Why a Broad IR Band Contains Different Information from a Sharp Peak
Learn how broad and sharp IR bands reveal different molecular environments, particularly the effects of hydrogen bonding on O-H vibrations. Includes spectral comparisons, worked calculations, and HSC-style practice questions with solutions.
A liquid alcohol can show a broad O-H absorption, while the same alcohol, measured when its molecules are mostly isolated, can show a much sharper band. The molecule has not gained or lost an O-H bond. What changed is the environment around that bond.
That is why a broad band contains different information from a sharp peak: its width can reveal how much the vibrating bond’s surroundings vary, often because of hydrogen bonding. A sharp peak indicates a narrower range of vibrational frequencies. The position helps identify the type of vibration, while the width provides evidence about the chemical environment.
01Start with the puzzle: one bond, two different spectra
Imagine measuring the infrared spectrum of an alcohol twice. First, you measure a very dilute solution, where most alcohol molecules are separated from one another. Then you measure the same alcohol as a concentrated liquid, where its molecules interact frequently.
Predict what will happen to the O-H absorption.
You might expect the same molecule to produce the same peak in both measurements. After all, its molecular formula and covalent bonds have not changed.
But the spectra can look quite different:
- In the dilute solution, the O-H absorption can be relatively sharp and appear at a higher wavenumber.
- In the concentrated liquid, the O-H absorption is generally broader and appears across a lower range of wavenumbers.
The important change is intermolecular hydrogen bonding. In the liquid, different O-H bonds experience different strengths of attraction to neighbouring molecules. Their vibrations therefore absorb infrared radiation across a wider range of frequencies.
02What an IR spectrum actually measures
An infrared (IR) spectrometer measures how much infrared radiation a sample absorbs at different wavenumbers.
Molecules are constantly vibrating. Their covalent bonds can stretch and bend. When infrared radiation has the appropriate frequency, it can transfer energy to a molecular vibration, provided that vibration produces a change in dipole moment.
The instrument detects the radiation that passes through the sample. It then processes the measurements to produce an IR spectrum.
Follow the radiation through the instrument
In a simplified Fourier-transform infrared (FTIR) experiment, the radiation follows this path:
- An infrared source produces radiation across a range of wavenumbers.
- An interferometer processes the radiation before it reaches the sample.
- The radiation passes through a sample held in a suitable cell.
- A detector measures the radiation that emerges.
- A computer processes the signal and produces the spectrum.
A background measurement helps account for absorption from the instrument, the atmosphere, and the sample holder or solvent.
What should happen if the sample contains O-H bonds?
The detector should receive less radiation at wavenumbers absorbed by O-H stretching vibrations. The resulting spectrum will contain an absorption feature in that region.
However, the instrument does not directly count O-H bonds. It records how the sample absorbs radiation across a range of wavenumbers.
That distinction is important when interpreting band shape.
Read both axes before interpreting the shape
The horizontal axis normally shows wavenumber, measured in reciprocal centimetres, \(\mathrm{cm}^{-1}\).
A typical organic IR spectrum decreases from approximately \(4000\ \mathrm{cm}^{-1}\) on the left to \(500\ \mathrm{cm}^{-1}\) on the right.
For example:
- \(3600\ \mathrm{cm}^{-1}\) appears towards the left.
- \(3000\ \mathrm{cm}^{-1}\) appears further right.
- \(1700\ \mathrm{cm}^{-1}\) appears further right again.
The vertical axis can show either percentage transmittance or absorbance.
| Vertical axis | Appearance of absorption | Meaning |
|---|---|---|
| Percentage transmittance (%T) | Downward trough | Less radiation passes through the sample |
| Absorbance | Upward peak | More radiation is absorbed |
A downward trough on a transmittance spectrum represents strong absorption, not weak absorption.
There are three separate things to examine:
- Position: where the absorption occurs on the wavenumber axis.
- Width: how far the absorption spreads horizontally.
- Intensity: how strongly the sample absorbs, represented by peak height or trough depth.
Don’t confuse them. A tall, narrow peak and a shorter, broad peak provide different information. The height does not tell you the width.
