the general structure of a carboxylic acid:
\[ \ce{R-C(=O)-OH} \]
Here, \(R\) represents the remainder of the carbon chain.
The two oxygen-containing groups have different roles.
1. The hydroxyl group is the hydrogen-bond donor.
Oxygen is more electronegative than hydrogen. It attracts the shared electrons in the O-H covalent bond towards itself.
This creates partial charges:
- Oxygen becomes partially negative, \(\delta^-\).
- Hydrogen becomes partially positive, \(\delta^+\).
The partially positive hydrogen can be attracted to an oxygen atom on another molecule.
2. The carbonyl oxygen is the hydrogen-bond acceptor.
The carbonyl oxygen has lone pairs of electrons and a partial negative charge.
These lone pairs allow it to attract the partially positive hydrogen from another molecule’s O-H group.
So, one carboxylic acid molecule has:
| Part of molecule | Role |
|---|---|
| O-H group | Hydrogen-bond donor |
| Carbonyl oxygen, C=O | Hydrogen-bond acceptor |
This combination is the key to dimer formation.
How two molecules connect
Imagine two carboxylic acid molecules, labelled A and B.
- Molecule A donates its O-H hydrogen to the carbonyl oxygen of molecule B.
- Molecule B donates its O-H hydrogen to the carbonyl oxygen of molecule A.
Both molecules donate one hydrogen bond and accept one hydrogen bond.
Two hydrogen bonds now connect the same pair of molecules.
The arrangement closes into a ring-like structure, which is why it is called a cyclic dimer.
The word dimer simply means a pair of two molecules. The word cyclic describes the closed arrangement formed by the two hydrogen bonds.
Think of it as a two-way handshake. Each molecule offers something the other can accept, and both interactions happen at once.
The analogy has a limit: hydrogen bonds are electrostatic attractions between partially charged atoms, not literal connections between two objects. Their geometry and electron distributions determine whether the arrangement is favourable.
Why two hydrogen bonds matter
A single hydrogen bond could connect two carboxylic acid molecules at one point.
However, each molecule has another suitable site available. Forming a second reciprocal hydrogen bond creates an additional attractive interaction and closes the cyclic arrangement.
This makes the paired structure particularly favourable.
The association can be represented schematically as:
\[ 2\,\mathrm{RCOOH} \rightleftharpoons (\mathrm{RCOOH})_2 \]
The symbol \(\rightleftharpoons\) indicates that the association is reversible.
The two molecules are not joined by new covalent bonds. They remain separate molecules held together by intermolecular hydrogen bonding.
There is another important qualification: a liquid carboxylic acid is not necessarily made up of permanent, isolated pairs. Molecules are constantly moving, and hydrogen bonds can break and reform. The cyclic dimer is a particularly useful model for understanding their strong molecular association.
02Why having a carbonyl group is not enough
Here’s a common misconception.
A student sees two molecules containing carbonyl groups and predicts that they should both form hydrogen-bonded dimers.
That prediction misses the need for a hydrogen-bond donor.
Compare propanoic acid with propanone.
| Structural feature | Propanoic acid | Propanone |
|---|---|---|
| Molecular formula | \(\ce{C3H6O2}\) | \(\ce{C3H6O}\) |
| Carbonyl group | Yes | Yes |
| O-H group | Yes | No |
| Can donate a hydrogen bond to another identical molecule? | Yes | No |
| Can accept a hydrogen bond? | Yes | Yes |
| Can form the same reciprocal cyclic dimer? | Yes | No |
Propanone contains a carbonyl oxygen, so it can accept hydrogen bonds from suitable molecules.
However, it has no hydrogen directly bonded to oxygen. It therefore cannot donate the hydrogen bond needed to complete the reciprocal arrangement with another propanone molecule.
This does not mean propanone has no intermolecular forces. It still experiences London dispersion forces and dipole-dipole attractions.
It can also accept hydrogen bonds from other substances, such as water or carboxylic acids.
The important distinction is between being able to accept a hydrogen bond and being able to both donate and accept one.
This is why you should examine the whole molecular structure rather than simply count oxygen atoms.
