Difficulty: Intermediate | Prerequisites: Chapters 1–13 (bonding, functional groups, reaction mechanisms)
Infrared spectroscopy sits at the centre of the structure-determination toolkit you will use for the rest of organic chemistry. This chapter teaches you how molecules absorb IR light, which bonds absorb where, and how to read a spectrum to identify functional groups. You need a solid handle on bond polarity, hybridisation, and common functional groups before diving in. If you are shaky on carbonyl chemistry or the difference between sp, sp2, and sp3 carbon, revisit those chapters first.
IR spectroscopy identifies functional groups by measuring which frequencies of infrared light a molecule absorbs. Stronger, lighter bonds absorb at higher wavenumbers. The carbonyl (C=O) stretch near 1700 cm⁻¹ and the broad O–H stretch near 3300 cm⁻¹ are the two most diagnostic peaks you will encounter on exams.
Infrared (IR) spectroscopy
An analytical technique that passes infrared radiation through a sample and records which frequencies are absorbed. The absorption pattern reveals which functional groups are present. Think of it as a fingerprint scanner for bonds.
Wavenumber (cm⁻¹)
The unit used on the x-axis of an IR spectrum, equal to 1/wavelength in centimetres. Higher wavenumber means higher energy. In simple terms, wavenumber tells you how energetically a bond vibrates.
Functional group region
The portion of the IR spectrum from roughly 1500 to 4000 cm⁻¹ where most diagnostic stretches appear (O–H, N–H, C–H, C=O, C≡C, C≡N). This is where you look first.
Fingerprint region
The portion from roughly 400 to 1500 cm⁻¹. The pattern here is unique to each molecule but difficult to assign to individual bonds. Useful for confirming identity by comparison with a reference spectrum, less useful for identifying functional groups from scratch.
Transmittance (%T)
The y-axis on most IR spectra. 100% means all light passed through; a dip (trough) means absorption occurred. Peaks in IR point downward, which is the opposite of UV-Vis.
Dipole moment change
A bond must undergo a change in dipole moment during vibration to absorb IR radiation. Symmetric stretches of nonpolar bonds (such as the C≡C in a symmetric internal alkyne) give weak or absent absorptions. Think of it as: if the charge distribution does not shift during vibration, there is nothing for the IR photon to interact with.
Absorption intensity
How strong (deep) a peak appears. Bonds with larger dipole moment changes give stronger absorptions. C=O is typically the strongest organic absorption; C≡C is often weak.
What IR spectroscopy tells you (and what it does not)
IR identifies functional groups present in a compound.
It does not tell you molecular weight, the arrangement of C and H atoms, or conjugated π systems. Those come from mass spectrometry, NMR, and UV-Vis respectively.
Type of radiation
IR radiation causes vibrational excitation of bonds. This is distinct from UV/Visible (electronic excitation), microwave (rotational excitation), and X-ray (core electron excitation).
The two main regions of the IR spectrum
Functional group region: 1500–4000 cm⁻¹. Contains the stretches you will be asked to identify.
Fingerprint region: 400–1500 cm⁻¹. Complex pattern, useful for matching against known spectra but rarely assigned to specific bonds on an exam.
What determines wavenumber
Two factors control where a bond absorbs:
Bond strength (force constant): stronger bonds absorb at higher wavenumber. Triple bonds > double bonds > single bonds.
Atom mass: lighter atoms vibrate faster. C–H stretches appear at higher wavenumber than C–C stretches because hydrogen is lighter than carbon.
Ranking bond stretches by wavenumber (highest to lowest)
sp C–H (alkyne, around 3300 cm⁻¹) > sp2 C–H (alkene/aromatic, around 3020–3080 cm⁻¹) > sp3 C–H (alkane, around 2850–2960 cm⁻¹). The more s-character in the hybrid orbital, the stronger the bond and the higher the wavenumber.
What determines absorption intensity
Greater change in dipole moment during the vibration produces a stronger (deeper) absorption.
C=O has the strongest absorption among common organic bonds because the large electronegativity difference between C and O creates a large oscillating dipole.
C≡C is often weak (especially in symmetric internal alkynes) because the dipole change is small.