03Why hydrogen bonding produces a broad band
Now we can explain the difference between a broad and a sharp O-H absorption.
An alcohol contains an O-H bond. Oxygen attracts the shared electrons more strongly than hydrogen, making the bond polar.
This allows the hydrogen atom to form an intermolecular hydrogen bond with an oxygen atom on a neighbouring molecule.
Consider two alcohol molecules:
\[ \ce{R-O-H...O-R} \]
The dotted interaction represents a hydrogen bond between molecules. It is not another covalent O-H bond.
One bond, many slightly different environments
Imagine a liquid alcohol containing thousands of molecules.
Some O-H bonds experience relatively weak hydrogen bonding. Others experience stronger interactions. Some molecules may participate in several neighbouring interactions.
Consequently, the O-H bonds do not all experience precisely the same environment.
Hydrogen bonding changes the effective strength of the O-H stretching vibration. Stronger hydrogen bonding generally weakens the effective O-H bond, lowering its stretching frequency and wavenumber.
The different environments therefore produce a range of vibrational frequencies.
Instead of one narrow absorption, the spectrometer records a broad band.
A useful analogy is a spring.
Imagine that an O-H bond behaves approximately like a spring connecting two atoms. Changing the interactions around that bond changes its effective stiffness, which changes its vibration.
The analogy helps explain why hydrogen bonding shifts the absorption. However, a real molecular vibration is more complicated than a mechanical spring, and the spectrum represents quantum-mechanical vibrational transitions.
For alcohols, a broad O-H absorption commonly appears around:
\[ 3230-3550\ \mathrm{cm}^{-1} \]
The width reflects the range of hydrogen-bonding environments.
Why can the same O-H bond produce a sharp peak?
Now consider the same alcohol in a very dilute, non-polar solution.
Most alcohol molecules are separated from one another. There are fewer opportunities for intermolecular hydrogen bonding.
The O-H bonds therefore experience more similar environments.
Their stretching absorptions become concentrated within a narrower range of wavenumbers, producing a sharper band.
Free O-H stretching absorptions can occur around:
\[ 3620-3670\ \mathrm{cm}^{-1} \]
These values are useful for understanding the physical effect of hydrogen bonding. They are not a replacement for the characteristic ranges supplied in your HSC Chemistry data sheet.
The comparison is the important part:
| Feature | Relatively free O-H | Hydrogen-bonded O-H |
|---|---|---|
| Molecular environment | More isolated molecules | Interacting neighbouring molecules |
| Hydrogen bonding | Little or none between alcohol molecules | Present to varying degrees |
| Effective O-H bond | Less affected by hydrogen bonding | Generally weakened by hydrogen bonding |
| Wavenumber | Generally higher | Generally lower |
| Band shape | Relatively sharp | Broad |
A sharp band does not mean there is only one O-H bond in the entire sample. Thousands of equivalent molecules can produce a sharp absorption if their bonds experience similar environments.
Likewise, a broad band does not necessarily mean the molecule contains several different types of O-H bonds.
Band width tells you about the spread of vibrational environments, not simply the number of bonds.
04How an experiment separates the two explanations
Suppose a chemist observes a broad O-H band in an unknown alcohol sample.
There are two possible explanations:
- The molecule contains several different O-H environments.
- The O-H bonds experience different intermolecular hydrogen-bonding environments.
A single spectrum cannot always distinguish these explanations.
The chemist needs to change one condition while keeping the molecular identity constant.
The revised experiment: change concentration
A useful investigation measures the IR spectra of the same alcohol at several concentrations in a suitable dry, non-polar solvent.
Keep the following conditions controlled:
- Temperature.
- Sample cell and path length.
- Solvent.
- Instrument resolution.
- Background correction.
Compare the position and shape of the O-H absorption, rather than relying on raw peak height.
What should happen if intermolecular hydrogen bonding is responsible?
At low concentration, most alcohol molecules are separated. A relatively sharp, higher-wavenumber free O-H absorption should dominate.
As concentration increases, alcohol molecules encounter one another more frequently. Hydrogen bonding becomes more significant, producing a broader absorption at lower wavenumbers.
