How to Distinguish Esters and Carboxylic Acids Using IR

A carboxylic acid and an ester can both produce a strong absorption band in the carbonyl region of an IR spectrum. So, if you spot a strong peak around 1700-1750 cm⁻¹, you haven't identified the functional group yet. You've only established that a carbonyl group is likely present.

The most reliable way to distinguish them is to look for a combination of features:

- **Carboxylic acid:** a strong C=O absorption, usually around 1680-1725 cm⁻¹, together with a very broad O-H absorption extending approximately from 2500 to 3300 cm⁻¹.
- **Ester:** a strong C=O absorption, usually around 1735-1750 cm⁻¹, together with strong C-O absorptions in the fingerprint region, approximately 1050-1300 cm⁻¹. It does not have the characteristic broad carboxylic acid O-H band.

The key is to identify the pattern, not to make a decision from one peak.

## 1. Start with the functional groups

Before interpreting the spectrum, look at what is actually bonded to the carbonyl carbon.

A carboxylic acid contains the group -COOH. An ester contains -COO-, with the carbonyl carbon bonded to an oxygen that is also attached to another carbon-containing group.

[IMAGE: Side-by-side structural drawings of a carboxylic acid R-C(=O)-OH and an ester R-C(=O)-O-R′. Show the carbonyl and hydroxyl bonds clearly, with the different bonding arrangements highlighted.]

Both structures contain a C=O bond. This explains why both produce a strong carbonyl absorption.

However, only the carboxylic acid has an O-H bond directly attached to the carboxyl group. That gives us a much more useful distinguishing feature.

Think about what you would predict before looking at the spectra:

- If the unknown is a carboxylic acid, you should find evidence of both C=O and O-H.
- If it is an ester, you should find evidence of C=O and C-O, but not the characteristic carboxylic acid O-H absorption.

This gives us our first identification rule: **look for the O-H band before relying on the exact position of the carbonyl band.**

## 2. Know what to look for in the IR spectrum

An IR spectrum shows how strongly a sample absorbs infrared radiation at different wavenumbers.

Wavenumber is measured in reciprocal centimetres (cm⁻¹). It is commonly plotted from approximately 4000 cm⁻¹ on the left to 400 cm⁻¹ on the right.

The vertical axis is usually either:

- **Transmittance (%):** how much infrared radiation passes through the sample. Absorption bands appear as downward dips.
- **Absorbance:** how much radiation the sample absorbs. Absorption bands appear as upward peaks.

Always check the axis label before interpreting a spectrum. A downward feature on a transmittance spectrum is not evidence of weak absorption. It indicates that less radiation has passed through the sample.

### The important absorption ranges

The following are approximate guide ranges for common organic compounds. Actual positions can vary with molecular structure, hydrogen bonding, and measurement conditions.

| Bond or feature | Approximate wavenumber (cm⁻¹) | What to look for |
|---|---:|---|
| Carboxylic acid O-H | 2500-3300 | Very broad absorption |
| Carboxylic acid C=O | 1680-1725 | Strong, relatively sharp absorption |
| Ester C=O | 1735-1750 | Strong, relatively sharp absorption |
| Ester C-O | 1050-1300 | One or more strong absorptions |
| Alkyl C-H | 2850-3000 | Several smaller absorptions |

The two most useful differences are the broad acid O-H band and the combination of ester C=O and C-O absorptions.

[IMAGE: An annotated comparison of two illustrative IR transmittance spectra, one for a carboxylic acid and one for an ester. Use a reversed wavenumber axis from 4000 on the left to 400 on the right. Show downward absorption bands. The acid spectrum must have a very broad O-H feature extending from approximately 2500 to 3300 cm⁻¹ and a strong carbonyl band near 1710 cm⁻¹. The ester spectrum must have no broad acid O-H feature, a strong carbonyl band near 1740 cm⁻¹, and strong C-O absorptions in the 1050-1300 cm⁻¹ region. Label the approximate regions and clearly identify the spectra as illustrative, not experimental measurements.]

