Infrared Spectroscopy of Organic Compounds: HSC Chemistry Guide

Imagine you are given two clear, colourless organic liquids. Both have the same molecular formula, \( \ce{C3H6O} \). One is propanal, an aldehyde. The other is propanone, a ketone. Their atoms are the same, but their structures are different.

How could you tell them apart without breaking the molecules apart?

A student might predict that the answer should come from the molecular formula. After all, the formula tells you how many carbon, hydrogen, and oxygen atoms are present. The problem is that a formula does not tell you how those atoms are arranged. Organic chemistry often comes down to identifying the arrangement, not just counting the atoms.

Infrared spectroscopy gives us a way to detect the bonds inside a molecule by measuring how they interact with infrared radiation.

01The basic idea: molecules have bond vibrations

Picture a bond between two atoms as a tiny spring. The atoms are not sitting perfectly still. They vibrate back and forth.

A bond can:

• stretch, where the atoms move closer together and further apart
• bend, where the atoms change angle

Different bonds behave differently because different atoms have different masses and different bond strengths.

A strong bond between light atoms vibrates differently from a weaker bond between heavier atoms.

When infrared radiation passes through a sample, some wavelengths can transfer energy to these vibrations. The molecule absorbs those wavelengths, and the missing energy appears as a peak in the infrared spectrum.

The key question is:

If every bond can vibrate, why does infrared spectroscopy not just produce a confusing mess of peaks?

The answer is that only certain vibrations absorb infrared radiation strongly. A vibration must involve a change in the molecule’s dipole moment, which is the separation of positive and negative charge.

For example, the \( \ce{C=O} \) bond in a carbonyl group has a large change in charge distribution as it stretches, so it produces a strong absorption.

02Building a mental model before the numbers

Think of infrared spectroscopy like listening to a group of people walking.

If everyone walks at exactly the same pace, the sound is simple. But if one person is wearing heavy boots, another is running, and another is dragging their feet, the pattern changes.

A bond’s “walking style” depends on:

• the atoms involved
• the bond strength
• the surrounding atoms in the molecule

The analogy helps because each bond has a characteristic vibration. However, it is not perfect. Molecules are not literally independent springs. Vibrations can interact with each other, and the surrounding structure can shift absorption positions slightly.

So infrared spectroscopy does not identify a molecule by finding one magical peak. It works by recognising patterns.

03Reading an infrared spectrum

Infrared spectra are usually displayed with:

• wavenumber on the horizontal axis, measured in \( \text{cm}^{-1} \)
• percentage transmittance or absorbance on the vertical axis

Wavenumber describes how many waves fit into a centimetre of space. In infrared spectroscopy, it is commonly used instead of wavelength.

A peak at a particular wavenumber tells you that a bond vibration absorbed infrared radiation at that energy.

The spectrum is often divided into two useful regions:

For HSC Chemistry, the functional group region usually does most of the early identification work.

04The important HSC absorption bands

You do not need to memorise every possible peak. Focus on the patterns that distinguish common organic functional groups.

The word “broad” matters. An O-H absorption spreads over a range of wavenumbers because hydrogen bonding affects the vibration.

A common mistake is to see a peak near \(1700\ \text{cm}^{-1}\) and immediately announce “ketone”. That is too quick.

A carbonyl peak tells you a \( \ce{C=O} \) group is present. It does not, by itself, tell you whether the molecule is a ketone, aldehyde, ester, or carboxylic acid.

You need other evidence.

05Using infrared data to support a structure

Infrared spectroscopy is usually a piece of evidence, not the entire answer.

A good identification process is:

1. Identify strong peaks that indicate important functional groups.
2. Compare the spectrum with possible structures.
3. Check whether the proposed structure explains all major peaks.
4. Use other evidence if needed, such as chemical tests or NMR spectroscopy.

For example, if a compound has:

• a strong peak near \(1700\ \text{cm}^{-1}\)
• no broad O-H peak

then it likely contains a carbonyl group, but it is not likely to be an alcohol or carboxylic acid.

Infrared spectroscopy works especially well alongside techniques such as carbon-13 NMR spectroscopy and proton NMR spectroscopy, which provide different information about the structure of the molecule:

• Carbon-13 NMR Spectroscopy: Interpreting HSC Chemistry Data
• Proton NMR Spectroscopy: How to Interpret HSC Chemistry Data

08The fingerprint region: useful, but not the first clue

The fingerprint region below \(1500\ \text{cm}^{-1}\) contains many complicated absorptions caused by bending and combined vibrations.

Students often try to memorise the fingerprint region peak-by-peak. This is usually inefficient.