For a related comparison involving the same structural issue, see Carboxylic Acids vs Esters: Structure, Properties and Reactions.
03How experiments provide evidence for dimer formation
Knowing that a cyclic dimer is structurally possible is one thing. How can we investigate whether carboxylic acid molecules actually associate this way?
One useful technique is infrared (IR) spectroscopy.
The idea is straightforward: molecular bonds vibrate, and they absorb infrared radiation at characteristic frequencies. Changes in the environment around a bond can change the position and shape of its absorption.
For carboxylic acids, the O-H stretching absorption provides useful evidence of hydrogen bonding.
The experimental setup
An infrared spectrometer contains an infrared source, a sample compartment, and a detector.
The measurement follows this sequence:
- The infrared source produces radiation across a range of frequencies.
- The radiation passes through a sample containing the carboxylic acid.
- The sample absorbs particular frequencies that match its molecular vibrations.
- The remaining radiation reaches a detector.
- The instrument produces a spectrum showing absorption against wavenumber.
The question is what should happen to the O-H absorption if the acid molecules are hydrogen-bonded to one another.
If the O-H groups are strongly interacting with other molecules, their vibrations should differ from those of relatively isolated O-H groups.
What the measurements show
Compare propanoic acid under two different conditions.
| Sample conditions | O-H stretching absorption | Interpretation |
|---|---|---|
| Liquid or relatively concentrated acid | Strong, very broad absorption, approximately \(2500\)-\(3300\ \mathrm{cm^{-1}}\) | Extensive hydrogen bonding between acid molecules |
| Very dilute acid in a non-polar solvent | Sharper absorption at a higher wavenumber, around \(3500\ \mathrm{cm^{-1}}\) | More isolated acid molecules and less acid-acid association |
| Low-pressure acid vapour | Dimer-associated absorption is greatly reduced | Monomers become more prominent |
These are representative spectral observations, not universal numerical boundaries. Exact absorption positions depend on molecular structure and experimental conditions.
The broad absorption in condensed carboxylic acids is characteristic of strongly hydrogen-bonded O-H groups. Reference IR spectra also show that the dimer-associated features diminish in low-pressure vapour. Organic Chemistry Data: Infrared Spectroscopy【turn164156view0】【turn787013view0】
Why does the absorption change?
Hydrogen bonding affects the O-H covalent bond. The attraction to the neighbouring carbonyl oxygen changes the electron distribution around the O-H bond, reducing its effective bond strength.
The O-H stretching vibration therefore occurs at a lower frequency. A range of different hydrogen-bonding environments also contributes to the broad absorption.
When the acid is highly diluted in a non-polar solvent, acid molecules are less likely to encounter one another. More O-H groups remain relatively free, producing a sharper absorption at a higher wavenumber.
What the experiment can and cannot prove
The first experiment might simply measure the IR spectrum of pure liquid acid.
That reveals extensive hydrogen bonding, but it does not independently establish that every association is a cyclic dimer. Other hydrogen-bonded arrangements could also affect the spectrum.
A better investigation compares samples at different concentrations, using a non-polar solvent that does not strongly compete for hydrogen bonds.
If dilution reduces the broad absorption and increases the free O-H absorption, this supports the interpretation that the interactions occur between separate acid molecules.
Combining this evidence with the molecular structure gives a convincing explanation for the cyclic dimer: each molecule has a suitable donor and acceptor site, allowing two reciprocal hydrogen bonds to form.
One limitation remains. An IR spectrum is evidence about molecular vibrations and their environments. It is not a photograph of individual dimers, and it does not establish that every molecule exists in one permanent pair.
Worked example: Convert an IR wavenumber into wavelength
A broad O-H absorption associated with hydrogen-bonded carboxylic acids is observed near \(3000\ \mathrm{cm^{-1}}\).
Estimate the wavelength of the infrared radiation at this wavenumber, in micrometres.