### Why is the carboxylic acid O-H band so broad?

You might expect an O-H bond to produce a fairly narrow absorption. That would be a reasonable prediction if all the O-H bonds experienced the same molecular environment.

Carboxylic acids, however, form hydrogen bonds with neighbouring molecules. In a liquid or concentrated sample, this can produce associated molecules, including hydrogen-bonded dimers.

The O-H bonds experience a range of hydrogen-bonding strengths. Their absorption frequencies are therefore spread over a wide range, producing the characteristic broad band.

[IMAGE: Two carboxylic acid molecules forming a hydrogen-bonded cyclic dimer. Show the two C=O and O-H groups, with dashed lines between the hydroxyl hydrogen of each molecule and the carbonyl oxygen of the other. Distinguish covalent bonds from intermolecular hydrogen bonds.]

This is an important distinction: the broad feature is not simply a collection of unrelated O-H peaks. It reflects the range of hydrogen-bonding environments in the sample.

An ester does not have this carboxylic acid O-H bond, so it does not produce this characteristic broad absorption.

Be careful, though. An ester could contain an additional alcohol O-H group elsewhere in its structure. In that case, an O-H absorption does not automatically make the compound a carboxylic acid. You must interpret the entire pattern.

## 3. Why the carbonyl peak alone is not enough

Both functional groups contain a carbonyl bond, so both produce a strong absorption in the 1700-1750 cm⁻¹ region.

An ester commonly absorbs at a higher wavenumber than a simple, saturated carboxylic acid. For example, an illustrative acid carbonyl band might occur at 1710 cm⁻¹, while an ester carbonyl band might occur at 1740 cm⁻¹.

But these values are not fixed identification boundaries.

Conjugation, molecular structure, and hydrogen bonding can shift the position of a carbonyl absorption. A conjugated ester, for instance, can absorb at a lower wavenumber than an ordinary saturated ester.

This means that a peak at 1720 cm⁻¹ should not automatically be labelled a carboxylic acid. The O-H region and the fingerprint region provide the supporting evidence.

[IMAGE: Two simplified carbonyl absorption bands plotted against a shared wavenumber axis, showing an illustrative carboxylic acid band near 1710 cm⁻¹ and ester band near 1740 cm⁻¹. Include a note that the ranges can overlap and shift with molecular structure. Do not represent the positions as universal boundaries.]

### The C-O region is useful, but not decisive

There is another feature worth checking: the C-O absorption.

Esters commonly produce strong C-O absorptions between approximately 1050 and 1300 cm⁻¹. You may see more than one absorption in this region.

However, a carboxylic acid also contains a C-O bond. It can therefore produce absorption in a similar region.

This is where a common mistake occurs:

> The spectrum contains a C-O absorption, so the compound must be an ester.

Not necessarily. Both functional groups contain C-O bonds.

Use the C-O region as supporting evidence. The broad acid O-H band is generally the more useful feature for distinguishing these two functional groups.

### A note about the different O-H ranges

You might notice that different chemistry references give slightly different upper limits for carboxylic acid O-H absorption.

For example:

- Some HSC examination material describes the broad region as approximately 2500-3000 cm⁻¹.
- Broader spectroscopy references commonly extend the range to approximately 3100-3300 cm⁻¹.

These ranges are not contradictory. The absorption is exceptionally broad, and its shape and extent depend on the sample's hydrogen-bonding environment.

NESA's 2022 Chemistry marking guidelines, for example, explicitly use the absence of a broad O-H stretch between 2500 and 3000 cm⁻¹ as evidence against a carboxylic acid. Use the reference range supplied in your question, but pay attention to the overall shape rather than treating the upper limit as an exact boundary. 

## 4. How the IR experiment produces the spectrum

It helps to understand what the instrument actually measures. Otherwise, it is easy to treat the spectrum as a collection of mysterious peaks.

A common modern instrument is a Fourier transform infrared (FTIR) spectrometer.

Consider a transmission arrangement using a liquid sample.

[IMAGE: Scientifically accurate side-view schematic of a transmission FTIR spectrometer. Show an infrared source at the left, followed by a Michelson interferometer containing a beam splitter, one fixed mirror and one moving mirror. Show the recombined beam travelling through a thin liquid sample between IR-transparent plates, then reaching a detector. Connect the detector signal to a computer displaying an IR spectrum. Use arrows to show the actual beam path and distinguish the two interferometer paths.]

