IR spectroscopy: reading the functional-group region
An IR spectrum is a plot of transmittance (or absorbance) versus wavenumber, running from about 4000 cm⁻¹ on the left to 400 cm⁻¹ on the right. Each dip is a vibrational absorption — a bond that stretches or bends at exactly the wavenumber of the incoming photon. The trick to reading IR is to split the spectrum into two zones with different jobs: 4000-1500 cm⁻¹ names the functional groups; 1500-400 cm⁻¹ is a compound-specific fingerprint. This guide walks the diagnostic bands, the position/intensity/shape rules that name each one, and the four questions that lead from a spectrum to a functional-group list.
Prerequisites: functional-group vocabulary. If reading an NMR spectrum is comfortable, IR is easier — fewer axes, coarser resolution, faster reading.
Where the numbers come from
A bond stretches like a spring, and Hooke’s law plus the harmonic-oscillator quantisation gives the fundamental vibrational frequency:
where k is the force constant (bond stiffness) and μ is the reduced mass of the two atoms. Two consequences:
- Stiffer bonds vibrate at higher wavenumber. Triple bonds (2200-2100 cm⁻¹) > double bonds (1750-1600) > single bonds (1500 and below).
- Lighter atoms vibrate at higher wavenumber. X-H stretches sit at 3600-2800 cm⁻¹ — highest wavenumber region — because μ is tiny when one atom is hydrogen.
The two effects combine cleanly: a C=O stretch is above a C=C stretch (bigger k) and both are above a C-C stretch. An O-H stretch is above a C-H stretch (smaller reduced mass with H, plus stiffer bond due to O’s electronegativity).
The two zones
Functional-group zone (4000-1500 cm⁻¹). Fewer bands. Each band position, intensity, and shape reliably diagnoses a specific bond type. Read this region first.
Fingerprint zone (1500-400 cm⁻¹). Many overlapping bands corresponding to C-C, C-O, and C-N single-bond stretches plus bending modes. Impossible to assign individually, but the pattern is unique to a specific molecule — the compound’s “fingerprint”. Match to a reference spectrum (NIST WebBook) to confirm identity.
Do not try to identify functional groups by looking in the fingerprint region. Do not try to match the whole spectrum without first extracting functional-group information from the top region.
The diagnostic band table
Committed to memory, this table converts any IR spectrum to a functional-group list in under a minute (Silverstein 8e Ch. 2, Coates 2000):
| Band (cm⁻¹) | Intensity | Shape | Assignment |
|---|---|---|---|
| 3600-3200 | m-s | Broad | O-H stretch (H-bonded alcohol/acid) |
| 3500 | m | Sharp | O-H stretch (free, dilute) |
| 3500-3300 | m | Two bands | N-H stretch (primary amine, 2 bands) |
| 3400-3200 | m | One band | N-H stretch (secondary amine) |
| 3300 | s | Sharp | ≡C-H stretch (terminal alkyne) |
| 3100-3000 | m | =C-H stretch (alkene, aromatic) | |
| 3000-2850 | s | Several | C-H stretch (alkane) |
| 2830, 2720 | m | Two bands | Aldehyde C-H (Fermi resonance) |
| 2260-2210 | m | C≡N stretch (nitrile) | |
| 2260-2100 | w | C≡C stretch (alkyne) | |
| 2260-2100 | s | X=C=Y (isocyanate, azide) | |
| 1815-1770 | s | C=O acid chloride | |
| 1810-1750 | s | C=O anhydride (2 bands) | |
| 1750-1720 | s | C=O ester | |
| 1735-1720 | s | C=O aldehyde | |
| 1720-1680 | s | C=O ketone, acid | |
| 1690-1630 | s | C=O amide | |
| 1680-1620 | m | C=C alkene | |
| 1600, 1500 | m | C=C aromatic (2 bands) | |
| 1560, 1380 | s | NO₂ stretches (nitro) | |
| 1300-1000 | s | C-O stretch (alcohol, ether, ester) |
Reading the intensity codes: s = strong (over 60% absorption); m = medium (30-60%); w = weak (under 30%); v = variable.