At intermediate concentrations, both features can appear together.
Illustrative experimental data
The following values are constructed to demonstrate the expected pattern. They are not measurements from a particular experiment.
| Alcohol concentration | O-H absorption observed | Interpretation |
|---|---|---|
| 0.01 mol L⁻¹ | Relatively sharp band near 3650 cm⁻¹ | Most O-H groups are relatively free |
| 0.10 mol L⁻¹ | Sharp band near 3650 cm⁻¹ and broad band around 3370-3520 cm⁻¹ | Both free and hydrogen-bonded environments are present |
| 1.00 mol L⁻¹ | Broad band around 3260-3500 cm⁻¹ dominates | Hydrogen-bonded environments are more significant |
The observation supports intermolecular hydrogen bonding as the explanation for the broad band.
Notice something particularly useful: the chemical identity of the alcohol has not changed. Its molecular structure remains the same throughout the investigation.
The experiment therefore demonstrates that band shape can provide information about the physical and intermolecular environment of a molecule, not just its functional groups.
There is one experimental limitation worth remembering. Increasing concentration also changes the total amount of absorbing material in the sample. This can change absorption intensity independently of band shape. Spectra should therefore be appropriately corrected and normalised before comparing the relative contributions of different O-H environments.
05Broad bands and sharp peaks: what can you actually conclude?
Let’s bring the reasoning together.
| Spectral observation | What it supports | What it does not establish |
|---|---|---|
| Broad O-H band | A range of O-H vibrational environments, often due to hydrogen bonding | That the molecule contains several O-H bonds |
| Sharp O-H band | A relatively uniform O-H environment, potentially with little hydrogen bonding | That only one O-H bond is present |
| Strong, narrow C=O absorption | A carbonyl stretching vibration | Which carbonyl-containing functional group is present |
| Broad band with smaller peaks superimposed | A broad absorption overlapping other vibrations | That every visible feature belongs to the same bond type |
A contrasting case: the broad band can contain overlapping absorptions
Carboxylic acids provide a useful example.
Their O-H absorption is very broad, commonly appearing around:
\[ 2500-3000\ \mathrm{cm}^{-1} \]
They also contain a carbonyl bond, producing a strong absorption around:
\[ 1680-1750\ \mathrm{cm}^{-1} \]
The acid O-H band can overlap with the C-H stretching region.
This means that several smaller, relatively narrow absorptions may appear within the broad O-H envelope.
A student might interpret these smaller features as separate O-H absorptions. That is not necessarily correct. They may be C-H vibrations superimposed on the broad O-H absorption.
The distinction is useful because it shows why band shape must be interpreted alongside position and other spectral evidence.
If you want to develop this comparison further, read How to Distinguish Alcohol and Carboxylic-Acid O-H Bands.
A reliable decision process
When you encounter a broad or sharp absorption in an HSC question, work through these steps.
- Read the axes. Identify wavenumber and determine whether the vertical axis shows absorbance or percentage transmittance.
- Locate the absorption. Use its wavenumber to identify the likely bond vibration.
- Examine the width. Decide whether the absorption is relatively broad or sharp.
- Consider the molecular environment. For O-H, investigate whether hydrogen bonding could explain the width and position.
- Check other absorptions. Look for overlapping vibrations and additional functional-group evidence.
- Limit your conclusion. Explain what the spectrum supports without claiming more than the evidence establishes.
The most useful habit is to describe position, width, and intensity separately before combining them into a chemical conclusion.
06Worked examples
Worked example: Comparing the widths of two O-H bands
Two absorbance spectra are obtained from the same alcohol under different conditions.
- Sample A has a relatively sharp O-H band extending from \(3635\) to \(3665\ \mathrm{cm}^{-1}\).
- Sample B has a broad O-H band extending from \(3260\) to \(3500\ \mathrm{cm}^{-1}\).
Calculate the width of each band and determine which sample provides evidence of more extensive hydrogen bonding.
Step 1
Band width is the difference between the higher and lower wavenumber boundaries:
\[ \Delta\tilde{\nu} = \tilde{\nu}_{\mathrm{high}} - \tilde{\nu}_{\mathrm{low}} \]
Here, \(\Delta\tilde{\nu}\) represents the band width in \(\mathrm{cm}^{-1}\).