The measurement proceeds in a particular order.

**Step 1: The infrared source produces radiation.**

The source emits infrared radiation containing a range of wavelengths.

**Step 2: The interferometer processes the beam.**

Inside the Michelson interferometer, a beam splitter divides the incoming radiation into two paths.

One part travels towards a fixed mirror. The other travels towards a moving mirror.

The reflected beams return to the beam splitter and recombine. As the moving mirror changes position, the difference between the two optical path lengths changes. This produces a changing interference signal.

**Step 3: The beam passes through the sample.**

The recombined beam passes through the sample.

Some infrared frequencies are absorbed by the molecules because their energies match particular molecular vibrations.

A bond can absorb infrared radiation when its vibration produces a change in the molecule's dipole moment.

Think of a bond as two masses connected by a spring. A vibration makes the atoms move relative to each other, and different bonds have different preferred vibration frequencies.

The spring analogy is useful for understanding why different bonds absorb at different frequencies, but real molecular vibrations are quantum mechanical and can involve several atoms moving together.

**Step 4: The detector measures the radiation.**

The detector measures the radiation reaching it.

At frequencies absorbed by the sample, less radiation reaches the detector.

**Step 5: The computer produces the spectrum.**

The instrument records the interference signal, called an interferogram. A mathematical operation called a Fourier transform converts this signal into the familiar spectrum of intensity against wavenumber.

The instrument also normally takes a background scan without the sample. Comparing the sample scan with this background helps account for the instrument response and contributions from the surrounding atmosphere.

This matters because water vapour and carbon dioxide in the air can also absorb infrared radiation. Without a suitable background correction, atmospheric absorptions can complicate the interpretation. 

### A small calculation: converting wavenumber into wavelength

IR spectra usually use wavenumber rather than wavelength.

Wavenumber, represented by \(\tilde{\nu}\), is the reciprocal of wavelength:

\[
\tilde{\nu}=\frac{1}{\lambda}
\]

Here:
- \(\tilde{\nu}\) is wavenumber in cm⁻¹.
- \(\lambda\) is wavelength in centimetres.

Suppose a strong carbonyl absorption occurs at 1710 cm⁻¹.

\[
\begin{aligned}
\lambda &= \frac{1}{1710}\\
&=5.85\times10^{-4}\ \mathrm{cm}
\end{aligned}
\]

Since \(1\ \mathrm{cm}=10^4\ \mathrm{\mu m}\):

\[
\lambda=5.85\ \mathrm{\mu m}
\]

The absorption therefore corresponds to an infrared wavelength of approximately 5.85 micrometres.

You generally won't need to convert every absorption in an HSC question. The important point is understanding that higher wavenumber means higher frequency and shorter wavelength.

## 5. A reliable method for comparing an ester and a carboxylic acid

When you receive an unfamiliar spectrum, follow this sequence.

| Step | What to examine | What it tells you |
|---|---|---|
| 1 | Horizontal and vertical axes | Whether you're reading wavenumber correctly and whether absorptions appear as peaks or dips |
| 2 | C=O region, approximately 1680-1750 cm⁻¹ | Whether a carbonyl group is present |
| 3 | O-H region, particularly 2500-3300 cm⁻¹ | Whether the characteristic very broad carboxylic acid absorption is present |
| 4 | C-O region, approximately 1050-1300 cm⁻¹ | Provides supporting evidence for an oxygen-containing functional group |
| 5 | Overall pattern | Whether all observations are consistent with the proposed structure |

A useful decision process is:

[IMAGE: A scientific decision diagram beginning with a strong C=O absorption. Branch to examination of the 2500-3300 cm⁻¹ region. One branch shows a very broad O-H absorption and leads to carboxylic acid as the likely assignment. The other shows no broad acid O-H absorption in a complete, good-quality spectrum and leads to ester as the more likely assignment, supported by C-O absorptions. Include a caution that additional O-H groups and incomplete spectra require further investigation.]

