You are viewing a free preview of this lesson.
Subscribe to unlock all 10 lessons in this course and every other course on LearningBro.
Now that you understand how IR spectroscopy works, the real skill lies in looking at a spectrum and deducing which functional groups are present. In Edexcel A-Level Chemistry, you are expected to identify functional groups from their characteristic IR absorptions and to distinguish between different types of compounds using IR data.
The following table contains the absorptions you must be able to identify. These values should be committed to memory — they are tested repeatedly in exams.
| Bond | Wavenumber range (cm⁻¹) | Appearance | Found in |
|---|---|---|---|
| O–H (alcohol) | 3200–3600 | Broad | Alcohols, phenols |
| O–H (carboxylic acid) | 2500–3300 | Very broad, often overlaps C–H | Carboxylic acids |
| N–H | 3300–3500 | Medium, sometimes double peak | Amines, amides |
| C–H | 2850–3100 | Medium to strong | Most organic compounds |
| C≡N | 2200–2260 | Medium, sharp | Nitriles |
| C≡C | 2100–2260 | Weak to medium | Alkynes |
| C=O | 1680–1750 | Strong, sharp | Aldehydes, ketones, carboxylic acids, esters, amides |
| C=C | 1620–1680 | Medium | Alkenes, aromatics |
| C–O | 1000–1300 | Strong | Alcohols, esters, ethers, carboxylic acids |
The real power of IR comes from combining multiple absorptions to narrow down the identity of a compound.
Alcohols show a broad O–H stretch at 3200–3600 cm⁻¹ and a C–O stretch at 1000–1300 cm⁻¹. The broadness of the O–H peak is due to hydrogen bonding between alcohol molecules, which causes a range of slightly different O–H bond strengths and therefore a broad absorption.
A sharp O–H peak (instead of broad) suggests a dilute solution or a gas-phase spectrum where hydrogen bonding is absent.
Carboxylic acids show a very broad O–H absorption at 2500–3300 cm⁻¹ — so broad that it often overlaps with and obscures the C–H stretches. This extreme breadth is caused by very strong hydrogen bonding in carboxylic acid dimers.
In addition, carboxylic acids show a strong C=O absorption at 1700–1725 cm⁻¹ and a C–O stretch at 1000–1300 cm⁻¹.
The combination of very broad O–H and strong C=O is diagnostic for carboxylic acids — no other common functional group produces this pattern.
Both show a strong C=O absorption at 1680–1750 cm⁻¹, but neither shows an O–H absorption (unless they also contain an OH group).
Esters show a strong C=O absorption at 1730–1750 cm⁻¹ (typically slightly higher wavenumber than ketones or acids) and a strong C–O stretch at 1000–1300 cm⁻¹. Crucially, they do not show an O–H absorption. The combination of C=O and C–O without O–H points to an ester.
Primary amines (RNH₂) show two N–H stretch peaks at 3300–3500 cm⁻¹ (one for symmetric and one for asymmetric stretching). Secondary amines (R₂NH) show one N–H peak in the same region. Tertiary amines (R₃N) show no N–H peak.
The N–H peaks are narrower than O–H peaks and typically appear as moderate-intensity absorptions.
Amides (RCONH₂ or RCONHR') show both the C=O stretch at 1630–1680 cm⁻¹ (lower than other C=O groups due to resonance with nitrogen) and N–H stretch at 3300–3500 cm⁻¹. Primary amides show two N–H peaks; secondary amides show one.