Step 2
\[ \begin{aligned} \Delta\tilde{\nu}_{\mathrm{A}} &=3665-3635\\ &=30\ \mathrm{cm}^{-1} \end{aligned} \]
Sample A has a relatively narrow absorption.
Step 3
\[ \begin{aligned} \Delta\tilde{\nu}_{\mathrm{B}} &=3500-3260\\ &=240\ \mathrm{cm}^{-1} \end{aligned} \]
Sample B has a much broader absorption.
Step 4
\[ \frac{\Delta\tilde{\nu}_{\mathrm{B}}} {\Delta\tilde{\nu}_{\mathrm{A}}} = \frac{240}{30} =8 \]
Sample B’s band is eight times wider than Sample A’s band in this illustrative comparison.
Conclusion: Sample B provides stronger evidence of a wider distribution of hydrogen-bonding environments. Its absorption also occurs at lower wavenumbers, consistent with hydrogen bonding weakening the effective O-H stretching vibration.
The width comparison is illustrative rather than a universal numerical test. Real band boundaries are not always sharply defined.
Worked example: Interpreting two overlapping O-H features
An alcohol is measured at three concentrations using the same sample cell and temperature. The background-corrected absorbance spectra show the following features:
| Concentration | Sharp absorption | Broad absorption |
|---|---|---|
| 0.01 mol L⁻¹ | 3650 cm⁻¹ | Not resolved |
| 0.10 mol L⁻¹ | 3650 cm⁻¹ | 3370-3520 cm⁻¹ |
| 1.00 mol L⁻¹ | Weak feature near 3650 cm⁻¹ | 3260-3500 cm⁻¹ |
Explain the changes and determine why the intermediate-concentration spectrum contains both features.
Step 1
The same alcohol is measured at every concentration. Its molecular structure and number of O-H bonds per molecule remain constant.
Therefore, the changing band shape cannot be explained simply by the alcohol gaining or losing O-H bonds.
Step 2
At \(0.01\ \mathrm{mol\,L^{-1}}\), the alcohol molecules are relatively separated.
Most O-H groups experience little intermolecular hydrogen bonding, producing the relatively sharp absorption near \(3650\ \mathrm{cm}^{-1}\).
Step 3
At \(1.00\ \mathrm{mol\,L^{-1}}\), more alcohol molecules interact through hydrogen bonding.
The different hydrogen-bonding environments produce a broad distribution of lower-wavenumber O-H absorptions.
Step 4
At \(0.10\ \mathrm{mol\,L^{-1}}\), some O-H groups are relatively free while others participate in hydrogen bonding.
The spectrum therefore contains both a sharp free O-H feature and a broad hydrogen-bonded envelope.
Conclusion: The two features represent different intermolecular environments of the same O-H functional group. They do not automatically indicate two chemically different types of O-H bonds.
07Questions and solutions
Question 1
An absorbance IR spectrum has a horizontal axis decreasing from \(4000\) to \(500\ \mathrm{cm}^{-1}\).
It contains:
- A broad absorption between \(3270\) and \(3510\ \mathrm{cm}^{-1}\).
- A relatively sharp absorption at \(1720\ \mathrm{cm}^{-1}\).
(a) Which absorption is broader?
(b) Calculate the approximate width of the broad absorption.
(c) Explain what the broad absorption suggests about the O-H bonding environment.
(d) Explain why the height of the sharp absorption does not tell you whether it is broad or narrow.
Solution 1
(a) The absorption between \(3270\) and \(3510\ \mathrm{cm}^{-1}\) is broader.
It covers a larger horizontal range of wavenumbers than the relatively sharp feature at \(1720\ \mathrm{cm}^{-1}\).
(b) Its approximate width is \(240\ \mathrm{cm}^{-1}\).
