Master AQA A Level IR Spectroscopy with this complete revision guide. Learn characteristic peaks, fingerprint region, functional groups, exam tips, worked examples and common mistakes to boost your Chemistry grade.
INTRODUCTION
Infrared spectroscopy is certainly one of the most important analytical techniques available to today’s scientists. It helps to decode the type of bonds present in a molecule, basically it helps to do functional group analysis because every molecule has a specific vibration and it’s exclusive to them.
In AQA A Level Chemistry, IR spectroscopy is mainly used to identify functional groups such as:
- Alcohols
- Carboxylic acids
- Aldehydes
- Ketones
- Esters
- Amines
- Alkenes
- Nitriles
What Is Infrared Radiation?
Infrared radiation is a type of electromagnetic radiation.
The electromagnetic spectrum includes:
- Gamma rays
- X-rays
- Ultraviolet (UV)
- Visible light
- Infrared (IR)
- Microwaves
- Radio waves
Infrared radiation has a lower frequency than visible light but a higher frequency than microwaves.
Unlike visible light, infrared radiation is invisible to the human eye. However, we experience it every day as heat.
For example:
- Sunshine warms your skin.
- A toaster glowing red while producing heat.
These all emit infrared radiation.
Exam Tip- Do not write “IR spectroscopy uses visible light.”This is incorrect.Always state that IR spectroscopy uses infrared radiation.
How Infrared Spectroscopy Works?
Before going into working of Infrared spectroscopy we need to go through few assumptions of Atom model –
- Atoms are treated as massless points
- Covalent bonds are not treated as rigid bonds but springs capable of vibrating at certain frequencies.
Few vibrations include asymmetric stretching , symmetric stretching , bending , wagging etc
Naturally molecules are always vibrating at certain frequencies When infrared radiation passes through a sample of these molecules , the energy in the form of an infrared is absorbed by these molecules, causing a change in their vibrations. This change is detected by a spectrometer therefore producing the characteristic peaks seen on an IR spectrum
Why Do Different Bonds Absorb Different Frequencies?
Every molecule has a distinguished vibration because they vary in bond strength and polarity difference and atomic mass of both atoms in a bond, that is why C – O bond is different from
C – H bond.Each functional group has its own characteristic absorption range.This allows chemists to identify compounds quickly and accurately.
KEY AQA TERMINOLOGIES TO REMEMBER
- Bonds absorb IR radiation and undergo vibrational excitation
- Each type of bond (e.g., C=O, O–H, C–H) absorbs at a characteristic wavenumber because different bonds have different strengths and different masses of atoms involved
- The wavenumber is proportional to the frequency of vibration: higher wavenumber = higher frequency = stronger bond or lighter atoms
- The spectrum shows transmittance — the percentage of IR radiation that passes through the sample — so a peak represents low transmittance, meaning high absorption
What Is an IR Spectrum?
An IR spectrum is a graph showing how much infrared radiation a compound absorbs at different wavenumbers.
| Axis | What it shows |
| Y-axis | % Transmittance (how much IR passes through) |
| X-axis | Wavenumber / cm⁻¹ (decreasing left to right: 4000 → 400) |
EXAM TIP – Higher wavenumber = higher frequency = higher Energy
Decoding the IR Spectra
Every downward peak represents a bond absorbing infrared radiation.
A strong peak indicates strong absorption & A weak peak indicates weaker absorption.
Both the position and shape of the peak are important as they correspond to some specific function groups.
As far position is concerned the graph can be classified into 2 regions –
Region 1
Functional Group Region – Here you will find peaks of all the prominent functional groups like O–H etc. This region lies above 1500 cm-1.
Region 2
Fingerprint Region – This region is characteristic to each molecule existing in the universe. Like every human has its own DNA similarly every molecule has its own finger print region. This region lies below 1500cm-1
Shape of Peaks –
1)Broad peaks – They are across a wide range of wavenumber like they don’t have an exact value. The most common example is the O–H bond. This is mainly due to hydrogen bonds present in the molecule.
