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Organic Chemistry 2 · Spectroscopy

Infrared Spectroscopy

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On this page 7 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Worked example
  6. Key takeaway
  7. Study tools

In 30 seconds

Infrared (IR) spectroscopy measures which frequencies of infrared light a molecule absorbs as its bonds stretch and bend. Each functional group absorbs at a characteristic wavenumber, so the spectrum acts like a fingerprint of the bonds present. A carbonyl near 1700 cm⁻¹, a broad O–H near 3200–3600 cm⁻¹, and C≡C/C≡N near 2100–2260 cm⁻¹ are among the most diagnostic signals. IR confirms functional groups but does not give a complete structure.

Why this matters

identifies pharmaceuticals by matching a drug's spectrum against a pharmacopeial reference before release. Clinically, breath and blood analyzers use IR absorption (e.g., CO₂, ethanol) for rapid measurement, and near-IR underlies pulse oximetry's blood-oxygen monitoring.

The college version

1. Molecular vibrations and wavenumber

IR light (~4000–400 cm⁻¹) is too weak to break bonds but strong enough to set them vibrating. are (atoms moving along the bond axis) and (angle changes — scissoring, rocking, wagging, twisting). The x-axis is wavenumber (cm⁻¹), proportional to frequency and energy.

2. Hooke's law and the stretching frequency

The models a bond as a spring. Stretching wavenumber rises with (stiffer spring) and falls with (heavier atoms): ν̃ ∝ kμ,   μ= m1 m2m1 + m2 So triple bonds (highest k) absorb highest (C≡C ~2100, C≡N ~2260 cm⁻¹), double bonds next (C=C ~1650, C=O ~1700 cm⁻¹), and single bonds lowest (C–C ~1200 cm⁻¹). It also explains why C–H (~3000) absorbs far higher than C–Cl (~700): hydrogen is much lighter than chlorine.

3. Diagnostic and fingerprint regions

The (~4000–1400 cm⁻¹) holds functional-group stretches that are easy to assign. The (~1400–400 cm⁻¹) holds complex bending and single-bond patterns unique to each molecule — used to match a compound against a reference, not to assign groups one by one.

How it works

  1. Infrared light of many frequencies passes through (or reflects off) the sample.
  2. Each bond absorbs only at frequencies matching its natural vibrations.
  3. A detector compares transmitted light to a reference beam and records the missing frequencies.
  4. Absorbance is plotted against .
  5. The chemist assigns diagnostic-region peaks and matches the fingerprint region.

Common confusions

Do not confuseWithDifference
StretchingBendingBond length vs. angle changes
Wavenumber (cm⁻¹)Wavelength (nm/µm)Wavenumber ∝ frequency; they move oppositely
Diagnostic regionFingerprint regionAssignable group stretches vs. unique matching pattern
Broad O–HSharp N–HO–H strongly H-bonded; N–H weaker
C=C stretchC=O stretchC=O stronger, higher; C=C may be absent
"No peak""No group"Symmetric/weak absorptions may not appear

Memory aids

"Heavy and Stiff move Higher; Light and Weak move Lower" — a bond absorbs at higher wavenumber when it is stronger and its atoms are lighter. And "ONE-C" for the scan order: O–H/N–H, Nitrile/alkyne (triple), Ester/ketone (C=O), C=C — then the fingerprint.

Quick review

Topic Recap

Infrared spectroscopy excites bond stretching and bending and records absorbed frequencies as wavenumbers. The Hooke's-law model explains why absorption wavenumber rises with bond strength and falls with atomic mass (triple > double > single). The diagnostic region reveals functional groups — broad O–H, sharp N–H, strong C=O, and the C=C/C≡C/C≡N family — while the fingerprint region confirms identity by matching. IR is fast and group-specific but cannot supply a formula, skeleton, or stereochemistry.

Knowledge Check

  1. Why does a C≡C bond absorb at higher wavenumber than a C=C bond?
  2. A spectrum shows a very broad absorption near 3300 cm⁻¹ and a strong peak at 1710 cm⁻¹. Which functional group?
  3. Why is C–H (~3000 cm⁻¹) so much higher than C–Cl (~700 cm⁻¹)?
  4. What does the fingerprint region tell you that the diagnostic region cannot?
  5. A molecule has an internal symmetric C=C but no clear peak near 1650 cm⁻¹. Explain.