Use:
\[ \Delta\tilde{\nu} = \tilde{\nu}_{\mathrm{high}} - \tilde{\nu}_{\mathrm{low}} \]
Substitute the boundaries:
\[ \begin{aligned} \Delta\tilde{\nu} &=3510-3270\\ &=240\ \mathrm{cm}^{-1} \end{aligned} \]
(c) The broad absorption is consistent with an O-H bond experiencing a range of hydrogen-bonding environments.
Different intermolecular interactions alter the effective O-H stretching vibration, producing absorptions across a range of wavenumbers.
(d) Peak height and band width describe different properties.
Height represents absorption intensity, whereas width describes the horizontal range of wavenumbers over which absorption occurs.
A sharp peak can be very intense, and a broad band can have a relatively low maximum absorbance. One does not determine the other.
Question 2
Two samples contain the same alcohol.
Sample A is measured in a very dilute, non-polar solution and produces a sharp O-H absorption from \(3630\) to \(3670\ \mathrm{cm}^{-1}\).
Sample B is measured as a concentrated liquid and produces a broad O-H absorption from \(3260\) to \(3500\ \mathrm{cm}^{-1}\).
(a) Calculate the width of each absorption.
(b) Calculate how many times wider Sample B’s absorption is.
(c) Explain why the absorption shifts towards lower wavenumbers in Sample B.
(d) Does this result mean that Sample B contains a different alcohol? Explain.
Solution 2
(a) Sample A has a width of \(40\ \mathrm{cm}^{-1}\), while Sample B has a width of \(240\ \mathrm{cm}^{-1}\).
For Sample A:
\[ \begin{aligned} \Delta\tilde{\nu} &=3670-3630\\ &=40\ \mathrm{cm}^{-1} \end{aligned} \]
For Sample B:
\[ \begin{aligned} \Delta\tilde{\nu} &=3500-3260\\ &=240\ \mathrm{cm}^{-1} \end{aligned} \]
(b) Sample B’s absorption is six times wider.
\[ \frac{240}{40}=6 \]
(c) Hydrogen bonding lowers the effective O-H stretching frequency.
In concentrated liquid alcohol, neighbouring molecules interact through intermolecular hydrogen bonds.
These interactions weaken the effective O-H bond, shifting its stretching absorption towards lower wavenumbers. Different hydrogen-bonding strengths also produce a wider range of absorption frequencies.
(d) No. Both samples contain the same alcohol.
The molecular structure remains unchanged. The difference arises from the intermolecular environment.
This demonstrates why a change in band position and width does not necessarily indicate a change in molecular identity.
Question 3
An unknown organic compound produces a percentage-transmittance IR spectrum with the following features:
- A very broad downward absorption extending from \(2500\) to \(3000\ \mathrm{cm}^{-1}\).
- Two smaller, relatively narrow downward absorptions at \(2860\) and \(2930\ \mathrm{cm}^{-1}\), appearing within the broad feature.
- A strong downward absorption at \(1712\ \mathrm{cm}^{-1}\).
A student suggests that the three absorptions between \(2500\) and \(3000\ \mathrm{cm}^{-1}\) represent three different O-H bonds.
Evaluate this interpretation and identify the most likely functional group.
Solution 3
The most likely functional group is a carboxylic acid, and the student’s interpretation is incorrect.
The broad absorption and the two narrower features provide different information.
Step 1: Interpret the broad absorption.
The very broad feature between \(2500\) and \(3000\ \mathrm{cm}^{-1}\) is characteristic of a carboxylic-acid O-H absorption.
Strong hydrogen bonding between carboxylic acid molecules produces a wide range of O-H stretching frequencies.
Step 2: Interpret the smaller absorptions.
The absorptions at \(2860\) and \(2930\ \mathrm{cm}^{-1}\) fall within the typical C-H stretching region.
They can therefore represent C-H vibrations superimposed on the broad acid O-H envelope.
They should not automatically be assigned to additional O-H bonds.
Step 3: Examine the carbonyl absorption.
The strong feature at \(1712\ \mathrm{cm}^{-1}\) falls within the characteristic carbonyl range of approximately \(1680\) to \(1750\ \mathrm{cm}^{-1}\).
This supports the presence of a \(\ce{C=O}\) bond.
Step 4: Combine the evidence.
