Infrared (IR) Spectroscopy
Every bond in a molecule jiggles at its own frequency. Shine infrared light through, and the bonds "steal" their own frequencies, leaving a spectrum of dips that reveals which functional groups are present. You read the key bands straight off the NESA data sheet.
An IR spectrum shows a broad band at about 3350 cm⁻¹, a C–H band near 2950 cm⁻¹, a band near 1050 cm⁻¹, and no band near 1700 cm⁻¹.
- Which functional group does the broad band near 3350 cm⁻¹ suggest?
- What does the absence of a band near 1700 cm⁻¹ rule out?
Know
- That IR radiation causes covalent bonds to vibrate (stretch/bend)
- That functional groups absorb at characteristic wavenumbers (read off the NESA data sheet)
Understand
- Why the x-axis is wavenumber, and how bond strength and atomic mass shift a band
- Why a carboxylic acid O–H is "very broad"
Can Do
- Assign key absorption bands to functional groups using the data-sheet table
- Use IR to distinguish an alcohol from a carboxylic acid from a ketone
Core Content
How IR Works: Bonds Vibrate
+5 XPAbsorb only when frequency matches and the dipole changes
Infrared radiation has the right energy to make covalent bonds vibrate stretching (the bond length changes) and bending (the bond angle changes).
A bond absorbs IR only when the radiation frequency matches the bond's natural vibration frequency and the vibration changes the molecule's dipole moment. Perfectly symmetric vibrations, such as the C=C stretch in symmetrical ethene, produce little or no IR signal; they are "IR-inactive."
IR makes bonds stretch and bend. A bond absorbs only if the IR frequency matches its vibration AND the vibration changes the dipole moment. Perfectly symmetric vibrations (e.g. C=C in ethene) are IR-inactive.
Pause, copy the highlighted absorption condition into your book.
What Sets the Wavenumber
+5 XPStronger and lighter bonds vibrate faster
We just saw that bonds absorb at their own vibration frequency. That raises a question: what makes one bond absorb at a higher wavenumber than another? This card answers it → bond strength and atomic mass.
- Stronger / shorter bonds vibrate faster → higher wavenumber. A C=O bond (~1700 cm⁻¹) absorbs at a higher wavenumber than a C–O single bond (~1000–1300 cm⁻¹).
- Lighter atoms vibrate faster → higher wavenumber. A C–H bond (~2850–3300 cm⁻¹) is higher than a C–Cl bond, because hydrogen is much lighter than chlorine.
- Band intensity depends on bond polarity: very polar bonds (C=O, O–H) give strong, deep dips; weakly polar or symmetric bonds give weak or no bands.
Higher wavenumber = stronger/shorter bond OR lighter atoms (C=O > C–O; C–H > C–Cl). Band intensity rises with bond polarity (C=O and O–H are strong; symmetric bonds are weak/absent).
Pause, copy the highlighted wavenumber rules into your book.
The Characteristic Absorption Table
+5 XPThe ranges you read straight off the NESA data sheet
We just saw what sets a band's position. That raises a question: which exact ranges should you use in the exam? This card answers it → these are the IR absorption ranges printed on the NESA Chemistry data sheet.
You do not memorise these, they are printed on the data sheet and you read bands off them. The HSC data-sheet ranges are:
| Bond | Wavenumber (cm⁻¹) | Note |
|---|---|---|
| N–H (amines) | 3300–3500 | |
| O–H (alcohols) | 3230–3550 | broad |
| C–H | 2850–3300 | |
| O–H (acids) | 2500–3000 | very broad |
| C≡N | 2220–2260 | |
| C=O | 1680–1750 | strong |
| C=C | 1620–1680 | |
| C–O | 1000–1300 | |
| C–C | 750–1100 |
NESA IR ranges (cm⁻¹): N–H 3300–3500; O–H alcohols 3230–3550 (broad); C–H 2850–3300; O–H acids 2500–3000 (very broad); C≡N 2220–2260; C=O 1680–1750; C=C 1620–1680; C–O 1000–1300; C–C 750–1100.
Pause, copy the highlighted data-sheet ranges into your book.
