Organic Chemistry · Structure Determination: Mass Spectrometry and Infrared Spectroscopy
Infrared Spectra of Some Common Functional Groups
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In 30 seconds
This topic is the practical payoff of Chapter 12: a catalog of IR absorptions that identify common functional groups. Each group has a Diagnostic band An absorption whose position identifies a functional group Full entry → — a stretch whose position, shape, and intensity announce its presence. Band positions follow from bond strength and atomic masses (Hooke's law), fine-tuned by Conjugation Alternating single/double bonds sharing π electrons Full entry → (lowers C=O/C=C ~20–30 cm⁻¹), Ring strain Angle compression in small rings Full entry → (raises C=O), and Resonance Electron delocalization, e.g., in amides Full entry → (weakens bonds; amide C=O sits lower than ketone C=O).
The five most important checks cover most organic unknowns: O–H (~3300, broad), N–H (~3300–3500, sharper), C–H (2850–3100), C≡N (~2250, medium-sharp), and C=O (~1715, strong). Add the C=C, aromatic, and C–O bands and you can identify the functional-group content of nearly any molecule from one spectrum.
Why this matters
These band positions are the working vocabulary of spectrum-reading chemists. When a synthesis produces a white solid, the first question is "did I make the target?" — usually answered by checking for the target's signature bands (and absence of the starting material's) in a two-minute IR scan. The same patterns run through pharmaceutical QC, polymer identification, food-safety screening, and forensics. On exams, "which compound matches this spectrum" questions directly test this table, so learn the logic rather than memorizing numbers.
The college version
Core Concepts
The master absorption table
| Functional group | Diagnostic band (cm⁻¹) | Shape / intensity | Notes |
|---|---|---|---|
| Alcohol O–H | 3200–3600 | Broad, strong | H-bonding broadens it |
| Carboxylic acid O–H | 2500–3300 | Very broad | Pairs with C=O near 1710 |
| Amine N–H | 3300–3500 | Sharper than O–H; 1 or 2 bands | Two bands = primary amine; one = secondary |
| Alkane C–H | 2850–2960 | Strong | sp³ C–H only |
| Alkene / arene C–H | 3000–3100 | Medium | "Above 3000" = the test |
| Terminal alkyne C–H | ~3300 | Sharp, medium | Distinct from the O–H hump |
| Aldehyde C–H | ~2720 and ~2820 | Weak doublet | The aldehyde signature |
| Nitrile C≡N | 2210–2260 | Medium, sharp | Unmistakable |
| Alkyne C≡C | 2100–2260 | Weak | Often barely visible |
| Ketone C=O | 1705–1725 | Strong | The anchor band |
| Ester C=O | 1735–1750 | Strong | Raised by the attached O |
| Carboxylic acid C=O | 1700–1725 | Strong | Companion: very broad O–H |
| Amide C=O | 1630–1690 | Strong | Lowered by resonance |
| Alkene C=C | 1640–1680 | Weak–medium | Conjugation lowers it |
| Aromatic C=C | 1450–1600 | Medium, 2–3 bands | Out-of-plane C–H at 690–900 |
| C–O (alcohol/ether/ester) | 1000–1300 | Strong | Confirms O-containing groups |
Carbonyl position tells the story
The C=O stretch is strong and isolated; position identifies the class: ketone ~1715, aldehyde ~1725, ester ~1735, carboxylic acid ~1710, amide ~1650, acid chloride ~1800. Two electronic effects explain the spread: Resonance (amide): nitrogen lone-pair donation gives the C–N bond partial double-bond character, weakening the C=O and lowering the stretch to ~1650 cm⁻¹. Inductive withdrawal (ester, acid chloride): an attached electronegative O or Cl stiffens the C=O, raising the stretch. Conjugation (C=O attached to a ring or C=C) lowers the band ~20–30 cm⁻¹; ring strain (small rings) raises it.
O–H and N–H hydrogen bonding
Hydrogen bonding O–H / N–H groups associating via H-bonds Full entry → stretches the O–H bond and spreads its energy, producing the classic broad hump at 3200–3600 cm⁻¹ (alcohols) or 2500–3300 cm⁻¹ (acids). N–H bands are narrower; primary amines show two N–H bands, secondary one, tertiary none — a quick branching test. Shape: O–H broad, N–H sharp.
Aromatic ring absorptions
Arenes show three signatures: C–H stretches at 3030–3100 cm⁻¹, ring C=C bands at 1450–1600 cm⁻¹, and out-of-plane C–H bending at 690–900 cm⁻¹ whose pattern reflects the substitution pattern — a classic exam question.
Elimination reasoning with the table
Interpretation becomes elimination: is the candidate band present, and are its companion bands? A strong C=O at 1735 plus strong C–O bands near 1240 and 1100 cm⁻¹ with no O–H → ester; the same C=O with a very broad 2500–3300 hump → acid; the same C=O with N–H near 3300 and a lower position near 1650 → amide. Each group has a companion-band fingerprint that prevents misassignment.
