How to Choose Between Organic Structures Using Two IR Features

Three candidate structures all contain a carbonyl group: a carboxylic acid, an ester, and a hydroxyketone. An IR spectrum shows a strong absorption near \(1715\ \mathrm{cm}^{-1}\). Can you identify which structure is present?

Not yet. All three candidates contain a \(\ce{C=O}\) bond, so that absorption doesn’t distinguish them. However, if the spectrum also contains a very broad absorption between \(2500\) and \(3000\ \mathrm{cm}^{-1}\), the carboxylic acid becomes the clear favourite.

This is the key to choosing between several structures using IR spectroscopy: combine two diagnostic features and check which candidate explains both. Don’t simply identify the two largest peaks and assume they reveal the structure.

01Start by reading the IR spectrum correctly

Before interpreting any absorption bands, check the axes. Misreading the horizontal or vertical axis can turn a correct chemical observation into the wrong structural identification.

The horizontal axis: wavenumber

The horizontal axis usually shows wavenumber, measured in reciprocal centimetres, \(\mathrm{cm}^{-1}\).

A typical IR spectrum runs from approximately \(4000\ \mathrm{cm}^{-1}\) on the left to \(500\ \mathrm{cm}^{-1}\) on the right.

Notice something unusual: the wavenumber decreases as you move from left to right.

This means:

• \(3400\ \mathrm{cm}^{-1}\) appears to the left of \(1700\ \mathrm{cm}^{-1}\).
• \(1700\ \mathrm{cm}^{-1}\) appears to the left of \(1100\ \mathrm{cm}^{-1}\).
• The fingerprint region is towards the right-hand side.

Wavenumber is inversely related to wavelength. A higher wavenumber corresponds to a shorter wavelength and higher-frequency radiation.

The vertical axis: transmittance or absorbance?

This is where students often make an avoidable mistake.

IR spectra can display either percentage transmittance or absorbance. The same absorption appears differently depending on which quantity is plotted.

On a percentage transmittance spectrum, a downward dip means less infrared radiation passes through the sample because more has been absorbed.

On an absorbance spectrum, a taller upward peak represents greater absorption.

Always check the vertical axis before interpreting a feature. A downward trough on a transmittance spectrum is not evidence of weak absorption. It is the opposite.

02Choose two features that distinguish the candidates

The NSW Chemistry data sheet provides characteristic ranges for several important infrared absorptions.

You don’t need to memorise every possible organic absorption. Focus on the features that help distinguish the structures in the question.

Important IR absorption ranges

The ranges are approximate. Molecular environment, hydrogen bonding, and experimental conditions can affect the precise position and shape of an absorption.

Think of these ranges as postcodes rather than exact street addresses. A band falling within a range supports a particular bond assignment, but its precise position does not automatically identify the entire molecule.

Feature 1: The carbonyl absorption

A carbonyl group contains a carbon-oxygen double bond, \(\ce{C=O}\).

It produces a characteristic, usually strong absorption between approximately \(1680\) and \(1750\ \mathrm{cm}^{-1}\).

This is useful because it tells you that a carbonyl group is probably present.

However, several organic functional groups contain carbonyl bonds:

• Aldehydes
• Ketones
• Carboxylic acids
• Esters

Consequently, a strong absorption near \(1715\ \mathrm{cm}^{-1}\) does not prove that a compound is a ketone or a carboxylic acid.

You need another feature to narrow the possibilities.

Feature 2: The shape and position of the O-H absorption

This is often the most useful second feature when comparing alcohols, carboxylic acids, and esters.

Alcohols and carboxylic acids both contain O-H bonds, but their IR absorptions have different characteristic ranges and shapes.

Alcohol O-H:

• Appears approximately between \(3230\) and \(3550\ \mathrm{cm}^{-1}\).
• Usually produces a broad absorption.
• Supports the presence of an alcohol functional group.

Carboxylic acid O-H:

• Appears approximately between \(2500\) and \(3000\ \mathrm{cm}^{-1}\) on the NSW Chemistry data sheet.
• Produces a very broad absorption.
• Can overlap with the C-H region, making the spectrum look less tidy.

The broadness is important. Hydrogen bonding affects the O-H bond’s vibrational frequency, spreading the absorption across a wider range of wavenumbers.

An ester contains no O-H bond, so a pure ester should not produce the characteristic alcohol or carboxylic acid O-H absorption.

Why the C-O region needs careful interpretation

A C-O absorption between approximately \(1000\) and \(1300\ \mathrm{cm}^{-1}\) can support the presence of oxygen-containing functional groups.