Telling Groups Apart
+5 XPCombinations of bands, not single bands
We just saw the data-sheet ranges. That raises a question: how do you separate groups that share a band, like alcohols and acids that both have O–H? This card answers it → look at the combination and the shape.
- Alcohol vs carboxylic acid: both have O–H, but an acid's O–H is very broad, 2500–3000 cm⁻¹ (often overlapping the C–H region) and the acid also shows a strong C=O ~1680–1750. An alcohol shows a broad O–H higher up (3230–3550) and no C=O.
- Ketone/aldehyde vs ester vs acid: all show C=O 1680–1750. Esters and acids additionally show C–O (1000–1300); the acid adds the very-broad O–H. Ketones and aldehydes are hard to tell apart by IR alone (need NMR/MS).
- Why the acid O–H is broader: extensive hydrogen bonding (acids dimerise, forming two H-bonds) stretches and weakens the O–H over a wider range of bond strengths → a broader band at lower wavenumber.
Alcohol = broad O–H 3230–3550, no C=O. Carboxylic acid = very broad O–H 2500–3000 PLUS strong C=O 1680–1750. Acid O–H is broader because dimer hydrogen bonding weakens the O–H over a wider range.
Pause, copy the highlighted diagnostic patterns into your book.
Reading a Real Spectrum
+5 XPA routine that works every time
We just saw how band combinations separate groups. That raises a question: what order should you scan a spectrum in? This card answers it → a fixed routine through the diagnostic regions.
- Ethanol: broad O–H ~3300 + sharp C–H ~2950 + C–O ~1050. No C=O → an alcohol.
- Ethanoic acid: very broad O–H 2500–3000 + strong C=O ~1700 + C–O. The very-broad O–H plus C=O = a carboxylic acid.
- Propanone (a ketone): strong C=O ~1700, C–H ~3000, and no O–H and no C–O → a ketone.
Scanning routine: (1) O–H/N–H region 3230–3550; (2) the acid's very-broad O–H 2500–3000; (3) C–H ~3000; (4) C=O ~1700; (5) C–O 1000–1300, then assign.
Ethanoic acid: a very broad O–H dip across 2500–3000 plus a strong C=O dip near 1700 (and a C–O dip near 1100), the signature of a carboxylic acid.
Scan: O–H/N–H (3230–3550) → very-broad acid O–H (2500–3000) → C–H (~3000) → C=O (~1700) → C–O (1000–1300). Ethanol = broad O–H + C–O, no C=O. Ethanoic acid = very broad O–H + C=O. Propanone = C=O only.
Pause, copy the highlighted scanning routine into your book.
The very broad band across 2500–3000 cm⁻¹ is the O–H of a carboxylic acid (data-sheet O–H acids 2500–3000, very broad), and the strong dip near 1710 cm⁻¹ is the C=O carbonyl (data-sheet C=O 1680–1750). The combination of a very-broad O–H with a strong C=O identifies a carboxylic acid an alcohol would show its O–H higher (3230–3550) with no C=O.
Complete the Learn phase to unlock Practice.
Activities
Use the data-sheet ranges to assign each band.
1. A spectrum shows a strong dip at 1715 cm⁻¹ and a dip at 1100 cm⁻¹, with no O–H. Assign each band and suggest the type of compound.
2. Why does a carboxylic acid O–H appear lower and broader than an alcohol O–H?
3. Why can IR struggle to distinguish a ketone from an aldehyde?
Activity 2, Match the Spectrum
Reason from bands to identity.