Answers and Rationales

  1. Stronger bond. The triple bond has a higher force constant k; since ν̃ ∝ k/μ, it vibrates at higher wavenumber.
  2. A carboxylic acid. Very broad O–H (2500–3300 cm⁻¹) plus C=O near 1710 cm⁻¹ is the classic acid signature.
  3. Lighter atom. H is far lighter than Cl, so the reduced mass μ for C–H is smaller, raising the wavenumber.
  4. Identity. Its unique pattern confirms identity by matching a reference; the diagnostic region only names individual groups.
  5. Weak/inactive vibration. A symmetric C=C stretch barely changes the dipole moment, so it absorbs IR only weakly — absence is not proof of absence.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Think of a molecule as tiny springs connecting balls (the atoms). Push a swing at its natural rhythm and it moves strongly; push at any other speed and it barely moves. Infrared light is that push — when its frequency matches a bond's natural vibration, the molecule absorbs it and we record a peak. Stretching is the bond getting longer and shorter; bending is the angle opening and closing.

The comparison "where it stops being exact": treating bonds as ideal springs predicts roughly where a vibration appears, but real bonds also bend, wag, and twist, and neighboring groups shift every absorption. So IR reliably tells you "there is a C=O here," but not which carbonyl compound you have, nor how the atoms connect in three dimensions.

Simple Example

Water absorbs IR strongly where its O–H bonds stretch and the H–O–H angle bends. That is why water vapor is a greenhouse gas and why wet samples show a big O–H peak.

Worked example

Assigning functional groups from an IR spectrum:

  1. O–H/N–H zone (3600–3100 cm⁻¹). A broad absorption near 3300 cm⁻¹ indicates O–H (alcohol or acid); acid O–H is very broad (2500–3300 cm⁻¹). N–H appears sharp(er), often as two spikes for a primary amine.
  2. Triple-bond zone (2260–2100 cm⁻¹). A sharp peak near 2260–2220 cm⁻¹ is C≡N; near 2260–2100 cm⁻¹ is C≡C (weak or absent for symmetric alkynes).
  3. Carbonyl zone (1850–1650 cm⁻¹). A strong C=O near 1715 (ketone/aldehyde), 1735 (ester), or 1680–1690 cm⁻¹ (amide) is the single most diagnostic feature.
  4. Double-bond zone (1680–1600 cm⁻¹). A C=C near 1650 cm⁻¹ (weak/absent for symmetric alkenes); aromatic C=C gives a pair near 1600 and 1500 cm⁻¹.
  5. Confirm in the fingerprint region. Match the full 1400–400 cm⁻¹ pattern against a reference to verify identity.
  6. State the limits. IR gives functional groups, not formula, skeleton, or stereochemistry; a missing weak peak (e.g., symmetric C=C) is not proof the group is absent.

Key takeaways

  • High yield: The C=O stretch near 1700 cm⁻¹ is the single most useful IR signal — strong and reliable.
  • High yield: O–H is broad (~3200–3600 cm⁻¹) from H-bonding; N–H is sharper, often two peaks for primary amines.
  • High yield: Triple bonds (C≡C, C≡N) absorb ~2100–2260 cm⁻¹, a region almost nothing else occupies.
  • High yield: Ordering follows bond strength and mass: triple > double > single; C–H > C–C because H is light.
  • High yield: A symmetric C=C can be weak or invisible — absence of a peak is not proof of absence.
  • Carboxylic-acid O–H is the broadest (2500–3300 cm⁻¹) and overlaps the C–H region.
  • The fingerprint region proves identity only by matching a reference.

Keep learning

Ready to build on this? Continue to the next lesson.

Study tools & related lessonsYou’ll learn to · Key vocabulary · Related

You’ll learn to

  • Explain how infrared light excites molecular vibrations (stretching and bending) to produce an absorption spectrum.
  • Relate absorption wavenumber to bond strength and atomic mass using the Hooke's-law model.
  • Identify O–H, N–H, C=O, C=C, C≡C, and C≡N absorptions in the diagnostic region and use the fingerprint region appropriately.
  • Apply a functional-group identification workflow and state the limitations of IR spectroscopy.

Key vocabulary

IR spectroscopy
Measures absorbed infrared frequencies
Molecular vibrations
Stretching and bending motions
Stretching
Atoms moving along the bond axis
Bending
Atoms changing bond angles
Wavenumber (cm⁻¹)
Energy/frequency unit on the x-axis
Hooke's law relationship
ν̃ ∝ k/μ
Bond strength
Force constant k
Atomic mass
Reduced mass μ of bonded atoms
Diagnostic region
4000–1400 cm⁻¹
Fingerprint region
1400–400 cm⁻¹
Broad vs. sharp
Peak width from H-bonding
O–H identification
Broad ~3200–3600 cm⁻¹
N–H identification
Sharp ~3300–3500 cm⁻¹, often double
C=O identification
Strong ~1650–1850 cm⁻¹
C=C identification
~1600–1680 cm⁻¹
C≡C identification
~2100–2260 cm⁻¹
C≡N identification
~2220–2260 cm⁻¹
Functional-group identification workflow
O–H/N–H → triple → C=O → C=C → fingerprint
IR limitations
What IR cannot tell you

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