When presented with an IR spectrum, work through the following steps:
flowchart TD
A["Start: Examine IR Spectrum"] --> B{"Check 2500-3600 cm⁻¹\nBroad absorption?"}
B -->|"Very broad 2500-3300"| C["Carboxylic acid\n(confirm with C=O ~1710)"]
B -->|"Broad 3200-3600"| D["Alcohol or phenol\n(confirm with C-O)"]
B -->|"Medium peaks 3300-3500"| E{"One or two peaks?"}
B -->|"No broad absorption"| F{"Check 1600-1800 cm⁻¹\nStrong sharp peak?"}
E -->|"Two peaks"| G["Primary amine or\nprimary amide"]
E -->|"One peak"| H["Secondary amine or\nsecondary amide"]
F -->|"Yes, C=O present"| I{"C=O wavenumber?"}
F -->|"No C=O"| J["Check 2100-2260 cm⁻¹\nfor C≡N or C≡C"]
I -->|"1730-1750"| K["Ester\n(confirm: C-O, no O-H)"]
I -->|"1720-1740"| L["Aldehyde\n(confirm: 2720/2820 C-H)"]
I -->|"1705-1725"| M["Ketone or acid\n(check for O-H)"]
I -->|"1630-1680"| N["Amide\n(confirm: N-H present)"]
Step 1: Look at the 2500–3600 cm⁻¹ region. Is there a broad absorption? If very broad (2500–3300), think carboxylic acid. If broad (3200–3600), think alcohol. If medium/sharp peaks at 3300–3500, think amine or amide.
Step 2: Look at the 1600–1800 cm⁻¹ region. Is there a strong, sharp absorption? If so, a C=O is present. Note the exact position:
Step 3: Look at the 2100–2260 cm⁻¹ region for C≡N (nitrile) or C≡C (alkyne).
Step 4: Look at the 1000–1300 cm⁻¹ region for C–O stretches, which are present in alcohols, esters, ethers, and carboxylic acids.
Step 5: Combine your observations to determine the functional group.
An IR spectrum shows:
Analysis:
Three compounds, all with molecular formula C₃H₆O₂, have the following IR spectra:
Compound X: Very broad absorption 2500–3300 cm⁻¹, strong C=O at 1712 cm⁻¹, C–O at 1200 cm⁻¹.
Compound Y: No O–H, strong C=O at 1745 cm⁻¹, strong C–O at 1240 cm⁻¹.
Compound Z: Broad O–H at 3400 cm⁻¹, no C=O, strong C–O at 1050 cm⁻¹.
Analysis:
| If you see... | It is probably... | Confirm by checking... |
|---|---|---|
| Very broad O–H (2500–3300) + C=O | Carboxylic acid | C–O at 1000–1300 |
| Broad O–H (3200–3600) + no C=O | Alcohol | C–O at 1000–1300 |
| C=O (1730–1750) + C–O + no O–H | Ester | No N–H either |
| C=O (1720–1740) + 2720/2820 peaks | Aldehyde | No broad O–H |
| C=O (1705–1725) + no O–H + no 2720 | Ketone | No C–O absorption |
| C=O (1630–1680) + N–H (3300–3500) | Amide | Lower C=O than other carbonyls |
| Two N–H peaks + no C=O | Primary amine | Peaks narrower than O–H |
| Sharp peak 2200–2260 | Nitrile (C≡N) | No broad absorptions |
Edexcel specifically requires you to understand that greenhouse gases absorb IR radiation because their bonds vibrate at frequencies that match the IR emitted by the Earth. Increasing concentrations of CO₂, CH₄, and other greenhouse gases means more IR is absorbed and re-emitted, trapping thermal energy in the atmosphere.
CFCs (chlorofluorocarbons) are particularly effective greenhouse gases because they absorb IR strongly in regions of the spectrum where neither CO₂ nor H₂O absorb — known as atmospheric windows. A single CFC molecule can have thousands of times the warming effect of a single CO₂ molecule.
Mistake 1: Confusing a broad O–H with an N–H. N–H absorptions are narrower and often appear as a pair of medium-intensity peaks (for primary amines), while O–H absorptions are characteristically broad.
Mistake 2: Forgetting that conjugation lowers C=O frequency. A C=O next to a C=C or aromatic ring absorbs at a lower wavenumber than an isolated C=O. An amide C=O at 1650 cm⁻¹ is lower than a ketone C=O at 1715 cm⁻¹ because nitrogen donates electron density into the C=O through resonance.