Functional Group | Wavenumber Range (cm⁻¹) | Appearance | Notes |
|---|---|---|---|
O–H (alcohol) | 3200–3600 | Broad, strong | Broadness from hydrogen bonding |
O–H (carboxylic acid) | 2500–3300 | Very broad, strong | Even broader than alcohol O–H; overlaps C–H region |
N–H (primary amine) | 3300–3500 | Two peaks ("rabbit ears") | Two N–H stretches for NH₂ |
N–H (secondary amine) | 3300–3500 | One peak | Single N–H stretch |
sp C–H (terminal alkyne) | ~3300 | Sharp | Appears alongside C≡C stretch |
sp2 C–H | 3020–3080 | Medium | Alkene and aromatic C–H |
sp3 C–H | 2850–2960 | Strong | Present in nearly all organic compounds |
C≡N (nitrile) | 2100–2300 | Medium, sharp | Appears in the "quiet" triple-bond region |
C≡C (alkyne) | 2100–2300 | Weak to medium | Weak or absent if internal and symmetric |
C=O (carbonyl) | 1600–1850 | Strong, sharp | The most recognisable IR absorption |
C=C (alkene) | ~1650 | Medium | Weaker than C=O |
C–O (ether, alcohol) | 1000–1260 | Strong | In the fingerprint region; hard to use alone |
Carbonyl wavenumber and conjugation
Conjugation with adjacent π systems lowers the carbonyl stretching frequency. Resonance delocalises electron density into the C=O bond, giving it partial single-bond character. Among carbonyls:
Amides absorb lowest (around 1650 cm⁻¹) because nitrogen's lone pair donates strongly into the carbonyl.
Carboxylic acids and esters absorb higher.
Ketones and aldehydes absorb in the 1700–1740 cm⁻¹ range.
Acid chlorides absorb highest (around 1800 cm⁻¹).
Conjugation with a C=C or aromatic ring also lowers the carbonyl stretch by 20–40 cm⁻¹ compared to the unconjugated analogue.
How to read an IR spectrum (a systematic approach)
Check the O–H / N–H region (3200–3600 cm⁻¹). A broad absorption here means alcohol, carboxylic acid, or amine. The breadth and shape narrow the options.
Check for C=O (1600–1850 cm⁻¹). If both a broad O–H and a carbonyl are present, suspect a carboxylic acid. A carbonyl without broad O–H suggests a ketone, aldehyde, or ester.
Check for C≡C or C≡N (2100–2300 cm⁻¹). This region is often empty, so any peak there is distinctive.
Look at the sp C–H region (~3300 cm⁻¹). A sharp peak here alongside a C≡C peak confirms a terminal alkyne.
Distinguish aldehyde from ketone: aldehydes show two small C–H stretches near 2720 and 2820 cm⁻¹ (Fermi resonance doublet). Ketones do not.
Distinguishing functional groups by their IR signatures
Alcohol: broad O–H stretch (3200–3600), no carbonyl.
Ketone: strong C=O near 1715 cm⁻¹, no broad O–H, no aldehyde C–H doublet.
Aldehyde: strong C=O near 1725 cm⁻¹, plus two small peaks near 2720/2820 cm⁻¹.
Carboxylic acid: very broad O–H (2500–3300 cm⁻¹) plus strong C=O near 1710 cm⁻¹.
Primary amine: two N–H stretches (3300–3500 cm⁻¹), no carbonyl.
Secondary amine: one N–H stretch (3300–3500 cm⁻¹), no carbonyl.
Predicting IR changes from a reaction
Exam questions frequently show a reaction and ask what changes in the IR spectrum. The approach:
Identify what bonds are present in the starting material and which are in the product.
Any bond in the starting material but not the product: its absorption should disappear.
Any bond in the product but not the starting material: a new absorption should appear.
Example: a terminal alkyne converted to a ketone (via Markovnikov hydration with HgSO₄/H₃O⁺). The sp C–H stretch (~3300 cm⁻¹) and C≡C stretch (~2150 cm⁻¹) both disappear, and a new C=O stretch (~1720 cm⁻¹) appears.
Example: an alcohol converted to an ether (via Williamson ether synthesis with NaH then alkyl halide). The broad O–H stretch (3200–3600 cm⁻¹) disappears. The C–O stretch remains but shifts.
IR spectroscopy is how chemists confirm whether a reaction worked before running more expensive tests. If you reduce a ketone to an alcohol, the carbonyl peak should vanish and a broad O–H should appear. Quality control labs in pharmaceutical and polymer manufacturing run IR scans routinely to verify that raw materials and products match their expected spectra.
