Distinguishing Carbonyl Compounds by IR

Introduction

A strong band near 1700 cm⁻¹ tells you there is a carbonyl group, but not which kind. Ketones, aldehydes, carboxylic acids, esters, amides and acyl chlorides all contain C=O. Fortunately, the atom attached to the carbonyl carbon nudges the band up or down in a predictable way, and each class has its own supporting bands elsewhere in the spectrum. Together, these clues usually pin down the class with confidence.

Core explanation

Typical C=O positions for simple saturated compounds:

Class General formula C=O / cm⁻¹ Supporting bands --- --- --- --- Acyl chloride RCOCl about 1800 none above 3000 except C–H Acid anhydride (RCO)₂O about 1820 and 1760 (two bands) C–O 1000–1300 Ester RCOOR′ 1735–1750 strong C–O 1000–1300 Aldehyde RCHO 1720–1740 C–H about 2720 and 2820 Ketone RCOR′ about 1715 no O–H, no aldehyde C–H Carboxylic acid RCOOH 1700–1725 very broad O–H 2500–3300 Amide RCONH₂ 1630–1690 N–H 3100–3500

Explaining the order. Two competing effects operate on the atom attached to the carbonyl carbon:

- Inductive withdrawal by an electronegative atom pulls electron density away from carbon through the σ bond. This reduces the contribution of the charge-separated C⁺–O⁻ form and strengthens the C=O bond, raising its wavenumber. - Resonance donation by a lone pair on the attached atom delocalises into the C=O π system, giving the bond more single-bond character and lowering its wavenumber.

In an acyl chloride, chlorine is strongly electron-withdrawing and its lone pairs overlap poorly with carbon's 2p orbital (a 3p–2p mismatch), so the inductive effect wins and C=O absorbs very high, near 1800 cm⁻¹. In an ester, the alkoxy oxygen withdraws inductively a little more than it donates, so the C=O is somewhat higher than in a ketone. In an amide, nitrogen is less electronegative and a much better lone-pair donor, so resonance wins decisively and the band falls to 1630–1690 cm⁻¹. Carboxylic acids sit slightly below ketones because hydrogen bonding in their dimers weakens C=O.

Supporting bands are essential. The C=O ranges overlap, so a single wavenumber is rarely conclusive. An acid is confirmed by its very broad O–H band; an amide by N–H bands; an ester by strong C–O bands near 1000–1300 cm⁻¹ with no O–H; an aldehyde by the weak C–H pair near 2720 and 2820 cm⁻¹. A ketone is identified largely by elimination: a C=O band near 1715 cm⁻¹ with none of these extra features.

Step-by-step reasoning

To classify a carbonyl compound:

1. Confirm a strong C=O band and note its wavenumber. 2. Check 2500–3300 cm⁻¹ for a very broad O–H: acid. 3. Check 3100–3500 cm⁻¹ for N–H bands with low C=O: amide. 4. Check 2720 and 2820 cm⁻¹ for aldehyde C–H. 5. Check 1000–1300 cm⁻¹ for strong C–O with C=O near 1740: ester. 6. If none apply and C=O is near 1715, suspect a ketone; near 1800, an acyl chloride.

Visual explanation

Imagine a ladder of C=O wavenumbers, with the acyl chloride on the top rung near 1800 cm⁻¹, then esters, aldehydes, ketones and acids clustered in the middle between 1700 and 1750 cm⁻¹, and amides on the bottom rung near 1650 cm⁻¹.

Real-world analogy

Identifying a carbonyl class is like identifying a car from its engine note alone: many models sound similar. But add the colour, the badge and the number of doors, and the identity becomes clear. The supporting bands are the extra details that confirm what the C=O band only suggests.

Real-world example

Aspirin (2-ethanoyloxybenzoic acid) contains both an ester and a carboxylic acid. Its spectrum shows two C=O bands, one near 1750 cm⁻¹ (ester) and one near 1690 cm⁻¹ (acid, conjugated with the ring), plus a very broad O–H band. As aspirin hydrolyses on storage to salicylic acid, the ester band weakens, so IR can reveal degraded tablets.

Why?

Why does an amide absorb so much lower than an acyl chloride when both have an electronegative atom next to C=O? Nitrogen's lone pair sits in a 2p orbital that overlaps well with carbon, so it donates strongly into the C=O π system. Chlorine's 3p lone pairs overlap poorly and chlorine is more electronegative, so withdrawal dominates.

Common misconception

"Esters show an O–H band because they contain two oxygen atoms." Esters have no O–H bond. Their second oxygen appears as strong C–O stretches in the fingerprint region, and the absence of a broad O–H band helps distinguish them from acids.

Worked example

Question: An isomer of C₄H₈O₂ shows a strong band at 1740 cm⁻¹ and strong bands at 1240 and 1050 cm⁻¹, with nothing broad above 3000 cm⁻¹. Another isomer shows a band at 1710 cm⁻¹ and a very broad band from 2500 to 3300 cm⁻¹. Classify each.

Reasoning: The first has a C=O at the ester position and strong C–O bands, with no O–H. The second has C=O near 1710 cm⁻¹ and the characteristic acid O–H.

Answer: The first is an ester, such as ethyl ethanoate; the second is a carboxylic acid, butanoic acid or 2-methylpropanoic acid.

Quick check

1. Which supporting band distinguishes an aldehyde from a ketone in an IR spectrum? Answer: The weak aldehyde C–H stretches near 2720 and 2820 cm⁻¹, absent in ketones.

Exam focus

Examiners often give isomers, such as an acid and an ester with the same formula, and ask how IR distinguishes them. Always name at least two pieces of evidence: the O–H band for the acid, and the C–O bands and slightly higher C=O for the ester.

Advanced insight

Acid anhydrides show two C=O bands because the two carbonyl groups share an oxygen and their stretches couple into symmetric and asymmetric modes, near 1820 and 1760 cm⁻¹. This is the carbonyl analogue of the two N–H bands of a primary amine, and it makes anhydrides easy to recognise.

Summary

Carbonyl classes absorb at characteristic C=O positions: acyl chlorides near 1800, esters 1735–1750, aldehydes 1720–1740, ketones about 1715, acids 1700–1725 and amides 1630–1690 cm⁻¹. Inductive withdrawal raises the wavenumber and lone-pair resonance lowers it. Because ranges overlap, supporting bands such as O–H, N–H, C–O and aldehyde C–H are needed for a firm identification.

Practice questions

1. Arrange an amide, a ketone and an acyl chloride in order of increasing C=O wavenumber. Answer: Amide (1630–1690) < ketone (about 1715) < acyl chloride (about 1800 cm⁻¹). 2. How could IR distinguish propanal from propanone? Answer: Propanal shows aldehyde C–H bands near 2720 and 2820 cm⁻¹ and a C=O band near 1730 cm⁻¹; propanone lacks the aldehyde C–H bands and absorbs near 1715 cm⁻¹. 3. Explain why the C=O of an amide absorbs at a lower wavenumber than that of a ketone. Answer: The nitrogen lone pair is delocalised into the C=O π system, giving the carbonyl bond partial single-bond character and a lower force constant. 4. A compound shows C=O at 1745 cm⁻¹, strong bands at 1200 cm⁻¹ and no absorption above 3100 cm⁻¹. Suggest its class. Answer: An ester, indicated by the C=O position, strong C–O bands and the absence of O–H or N–H.