IR Spectroscopy Calculator

Convert between IR wavenumber, wavelength, and frequency. Look up characteristic absorption bands.

Mode

Wavenumber ṽ (cm⁻¹) (cm⁻¹)

C=O: 1700-1750, C-H: 2850-3000

Functional group for lookup

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What Determines Absorption Frequency

An IR band appears where the bond's vibrational frequency matches the photon energy. Two factors set that frequency, and they work in opposite directions.

ν ∝ √(k/μ) μ = m₁m₂/(m₁+m₂)

Factor	Effect on wavenumber	Example
Stronger bond (higher k)	Higher	C≡C 2200 > C=C 1650 > C–C 1000
Lighter atoms (lower μ)	Higher	C–H 3000 > C–C 1000
Deuteration	Lower by ~√2	C–D near 2200 vs C–H 3000

This is why C–H stretches always sit around 3000 cm⁻¹ regardless of the molecule — hydrogen's tiny mass dominates the reduced mass term.

Group	Wavenumber (cm⁻¹)	Appearance
O–H (alcohol)	3200–3600	Broad — hydrogen bonding
N–H	3300–3500	Sharper; two peaks for NH₂
C–H	2850–3100	Above 3000 indicates sp²
C≡N	2220–2260	Sharp, distinctive
C=O	1650–1780	Strong and sharp
C=C	1620–1680	Weak
Fingerprint region	600–1400	Unique to each compound

Carbonyl Position Is Diagnostic

The exact C=O wavenumber identifies which carbonyl you have. Ester at 1735–1750 sits above ketone at 1715 because the ester oxygen withdraws electron density inductively, strengthening the C=O bond. Conjugation lowers it — an aryl ketone drops to about 1685 as the double bond character is partly delocalised. Amides sit lowest, near 1650.

A vibration is only IR-active if it changes the molecular dipole moment. Symmetric stretches in symmetric molecules — N₂, O₂, the symmetric stretch of CO₂ — are invisible in IR but appear in Raman, which is why the two techniques are complementary.

Worked Examples

Example 1: C=O in ketone: 1715 cm⁻¹

λ=10000/1715=5.83μm

Result: ν=1715×3e10=5.15×10¹³ Hz

Carbonyl region 1650-1850 diagnostic

Example 2: Lookup: C=O carbonyl

Aldehyde vs ketone vs ester?

Result: Different carbonyl environments shift C=O by 10-30 cm⁻¹

Use position to distinguish compound classes

Example 3: Distinguishing ester from ketone

C=O at 1740 versus 1715 cm⁻¹

Result: Ester versus ketone

The ester oxygen withdraws electron density inductively, strengthening the C=O bond and raising its frequency by about 25 cm⁻¹.

Example 4: Effect of conjugation

Acetophenone C=O at 1685 rather than 1715

Result: Conjugation with the ring

Delocalisation reduces double bond character, weakening the bond and lowering the wavenumber by roughly 30 cm⁻¹.

Common Mistakes

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Ignoring peak shape

A broad band near 3300 indicates hydrogen-bonded O–H; a sharp one at similar position suggests N–H. Shape carries as much information as position.

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Treating all carbonyls as one peak

Position distinguishes them: amide ~1650, aryl ketone ~1685, ketone ~1715, ester ~1740, anhydride ~1800. A 30 cm⁻¹ shift is meaningful.

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Expecting every bond to absorb

Only vibrations that change the dipole moment are IR-active. Symmetric molecules such as N₂ and O₂ show nothing at all.

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Trying to assign every fingerprint peak

Below 1400 cm⁻¹ the pattern is a molecular signature rather than a set of assignable bands. Use it for comparison, not interpretation.

Frequently Asked Questions

Fingerprint region?

600-1500 cm⁻¹: many overlapping C-C, C-N, C-O single bonds. Unique to each molecule (like fingerprint). Used for compound identification by library comparison, not individual bond assignment.

IR vs Raman?

IR active: vibrations that change dipole moment. Raman active: vibrations that change polarizability. Symmetric vibrations: often IR inactive but Raman active. Centrosymmetric molecules (CO2, benzene): mutually exclusive rule. Both techniques are complementary.

Why do C–H stretches always appear near 3000?

Because hydrogen's small mass dominates the reduced mass term. Frequency scales with the inverse square root of reduced mass, so C–H sits high regardless of the rest of the molecule.

Why is the O–H peak broad?

Hydrogen bonding creates a range of slightly different O–H environments, each absorbing at a slightly different frequency. The bands overlap into one broad envelope.

How do I tell an ester from a ketone?

By carbonyl position. Esters absorb near 1740 cm⁻¹ and ketones near 1715, because the ester oxygen inductively strengthens the C=O bond.

Why are some bonds IR-inactive?

A vibration must change the molecular dipole moment to absorb infrared light. Symmetric stretches in symmetric molecules produce no dipole change and appear only in Raman.

Formula Explorer connections

Interpretation: This analytical relationship converts an instrument signal, separation measure or optical response into concentration, identity or performance. Assumption: Calibration, blank correction, linear range, path length, matrix effects and instrument settings must match the sample and method.

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