However, it isn’t unique to esters.

Alcohols, ethers, esters, and carboxylic acids can all contain C-O bonds.

This means you shouldn’t write:

There is a C-O absorption, therefore the compound is an ester.

Instead, consider the other evidence.

For example:

• C=O and C-O, with no O-H, can support an ester over a ketone.
• C=O and a very broad acid O-H support a carboxylic acid.
• C=O and a broad alcohol O-H support a compound containing both a carbonyl group and an alcohol group.

The candidate structures determine which pair of features is most useful.

03A practical decision process

When you’re given several possible structures, use this sequence.

Step 1: List the functional groups in each candidate.

Don’t start by assigning every absorption. First work out what bonds each proposed structure should contain.

Step 2: Identify the strongest diagnostic feature.

Look for a strong C=O absorption, a broad O-H band, or another characteristic feature relevant to the candidates.

Step 3: Choose a second feature that separates the remaining candidates.

For example, if every candidate contains C=O, that band cannot distinguish them. Investigate the O-H region instead.

Step 4: Check the combination against every candidate.

A good structure should explain both observed features and should not predict a strong diagnostic absorption that is clearly absent from a complete, good-quality spectrum.

Step 5: Decide whether the evidence is sufficient.

If two structures contain the same functional groups, two IR features may not distinguish their exact arrangements. Don’t force a unique answer when the evidence doesn’t support one.

06A common mistake: treating one absorption as proof

Suppose an unknown has:

• A strong C=O absorption.
• A clear C-O absorption.
• No obvious O-H absorption.

A student concludes that the unknown must be an ester.

That conclusion might be correct, but it isn’t automatically justified.

A carboxylic acid also contains C=O and C-O bonds. The crucial difference is its very broad O-H absorption.

If the spectrum clearly covers the \(2500\)-\(3000\ \mathrm{cm}^{-1}\) region and shows no acid O-H band, the ester becomes more likely.

If that region is missing, cropped, or obscured, you cannot confidently eliminate the carboxylic acid.

This leads to an important rule:

An absent absorption is only useful evidence when the spectrum actually gives you a reliable opportunity to observe it.

Also, don’t rely on a tiny difference in carbonyl position to identify a functional group when the data-sheet ranges overlap. A carbonyl absorption at \(1715\ \mathrm{cm}^{-1}\), for example, is not a unique fingerprint for one particular type of carbonyl compound.

07Questions and solutions

The following questions use illustrative IR data, not experimental measurements. Read the axes, compare the band shapes, and use the candidate structures rather than guessing from individual numbers.

Question 1

An unknown compound is one of the following:

• Propanoic acid
• Propanone
• Propan-1-ol

Its percentage-transmittance IR spectrum contains:

• A strong, narrow downward absorption at \(1715\ \mathrm{cm}^{-1}\).
• A very broad downward absorption between \(2500\) and \(3000\ \mathrm{cm}^{-1}\).

Identify the compound and explain how both features support your answer.

Solution 1

The compound is propanoic acid.

The strong absorption at \(1715\ \mathrm{cm}^{-1}\) indicates a carbonyl group.

This eliminates propan-1-ol, which contains no C=O bond.

However, both propanone and propanoic acid contain carbonyl groups, so the first feature cannot distinguish between them.

The second feature provides the distinction. The very broad absorption between \(2500\) and \(3000\ \mathrm{cm}^{-1}\) is characteristic of the O-H bond in a carboxylic acid.

Propanone has no O-H bond, so it cannot explain this absorption.

Therefore, the combination of C=O and very broad acid O-H identifies propanoic acid.

Question 2

An unknown has the molecular formula \(\ce{C4H8O2}\).

Three possible structures are:

• Butanoic acid, \(\ce{CH3CH2CH2COOH}\)
• Methyl propanoate, \(\ce{CH3CH2COOCH3}\)
• 4-hydroxybutan-2-one, \(\ce{CH3COCH2CH2OH}\)

Its illustrative IR spectrum shows:

• A strong C=O absorption at \(1725\ \mathrm{cm}^{-1}\).
• A broad O-H absorption between \(3230\) and \(3550\ \mathrm{cm}^{-1}\).
• No very broad absorption extending through the \(2500\)-\(3000\ \mathrm{cm}^{-1}\) region.

Identify the most likely structure. Explain why the carbonyl absorption alone cannot distinguish the three candidates.

Solution 2

The most likely structure is 4-hydroxybutan-2-one, \(\ce{CH3COCH2CH2OH}\).

