IR Group Frequencies: The Key Table

Introduction

An infrared spectrum can look like a confusing jumble of dips and troughs, yet chemists read it within seconds. The secret is that particular bonds absorb in particular regions, almost regardless of which molecule they sit in. A C=O bond in a ketone and a C=O bond in a steroid both absorb close to 1700 cm⁻¹. These characteristic positions are called group frequencies , and a short table of them is the single most useful tool in IR interpretation.

Core explanation

Why bonds have their own frequencies. Treating a bond as a spring, the wavenumber of its stretching vibration depends on two things: the bond strength (force constant, k) and the reduced mass (μ) of the two atoms. Stronger bonds and lighter atoms vibrate at higher wavenumbers. Because a given bond type has a fairly consistent strength and involves the same atoms, it absorbs within a fairly narrow range wherever it occurs.

The key table. The values below are the standard textbook ranges used at advanced level:

Bond Where found Wavenumber / cm⁻¹ Typical appearance --- --- --- --- O–H alcohols, phenols 3200–3550 strong, broad O–H carboxylic acids 2500–3300 very broad N–H amines, amides 3300–3500 medium, sharper than O–H C–H alkanes, alkenes, arenes 2850–3100 medium to strong C≡N nitriles 2220–2260 medium, sharp C≡C alkynes 2100–2260 weak to medium C=O aldehydes, ketones, acids, esters, amides 1630–1820 strong, sharp C=C alkenes 1620–1680 weak to medium C=C arenes about 1450–1600 several medium bands C–O alcohols, esters, ethers, acids 1000–1300 strong C–Cl chloroalkanes 600–800 strong

Reading the table as a map. The spectrum can be divided into zones. Above 2500 cm⁻¹ lie the stretches of bonds to hydrogen (O–H, N–H, C–H), which are high because hydrogen is so light. Between about 2000 and 2300 cm⁻¹ sit triple bonds, which are very strong. Double bonds (C=O and C=C) fall between about 1600 and 1850 cm⁻¹. Below 1500 cm⁻¹ single bonds between heavier atoms and many bending vibrations crowd together in the fingerprint region.

Intensity matters too. Position tells you which bond; intensity and shape add confirmation. A very polar bond such as C=O changes its dipole moment a great deal as it stretches, so its band is strong. A symmetrical C=C bond changes its dipole little, so its band is often weak. Hydrogen bonding makes O–H bands broad.

Formulae

wavenumber (cm⁻¹) = 1 ÷ wavelength (cm). Higher wavenumber means higher energy: E = hc × wavenumber. For a bond modelled as a spring, wavenumber ∝ √(k ÷ μ).

Step-by-step reasoning

To use the table on an unknown absorption:

1. Read the wavenumber of the band's minimum in transmittance. 2. Decide which zone it is in: bonds to hydrogen, triple bonds, double bonds or the fingerprint region. 3. Find every table entry whose range includes that value. 4. Use shape and intensity to decide between overlapping entries. 5. Look for supporting bands elsewhere before drawing a conclusion.

Visual explanation

Imagine the horizontal axis of an IR spectrum from 4000 cm⁻¹ on the left to 500 cm⁻¹ on the right, divided into coloured strips: X–H stretches at the far left, a narrow triple-bond strip near 2200, a double-bond strip around 1700, and a crowded fingerprint strip on the right.

Real-world analogy

Group frequencies are like the notes produced by strings on a guitar. A thick, slack string always gives a low note and a thin, tight string a high one, whichever guitar they are fitted to. You can recognise the string from the note without seeing the instrument.

Real-world example

Quality-control laboratories in the pharmaceutical industry check incoming raw materials with IR. A technician can confirm in under a minute that a delivered drum of ethyl ethanoate shows a strong C=O band near 1740 cm⁻¹ and strong C–O bands near 1240 cm⁻¹, with no broad O–H band indicating contamination by water or acid.

Why?

Why do bonds to hydrogen absorb at the highest wavenumbers even though they are not the strongest bonds? Because the reduced mass of an X–H pair is close to the mass of hydrogen itself, which is very small. The low reduced mass raises the vibrational frequency more than the modest bond strength lowers it.

Common misconception

"Each bond absorbs at one exact wavenumber." In reality each bond absorbs across a range, because neighbouring atoms, conjugation, ring strain and hydrogen bonding all shift the band slightly. The table gives ranges, not single values.

Worked example

Question: A compound shows a strong, sharp band at 1715 cm⁻¹ and a medium band at 2960 cm⁻¹, but nothing between 3200 and 3550 cm⁻¹. Which bonds are present?

Reasoning: 1715 cm⁻¹ falls in the C=O range and the band is strong, as expected for a carbonyl. 2960 cm⁻¹ is a C–H stretch. The absence of a broad band above 3200 cm⁻¹ rules out an alcohol O–H.

Answer: C=O and C–H are present; there is no alcohol O–H. The compound could be a ketone such as propanone.

Quick check

1. In which range would you expect the stretching band of a C≡N bond in a nitrile to appear? Answer: Between about 2220 and 2260 cm⁻¹, in the triple-bond region.

Exam focus

Exam papers usually supply a data table, but you must quote the range you used and name the bond, not just the functional group. Write, for example, "absorption at 1720 cm⁻¹ shows C=O". Be precise about broad versus sharp and strong versus weak when a question asks you to justify a choice.

Advanced insight

Group frequencies work because many vibrations are localised in one bond. When two bonds of similar frequency are adjacent, their vibrations couple and split into symmetric and asymmetric modes, as in the two C=O bands of acid anhydrides near 1820 and 1760 cm⁻¹. Coupling is why the table must be applied with care.

Summary

Each bond type absorbs IR radiation in a characteristic range, its group frequency, set by bond strength and the masses of the atoms. Bonds to hydrogen absorb highest, then triple bonds, then double bonds, with single bonds to heavy atoms in the fingerprint region. Band position identifies the bond, while intensity and shape confirm it.

Practice questions

1. Explain why an O–H stretch appears at a higher wavenumber than a C–O stretch. Answer: Hydrogen is much lighter than carbon, so the reduced mass of O–H is far smaller, and O–H is also a stronger bond; both factors raise the vibrational wavenumber. 2. Which bond is responsible for a strong band at 1050 cm⁻¹ in the spectrum of ethanol? Answer: The C–O single bond, which absorbs between 1000 and 1300 cm⁻¹. 3. Why is the C=C stretch of an alkene usually weaker than the C=O stretch of a ketone? Answer: C=C is almost non-polar, so its dipole moment changes little when it stretches, whereas the highly polar C=O bond changes its dipole greatly. 4. A band appears at 2240 cm⁻¹. Name two bonds that might cause it. Answer: A C≡N bond in a nitrile or a C≡C bond in an alkyne.