Mistake 3: Concluding "no functional group" because there is no obvious absorption. Every organic compound has C–H absorptions. The absence of O–H, N–H, and C=O does not mean the compound has no structure — it means it is likely an alkane, alkene, halogenoalkane, or ether.
Mistake 4: Not considering that impurities (especially water) can give misleading O–H absorptions. A broad peak around 3400 cm⁻¹ could be water contamination rather than an alcohol group. Always consider sample purity.
IR spectroscopy is an essential tool for identifying functional groups. The key is to learn the characteristic absorption ranges and to use a systematic approach: check for O–H/N–H first, then look for C=O, then check for C≡N/C≡C, and finally look for C–O. Combining these observations allows you to determine which functional group is present, and when combined with other data (mass spectrometry, NMR), you can identify the complete structure.
Edexcel 9CH0 specification, Topic 18 — Modern Analytical Techniques I, sub-strand 18.2 requires candidates to interpret IR spectra of organic compounds: distinguishing alcohols from carbonyl compounds from carboxylic acids; distinguishing aldehydes from ketones (with chemical test support); identifying esters; and using the presence and absence of key absorptions to differentiate between candidate structures (refer to the official specification document for exact wording). The data booklet provides the absorption ranges; the candidate must apply them. Examined on Paper 2 (Core Organic and Physical Chemistry) and routinely on Paper 3 (General and Practical Principles) in synoptic combined-data questions where IR is presented alongside MS, ¹H NMR, ¹³C NMR, and chemical-test results.
Question (8 marks):
Three colourless liquids — A, B, and C — have molecular formula C₃H₆O₂, C₃H₈O₃, and C₃H₆O respectively. Their IR spectra show:
(a) Identify the functional groups in each compound. (3)
(b) Suggest a structural formula for each, justifying your choice. (4)
(c) Predict whether C would give a positive Tollens' test, and use IR evidence to justify. (1)
Solution with mark scheme:
(a) A has broad 2500–3300 cm⁻¹ + sharp 1710 cm⁻¹ → carboxylic acid (broad O–H of acid + C=O). M1. B has very broad 3200–3550 cm⁻¹ but no C=O → polyol (multiple O–H, one or more alcohols). M1. C has C=O at 1715 cm⁻¹ but no broad O–H above 3000 → aldehyde or ketone. M1.
(b) A is propanoic acid CH₃CH₂COOH (Mr = 74, fits C₃H₆O₂). M1. B is propane-1,2,3-triol (glycerol) HOCH₂CH(OH)CH₂OH (Mr = 92, fits C₃H₈O₃) — three O–H groups give the very broad band, no C=O present. M1. C is propanal CH₃CH₂CHO (Mr = 58, fits C₃H₆O) or propanone (CH₃)₂CO (also Mr = 58, also C₃H₆O). M1. M1 for noting that C cannot be definitively assigned from IR alone (aldehyde vs ketone overlap in C=O range); chemical test or NMR needed. Or candidates may cite a weak C–H aldehyde absorption near 2820 cm⁻¹ (the aldehyde "Bohlmann band"); not all data booklets list this.
(c) Tollens' test is positive only for aldehydes. From IR alone, propanal (C = aldehyde) gives positive Tollens'; propanone (C = ketone) does not. A1 for noting that IR cannot definitively distinguish aldehyde from ketone, so chemical test would be required.
Total: 8 marks (M7 A1).
Question (6 marks): An unknown organic compound X is known to be either ethanol (C₂H₆O) or ethanoic acid (C₂H₄O₂). Its IR spectrum shows a broad absorption from 2500–3300 cm⁻¹ and a strong sharp peak at 1710 cm⁻¹.
(a) State, with reasoning from each peak, which compound X must be. (4)
(b) Predict the IR features that ethanol would show instead, and how they differ. (2)
Mark scheme decomposition by AO:
Subscribe to continue reading
Get full access to this lesson and all 10 lessons in this course.