1. Spectrum P: broad O–H ~3350, C–H ~2950, C–O ~1050, no C=O. Identify the functional group.
2. Spectrum Q: strong C=O ~1715, C–H ~3000, no O–H, no C–O. Identify the functional group.
3. Explain the scanning routine you would use to assign an unknown IR spectrum.
Check Your Understanding
Multiple Choice
1. A strong absorption near 1700 cm⁻¹ in an IR spectrum is characteristic of:
2. Which feature best distinguishes a carboxylic acid from an alcohol in an IR spectrum?
3. Why does a C=O bond absorb at a higher wavenumber than a C–O bond?
4. The symmetric C=C stretch in ethene gives little or no IR absorption because:
5. An IR spectrum shows a broad band ~3350 cm⁻¹, C–H ~2950 cm⁻¹, a band ~1050 cm⁻¹, and no band near 1700 cm⁻¹. The compound is most likely:
Short Answer
1. Two unlabelled IR spectra are obtained, one for ethanol and one for ethanoic acid. Describe the key bands you would use to assign each spectrum, and explain why the O–H band differs between the two. (4 marks)
2. Explain, using the idea of bond vibration, why a C–H bond absorbs at a higher wavenumber than a C–Cl bond. (3 marks)
3. An unknown shows a strong band at 1715 cm⁻¹ but no broad O–H band anywhere. Explain what this tells you and why IR alone may not identify the exact compound. (5 marks)
Show All Answers
Activity 1
1. 1715 = C=O (carbonyl, 1680–1750); 1100 = C–O (1000–1300). With no O–H, this fits an ester (C=O + C–O, no acid O–H).
2. Carboxylic acids hydrogen-bond extensively (dimerise), weakening the O–H over a wide range of bond strengths, which broadens and lowers the band to 2500–3000.
3. Both ketones and aldehydes show only a strong C=O (1680–1750) with no distinguishing band on the data-sheet table, so IR alone cannot separate them, NMR/MS is needed.
Activity 2
1. P is an alcohol: broad O–H 3230–3550 and C–O, with no C=O.
2. Q is a ketone (or aldehyde): a strong C=O with no O–H and no C–O.
3. Scan in order: O–H/N–H (3230–3550) → very-broad acid O–H (2500–3000) → C–H (~3000) → C=O (~1700) → C–O (1000–1300), then assign the groups present.
Multiple Choice
1. B a strong band near 1700 cm⁻¹ is the C=O carbonyl.
2. B a very broad O–H (2500–3000) plus a C=O identifies a carboxylic acid.
3. B the C=O bond is stronger and shorter, so it vibrates at a higher frequency.
4. B a symmetric C=C stretch produces no dipole change, so it is IR-inactive.
5. C broad O–H + C–O with no C=O identifies an alcohol.
Short Answer Model Answers
Q1 (4 marks): Ethanoic acid shows a very broad O–H band at ~2500–3000 cm⁻¹ plus a strong C=O band near 1680–1750 and a C–O band 1000–1300; this O–H/C=O combination identifies the carboxylic acid. Ethanol shows a broad O–H band higher up at ~3230–3550 cm⁻¹, a C–H band near 3000, a C–O band ~1050, and no C=O, identifying the alcohol. The acid's O–H band is broader and at lower wavenumber because carboxylic acids hydrogen-bond more extensively (dimerise), stretching and weakening the O–H across a wider range of strengths.
Q2 (3 marks): Wavenumber increases with the vibration frequency of a bond, which is higher when the atoms are lighter (and the bond stronger). Hydrogen is much lighter than chlorine, so the C–H bond vibrates at a higher frequency than C–Cl and therefore absorbs IR at a higher wavenumber.
Q3 (5 marks): A strong band at 1715 cm⁻¹ is a C=O carbonyl (data-sheet 1680–1750), so the compound contains a carbonyl group. The absence of a broad O–H rules out a carboxylic acid and an alcohol, narrowing it to a ketone, aldehyde or ester. However, IR alone cannot decide between these: ketones and aldehydes both show only the C=O, and distinguishing them (or locating the group) needs the molecular mass from MS and the carbon/hydrogen environments from NMR. IR therefore identifies the functional-group class but not the full structure.
Return to Think First
Return to the spectrum with a broad band ~3350, C–H ~2950, a band ~1050, and no band near 1700.
- Which group does the broad 3350 band identify, and which class is ruled out by the missing 1700 band?
- What is the most likely compound type, and which data-sheet ranges justify it?
Review
When does a bond absorb infrared radiation?
What makes a bond absorb at a higher wavenumber?
Which two bands identify a carboxylic acid?
How do you tell an alcohol from a carboxylic acid by IR?
What does IR not tell you on its own?