A carboxylic acid contains both an O-H bond and a carbonyl bond within the same functional group:
\[ \ce{R-C(=O)-OH} \]
The very broad O-H absorption and strong C=O absorption are consistent with this structure.
The important lesson is that a broad band can contain smaller overlapping absorptions. You must interpret the whole envelope rather than counting every visible trough as a separate functional group.
Question 4
A chemist measures the same alcohol at two concentrations using the same sample cell.
The following illustrative absorbance data are obtained:
| Measurement | Concentrated sample | Dilute sample |
|---|---|---|
| Concentration | 1.00 mol L⁻¹ | 0.10 mol L⁻¹ |
| Maximum absorbance in O-H region | 0.90 | 0.18 |
| Broad O-H envelope | 3260-3500 cm⁻¹ | 3390-3540 cm⁻¹ |
| Sharp feature near 3650 cm⁻¹ | Weak | More prominent |
A student concludes:
“The concentrated sample must contain five times as many O-H bonds per molecule because its maximum absorbance is five times greater.”
(a) Explain why this conclusion is incorrect.
(b) Calculate the width of the broad absorption in each sample.
(c) Explain what the changes in band shape suggest about the intermolecular environment.
(d) Describe how the chemist should compare the spectra more reliably.
Solution 4
The conclusion is incorrect because absorbance intensity depends on the amount of absorbing material and the measurement conditions, not simply the number of O-H bonds in each molecule.
(a) Both samples contain the same alcohol, so each molecule has the same number of O-H bonds.
The concentrated sample contains more alcohol molecules per unit volume. It can therefore produce a greater overall absorbance without having more O-H bonds per molecule.
This is a concentration effect, not evidence of a structural change.
(b) The concentrated sample has a width of \(240\ \mathrm{cm}^{-1}\), while the dilute sample has a width of \(150\ \mathrm{cm}^{-1}\).
For the concentrated sample:
\[ \begin{aligned} \Delta\tilde{\nu} &=3500-3260\\ &=240\ \mathrm{cm}^{-1} \end{aligned} \]
For the dilute sample:
\[ \begin{aligned} \Delta\tilde{\nu} &=3540-3390\\ &=150\ \mathrm{cm}^{-1} \end{aligned} \]
The concentrated sample therefore has a broader O-H absorption.
(c) The concentrated sample contains a greater distribution of hydrogen-bonded environments.
At higher concentration, alcohol molecules interact more frequently.
This produces a broader absorption at lower wavenumbers. When diluted, fewer molecules participate in intermolecular hydrogen bonding, so the sharp, higher-wavenumber free O-H feature becomes more prominent.
(d) The chemist should compare corrected and appropriately normalised spectra.
The spectra should be background-corrected and compared under consistent measurement conditions.
Normalising the spectra allows the chemist to examine differences in their relative band shapes without confusing these differences with the overall amount of absorbing material.
The position and width of the bands provide evidence about the molecular environment, while raw absorbance intensity must be interpreted with concentration and path length in mind.
Question 5
A chemist investigates a pure organic compound known to contain exactly one O-H bond per molecule.
Three IR spectra are obtained at different concentrations in the same dry, non-polar solvent.
The temperature, sample cell, path length, and instrument resolution are kept constant. The spectra are background-corrected and normalised to the same total O-H absorption area.
The illustrative results are:
| Concentration | Observed O-H absorption |
|---|---|
| 0.005 mol L⁻¹ | A sharp band at 3655 cm⁻¹, approximately 30 cm⁻¹ wide |
| 0.10 mol L⁻¹ | A sharp band near 3650 cm⁻¹ and a broad envelope from 3370-3520 cm⁻¹ |
| 1.00 mol L⁻¹ | A dominant broad envelope from 3260-3500 cm⁻¹, with a much weaker sharp feature |
A student makes the following argument:
“The broad envelope must represent several chemically different O-H bonds. The sharp peak at 3655 cm⁻¹ must be an impurity because it appears separately from the broad band.”
Evaluate the student’s argument.
In your response:
(a) Explain why the broad and sharp features can belong to the same O-H functional group.
