Carbonyl Peaks in FTIR Spectra

The carbonyl (C=O) stretch is the single most diagnostic absorption in infrared spectroscopy. Strong, sharp, and predictable, it appears between 1650–1800 cm⁻¹ — and its exact position tells you the compound class. Whether you are distinguishing a ketone from an aldehyde or confirming an ester linkage, the carbonyl region is where you start. Use the full FTIR spectrum table to cross-reference other peaks once you have identified the C=O band.

Quick rule of thumb

Higher C=O frequency = stronger bond = more electron withdrawal. Acyl halides sit highest (~1800), amides sit lowest (~1650).

Carbonyl C=O Stretch Positions by Compound Type

The table below ranks carbonyl-containing compound classes from highest to lowest C=O stretching frequency. Higher frequency means a stronger, stiffer C=O bond.

Compound Type	Typical Range (cm⁻¹)	Example	Notes
Acid anhydrides	1800–1830, 1740–1775	Acetic anhydride	Two C=O bands (symmetric + asymmetric)
Acyl halides	1770–1815	Acetyl chloride	Highest frequency due to electron withdrawal
Esters / Lactones	1735–1750	Ethyl acetate	Conjugation lowers position
Aldehydes	1720–1740	Acetaldehyde	Also shows C–H stretch at 2720–2850 (doublet)
Ketones	1705–1725	Acetone	Most "standard" C=O position
Carboxylic acids	1700–1725	Acetic acid	Very broad O–H at 2500–3300 confirms
Amides (primary)	1630–1680	Acetamide	Lowest due to N lone pair resonance
Amides (secondary)	1630–1680	N-methylacetamide	Also called "Amide I band"
Carboxylate salts	1550–1610	Sodium acetate	Asymmetric COO⁻ stretch

How to Narrow Down the Carbonyl Type

Once you spot a strong absorption in the carbonyl region, use the following decision-tree approach to zero in on the compound class. Each step narrows the possibilities based on the exact wavenumber and supporting peaks visible in the spectrum.

Is the C=O above 1750 cm⁻¹? You are looking at an acid anhydride or an acyl halide. Anhydrides show two C=O bands; acyl halides show one.
C=O near 1735 cm⁻¹? This is the classic esterposition. Look for a strong C–O stretch near 1000–1300 cm⁻¹ in the fingerprint region to confirm.
C=O near 1715–1725 cm⁻¹? Either a ketone or an aldehyde. Check for the telltale aldehyde C–H doublet at 2720–2850 cm⁻¹. If those two weak bands are absent, you have a ketone.
C=O near 1710 cm⁻¹plus a broad O–H? A very broad absorption spanning 2500–3300 cm⁻¹ alongside a carbonyl near 1710 cm⁻¹ is the hallmark of a carboxylic acid. The breadth of the O–H band distinguishes it from alcohols.
C=O below 1700 cm⁻¹? You are in amideterritory. Check for N–H stretches near 3100–3500 cm⁻¹: primary amides show two bands (symmetric and asymmetric N–H), while secondary amides show one.

For a complete walkthrough of the interpretation process, see How to Read FTIR Spectra.

Factors That Shift Carbonyl Position

The ranges above are guidelines for simple, unconjugated compounds in dilute solution. Several structural and environmental factors can shift the C=O frequency up or down from its expected position.

Conjugation

When a carbonyl is conjugated with a C=C double bond or an aromatic ring, electron delocalization weakens the C=O bond. This lowers the stretching frequency by roughly 20–30 cm⁻¹. For example, acetophenone absorbs near 1682 cm⁻¹ rather than the typical ketone value of 1715 cm⁻¹.

Ring Strain

Incorporating a carbonyl into a small ring raises its frequency. The compressed bond angle forces more s-character into the C=O bond, stiffening it. Cyclopentanone absorbs at 1745 cm⁻¹, compared to 1715 cm⁻¹ for cyclohexanone. Four-membered-ring lactones (beta-lactones) can reach 1800 cm⁻¹ or higher.

Hydrogen Bonding

Intermolecular hydrogen bonding to the carbonyl oxygen weakens the C=O bond and lowers the stretching frequency. This is why carboxylic acids in concentrated samples or neat liquids often absorb 10–20 cm⁻¹ lower than the dilute-solution value. The effect is strongest when the carbonyl acts as the hydrogen-bond acceptor.

Electron-Withdrawing Groups

Substituents that pull electron density away from the carbonyl carbon strengthen the C=O bond through increased double-bond character. This raises the stretching frequency. Acyl halides are the most extreme example — the highly electronegative chlorine in acetyl chloride pushes the C=O stretch up to 1800 cm⁻¹. Alpha-halogenation of ketones and esters produces a similar but smaller upward shift.

Confirming Carbonyl Assignments

A carbonyl peak alone tells you a C=O is present, but not which functional group it belongs to. Always look for corroborating peaks to lock down the assignment. The table below lists the key supporting absorptions for each carbonyl type. For peaks in the fingerprint region, pattern-matching against reference spectra becomes essential.

Carbonyl Type	C=O Position	Corroborating Peaks
Acid anhydrides	1800–1830 + 1740–1775	Two C=O bands always present; strong C–O stretch near 1000–1300 cm⁻¹
Acyl halides	1770–1815	No broad O–H or N–H; C–Cl stretch near 550–850 cm⁻¹ (weak)
Esters	1735–1750	Strong C–O stretch at 1000–1300 cm⁻¹; no O–H or N–H bands
Aldehydes	1720–1740	Two weak C–H stretches at 2720 and 2850 cm⁻¹ (Fermi resonance doublet)
Ketones	1705–1725	No aldehyde C–H doublet; no O–H or N–H bands
Carboxylic acids	1700–1725	Very broad O–H stretch from 2500–3300 cm⁻¹; O–H bend near 1420 cm⁻¹
Amides (primary)	1630–1680	Two N–H stretches at 3350 and 3180 cm⁻¹; N–H bend (Amide II) near 1620 cm⁻¹
Amides (secondary)	1630–1680	One N–H stretch near 3300 cm⁻¹; Amide II band near 1550 cm⁻¹
Carboxylate salts	1550–1610	Symmetric COO⁻ stretch near 1400 cm⁻¹; no broad O–H

When in doubt, return to the FTIR spectrum table and search by wavenumber to see every functional group that absorbs in the region you are investigating.

Frequently Asked Questions

What wavenumber range do carbonyl peaks appear in FTIR?

Carbonyl (C=O) stretching absorptions appear between 1550 and 1830 cm⁻¹. The exact position depends on the compound class: anhydrides and acyl halides absorb highest (1770–1830 cm⁻¹), esters near 1735–1750 cm⁻¹, ketones and aldehydes near 1705–1740 cm⁻¹, and amides lowest (1630–1680 cm⁻¹).

How do I distinguish a ketone from an aldehyde in FTIR?

Both ketones and aldehydes absorb near 1715–1725 cm⁻¹, but aldehydes produce a distinctive doublet of weak C–H stretches at 2720 and 2850 cm⁻¹ (the Fermi resonance doublet). If these two small peaks are absent, the compound is a ketone rather than an aldehyde.

Why do amide C=O peaks appear at lower wavenumbers than other carbonyls?

The nitrogen lone pair in amides donates electron density into the C=O bond through resonance, giving the bond partial single-bond character. This weakens the C=O bond and lowers its stretching frequency to 1630–1680 cm⁻¹ — the lowest of any common carbonyl type.