All three candidate structures contain a carbonyl group, so the strong absorption at \(1725\ \mathrm{cm}^{-1}\) is consistent with each one.

The O-H absorption provides the distinction.

The band between \(3230\) and \(3550\ \mathrm{cm}^{-1}\) is characteristic of an alcohol O-H bond.

Compare the three structures:

• Butanoic acid contains a carboxylic acid O-H bond, expected to produce a very broad absorption around \(2500\)-\(3000\ \mathrm{cm}^{-1}\).
• Methyl propanoate contains no O-H bond.
• 4-hydroxybutan-2-one contains an alcohol O-H bond, consistent with the observed broad absorption.

The absence of the very broad acid O-H feature also argues against butanoic acid.

Therefore, 4-hydroxybutan-2-one is the best fit.

The key is to recognise that the carbonyl identifies a feature shared by all three structures, while the O-H absorption identifies the particular type of oxygen-containing functional group.

Question 3

An unknown has one of the following structures:

• Propanoic acid
• 1-hydroxypropan-2-one
• Methyl ethanoate

An illustrative IR spectrum has absorbance on its vertical axis and wavenumber on its horizontal axis.

It shows:

• A broad upward absorption band centred near \(3380\ \mathrm{cm}^{-1}\).
• A strong upward absorption band near \(1715\ \mathrm{cm}^{-1}\).

A student makes the following claim:

“The band at \(3380\ \mathrm{cm}^{-1}\) is a carboxylic acid O-H absorption because it appears on the left-hand side of the spectrum.”

Explain what the student has misunderstood and identify the most likely compound.

Solution 3

The most likely compound is 1-hydroxypropan-2-one, \(\ce{HOCH2COCH3}\).

The student has confused the position of the band on the graph with the characteristic absorption range of the functional group.

The vertical axis is absorbance, so upward peaks represent absorption. Both observed features are therefore genuine absorptions.

The horizontal axis is wavenumber, which decreases from left to right. The band at \(3380\ \mathrm{cm}^{-1}\) appears towards the left because it has a higher wavenumber than the carbonyl absorption at \(1715\ \mathrm{cm}^{-1}\).

However, a carboxylic acid O-H absorption is expected around \(2500\)-\(3000\ \mathrm{cm}^{-1}\), not \(3380\ \mathrm{cm}^{-1}\).

The broad absorption near \(3380\ \mathrm{cm}^{-1}\) is consistent with an alcohol O-H bond.

The strong absorption at \(1715\ \mathrm{cm}^{-1}\) confirms a carbonyl group.

1-hydroxypropan-2-one contains both an alcohol O-H and a ketone C=O, explaining the two observed features.

Question 4

An unknown compound has the molecular formula \(\ce{C4H8O2}\) and is one of the following:

• Butanoic acid, \(\ce{CH3CH2CH2COOH}\)
• Methyl propanoate, \(\ce{CH3CH2COOCH3}\)

An illustrative, incomplete IR spectrum shows:

• A strong absorption at \(1735\ \mathrm{cm}^{-1}\).
• An absorption at \(1150\ \mathrm{cm}^{-1}\).
• No visible O-H absorption in the displayed spectrum.

However, the graph only contains two wavenumber intervals: \(4000\)-\(3000\ \mathrm{cm}^{-1}\) and \(1800\)-\(500\ \mathrm{cm}^{-1}\). The region between \(3000\) and \(1800\ \mathrm{cm}^{-1}\) has not been recorded.

A student selects methyl propanoate because there is no visible O-H absorption.

Is this conclusion justified? Explain what additional spectral information is required.

Solution 4

No. The available spectrum does not provide enough evidence to distinguish the two structures.

Both butanoic acid and methyl propanoate contain a C=O bond, so both can explain the strong absorption at \(1735\ \mathrm{cm}^{-1}\).

Both also contain C-O bonds, so the absorption at \(1150\ \mathrm{cm}^{-1}\) does not distinguish them.

The student’s mistake is assuming that an O-H absorption is absent simply because none is visible.

The missing interval is \(3000\)-\(1800\ \mathrm{cm}^{-1}\).

This includes the characteristic carboxylic acid O-H region of \(2500\)-\(3000\ \mathrm{cm}^{-1}\).

Therefore, the spectrum has not actually tested the region needed to eliminate butanoic acid.

To distinguish the candidates, the missing region must be measured.

If a complete, good-quality spectrum shows the very broad acid O-H absorption, butanoic acid is supported. If that absorption is genuinely absent while the carbonyl and C-O features remain, methyl propanoate becomes the stronger candidate.