There are two important qualifications.

First, an absent absorption is only useful evidence if the relevant region was actually measured. If the spectrum is cropped before 2500 cm⁻¹, you cannot use the absence of an acid O-H band to eliminate a carboxylic acid.

Second, a molecule can contain more than one functional group. An ester with an additional alcohol group can still produce an O-H absorption. You need to identify what type of O-H absorption is present.

Let's apply these ideas to two worked examples.

## 6. Worked examples

### Worked example 1: Identifying a simple carboxylic acid

An unknown organic compound has the molecular formula \(\ce{C4H8O2}\). It is known to be either butanoic acid or methyl propanoate.

An illustrative percentage-transmittance IR spectrum gives the following observations:

| Observation | Measured feature |
|---|---|
| Carbonyl absorption | Strong, narrow downward dip at 1714 cm⁻¹ |
| O-H region | Very broad downward trough extending from approximately 2500 to 3150 cm⁻¹ |
| C-O region | Absorptions near 1230 and 1280 cm⁻¹ |
| Remaining spectrum | Several smaller absorptions |

Identify the unknown compound and justify your answer using the IR evidence.

[IMAGE: Illustrative percentage-transmittance spectrum for the worked example. Show a complete reversed wavenumber axis from 4000 to 650 cm⁻¹. Include a strong narrow carbonyl dip at 1714 cm⁻¹, a very broad acid O-H trough from 2500 to 3150 cm⁻¹, and two fingerprint-region dips around 1230 and 1280 cm⁻¹. Label the supplied observations without adding any conclusions about the compound.]

**Step 1**: Identify the carbonyl group.

The strong absorption at 1714 cm⁻¹ is consistent with a C=O bond.

Both proposed compounds contain a carbonyl group, so this observation does not distinguish them.

**Step 2**: Examine the O-H region.

The spectrum contains a very broad absorption extending from approximately 2500 to 3150 cm⁻¹.

This is characteristic of a carboxylic acid O-H bond. The broadness is consistent with hydrogen bonding between neighbouring acid molecules.

Methyl propanoate has no O-H bond, so it cannot explain this feature.

**Step 3**: Consider the C-O absorptions.

The absorptions near 1230 and 1280 cm⁻¹ are consistent with C-O bonding.

However, both proposed structures contain C-O bonds. These absorptions support the presence of the relevant oxygen-containing functional group but do not independently establish which compound is present.

**Step 4**: Select the structure that explains the complete spectrum.

The unknown is **butanoic acid**, \(\ce{CH3CH2CH2COOH}\).

It accounts for:
- The strong C=O absorption.
- The very broad carboxylic acid O-H absorption.
- The C-O absorptions.

Methyl propanoate cannot account for the characteristic broad acid O-H feature.

The decisive evidence is the shape and position of the O-H absorption, not simply the presence of a carbonyl or C-O bond.

### Worked example 2: When an ester also contains an O-H bond

An unknown compound is one of the following:

- 2-hydroxypropanoic acid: \(\ce{CH3CH(OH)COOH}\)
- Ethyl 2-hydroxypropanoate: \(\ce{CH3CH(OH)COOCH2CH3}\)

An illustrative percentage-transmittance spectrum shows:

| Observation | Measured feature |
|---|---|
| Carbonyl absorption | Strong, narrow dip at 1742 cm⁻¹ |
| O-H absorption | Broad dip centred around 3450 cm⁻¹ |
| Low-frequency O-H region | No very broad trough extending through 2500-3000 cm⁻¹ |
| C-O region | Strong absorptions around 1160 and 1250 cm⁻¹ |

Identify the most likely compound. Explain why the O-H absorption does not automatically indicate a carboxylic acid.

[IMAGE: Illustrative percentage-transmittance spectrum for the hydroxy ester example. Show a strong narrow carbonyl absorption at 1742 cm⁻¹, a broad alcohol O-H absorption centred near 3450 cm⁻¹, no very broad trough extending through 2500-3000 cm⁻¹, and strong fingerprint absorptions around 1160 and 1250 cm⁻¹. Use the reversed wavenumber axis and show the complete 4000-650 cm⁻¹ region.]