A better approach is:

• use the functional group region to propose a structure
• use the fingerprint region to compare a sample against a known compound

The fingerprint region is like recognising a friend’s handwriting. A single letter may not prove who wrote it, but the whole pattern is distinctive.

09A common misconception: “A peak means the molecule definitely contains only that group”

This prediction feels reasonable. If a peak matches a functional group, why not stop there?

The problem is that many molecules share the same bond types.

For example:

• ketones contain \( \ce{C=O} \)
• aldehydes contain \( \ce{C=O} \)
• esters contain \( \ce{C=O} \)
• carboxylic acids contain \( \ce{C=O} \)

The peak tells you what bond is present, not the complete identity of the molecule.

A stronger chemist asks:

“Does this proposed structure explain all the important evidence?”

not:

“Can I find one peak that matches?”

10Questions and solutions

Question 1

A compound has a strong infrared absorption at \(1720\ \text{cm}^{-1}\). Which bond is most likely responsible for this absorption?

Solution 1

The absorption is most likely caused by a \( \ce{C=O} \) bond. A peak near \(1700\ \text{cm}^{-1}\) is characteristic of carbonyl stretching vibrations.

The tempting mistake is to name a specific compound, such as a ketone. The spectrum only proves that a carbonyl group is likely present. Additional evidence is needed to decide which carbonyl compound it is.

Question 2

An unknown compound shows:

• a broad absorption from \(3200-3600\ \text{cm}^{-1}\)
• no strong absorption near \(1700\ \text{cm}^{-1}\)
• C-H absorptions near \(2900\ \text{cm}^{-1}\)

What functional group is most likely present?

Solution 2

The compound most likely contains an alcohol functional group because the broad absorption from \(3200-3600\ \text{cm}^{-1}\) indicates an O-H stretch.

The absence of a strong \(1700\ \text{cm}^{-1}\) peak suggests there is no carbonyl group.

The C-H absorptions are expected in many organic molecules, so they support an organic compound but are not the key identifying feature.

Question 3

Two compounds have the same molecular formula, \( \ce{C3H6O} \). Compound A has a strong absorption near \(1715\ \text{cm}^{-1}\) but no O-H absorption. Compound B has a broad O-H absorption and no carbonyl peak. Explain why infrared spectroscopy can distinguish them.

Solution 3

Infrared spectroscopy can distinguish the compounds because they contain different functional groups.

Compound A contains a carbonyl group, shown by the strong \(1715\ \text{cm}^{-1}\) absorption. The lack of O-H absorption means it is not an alcohol.

Compound B contains an O-H group, shown by the broad absorption, and the absence of a carbonyl peak means it does not contain \( \ce{C=O} \).

The important idea is that infrared spectroscopy identifies bonds and functional groups, not molecular formulas alone.

Question 4

A student examines an infrared spectrum and sees a strong peak at \(1710\ \text{cm}^{-1}\). They conclude the compound must be a ketone. Evaluate this conclusion.

Solution 4

The conclusion is incomplete because a peak at \(1710\ \text{cm}^{-1}\) only confirms that a carbonyl group is likely present.

The student’s reasoning is tempting because ketones are common carbonyl compounds. However, aldehydes, esters, and carboxylic acids can also contain \( \ce{C=O} \).

To decide between these possibilities, the student must use additional peaks. For example, a broad O-H absorption would support a carboxylic acid, while other evidence may be needed to distinguish aldehydes and ketones.

Question 5

An unknown compound has molecular formula \( \ce{C4H8O2} \). A student proposes that it is an ester because the infrared spectrum shows a strong absorption near \(1740\ \text{cm}^{-1}\) and no broad O-H absorption.

Another student argues that it could be a carboxylic acid because both contain \( \ce{C=O} \).

Using only the infrared evidence provided, decide which explanation is stronger and explain the limitation of the evidence.

Solution 5

The ester explanation is stronger, but the infrared evidence alone does not prove the complete structure.

The strong absorption near \(1740\ \text{cm}^{-1}\) supports the presence of a carbonyl group. The absence of a broad O-H absorption makes a carboxylic acid less likely because carboxylic acids show a very broad O-H stretch.

The trap is assuming that every carbonyl compound is equally likely once a \( \ce{C=O} \) peak appears. The missing evidence matters. Here, the absence of O-H changes the interpretation.

However, infrared spectroscopy alone cannot identify the exact ester structure. Different esters can produce similar functional group peaks, so techniques such as proton NMR, carbon-13 NMR, or chemical tests may be needed to confirm the full structure.

Infrared spectroscopy gives you the first structural clues. The next step is learning how other techniques add different pieces of evidence until the molecule can be identified with confidence.