Organic Chemistry · Structure Determination: Mass Spectrometry and Infrared Spectroscopy
Infrared Spectroscopy
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In 30 seconds
Infrared (IR) spectroscopy probes the vibrations of covalent bonds. A bond behaves like a tiny spring with a weight at each end: it can Stretch Vibration that changes bond length Full entry → (the bond length changes rhythmically) and it can Bend Vibration that changes bond angle (scissor, rock, wag, twist) Full entry → (the bond angle changes). Each motion has a natural vibrational frequency set by the atomic masses and the bond strength. When infrared light of exactly that frequency shines on the molecule, the bond absorbs the photon and vibrates more vigorously — an event recorded as an absorption band.
The mid-infrared region, 4000 to 400 cm⁻¹, covers essentially all organic bond vibrations. Because every functional group contains characteristic bonds vibrating at characteristic frequencies, an IR spectrum is a quick readout of which groups a molecule contains: carbonyl (C=O), alcohol (O–H), nitrile (C≡N), alkene (C=C), and so on. IR does not give a complete structure by itself, but it is the fastest way to confirm the presence or absence of a functional group.
Why this matters
IR spectroscopy is the workhorse functional-group test in organic laboratories: confirming a reaction worked (did the starting C=O disappear?), checking product identity and purity, and identifying unknowns from plastics, inks, and pharmaceutical tablets. The same physics operates in the atmosphere: CO₂, water, and methane absorb infrared radiation, the molecular basis of the greenhouse effect. On exams, IR questions present a spectrum and ask which functional groups are present. IR radiation is non-ionizing, so routine sample handling needs no special shielding, unlike UV sources.
The college version
Core Concepts
Stretches and bends: two families of vibration
Bond vibrations fall into two families. Stretches change the bond length along the bond axis; a group with two identical bonds (like -CH2-) can stretch symmetrically (both C–H bonds lengthen together) or antisymmetrically (one lengthens as the other shortens). Bends change bond angles — scissoring, rocking, wagging, twisting. Bending is easier to excite than stretching, so bends appear at lower wavenumbers: C–H stretches sit near 2850–3000 cm⁻¹, C–H bends near 1450 and 1375 cm⁻¹.
Hooke's law: what sets a vibration's frequency
The vibrational frequency of a bond is approximated by the harmonic-oscillator (Hooke's law) expression:
ν̃ = 12πckμ
where k is the force constant (bond stiffness, larger for multiple bonds), c is the speed of light, and μ is the reduced mass:
μ= m1 m2m1 + m2
with m1 and m2 the atomic masses. Two practical rules fall out: stronger bonds vibrate at higher wavenumber (C≡C ≈ 2150 cm⁻¹ > C=C ≈ 1650 cm⁻¹ > C–C ≈ 1200 cm⁻¹), and lighter atoms vibrate at higher wavenumber (C–H ≈ 3000 cm⁻¹ versus C–C ≈ 1200 cm⁻¹). These trends let you predict where a band should appear — far more useful than memorizing numbers.
The instrument: FT-IR
A modern IR spectrometer is a Fourier-transform instrument (FT-IR Fourier-transform infrared spectrometer Full entry →). A broadband source emits all IR frequencies at once; the beam passes through an interferometer (typically a Michelson design with a moving mirror), then through the sample, then to a detector. The detector records an Interferogram Raw detector signal (intensity vs mirror position) Full entry → — intensity versus mirror position — and a Fourier transform converts it into the familiar spectrum. Collecting all frequencies simultaneously gives speed and signal-to-noise: an FT-IR records a good spectrum of a tiny sample in seconds.
Reading the raw spectrum
IR spectra plot wavenumber on the x-axis, running from about 4000 cm⁻¹ on the left to 400 cm⁻¹ on the right (energy decreases left to right), and percent Transmittance (%T) Fraction of IR light passing through the sample on the y-axis, with 100% at the top. Absorption appears as downward dips whose depth reflects how strongly the Vibration Rhythmic motion of atoms in a bond — stretching or bending Full entry → changes the bond's dipole moment.
What makes a vibration IR-active
A vibration absorbs infrared light only if it changes the molecule's dipole moment. Vibrations of polar bonds (C=O, O–H, N–H) are strong absorbers; nonpolar vibrations may be weak or invisible. The C≡C stretch of a symmetrically substituted alkyne (e.g., 3-hexyne) is a classic near-silent vibration, because the dipole moment stays zero during the stretch — a structural feature that can hide from IR. Complementary techniques (Raman spectroscopy) probe such vibrations.
Common Confusions
| Do not confuse | With | Difference |
|---|---|---|
| Stretching vibration | Bending vibration | Stretches change bond length and absorb at higher wavenumber; bends change angle at lower wavenumber |
| Symmetric stretch | Antisymmetric stretch | In -CH2-, symmetric = both C–H bonds lengthen together; antisymmetric = one lengthens as the other shortens |
| Direction of the wavenumber axis | Increasing energy left-to-right | Energy decreases left to right: 4000 cm⁻¹ (left) is high energy, 400 cm⁻¹ (right) is low |
| Absorption peaks | Emission peaks | In IR, absorption appears as downward dips in %T, not upward peaks |
| A vibration that exists | A vibration that absorbs IR | Only vibrations that change the dipole moment are IR-active; symmetric C≡C stretches can be silent |
| Weak IR bands | Unimportant bands | Weakness can be structural information (C≡C near 2150 cm⁻¹ is characteristically weak) |
Eli explains
The same idea, in plain words
Explain it like I’m 10
A molecule is like a set of springs with weights on the ends — stiffer springs and lighter weights wiggle faster. Infrared light is a box of wiggles: shine the exact speed a spring likes and it wiggles harder, swallowing that light. The list of swallowed speeds tells you what kinds of springs (bonds) are in the molecule.