The four-question workflow
Every IR spectrum answers four questions in order:
Question 1: Is there a C=O? Look for a strong, sharp band between 1830 and 1630 cm⁻¹. If yes, note the exact position; that position pins down the carbonyl subtype (acyl chloride high, amide low; see table).
Question 2: Is there an O-H or N-H? Look above 3200 cm⁻¹. O-H is broad and centred around 3400 (H-bonded alcohol) or 2500-3300 (H-bonded acid; overlaps with C-H). N-H is medium and sharp; primary amines/amides show two bands, secondary show one.
Question 3: Are there C-H bonds sp²/sp³/sp? Sp³ C-H at 2850-3000 (in every organic molecule with alkyl groups). Sp² C-H at 3000-3100 (alkene, aromatic — a small band just left of the alkane). Sp C-H at 3300 (terminal alkyne — sharp, distinctive).
Question 4: Are there aromatic C=C bands? Look for medium bands at 1600 and 1500 cm⁻¹ (sometimes 1450) and a set of overtones at 2000-1600 (weak). Aromatic rings show these consistently.
Answer these four in order and you have identified the functional-group class before touching the fingerprint region.
Worked example — identifying an unknown from IR
- Formula
- Substitute
- Note
The broad O-H stretch from 3300 to 2500 cm⁻¹ (extending across the C-H region) is the classic carboxylic-acid signature — hydrogen-bonded acid dimers have an O-H stretch this wide. Combined with the C=O at 1710 (right in the acid range), it is unambiguous. The molecular formula C₃H₆O₂ with a COOH accounts for CHO₂; the rest is C₂H₅. Propanoic acid. IR alone got the functional group; molecular formula narrowed to one structure.
- Result
Overtones, combination bands, and Fermi resonance
Real spectra are messier than the table suggests. Three complications:
Overtones are absorptions at approximately 2× (or 3×) a fundamental. Weak, but visible at high concentration. The C=O overtone at ~3400 cm⁻¹ can be confused with an O-H stretch if you’re not careful.
Combination bands are absorptions at the sum of two fundamental frequencies. The 2000-1600 cm⁻¹ region for aromatic rings shows an “aromatic overtone/combination fingerprint” that identifies mono-, di-, and tri-substituted benzenes — a small but useful diagnostic.
Fermi resonance happens when an overtone or combination band lies near an allowed fundamental and mixes with it, giving two bands where one was expected. The aldehyde C-H stretch shows this: instead of one band, aldehydes give two bands at 2830 and 2720 cm⁻¹ (from Fermi resonance between the C-H fundamental and the overtone of the C-H bend). Seeing that specific pair is the strongest aldehyde diagnostic in the IR.
C=O position — a table within a table
The carbonyl stretch is the most reactivity-informative band in IR. Its exact position within the 1830-1630 cm⁻¹ window is set by three effects:
- Electron-withdrawing groups raise the frequency (stiffer C=O bond). Acyl chlorides (1815-1770) > anhydrides (1810 and 1750) > esters (1750-1720) > aldehydes/ketones (1735-1710) > amides (1690-1630).
- Conjugation lowers the frequency (partial delocalisation weakens the C=O). Aryl ketone (1685) is ~30 cm⁻¹ below saturated ketone (1715); α,β-unsaturated (1680) is similar.
- Ring strain raises the frequency. Cyclobutanone (1780) > cyclopentanone (1750) > cyclohexanone (1715). Ring strain compresses the C-C-C angle, stiffening the C=O.
Given a C=O near 1740 cm⁻¹, three candidates fit — an ester, a strained ketone, or an aldehyde. The aldehyde C-H Fermi doublet at 2830/2720 distinguishes the aldehyde from the others.
Common pitfalls
Confusing O-H with N-H. Both appear above 3200 cm⁻¹, but O-H (H-bonded) is broad and rounded while N-H is medium and sharp. Primary N-H shows two bands (symmetric and asymmetric stretch); primary O-H shows one. Acid O-H is uniquely broad and extends into the C-H region.
Ignoring the fingerprint region entirely. The four-question workflow tells you the functional-group class, but two isomers can share the same functional groups. To distinguish 1-pentanol from 2-pentanol, the fingerprint region (specifically C-O bending around 1080 vs 1120) is what makes the call. Compare with a reference spectrum from NIST WebBook or Aldrich to confirm.