**Step 1**: Identify the carbonyl absorption.

The strong absorption at 1742 cm⁻¹ indicates a carbonyl group.

This is consistent with an ester, although carbonyl absorption alone is not sufficient to establish the functional group.

**Step 2**: Identify the type of O-H absorption.

The spectrum contains a broad absorption around 3450 cm⁻¹.

This is consistent with an alcohol O-H bond.

Notice that both candidate compounds contain an alcohol O-H bond. Therefore, this absorption does not distinguish them.

The important observation is that there is no very broad carboxylic acid O-H absorption extending through the 2500-3000 cm⁻¹ region.

**Step 3**: Examine the fingerprint region.

The strong absorptions around 1160 and 1250 cm⁻¹ support the presence of C-O bonds.

Together with the carbonyl absorption at 1742 cm⁻¹, this is consistent with an ester functional group.

**Step 4**: Compare the complete evidence.

The unknown is most likely **ethyl 2-hydroxypropanoate**.

It contains:
- An ester group, accounting for the strong C=O and C-O absorptions.
- An alcohol O-H group, accounting for the broad absorption near 3450 cm⁻¹.
- No carboxylic acid O-H bond, explaining the absence of the very broad low-frequency absorption.

The tempting mistake is to assume that any O-H absorption indicates a carboxylic acid. The spectrum shows why that conclusion is unreliable.

The position and shape of the O-H absorption matter just as much as its presence.

## 7. Questions and solutions

The following questions use original, illustrative spectral data. Unless stated otherwise, assume the samples are pure, the background has been corrected, and the stated spectrum covers the complete relevant wavenumber region.

### Question 1

An unknown compound is either butanoic acid or methyl propanoate.

Its percentage-transmittance IR spectrum contains:

- A strong carbonyl absorption at a wavelength of 5.75 μm.
- Strong C-O absorptions between 1050 and 1300 cm⁻¹.
- No broad O-H absorption between 2500 and 3300 cm⁻¹.

The spectrum covers 4000-650 cm⁻¹.

Calculate the carbonyl absorption's wavenumber in cm⁻¹ and identify the most likely compound. Explain your reasoning.

[IMAGE: Illustrative percentage-transmittance spectrum for methyl propanoate. Show a strong carbonyl dip at approximately 1740 cm⁻¹, strong C-O absorptions between 1050 and 1300 cm⁻¹, and no broad acid O-H absorption between 2500 and 3300 cm⁻¹. Include the complete reversed wavenumber axis.]

### Solution 1

The compound is **methyl propanoate**, with a carbonyl absorption at approximately 1740 cm⁻¹.

First, convert the wavelength into centimetres:

\[
\begin{aligned}
5.75\ \mathrm{\mu m}
&=5.75\times10^{-4}\ \mathrm{cm}
\end{aligned}
\]

Use the relationship between wavelength and wavenumber:

\[
\begin{aligned}
\tilde{\nu}&=\frac{1}{\lambda}\\
&=\frac{1}{5.75\times10^{-4}}\\
&=1739\ \mathrm{cm}^{-1}
\end{aligned}
\]

To three significant figures:

\[
\boxed{\tilde{\nu}=1.74\times10^3\ \mathrm{cm}^{-1}}
\]

This is consistent with an ester carbonyl absorption.

The C-O absorptions also support an ester structure. More importantly, the complete spectrum contains no very broad carboxylic acid O-H absorption.

Butanoic acid would be expected to produce that broad feature. Its absence, combined with the other observations, supports methyl propanoate.

### Question 2

A student receives an IR spectrum of an unknown compound.

The spectrum contains:

- A strong carbonyl absorption at 1720 cm⁻¹.
- A strong absorption at 1260 cm⁻¹.
- No visible O-H absorption.

The student concludes that the compound must be an ester.

However, the supplied spectrum only covers 1800-650 cm⁻¹.

Evaluate the student's conclusion. Can the compound be confidently identified as an ester?