Worked example
Example 1: Why C–H stretches at ~3000 cm⁻¹ but C–C stretches near 1200 cm⁻¹
Compute the reduced masses (atomic masses in amu):
μC–H = (12.0)(1.008)12.0 + 1.008 = 0.930 amu, μC–C = (12.0)(12.0)12.0 + 12.0 = 6.00 amu
Assuming similar force constants, the Hooke's law frequency ratio is:
ν̃C–Hν̃C–C = μC–CμC–H = 6.000.930 = 2.54
So the C–H stretch sits about 2.5× higher than the C–C stretch — indeed C–H appears near 3000 cm⁻¹ and C–C near 1200 cm⁻¹. The same logic explains why triple bonds (higher k) outrank double bonds, which outrank single bonds.
Example 2: The energy of an O–H stretching photon at 3600 cm⁻¹
First convert wavenumber to m⁻¹: 3600 cm-1 × 100 cm/m = 3.60 × 105 m-1. Then:
E = hcν̃ = (6.626 × 10-34 J·s)(2.998 × 108 m/s)(3.60 × 105 m-1) = 7.16 × 10-20 J
Per mole: (7.16 × 10-20 J)(6.022 × 1023 mol-1) ≈ 43 kJ/mol — about ten times too weak to break a typical C–H bond (~413 kJ/mol). IR merely intensifies vibration, which is why IR spectroscopy is a gentle, non-destructive analysis method.
Example 3: Converting percent transmittance to absorbance
A carbonyl band reads 25% T. Absorbance is:
A = 2 - log10(%T)
Substitute: A = 2 - log10(25) = 2 - 1.40 = 0.60. Absorbance scales linearly with concentration via the Beer–Lambert law, A = εb c, where ε is molar absorptivity (L·mol⁻¹·cm⁻¹), b the path length (cm), and c the concentration (mol/L). Dimensional analysis: L·mol-1·cm-1 × cm × mol/L = 1, so A is unitless. This conversion moves IR from qualitative to quantitative analysis.
Key takeaways
- Two families of vibration: stretches (higher wavenumber) and bends (lower wavenumber).
- Hooke's law rules: stronger bond → higher ν̃; lighter atoms → higher ν̃.
- Signature anchors: C–H ≈ 2850–3100 cm⁻¹; C=O ≈ 1715 cm⁻¹; C≡N ≈ 2250 cm⁻¹; O–H ≈ 3200–3600 cm⁻¹ (broad).
- FT-IR records all frequencies at once (interferogram + Fourier transform), giving fast, sensitive spectra.
- Axes: x = wavenumber (4000 → 400 cm⁻¹); y = % transmittance (100 top, 0 bottom); absorption = downward dips.
- A vibration is IR-active only if it changes the dipole moment; symmetric C≡C stretches in symmetric alkynes can be nearly invisible.
- IR photons deliver only ~20–40 kJ/mol — enough to vibrate bonds, not break them (non-ionizing).
Check yourself
5 review questions from the chapter. Try each one, then open the answer.
Rank these stretches by wavenumber, highest first: C–C, C=C, C≡C. Why?
Show answer
C≡C (≈2150 cm⁻¹) > C=C (≈1650 cm⁻¹) > C–C (≈1200 cm⁻¹), because force constant increases with bond order and ν̃ ∝ k.
Why does the C–H stretch appear near 3000 cm⁻¹ while the C–Cl stretch appears near 750 cm⁻¹?
Show answer
Hydrogen is far lighter than chlorine: μC–H = 0.930 amu vs μC–Cl = (12.0)(35.5)/(12.0+35.5) = 8.96 amu; lighter reduced mass → higher wavenumber.
A band has 50% transmittance. What is its absorbance?
Show answer
A = 2 - log10(50) = 2 - 1.70 = 0.30.
What is the advantage of FT-IR over older dispersive instruments?
Show answer
It records all frequencies simultaneously (interferogram + Fourier transform), so spectra are much faster with better signal-to-noise.
Why might the C≡C stretch of 3-hexyne be nearly invisible in its IR spectrum?
Show answer
The C≡C stretch in a symmetric alkyne does not change the dipole moment, so it is only weakly IR-active (often invisible).
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- Vibration
- Rhythmic motion of atoms in a bond — stretching or bending
- Stretch
- Vibration that changes bond length
- Bend
- Vibration that changes bond angle (scissor, rock, wag, twist)
- Force constant (k)
- A measure of bond stiffness
- Reduced mass (μ)
- m1m2/(m1+m2) for the two bonded atoms
- FT-IR
- Fourier-transform infrared spectrometer
- Interferogram
- Raw detector signal (intensity vs mirror position)
- Transmittance (%T)
- Fraction of IR light passing through the sample
- IR-active
- A vibration that changes the dipole moment and absorbs IR
- transmittance, T
- Fraction of incident light that passes through the sample (I/I0)
Sources & references
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