(b) Use the concentration data to identify the most likely molecular interaction responsible for the broad envelope.
(c) Predict what would happen if the concentrated sample were diluted further.
(d) Propose a second experimental change that could provide additional evidence for your explanation. Predict the expected observation.
(e) Explain what the spectra establish and what they cannot establish about the individual hydrogen bonds.
Solution 5
The student’s argument is incorrect. The results strongly support concentration-dependent intermolecular hydrogen bonding involving the same O-H functional group.
The key evidence is not simply that the spectrum contains two features. It is that their relative contributions change systematically as the concentration changes.
(a) The two features represent different environments of the same O-H bond.
The compound contains exactly one O-H bond per molecule.
At low concentration, most molecules are relatively isolated. Their O-H bonds experience similar environments, producing a sharp absorption near \(3655\ \mathrm{cm}^{-1}\).
At higher concentrations, more O-H groups interact with neighbouring molecules through hydrogen bonding.
These interactions produce a distribution of stretching frequencies, creating the broad, lower-wavenumber envelope.
The two features therefore represent relatively free and hydrogen-bonded populations of the same functional group.
There is no requirement for two chemically different types of O-H bonds.
(b) The most likely interaction is intermolecular hydrogen bonding.
Consider the progression:
- At \(0.005\ \mathrm{mol\,L^{-1}}\), the sharp free O-H band dominates.
- At \(0.10\ \mathrm{mol\,L^{-1}}\), both free and hydrogen-bonded environments are observed.
- At \(1.00\ \mathrm{mol\,L^{-1}}\), the broad hydrogen-bonded absorption dominates.
Increasing concentration brings alcohol molecules into closer contact and increases opportunities for intermolecular hydrogen bonding.
The broad band also shifts towards lower wavenumbers as the hydrogen-bonded environments become more significant.
This concentration-dependent change is strong evidence for intermolecular interactions rather than a fixed set of different covalent O-H bonds.
(c) Further dilution should increase the relative contribution of the sharp, higher-wavenumber absorption.
As the molecules become more separated, intermolecular hydrogen bonding becomes less significant.
The broad lower-wavenumber envelope should weaken relative to the sharp free O-H band.
At sufficiently low concentration, the sharp feature should dominate the spectrum.
This prediction follows directly from the proposed molecular model.
(d) Increasing the temperature provides another useful test.
The chemist could measure the concentrated sample at a modestly higher temperature while keeping the other experimental conditions controlled.
If intermolecular hydrogen bonding is responsible for the broad envelope, increasing thermal motion should generally reduce the population of hydrogen-bonded associations.
The expected observation is:
- A reduction in the relative contribution of the broad, lower-wavenumber absorption.
- An increase in the relative contribution of the sharper, higher-wavenumber free O-H absorption.
The change should be reversible when the sample returns to its original temperature, provided that the compound remains chemically unchanged and the measurement conditions are controlled.
This provides additional evidence that the broadness arises from intermolecular interactions rather than different covalent bonds.
(e) The spectra establish a distribution of O-H vibrational environments, but they do not reveal every individual hydrogen bond.
The concentration dependence supports intermolecular hydrogen bonding as the explanation for the broad absorption.
However, the spectrum does not tell the chemist the exact number of hydrogen bonds involving each molecule, the precise strength of every interaction, or the arrangement of every neighbouring molecule.
The broad envelope represents the combined absorption from many molecular environments.
The normalised spectra are particularly useful because they allow the chemist to compare the relative distribution of free and hydrogen-bonded O-H environments without treating a larger raw absorbance as evidence of more O-H bonds per molecule.
This is the deeper point: a broad band is not simply a less tidy version of a sharp peak. It can reveal how a molecule interacts with its surroundings, even when its covalent structure remains unchanged.
08Where to go next
Once you understand what band width reveals, the next step is using that information to identify an unknown organic compound.
Don’t rely on an O-H absorption alone. Combine its position and shape with other diagnostic features, particularly the carbonyl absorption and C-O region.
Continue with How to Choose Between Organic Structures Using Two IR Features to practise combining spectral evidence and deciding when an IR spectrum is not sufficient to identify a unique structure.