Step 1
Wavenumber, represented by \(\tilde{\nu}\), is the reciprocal of wavelength, \(\lambda\).
\[ \tilde{\nu}=\frac{1}{\lambda} \]
Therefore:
\[ \lambda=\frac{1}{\tilde{\nu}} \]
Step 2
\[ \lambda=\frac{1}{3000\ \mathrm{cm^{-1}}} \]
\[ \lambda=3.33\times10^{-4}\ \mathrm{cm} \]
Step 3
Since \(1\ \mathrm{cm}=10^{-2}\ \mathrm{m}\):
\[ \lambda=3.33\times10^{-6}\ \mathrm{m} \]
Step 4
Since \(1\ \mu\mathrm{m}=10^{-6}\ \mathrm{m}\):
\[ \boxed{\lambda=3.33\ \mu\mathrm{m}} \]
The infrared radiation has a wavelength of approximately \(3.33\ \mu\mathrm{m}\).
This is the wavelength associated with the absorption, not the distance between the two molecules or the length of the O-H bond.
Worked example: Interpret a change in an IR spectrum
A sample of propanoic acid is measured first as a liquid and then as a low-pressure vapour.
The liquid shows a broad O-H absorption between \(2500\) and \(3300\ \mathrm{cm^{-1}}\). The vapour shows a much sharper O-H absorption at a higher wavenumber, with the dimer-associated features greatly reduced.
Explain the observations.
Step 1
The chemical identity of propanoic acid has not changed. Its molecules still contain the same O-H and C=O covalent bonds.
Step 2
In the liquid, propanoic acid molecules are strongly associated through hydrogen bonding.
The O-H groups experience interactions with carbonyl oxygens on neighbouring molecules. These interactions alter the O-H stretching vibration, producing the broad absorption at lower wavenumbers.
Step 3
At low pressure, acid molecules are much less likely to remain associated with one another.
The proportion of monomers increases, so more O-H groups are relatively free from acid-acid hydrogen bonding.
Their O-H stretching absorption becomes sharper and occurs at a higher wavenumber.
Step 4
The spectral changes support the idea that carboxylic acid molecules form intermolecular hydrogen-bonded associations in the condensed phase.
They do not indicate that the covalent O-H bonds have broken or that the acid has decomposed.
04What happens to dimers when a carboxylic acid boils?
This is where students sometimes confuse two completely different types of bonding.
If carboxylic acids form dimers through hydrogen bonding, what happens when the liquid is heated until it boils?
Does the energy break the covalent bonds holding the atoms together?
No.
Boiling is a physical change. It involves overcoming intermolecular attractions sufficiently for molecules to escape from the liquid and enter the gas phase.
In a carboxylic acid, heating disrupts or rearranges the hydrogen bonds between neighbouring molecules.
However, the covalent bonds within each acid molecule remain intact.
Consider what happens to a pair of acid molecules.
Before heating, they may be associated through two hydrogen bonds.
As energy is supplied, molecular movement increases. The attractions between neighbouring molecules become less able to hold them together.
Eventually, molecules can escape into the gas phase.
The important distinction is:
| Interaction | What happens during ordinary boiling? |
|---|---|
| Hydrogen bonds between acid molecules | Disrupted or rearranged as molecules separate |
| Covalent O-H bonds within each molecule | Remain intact |
| Covalent C=O bonds within each molecule | Remain intact |
| Covalent bonds forming the carbon chain | Remain intact |
It is also important not to assume that every dimer must become two completely separate monomers at the boiling point.
Some carboxylic acid dimers can persist in the vapour, depending on temperature and pressure. The relative amounts of monomers and dimers change with the conditions.
This is why boiling should not be described as breaking the covalent bonds of the acid molecule.
The distinction is useful whenever you explain boiling points. Stronger intermolecular attractions generally mean that more energy is needed to separate molecules, resulting in a higher boiling point.
For a broader explanation of how intermolecular forces affect the physical properties of organic compounds, see HSC Chemistry: Carboxylic Acids, Amines and Amides Explained.
05Questions and solutions
Question 1
Ethanoic acid has the structural formula \(\ce{CH3COOH}\).
Explain how two ethanoic acid molecules can form a cyclic dimer. Identify the hydrogen-bond donor and acceptor on each molecule.