The important distinction is between an absorption that is absent and an absorption that was never measured.

Question 5

An unknown organic compound has the molecular formula \(\ce{C4H8O2}\).

Two possible structures are:

Structure A: 1-hydroxybutan-2-one

\[ \ce{HOCH2C(=O)CH2CH3} \]

Structure B: 4-hydroxybutan-2-one

\[ \ce{CH3C(=O)CH2CH2OH} \]

Both structures contain one ketone group and one alcohol group, but the position of the alcohol group differs.

An illustrative, high-quality IR spectrum shows:

• A strong C=O absorption at \(1718\ \mathrm{cm}^{-1}\).
• A broad O-H absorption centred near \(3370\ \mathrm{cm}^{-1}\).
• An absorption near \(1100\ \mathrm{cm}^{-1}\), consistent with C-O bonding.

A student argues that Structure A must be correct because the C=O absorption appears at \(1718\ \mathrm{cm}^{-1}\).

Evaluate this conclusion.

In your answer:

• Explain what the IR spectrum establishes.
• Explain whether the three observed features distinguish the position of the O-H group.
• Use the molecular formula to calculate the degree of unsaturation, assuming a neutral compound containing only carbon, hydrogen, and oxygen.
• Identify a suitable additional technique that could distinguish the two structures.

Solution 5

The IR evidence does not uniquely distinguish Structure A from Structure B. Both structures are consistent with the three observed absorptions.

Start with the molecular formula.

For a neutral organic compound containing only carbon, hydrogen, and oxygen, the degree of unsaturation (DBE) is:

\[ \mathrm{DBE}=\frac{2C+2-H}{2} \]

Oxygen is omitted from this calculation.

For \(\ce{C4H8O2}\):

\[ \begin{aligned} \mathrm{DBE} &=\frac{2(4)+2-8}{2}\\ &=\frac{2}{2}\\ &=1 \end{aligned} \]

The compound has one degree of unsaturation.

A carbonyl group contains one C=O double bond, accounting for this degree of unsaturation. Both candidate structures satisfy the molecular formula and this result.

Now consider the IR evidence.

The C=O absorption at \(1718\ \mathrm{cm}^{-1}\) confirms a carbonyl group. Both candidates contain a ketone, so both predict this absorption.

The broad O-H absorption near \(3370\ \mathrm{cm}^{-1}\) supports an alcohol group. Both candidates contain an alcohol, so both explain this feature.

The C-O absorption near \(1100\ \mathrm{cm}^{-1}\) is also consistent with both structures because each contains a C-O single bond.

The tempting mistake is to treat the exact carbonyl position as a unique identifier.

However, \(1718\ \mathrm{cm}^{-1}\) falls within the broad carbonyl range of \(1680\)-\(1750\ \mathrm{cm}^{-1}\). Both structures can produce a carbonyl absorption in this region. The precise position depends on the molecular environment and experimental conditions.

The spectrum identifies the functional groups, but the listed features do not establish where the alcohol group sits in the carbon skeleton.

A suitable next technique is proton NMR (\(^1\mathrm{H}\) NMR).

The two structures have different hydrogen environments:

• Structure A contains a \(\ce{HOCH2CO-}\) group.
• Structure B contains a \(\ce{CH3CO-}\) group and a terminal \(\ce{-CH2OH}\) group.

These differences produce different proton NMR signals, including different chemical shifts, integrations, and splitting patterns.

Carbon-13 NMR would provide another useful check through the different carbon environments.

The strongest justified conclusion is that the unknown is a hydroxyketone with molecular formula \(\ce{C4H8O2}\). The supplied IR features cannot establish which positional isomer is present.

This is an important limit of IR spectroscopy: identifying the functional groups is not always the same as identifying the complete molecular structure.

08Where to go next

When two candidate structures contain the same functional groups, stop trying to extract more certainty from the same two IR bands.

Move to evidence that reveals the arrangement of atoms.

• Mass Spectrometry, IR and NMR: How to Identify a Structure explains what each technique contributes and how to combine their evidence.
• How to Identify an Organic Structure Using MS, IR and NMR takes you through a complete unknown-structure problem.

The most useful habit to carry into an HSC question is simple: choose the IR features that eliminate the actual alternatives, check that your proposed structure explains both, and recognise when another technique is needed to finish the identification.

Reference: The approximate absorption ranges in this resource follow the infrared absorption data in the NSW Chemistry data sheet. Always use the data sheet supplied with your examination. NSW Chemistry formulae sheet and data sheet.