Students often think IR tells you the carbon skeleton or molecular weight. It does not. IR identifies functional groups only. For connectivity, you need NMR; for molecular weight, you need mass spectrometry.
Students confuse wavenumber with wavelength. Higher wavenumber means higher energy and shorter wavelength, not the other way around.
Students assume all C=O bonds absorb at the same wavenumber. They do not. Conjugation, ring strain, and neighbouring atoms shift the carbonyl stretch significantly (amide ~1650, acid chloride ~1800).
Students forget that a symmetric internal alkyne may show no C≡C peak at all, because there is no dipole moment change during the stretch.
⚠️ Know the wavenumber range for every major functional group in the table above. Multiple-choice questions frequently test whether you can match a functional group to its range.
⚠️ Be prepared to look at a spectrum and identify the compound class (alcohol vs. ketone vs. aldehyde vs. carboxylic acid vs. amine). The broad-versus-sharp O–H distinction is especially common.
⚠️ Reaction-to-IR-change questions are a favourite. Know which peaks appear and disappear for hydration of alkynes, Williamson ether synthesis, and oxidation/reduction of alcohols and carbonyls.
⚠️ The carbonyl conjugation trend (amide lowest, acid chloride highest) appears regularly.
True or false: IR spectroscopy can determine the molecular weight of a compound.
False. That is mass spectrometry.
Fill in the blank: Absorption of _______ radiation causes vibrational excitation of bonds.
Infrared (IR).
True or false: A symmetric internal alkyne will always show a strong C≡C stretch.
False. It may show a weak or absent peak because there is little change in dipole moment.
Fill in the blank: The fingerprint region spans from _______ to _______ cm⁻¹.
400 to 1500.
True or false: Conjugation raises the wavenumber of a carbonyl stretch.
False. Conjugation lowers the carbonyl stretching frequency.
Q: What information is primarily obtained from infrared spectroscopy?
A: Functional groups present in a compound.
Q: What information is primarily obtained from NMR spectroscopy?
A: The arrangement of carbon and hydrogen atoms in a compound.
Q: Which wavenumber range corresponds to the C=O double bond region?
A: 1600–1850 cm⁻¹.
Q: Which wavenumber range is the fingerprint region?
A: 400–1500 cm⁻¹.
Q: Among C=N, C≡C, C=O, sp2 C–H, and C–O, which bond has the strongest IR absorption?
A: C=O, because the large electronegativity difference produces the greatest change in dipole moment.
Q: Rank sp C–H, sp2 C–H, and sp3 C–H stretches from highest to lowest wavenumber.
A: sp C–H > sp2 C–H > sp3 C–H. More s-character means a stronger, shorter bond that vibrates at higher frequency.
Q: A terminal alkyne is hydrated with HgSO₄/H₃O⁺ to form a ketone. What changes would you expect in the IR spectrum?
A: The sp C–H stretch (~3300 cm⁻¹) and C≡C stretch (~2150 cm⁻¹) should both disappear, and a new C=O stretch (~1720 cm⁻¹) should appear.
Q: An alcohol is converted to an ether via Williamson ether synthesis (NaH, then alkyl halide). What changes in the IR?
A: The broad O–H absorption at 3200–3600 cm⁻¹ should disappear.
Q: A compound shows a broad absorption near 3300 cm⁻¹ and a sharp absorption at 1650 cm⁻¹. Which functional group combination is consistent with this?
A: A primary amide (N–H stretch near 3300 cm⁻¹ and C=O stretch at 1650 cm⁻¹), or possibly a compound containing both an amine/alcohol and a C=C. In the context of the practice set, the answer is an amide (compound with both NH₂ and C=O).
Q: Which carbonyl-containing compound class absorbs at the lowest wavenumber?
A: Amides, around 1650 cm⁻¹, because nitrogen’s lone pair donates into the C=O, giving it partial single-bond character.
IR spectroscopy connects directly to functional group chemistry (Chapters 1–13). Every reaction you have studied that creates or destroys a functional group produces a predictable change in the IR spectrum. This chapter also sets up the structure-determination strategy you will use in Chapter 15 and beyond, where you combine IR, NMR, and mass spectrometry data to identify unknown compounds.
The carbonyl absorption trends tie back to resonance and induction from Chapter 1–2. Understanding why an amide C=O absorbs lower than a ketone C=O requires the same resonance reasoning you used when comparing acidity and nucleophilicity.
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