2)Sharp peaks – They have an exact value and these help to determine the functional group precisely as they are well pointed. The common example include C=O ,C≡N,C–H.
EXAM TIP – These both properties are widely used in answering exam questions in A levels.
Ques1)Compound X has the molecular formula C5H12O and is a secondary alcohol. Identify one feature of this infrared spectrum of a pure sample of X that may be used to confirm that X is an alcohol.
Solution – Just look at a broad peak in the range 3230 to 3550 cm−1.
A broad peak around 3300 to 3500 cm-1.
What Is Wavenumber?
It is basically the reverse of wavelength. In IR spectroscopy it is usually represented in cm-1.
Wavenumber = 1wavelength(cm)
Characteristic Wavenumbers (AQA Data Sheet)
The AQA Chemistry Data Sheet provides the following infrared absorption data. You must learn to use this table in the exam.
Table
| Bond | Wavenumber / cm⁻¹ | Notes |
| N–H (amines) | 3300 – 3500 | Sharp peak |
| O–H (alcohols) | 3230 – 3550 | Strong, broad peak due to hydrogen bonding |
| C–H | 2850 – 3300 | Present in almost all organic compounds |
| O–H (carboxylic acids) | 2500 – 3000 | Very broad, often overlaps C–H region |
| C≡N (nitriles) | 2220 – 2260 | Sharp, distinctive peak |
| C=O (carbonyl) | 1680 – 1750 | Strong, sharp peak |
| C=C (alkenes/aromatics) | 1620 – 1680 | Medium intensity |
| C–O | 1000 – 1300 | Strong peak |
| C–C | 750 – 1100 | Fingerprint region |
For AQA Chemistry, you are not expected to calculate wavenumbers. Instead, you need to recognise characteristic absorption ranges from the data booklet and use them to identify functional groups.
Exam tip: Always write the wavenumber range from the data sheet and try to write a range , for eg- Write: “The peak at 3230–3550 cm⁻¹ indicates an O–H bond in an alcohol.”
Distinguishing Between Similar Compounds
| Compounds | How to Distinguish |
| Alcohol vs. Carboxylic Acid | Acid has very broad O–H (2500–3000) and C=O (~1700); alcohol has O–H (3230–3550) but no C=O |
| Aldehyde vs. Ketone | Aldehyde has weak C–H peaks at ~2720 and ~2820 cm⁻¹; ketone does not |
| Carboxylic Acid vs. Ester | Acid has broad O–H (2500–3000); ester has no O–H |
| Primary vs. Tertiary Amine | Primary has N–H stretch (3300–3500); tertiary has no N–H |
Detecting Impurities
IR spectroscopy can identify impurities in a sample.
For example:
- If a sample of ethanal is contaminated with unreacted ethanol, the IR spectrum will show an unexpected broad O–H peak at 3230–3550 cm⁻¹ (from the ethanol) alongside the expected C=O peak of ethanal.
- Water contamination often appears as a broad O–H peak around 3300 cm⁻¹.
PAST PAPER PRACTICE QUESTIONS
Ques1)The infrared spectra shown are those of three compounds.
- Compound A 1,4-dibromobutane
- Compound B butane-1,4-diol
- Compound C butanedioic acid
Identify the compound responsible for each spectrum by writing the correct letter, A, B or C, in the box next to each spectrum. You may find it helpful to refer to the Table on the Data Sheet.
Solution – Always start with looking at the most prominent peak in various spectra.
Spectra A shows a broad band around 3300 – 3500 cm-1 , which is surely an alcohol.
Hence the first spectra belongs to butane-1,4-diol.
For second spectra there is still a peak around 3300 + another prominent peak is around 1720 to 1750 and the peak is narrow which surely indicates a carbonyl
Combining both observations we will land on a carboxylic acid
Hence the second spectra belongs to butanedioic acid
The last spectra belongs to 1,4-dibromobutane.
Ques2)The infrared spectrum shown below is either that of butan-2-ol or that of butanone.
Identify the compound to which this infrared spectrum refers. Explain your answer.