[IMAGE: A cropped percentage-transmittance IR spectrum with a reversed wavenumber axis covering only 1800-650 cm⁻¹. Show a strong carbonyl dip at 1720 cm⁻¹ and a fingerprint-region absorption at 1260 cm⁻¹. Visually indicate that the 2500-3300 cm⁻¹ region is outside the recorded range, rather than showing it as a flat baseline.]

### Solution 2

The student's conclusion is not justified. The compound cannot be confidently identified as an ester from the supplied spectrum.

The strong absorption at 1720 cm⁻¹ indicates a carbonyl group. However, both carboxylic acids and esters contain C=O bonds.

The absorption at 1260 cm⁻¹ is consistent with C-O bonding, but this is also present in both functional groups.

The critical problem is the missing spectral region.

The carboxylic acid O-H absorption usually occurs between approximately 2500 and 3300 cm⁻¹. The supplied spectrum stops at 1800 cm⁻¹, so this feature cannot be assessed.

A carboxylic acid could therefore produce exactly the observed features in the recorded region.

The tempting argument is that no O-H absorption is visible. But the spectrum has not measured the region where the characteristic acid O-H absorption occurs.

To distinguish the compounds, a complete spectrum covering the O-H region is needed.

This illustrates an important experimental principle: **you cannot use an unmeasured region as evidence that a bond is absent.**

### Question 3

An unknown compound is known to be either:

- A conjugated carboxylic acid containing the structural group \(\ce{R-CH=CH-COOH}\).
- A saturated ester.

An illustrative percentage-transmittance IR spectrum shows:

- A strong carbonyl absorption at 1692 cm⁻¹.
- A very broad absorption extending from 2500 to 3200 cm⁻¹.
- A C-O absorption around 1270 cm⁻¹.

A student argues that the carbonyl absorption is too low to represent a carboxylic acid because the usual reference value is approximately 1710 cm⁻¹.

Calculate the difference between the observed carbonyl wavenumber and the reference value. Then determine which functional group is most likely and explain why the student's argument is incorrect.

[IMAGE: Illustrative spectrum of a conjugated carboxylic acid. Show a strong carbonyl dip at 1692 cm⁻¹, a very broad acid O-H trough from 2500 to 3200 cm⁻¹, and a C-O absorption near 1270 cm⁻¹. Include a small structural inset showing the conjugated arrangement C=C-C(=O)-OH.]

### Solution 3

The unknown is a **conjugated carboxylic acid**. Its carbonyl absorption is 18 cm⁻¹ lower than the approximate reference value, but this does not rule out an acid.

Calculate the difference:

\[
\begin{aligned}
\Delta\tilde{\nu}
&=1692-1710\\
&=-18\ \mathrm{cm}^{-1}
\end{aligned}
\]

The negative sign indicates that the observed absorption occurs at a lower wavenumber.

The reason is conjugation.

The C=C bond is directly adjacent to the carbonyl group. The electrons are delocalised across the connected system, changing the effective bonding within the carbonyl group.

This can reduce the carbonyl stretching frequency.

The second observation is much more decisive: the very broad absorption between 2500 and 3200 cm⁻¹ is characteristic of a carboxylic acid O-H bond.

The C-O absorption is also consistent with the proposed structure.

The student has treated 1710 cm⁻¹ as an exact boundary. It is only an approximate reference value, not a universal position for every carboxylic acid.

The complete pattern supports a conjugated carboxylic acid, despite its lower carbonyl wavenumber.

### Question 4

Two possible compounds are being compared:

- 2-hydroxypropanoic acid, \(\ce{CH3CH(OH)COOH}\).
- Ethyl 2-hydroxypropanoate, \(\ce{CH3CH(OH)COOCH2CH3}\).

Both compounds contain an O-H bond and a carbonyl group.

An illustrative IR spectrum of the unknown shows:

| Feature | Observation |
|---|---|
| C=O | Strong absorption at 1740 cm⁻¹ |
| O-H | Broad absorption centred around 3450 cm⁻¹ |
| Acid O-H region | No very broad absorption extending through 2500-3000 cm⁻¹ |
| C-O | Strong absorptions around 1170 and 1240 cm⁻¹ |

Identify the unknown.