Common Confusions
| Do not confuse | With | Difference |
|---|---|---|
| Ester C=O (~1735) | Ketone C=O (~1715) | Ester is raised by the adjacent O; confirm with C–O bands at 1000–1300 cm⁻¹ |
| Acid O–H (very broad) | Alcohol O–H (broad) | Acid O–H spans ~2500–3300 cm⁻¹ and pairs with C=O ~1710; alcohol O–H is 3200–3600, no C=O |
| Nitrile C≡N | Alkyne C≡C | C≡N is medium-sharp (~2240); C≡C is weak, often nearly invisible |
| Primary vs secondary amide N–H | Same region | Primary amide: two N–H bands; secondary: one; tertiary: none |
| Conjugation effect | Ring strain effect | Conjugation lowers C=O; ring strain raises it — opposite directions |
| "C=O at 1715 means any carbonyl" | Class-specific position | Position plus companion bands (O–H, N–H, C–O) determine the class |
Eli explains
The same idea, in plain words
Explain it like I’m 10
Each functional group wears a badge at a particular spot on the IR chart — C=O is a big strong mark near 1715, O–H a wide smudge near 3300, C≡N a sharp thin line near 2250. Reading a spectrum tells which badges the molecule wears. Conjugation, H-bonding, and ring strain move the badges slightly.
Worked example
Example 1: Identifying an alcohol — 1-butanol (C₄H₁₀O)
Spectrum: broad strong band near 3330 cm⁻¹; C–H stretches at 2960, 2935, 2875 (all below 3000); strong C–O near 1050 cm⁻¹; nothing in the carbonyl window.
Reasoning by elimination: The broad 3330 cm⁻¹ hump is the O–H stretch of a hydrogen-bonded alcohol (no C=O nearby → not an acid). The strong C–O band near 1050 cm⁻¹ confirms the alcohol C–O stretch (primary alcohols: 1040–1060 cm⁻¹). No C=O rules out ketone, aldehyde, ester, acid; no sharp N–H bands rule out amines. Conclusion: a saturated primary alcohol — consistent with 1-butanol, CH3CH2CH2CH2OH.
Example 2: Identifying a nitrile — butanenitrile (C₄H₇N)
Spectrum: medium-sharp band at 2245 cm⁻¹; C–H stretches at 2960 and 2875 cm⁻¹; no O–H; no carbonyl band.
Reasoning: The 2245 cm⁻¹ band sits in the triple-bond region where almost nothing else absorbs (C≡N 2210–2260). It is medium and sharp — typical of a nitrile, not the weak C≡C band of an alkyne. No O–H and no C=O eliminate oxygen groups; with nitrogen the only heteroatom, a C≡N stretch with no N–H bands is the nitrile signature. Conclusion: butanenitrile, CH3CH2CH2C≡N (SMILES: CCCC#N).
Example 3: Using the carbonyl ladder to tell ketone, ester, and amide apart
Three unknowns each show one strong C=O band, no O–H, and C–H stretches, at (a) 1715, (b) 1738, (c) 1655 cm⁻¹.
- (a) 1715 cm⁻¹ → ketone: reference position, no companions.
- (b) 1738 cm⁻¹ → ester: raised by the attached oxygen; confirm with C–O bands at 1000–1300 cm⁻¹.
- (c) 1655 cm⁻¹ → amide: resonance lowers the C=O; confirm with one N–H band near 3300 cm⁻¹ (secondary amide).
The ~20 cm⁻¹ class differences are decisive when combined with companion bands.
Key takeaways
- The big five: O–H broad ~3300; N–H sharp ~3300–3500 (two bands = primary amine); C–H 2850–3100 (3000 is the sp²/sp³ line); C≡N ~2250; C=O strong ~1715.
- Carbonyl ladder: acid chloride ~1800 > ester ~1735 > aldehyde ~1725 > ketone ~1715 > acid ~1710 > amide ~1650.
- Acid O–H (2500–3300, very broad) + C=O ~1710 = carboxylic acid; no O–H + C=O ~1735 + C–O = ester.
- Conjugation lowers C=O and C=C by ~20–30 cm⁻¹; ring strain raises C=O.
- Nitrile (medium-sharp ~2240) vs alkyne (weak ~2150): the C≡N band is the reliable one.
- Aromatic: C–H 3030–3100, ring C=C 1450–1600, out-of-plane C–H 690–900.
- Treat wavenumbers as ranges — solvent, state, and hydrogen bonding shift bands.
Check yourself
5 review questions from the chapter. Try each one, then open the answer.
A compound shows a very broad band from 2500–3300 cm⁻¹ and a strong C=O at 1710 cm⁻¹. What is it?
Show answer
A carboxylic acid — the exceptionally broad O–H (2500–3300 cm⁻¹) plus C=O near 1710 cm⁻¹.
Why does an amide C=O stretch appear lower (~1650 cm⁻¹) than a ketone's (~1715 cm⁻¹)?
Show answer
Resonance: the nitrogen lone pair donates into the carbonyl, giving the C–N bond partial double-bond character, weakening the C=O, and lowering the stretch to ~1630–1690 cm⁻¹.
How do you distinguish a primary amine from an alcohol in the 3200–3600 cm⁻¹ region?
Show answer
Shape and count: O–H is a broad hump; N–H is sharper, and a primary amine shows two N–H bands while an alcohol shows one.
A sharp, medium band appears at 2245 cm⁻¹ with no other bands above 3000 cm⁻¹. What group is most likely?
Show answer
A nitrile (C≡N, 2210–2260 cm⁻¹) — medium-sharp and isolated; a C≡C stretch would be much weaker.
Which two effects shift the C=O band, and in which directions?
Show answer
Conjugation lowers C=O by ~20–30 cm⁻¹; ring strain raises it.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- Diagnostic band
- An absorption whose position identifies a functional group
- Conjugation
- Alternating single/double bonds sharing π electrons
- Ring strain
- Angle compression in small rings
- Hydrogen bonding
- O–H / N–H groups associating via H-bonds
- Resonance
- Electron delocalization, e.g., in amides
- Fingerprint region
- Bands below ~1400 cm⁻¹
- Out-of-plane C–H bend
- Aromatic C–H bending at 690–900 cm⁻¹
Sources & references
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