Solution 1
Two ethanoic acid molecules form a cyclic dimer through two reciprocal hydrogen bonds.
Each molecule contains an O-H group that can donate a hydrogen bond and a carbonyl oxygen that can accept one.
The O-H hydrogen of the first molecule is attracted to the carbonyl oxygen of the second molecule. The second molecule forms an equivalent interaction in the opposite direction.
These two hydrogen bonds close into a cyclic arrangement.
The resulting dimer consists of two separate ethanoic acid molecules held together by intermolecular forces. No new covalent bonds are formed.
Question 2
Propanone, \(\ce{CH3COCH3}\), contains a polar carbonyl group.
A student claims that two propanone molecules should form the same cyclic hydrogen-bonded dimers as two propanoic acid molecules because both compounds contain a carbonyl oxygen.
Evaluate this claim. Explain the structural difference that determines whether the cyclic dimer can form.
Solution 2
The claim is incorrect. Two propanone molecules cannot form the same reciprocal, two-hydrogen-bond cyclic dimer as two propanoic acid molecules.
Both molecules contain a carbonyl oxygen, which can accept hydrogen bonds.
However, propanone has no O-H bond. It therefore has no hydrogen directly bonded to oxygen that can donate the required hydrogen bond to another propanone molecule.
Propanoic acid contains both the carbonyl oxygen and the hydroxyl hydrogen. This allows it to donate and accept hydrogen bonds simultaneously.
The distinction is not whether a molecule contains oxygen. It is whether its complete structure provides both a suitable hydrogen-bond donor and acceptor.
Propanone still experiences dipole-dipole attractions and London dispersion forces, and it can accept hydrogen bonds from other substances.
Question 3
A chemist records IR spectra of propanoic acid dissolved in a non-polar solvent at \(25^\circ\mathrm{C}\).
The following illustrative observations are obtained:
| Sample | Acid concentration | Main O-H spectral observation |
|---|---|---|
| A | \(0.10\ \mathrm{mol\,L^{-1}}\) | Strong, broad absorption between \(2500\) and \(3300\ \mathrm{cm^{-1}}\) |
| B | \(0.002\ \mathrm{mol\,L^{-1}}\) | Much weaker broad absorption and a stronger, sharper absorption near \(3500\ \mathrm{cm^{-1}}\) |
(a) Identify which sample is expected to contain a greater proportion of hydrogen-bonded acid dimers.
(b) Explain why dilution changes the O-H absorption.
(c) Explain why the spectral observations do not mean that the covalent O-H bonds have broken.
Solution 3
Sample A is expected to contain a greater proportion of hydrogen-bonded acid dimers.
(a) Sample A contains more dimers.
Its higher acid concentration means acid molecules are more likely to encounter one another and form intermolecular hydrogen bonds.
Sample B is much more dilute, so acid-acid associations become less common.
(b) Dilution reduces the frequency of interactions between acid molecules.
In sample A, many O-H groups participate in hydrogen bonding with carbonyl oxygens on neighbouring molecules. This produces a broad absorption at lower wavenumbers.
In sample B, a greater proportion of O-H groups are relatively free from acid-acid hydrogen bonding. Their stretching vibrations occur at higher wavenumbers, producing a sharper absorption.
The difference reflects a change in intermolecular association, not a change in the molecular formula.
(c) The covalent O-H bonds remain intact in both samples.
Hydrogen bonding changes the environment and effective strength of the O-H covalent bond, affecting its vibration and IR absorption.
A change in an absorption band does not automatically indicate that a covalent bond has been broken.
Also, the spectra establish a qualitative change in association. They do not provide an exact percentage of dimers without appropriate quantitative calibration and analysis.
Question 4
A sample of ethanoic acid is heated until it boils.
An IR spectrum of the resulting vapour shows a strong absorption associated with relatively free O-H groups, together with weaker dimer-associated features.
A student makes two claims:
- All hydrogen-bonded dimers must break into individual monomers during boiling.
- The covalent O-H bonds must also break because the molecules need to separate.
