The Carbonyl Stretch

Introduction

If you learn only one IR band, learn this one. The carbonyl stretch is usually the strongest band in the spectrum, it is sharp, and it sits in a region where few other bonds absorb, roughly 1650–1800 cm⁻¹. Its presence or absence immediately divides organic compounds into two groups, and its exact position gives clues about what is attached to the carbonyl carbon.

Core explanation

Why the band is strong. Infrared intensity depends on the change in dipole moment during a vibration. The C=O bond is highly polar, with a large δ+ on carbon and δ− on oxygen. When it stretches, the dipole moment changes greatly, so the bond absorbs infrared radiation very strongly. The resulting band often reaches close to zero transmittance.

Why the band is where it is. A double bond is stiffer than a single bond, so C=O stretches at a much higher wavenumber than C–O (1000–1300 cm⁻¹). But carbon and oxygen are much heavier than hydrogen, so C=O absorbs below the X–H stretches. A simple saturated ketone such as propanone absorbs at about 1715 cm⁻¹, and this is the usual reference value.

Typical range. Carbonyl bands appear between about 1630 and 1820 cm⁻¹ depending on the compound. Few other bonds absorb strongly here; C=C bands in the same region are usually much weaker.

Factors that shift the band. The C=O wavenumber follows the strength of the C=O bond, which is sensitive to its surroundings:

- Conjugation lowers it. When C=O is next to a C=C bond or benzene ring, π electrons are delocalised over both groups. The C=O gains some single-bond character and weakens, so the band drops by about 20–30 cm⁻¹. Phenylethanone (acetophenone) absorbs near 1685 cm⁻¹ compared with 1715 cm⁻¹ for propanone. - Ring strain raises it. Cyclohexanone absorbs near 1715 cm⁻¹, cyclopentanone near 1745 cm⁻¹ and cyclobutanone near 1780 cm⁻¹. Squeezing the ring angle alters the hybridisation at the carbonyl carbon and stiffens the C=O bond. - Electronegative substituents raise it. An electronegative atom such as Cl on the carbonyl carbon withdraws electrons inductively, strengthening C=O; acyl chlorides absorb near 1800 cm⁻¹. - Lone-pair donation lowers it. A nitrogen atom attached to the carbonyl carbon donates its lone pair by resonance, weakening C=O; amides absorb near 1650–1690 cm⁻¹. - Hydrogen bonding to the carbonyl oxygen lowers it slightly, as in carboxylic acid dimers.

These effects mean that the carbonyl band can reveal the type of carbonyl compound, which is developed further when comparing carbonyl classes.

Step-by-step reasoning

To use a carbonyl band:

1. Look for a strong, sharp band between 1630 and 1820 cm⁻¹. 2. If absent, the compound has no C=O group. 3. If present, compare its position with 1715 cm⁻¹ for a simple ketone. 4. A lower value suggests conjugation, an amide or hydrogen bonding. 5. A higher value suggests an ester, acyl chloride, anhydride or strained ring.

Visual explanation

Picture an IR spectrum as a landscape. Most bands are hills and valleys of modest depth, but in the middle, near 1700 cm⁻¹, a single narrow canyon plunges almost to the bottom of the chart. That canyon is the carbonyl stretch, visible even from across the room.

Real-world analogy

A bell rung hard gives a loud sound; a bell with a crack gives a lower, softer note. A carbonyl group that shares its electrons through conjugation is like the cracked bell: its bond is a little weaker, so it rings at a lower frequency.

Real-world example

The ageing of polymers such as polyethene in sunlight involves oxidation that introduces C=O groups into the chains. Engineers monitor the "carbonyl index", the size of the C=O band near 1715 cm⁻¹ compared with a reference band, to judge how much a plastic pipe or film has degraded.

Why?

Why does conjugation lower the C=O wavenumber? Delocalisation spreads the π electrons over the C=C–C=O system. A resonance form with a C–O single bond contributes to the structure, so the real carbonyl bond order is slightly below two. A weaker bond has a smaller force constant and vibrates at a lower wavenumber.

Common misconception

"Any strong band near 1700 cm⁻¹ means a ketone." Aldehydes, carboxylic acids, esters, amides, acyl chlorides and anhydrides all have C=O bands in this region. The exact position and supporting bands are needed to decide which.

Worked example

Question: Predict which absorbs at the higher wavenumber: butanone or but-3-en-2-one (CH₂=CH–CO–CH₃). Explain.

Reasoning: In but-3-en-2-one the C=O is conjugated with a C=C bond, which weakens the carbonyl bond. Butanone has an isolated C=O.

Answer: Butanone absorbs higher, near 1715 cm⁻¹, while the conjugated enone absorbs lower, around 1680–1690 cm⁻¹.

Quick check

1. Why is the C=O stretching band usually the most intense band in an IR spectrum? Answer: C=O is highly polar, so its dipole moment changes greatly when it stretches, giving strong absorption.

Exam focus

Quote the carbonyl range given in the data table (typically 1680–1750 cm⁻¹) and describe the band as strong and sharp. For higher marks, explain shifts using bond strength: conjugation weakens C=O and lowers the wavenumber; ring strain and electronegative substituents raise it.

Advanced insight

In metal carbonyl complexes such as Ni(CO)₄, the C=O stretch falls to about 2060 cm⁻¹ from 2143 cm⁻¹ in free carbon monoxide. The metal donates electrons into the π antibonding orbitals of CO, weakening the C≡O bond. Chemists use this shift to measure how electron-rich a metal centre is.

Summary

The C=O stretch is a strong, sharp band between about 1630 and 1820 cm⁻¹, with simple ketones near 1715 cm⁻¹. Its intensity comes from the high polarity of the bond. Its position reflects bond strength: conjugation, nitrogen lone-pair donation and hydrogen bonding lower it, while ring strain and electronegative substituents raise it.

Practice questions

1. State the approximate C=O stretching wavenumber of propanone. Answer: About 1715 cm⁻¹. 2. Explain why cyclobutanone absorbs at a higher wavenumber than cyclohexanone. Answer: Ring strain in the four-membered ring alters the bonding at the carbonyl carbon and stiffens the C=O bond, raising its wavenumber to about 1780 cm⁻¹. 3. Benzaldehyde absorbs near 1700 cm⁻¹, while ethanal absorbs near 1730 cm⁻¹. Explain the difference. Answer: In benzaldehyde the C=O is conjugated with the benzene ring, which delocalises the π electrons and weakens the C=O bond, lowering the wavenumber. 4. Why can a C=C band near 1650 cm⁻¹ usually be distinguished from a C=O band? Answer: The C=C band is usually weak because C=C is nearly non-polar, whereas the C=O band is strong.