Explain why the presence of an O-H absorption does not make it a carboxylic acid. Your answer should refer to the different O-H environments in the two candidate structures.

[IMAGE: Side-by-side structural diagrams of 2-hydroxypropanoic acid and ethyl 2-hydroxypropanoate. Show the alcohol O-H group in both molecules and the additional carboxylic acid O-H group only in the acid. Include a small accompanying illustrative spectral comparison showing the acid's very broad low-frequency O-H feature and the ester’s alcohol O-H feature near 3450 cm⁻¹.]

### Solution 4

The unknown is **ethyl 2-hydroxypropanoate**, which is an ester containing an additional alcohol group.

The important distinction is that the two compounds contain different types of O-H bonds.

2-hydroxypropanoic acid contains:
- An alcohol O-H bond attached to the second carbon.
- A carboxylic acid O-H bond within the -COOH group.

Ethyl 2-hydroxypropanoate contains:
- An alcohol O-H bond attached to the second carbon.
- An ester group, which has no carboxylic acid O-H bond.

The broad absorption around 3450 cm⁻¹ is consistent with the alcohol O-H bond present in both candidates.

However, the acid candidate should also produce the characteristic very broad carboxylic acid O-H absorption extending through the lower wavenumber region.

That feature is absent.

The strong carbonyl absorption at 1740 cm⁻¹ and the C-O absorptions around 1170 and 1240 cm⁻¹ provide further support for the ester.

The key is not simply counting the number of O-H absorptions. You need to identify which type of O-H bond is responsible for the observed band.

### Question 5

<badge color="danger">Extension</badge>

A chemist takes a pure sample of a carboxylic acid and records two IR spectra under different conditions.

The first spectrum is measured using the concentrated liquid sample. The second is measured after diluting the same compound in a dry, non-polar solvent.

The results are:

| Feature | Concentrated sample | Dilute sample |
|---|---|---|
| O-H absorption | Very broad, 2500-3300 cm⁻¹ | Much narrower, centred near 3550 cm⁻¹ |
| C=O absorption | 1712 cm⁻¹ | 1758 cm⁻¹ |
| C-O absorption | Around 1270 cm⁻¹ | Around 1270 cm⁻¹ |

Assume that:
- Both spectra are correctly background-corrected.
- The solvent does not contribute significant absorptions in the regions being interpreted.
- No chemical reaction occurs during dilution.
- Both spectra belong to the same pure compound.

A student examines the second spectrum and concludes:

"The compound must have changed into an ester. The broad acid O-H absorption has disappeared, and the carbonyl absorption has shifted into the ester region."

Evaluate this conclusion.

Your answer must explain:

1. Why the O-H absorption changes shape and position.
2. Why the carbonyl absorption shifts to a higher wavenumber.
3. Whether the functional group has actually changed.
4. Why comparing the two spectra without considering sample conditions could produce an incorrect identification.

[IMAGE: Side-by-side comparison of two illustrative IR transmittance spectra of the same carboxylic acid. The concentrated liquid spectrum must show a very broad O-H trough from 2500-3300 cm⁻¹ and a carbonyl dip at 1712 cm⁻¹. The dilute non-polar solution spectrum must show a much narrower O-H feature near 3550 cm⁻¹ and a carbonyl dip at 1758 cm⁻¹. Keep the C-O absorption near 1270 cm⁻¹ in both spectra. Add a small molecular inset comparing hydrogen-bonded carboxylic acid dimers in the concentrated sample with predominantly separated acid molecules in the dilute sample. Clearly indicate that the covalent molecular structure is unchanged.]

### Solution 5

The student's conclusion is incorrect. Both spectra are consistent with the same carboxylic acid measured under different hydrogen-bonding conditions.

The crucial idea is that changing the intermolecular environment can change an IR absorption without changing the covalent structure of the molecule.

**Step 1**: Explain the broad O-H absorption in the concentrated sample.

In the concentrated liquid, neighbouring carboxylic acid molecules form hydrogen bonds.

Two acid molecules can associate to form a hydrogen-bonded dimer. In this arrangement, each molecule interacts with the other through its carboxyl group.