Assigning “C-H stretches” as diagnostic. Everything organic has C-H stretches. Their exact position (sp³ below 3000 vs sp² above 3000) is diagnostic, but their presence is not.
Reading transmittance and absorbance backwards. Transmittance spectra: dips going DOWN are absorptions. Absorbance spectra: peaks going UP are absorptions. Most textbooks use transmittance; some journals use absorbance. Check the y-axis label.
Practice
Problem 1 Easy An IR spectrum shows a strong band at 1735 cm⁻¹ but no O-H or N-H stretch above 3100 cm⁻¹. What functional group is most likely present?
Answer: Ester (or possibly aldehyde or ketone). The C=O at 1735 is diagnostic of a carbonyl; the absence of O-H rules out a carboxylic acid; the absence of N-H rules out an amide. Between ester, aldehyde, and ketone: an ester's C=O typically sits at 1735-1740, an aldehyde at 1725-1735 (with Fermi doublet at 2830/2720), a saturated ketone at 1715-1720. Look for the aldehyde Fermi pair to distinguish; absence of it points to ester or ketone. C-O stretch bands at 1200 cm⁻¹ area confirm ester (strong 'ester C-O' band around 1200-1300).
Problem 2 Intermediate A molecule shows a sharp band at 3300 cm⁻¹ (medium intensity), a very weak band at 2110 cm⁻¹, and strong C-H bands below 3000 cm⁻¹. Identify the functional group and explain both diagnostic peaks.
Answer: Terminal alkyne. The 3300 cm⁻¹ band is the ≡C-H stretch (sharp because sp-hybridised C-H has a fixed local geometry and a stiffer bond than sp³ C-H). The 2110 cm⁻¹ band is the C≡C stretch — weak because the change in dipole moment during the vibration is small for a symmetric internal alkyne; even for a terminal alkyne it is only moderately intense. Together they are the alkyne signature. An internal alkyne (RC≡CR') would show only a weak or absent 2200 cm⁻¹ band and no 3300 stretch, making it hard to detect by IR alone.
Problem 3 Hard Two compounds have molecular formula C₄H₈O₂. Compound A shows a strong band at 1745 cm⁻¹ and no OH stretch. Compound B shows a broad OH from 3300 to 2500 cm⁻¹ and a strong C=O at 1710 cm⁻¹. Propose structures and rationalise the IR differences.
- Formula
- Substitute
- Note
Both isomers have one degree of unsaturation, one C=O, but differ in whether the OH is present. Methyl propanoate has C=O at 1740 cm⁻¹ (ester range) and no OH; butanoic acid has the classic acid signature. The wavenumber difference — 1745 vs 1710 — is a real 30-40 cm⁻¹ shift caused by the electron-donation from OR' (ester) vs OH (acid); the OR' donates less strongly, keeping the C=O stiffer. Between ethyl acetate (CH₃COOC₂H₅) and methyl propanoate (CH₃CH₂COOCH₃), IR alone won't distinguish — the fingerprint region or NMR would.
- Result
Where this leads
IR is one leg of the structure-elucidation tripod:
- Multi-technique elucidation — IR (functional groups), NMR (connectivity), and MS (mass + formula) together identify almost any organic compound.
- Quantitative IR — Beer’s law works for a single sharp band; see Beer-Lambert for the failure modes.
- Reaction monitoring — in-situ IR (ReactIR) tracks C=O disappearance during ketone reduction or ester formation in real time.
- Solid-state IR — ATR (attenuated total reflectance) accessories let you measure powders and thin films directly without KBr pellets.
Every one of them starts with the four-question workflow: C=O, O-H/N-H, C-H hybridisation, aromatic C=C.
Sources
- Silverstein, Webster & Kiemle, Spectrometric Identification of Organic Compounds, 8th ed. (2014), Ch. 2
- Pavia, Lampman, Kriz & Vyvyan, Introduction to Spectroscopy, 5th ed. (2015), Ch. 2
- Coates, J. (2000), 'Interpretation of Infrared Spectra' in Encyclopedia of Analytical Chemistry — canonical band table
- Socrates, G., Infrared and Raman Characteristic Group Frequencies, 3rd ed. (2001) — comprehensive reference
- NIST Chemistry WebBook — reference IR spectra for validation