Evaluate both claims.
Solution 4
Both claims are incorrect.
Boiling disrupts intermolecular hydrogen bonding, but it does not require every dimer to dissociate completely or break the covalent O-H bonds.
The strong absorption associated with relatively free O-H groups indicates that monomers are prominent in the vapour.
However, the weaker dimer-associated features indicate that some associated acid molecules remain.
The proportion of dimers depends on conditions such as temperature and pressure.
Most importantly, the energy required to separate molecules through intermolecular attractions is not the same as the energy required to break the covalent bonds within those molecules.
During ordinary boiling, the acid molecules remain chemically intact.
The correct explanation is that hydrogen bonds between molecules are disrupted or rearranged as the liquid becomes a gas, while the covalent structure of each acid molecule is preserved.
Question 5
A dilute solution of propanoic acid in a non-polar solvent contains a significant proportion of hydrogen-bonded acid dimers.
A large excess of propanone is then introduced. The temperature and propanoic acid concentration are kept constant.
After mixing, the IR spectrum shows:
- A substantial reduction in the absorption associated with acid-acid dimers.
- An O-H absorption that remains shifted and broadened compared with a relatively free O-H group.
Assume that the dominant new interaction is between the O-H group of propanoic acid and the carbonyl oxygen of propanone, and that no covalent chemical reaction occurs.
Explain:
(a) Why the acid-acid dimer absorption decreases.
(b) Why the O-H absorption remains shifted and broadened.
(c) How many hydrogen bonds can form in an idealised pair containing one propanoic acid molecule and one propanone molecule through this interaction.
(d) Why this new arrangement cannot reproduce the original cyclic acid dimer, even though hydrogen bonding is still occurring.
Solution 5
The propanoic acid molecules are forming hydrogen bonds with propanone instead of exclusively associating with other propanoic acid molecules.
This explains why the acid-acid dimer absorption decreases even though hydrogen bonding has not disappeared.
(a) The acid-acid dimer absorption decreases because propanone competes with other acid molecules for hydrogen bonding.
The carbonyl oxygen of propanone can accept a hydrogen bond from the hydroxyl hydrogen of propanoic acid.
As propanone becomes available in large excess, more acid molecules can interact with propanone rather than forming cyclic acid-acid dimers.
This reduces the proportion of the original dimer arrangement.
(b) The O-H absorption remains shifted and broadened because the acid’s hydroxyl hydrogen is still participating in hydrogen bonding.
Its interaction partner has changed.
Instead of being attracted to the carbonyl oxygen of another propanoic acid molecule, it is now attracted to the carbonyl oxygen of propanone.
Hydrogen bonding still changes the O-H stretching vibration, so the absorption does not return completely to the relatively free O-H position.
(c) One idealised propanoic acid-propanone pair forms one hydrogen bond through this interaction.
The O-H hydrogen of propanoic acid acts as the donor, while the carbonyl oxygen of propanone acts as the acceptor.
(d) The original cyclic dimer cannot form because propanone cannot donate a second hydrogen bond.
The propanoic acid molecule has an available carbonyl oxygen that could accept a hydrogen bond, but propanone has no O-H hydrogen to donate to it.
Consequently, the reciprocal two-hydrogen-bond arrangement cannot be completed.
This is the important distinction: hydrogen bonding can still occur without cyclic dimer formation.
A molecule’s ability to accept hydrogen bonds does not automatically mean it can form a reciprocal cyclic dimer with another molecule.
The example also shows why it is not enough to memorise that carboxylic acids form dimers. You need to examine which atoms are available to donate and accept hydrogen bonds, and which other molecules are competing for those sites.
The central idea is that carboxylic acids can form cyclic dimers because their carboxyl groups contain both a hydrogen-bond donor and a hydrogen-bond acceptor in a suitable arrangement. Two molecules can therefore form two reciprocal hydrogen bonds.
Once you understand that structure, the next useful step is applying the same donor-acceptor reasoning to mixtures, solvents, and other organic functional groups. That is how you move beyond simply recognising a functional group and start predicting the physical properties of unfamiliar molecules.