Solution – For this spectrum to be of Alcohol a broad peak around 3300 to 3400 should be there , but it is absent in this case ,which clearly means it belongs to butanone.The most common characteristic of a carbonyl is (Strong) absorption / peak at approximately 1700 (cm–1) / 1710 (cm–1) which is clearly present in this spectrum.
Ques3)Consider the four cyclic compounds, A, B, C, D
The infrared spectra of compounds A, B, C and D are shown below. Write the correct letter, A, B, C or D, in the box next to each spectrum.
Solution –
(i)It has a defined sharp peak around 1720 hence it is C.
(ii)Here no defined peak is present in functional group region , hence its an alkane, hence its compound A
(iii) Here again no defined peak is present in functional group region but a C=C stretching is present hence its compound D
(iv)It has broad band around 3400 which surely belong to an alcohol
Hence its compound B
Common Exam Mistakes to Avoid
| Mistake | Why It’s Wrong | Correct Approach |
| Calling any broad O–H peak an “alcohol” | Carboxylic acids also have O–H | Check position: alcohol O–H is 3230–3550; acid O–H is 2500–3000 and has C=O |
| Forgetting to mention what’s absent | You only get half the marks | Always state what peaks are missing to rule out other functional groups |
| Giving a single wavenumber | AQA expects ranges | Always quote the range from the data sheet |
| Confusing peak shapes | Shape gives crucial information | O–H = broad (H-bonding); C=O = sharp; N–H = sharp |
| Trying to assign fingerprint region peaks | This region is too complex | Only use it for comparison with database spectra |
Quick Revision Checklist
Before your exam, ensure you can:
- Explain that bonds absorb IR radiation at characteristic wavenumbers
- State that the fingerprint region (<1500 cm⁻¹) allows identification by comparison
- Use the AQA Data Sheet to identify particular bonds and functional groups
- Distinguish between alcohols, carboxylic acids, aldehydes, ketones, esters, amines, and nitriles
- Identify impurities from unexpected peaks
- Explain the link between IR absorption by CO₂, CH₄, and H₂O and global warming
- Quote wavenumber ranges, not single values
- Consider both present and absent peaks
Frequently Asked Questions
1. What is infrared spectroscopy used for in AQA A Level Chemistry?
Infrared (IR) spectroscopy is used to identify functional groups in organic compounds by detecting the characteristic wavenumbers at which bonds absorb infrared radiation. It can help identify groups such as O–H, C=O, C–O, N–H, C≡N and C=C.
2. What are the most important IR absorption ranges for AQA A Level Chemistry?
Some of the key AQA ranges include O–H (alcohol) at 3230–3550 cm⁻¹, O–H (carboxylic acid) at 2500–3000 cm⁻¹, C=O at 1680–1750 cm⁻¹, C≡N at 2220–2260 cm⁻¹, N–H at 3300–3500 cm⁻¹, and C–O at 1000–1300 cm⁻¹. Students should use the ranges provided in the AQA data sheet when answering exam questions.
3. What is the difference between the functional group region and fingerprint region in an IR spectrum?
The functional group region is generally above 1500 cm⁻¹ and contains characteristic absorptions that can help identify functional groups. The fingerprint region, below 1500 cm⁻¹, contains a complex pattern that is characteristic of a particular molecule and is mainly useful for comparison with a known spectrum.
4. How can IR spectroscopy distinguish an alcohol from a carboxylic acid?
An alcohol shows a broad O–H absorption at 3230–3550 cm⁻¹, whereas a carboxylic acid has a very broad O–H absorption around 2500–3000 cm⁻¹ as well as a strong C=O absorption around 1680–1750 cm⁻¹. Looking at both the presence and absence of these peaks helps distinguish the two functional groups.
5. What are common mistakes students make when interpreting IR spectra?
Common mistakes include identifying every broad O–H peak as an alcohol, ignoring the absence of peaks, giving a single wavenumber instead of the AQA range, confusing peak shapes, and trying to assign individual peaks in the fingerprint region. In exam answers, students should consider peak position, intensity, shape, and missing absorptions together.