These interactions weaken and alter the O-H bonding environment.

Different hydrogen-bonding strengths produce a range of O-H vibrational frequencies, giving the very broad absorption between 2500 and 3300 cm⁻¹.

**Step 2**: Explain what changes after dilution.

Dilution in a dry, non-polar solvent reduces the likelihood of acid molecules forming hydrogen-bonded dimers.

More molecules exist as individual acid molecules rather than strongly associated pairs.

Their O-H bonds now experience a more similar environment. The absorption becomes much narrower and moves towards a higher wavenumber, around 3550 cm⁻¹.

Notice that the O-H absorption has not necessarily vanished. It has changed shape and position.

**Step 3**: Explain the carbonyl shift.

In the concentrated sample, hydrogen bonding between carboxylic acid molecules also affects the carbonyl group.

The associated acid molecules commonly show a carbonyl absorption around 1710 cm⁻¹.

When the sample is diluted and the molecules become less associated, the carbonyl absorption shifts towards approximately 1760 cm⁻¹.

The observed change is:

\[
\begin{aligned}
\Delta\tilde{\nu}
&=1758-1712\\
&=+46\ \mathrm{cm}^{-1}
\end{aligned}
\]

The positive sign indicates that the carbonyl absorption has moved to a higher wavenumber.

This shift is consistent with a change from hydrogen-bonded acid molecules towards less-associated acid molecules.

**Step 4**: Decide whether the functional group changed.

It did not.

The molecule still contains the same carboxylic acid group. Dilution has changed the intermolecular attractions, not the covalent bonds.

The C-O absorption remaining around 1270 cm⁻¹ is also consistent with the same molecular structure.

**Step 5**: Evaluate the student's interpretation.

The student has used two observations that commonly distinguish acids and esters:

- The carbonyl absorption has shifted towards a typical ester region.
- The very broad acid O-H absorption is no longer visible in its original form.

However, the student has ignored the experimental conditions.

A carboxylic acid can produce different IR spectra depending on its concentration and the extent of hydrogen bonding. The narrower O-H absorption near 3550 cm⁻¹ is still evidence of an O-H bond.

The correct conclusion is that both spectra represent the same carboxylic acid under different conditions.

This is why IR identification must consider not only the absorption positions and shapes, but also the physical environment in which the sample was measured.

The deeper lesson is that an IR spectrum describes the vibrations of molecules in their actual environment. It does not provide a fixed set of peak positions that can be interpreted independently of experimental conditions.

## 8. What to remember when interpreting an exam spectrum

When comparing a carboxylic acid with an ester, the most useful distinction is the characteristic very broad carboxylic acid O-H absorption.

Keep these points in mind:

- Both compounds contain a C=O bond, so a strong carbonyl absorption alone does not distinguish them.
- A carboxylic acid produces a very broad O-H absorption, commonly described as extending from approximately 2500 to 3300 cm⁻¹.
- An ordinary ester has no carboxylic acid O-H bond.
- Esters commonly produce strong C-O absorptions between 1050 and 1300 cm⁻¹, but carboxylic acids also contain C-O bonds.
- Carbonyl wavenumbers are approximate and can shift with molecular structure and hydrogen bonding.
- Always check whether the spectrum actually covers the region you are using to eliminate a functional group.

For the structural differences between these compounds, including their intermolecular forces and chemical properties, read [Carboxylic Acids vs Esters: Structure, Properties and Reactions](https://tutorgum.com/learn/resources/carboxylic-acid-vs-ester-why-similar-formulas-can-produce-different-properties/).

Once you can distinguish these two functional groups, the next step is combining IR evidence with other analytical techniques. [How to Choose Between Organic Structures Using Two IR Features](https://tutorgum.com/learn/resources/hsc-chemistry-how-to-choose-between-several-structures-using-two-ir-features/) develops that broader skill.

In an HSC structural-identification question, IR can establish which functional groups are likely present. When several possible molecules contain the same functional groups, you'll need additional evidence, such as proton NMR, carbon-13 NMR, or mass spectrometry, to work out the complete molecular structure.