Peaks indicate bond types, not exact counts of bonds
Sharp, tall peaks mean strong absorption for that bond type; broad peaks give different clues
You won’t infer exact numbers of CH or OH bonds just from peak counts; you use peak positions, intensities, and shapes to deduce bond types and then compare the fingerprint region to known spectra
Three key parts of every IR peak
Wave number (the bottom axis): where the peak occurs; this is the most important descriptor
Intensity (how deep the peak goes): relates to how much light is absorbed at that wave number
Peak shape (broad vs sharp): provides clues about the bond environment (e.g., OH tends to be broad; CH stretches can be sharp)
The spectrum is interpreted using all three together, not in isolation
Where the wave number comes from (bond vibration basics)
The wave number \tilde{ν} is related to bond strength and masses via a harmonic oscillator model
The basic relation (for a diatomic approximation) is
ν~≈2πc1μf
where
f is the force constant (bond strength)
μ is the reduced mass, μ=m<em>1+m</em>2m<em>1m</em>2
c is the speed of light
Consequences you should know (without needing to calculate exact numbers):
Stronger bonds (larger f) give higher wave numbers
Heavier atom pairs (larger μ) give lower wave numbers
More massive atoms generally push the wave number down; lighter atoms push it up
Practical note from the instructor’s perspective:
You don’t need to memorize exact numbers for every bond (the exam provides an equation sheet and you’ll use it conceptually)
You should be able to predict trend and roughly locate peaks (high vs mid vs low regions) rather than memorize precise values
Where IR peaks commonly lie on the spectrum
The spectrum runs roughly from 4000 cm⁻¹ on the left to 400 cm⁻¹ on the right
The “left side” (higher wave numbers) is called the diagnostic region, where a given peak can correspond to multiple possible bonds
The “right side” (lower wave numbers) is the fingerprint region, where peaks are highly specific to a particular molecule or functional group
Typical guidance:
Diagnostic region: around ~1500–4000 cm⁻¹, where bond vibrations like C–H, C=O, N–H appear but aren’t unique to a single molecule
Fingerprint region: ~1500–400 cm⁻¹, complex pattern highly unique to a specific structure
Example note from the lecture: a peak near 1000 cm⁻¹ can arise from several different bonds (C–C, C–O, C–N, etc.) and by itself is not diagnostic
Analogy used: if you only have a fingerprint, you match it to a database to identify the molecule (like solving a crime with a fingerprint database)
Peak intensities and shapes: what they imply
Intensity describes how strongly a bond absorbs at its wave number
A very intense peak indicates strong absorption (low transmittance)
An example given: a peak with transmittance ~15% would correspond to absorbance ~85% (intense)
Shape (broad vs sharp) provides clues about the bond environment
OH stretches tend to be broad (often due to hydrogen bonding or multiple environments)
Most CH stretches are relatively sharp, but exact shapes depend on neighboring atoms and coupling
Size and breadth together help you guess the type of bond you’re looking at, but you still often need the fingerprint region to confirm identity
The OH vs CH regions and what to expect
OH bonds: broad absorption bands, often spanning a wide range, due to hydrogen bonding and interactions in the sample
CH bonds: typically sharper, more localized features
CH stretch location basics:
CH associated with a carbon involved in a double bond (alkene) tends to appear above ~3000 cm⁻¹, often around ~3100 cm⁻¹
CH stretches associated with a terminal alkyne (C≡C–H) can appear around ~3300 cm⁻¹ (also above 3000 cm⁻¹, but distinct from alkenes)
CH3 (and other alkyl) C–H stretches typically appear around ~2850–2960 cm⁻¹ (roughly below 3000 cm⁻¹)
Important qualitative rule from the lecture: stronger CH signals above 3000 cm⁻¹ can indicate vinyl (alkene) CH bonds, while CH stretches below ~3000 are often from alkyl CH
How to use the fingerprint region to identify a molecule
The left (diagnostic) region can tell you about the presence of certain bond types but is not definitive for identity
The right (fingerprint) region contains many small, unique peaks that together serve as a molecular fingerprint
Practical workflow:
Look for large, diagnostic peaks (e.g., OH broad band, strong C=O at ~1700 cm⁻¹, etc.) to infer functional groups
Compare the fingerprint region to known spectra in a database to confirm the exact molecule or distinguish between similar structures
Example discussed in class:
If you have a spectrum with a broad peak somewhere and a set of other peaks, you can suspect an alcohol (OH) or carboxylic acid depending on the exact pattern, then use the fingerprint to confirm by matching to a known compound such as butanol vs propanol
Conceptual exercise: predicting CH stretch regions for a set of molecules
Task: determine which CH stretches would appear above 3,000 cm⁻¹ for several molecules, ignoring OH bands (which are above ~3200–3600 cm⁻¹ but not the focus here)
Consider three example structures (described conceptually in the lecture):
Structure A: contains a carbon–carbon double bond (alkene) with a CH bonded directly to the alkene carbon (vinyl CH)
This CH is expected to show a CH stretch above 3,000 cm⁻¹, roughly near 3,100 cm⁻¹
Structure B: contains a C=C double bond and a C≡C triple bond, with the triple-bonded carbon not bearing a hydrogen (no CH on the triple-bond carbon)
CHs present elsewhere (e.g., from CH3 groups) will produce CH stretches around ~2900 cm⁻¹, but there may be no CH stretch above 3,000 cm⁻¹ associated with the triple bond itself
Structure C: neither double-bond carbons nor triple-bond carbons bear hydrogens (no vinyl CH at the alkene carbons and no terminal CH on the triple bond)
The CH stretches above 3,000 cm⁻¹ will be limited to any other CH groups (e.g., in substituents), but the characteristic vinyl or alkyne CH bands above 3,000 cm⁻¹ will be absent
Practical takeaway:
If you observe a peak around ~3,100 cm⁻¹, it likely indicates a vinyl CH attached to a C=C bond (Structure A)
If there is no CH stretch above ~3,000 cm⁻¹, you might be looking at Structures B or C where the high-lying CH stretch from an alkene CH is absent due to substitution patterns
The presence or absence of a peak near ~3,100–3,300 cm⁻¹ helps distinguish alkene-CH vs alkyne-CH environments, while CH stretches around ~2900 cm⁻¹ come from CH3/alkyl groups regardless of the multiple bonds present
Exam strategy and practical notes
The instructor emphasizes not memorizing every numeric value for bonds
You will be provided with an equation sheet that contains key constants and typical values; focus on the relationships rather than memorizing exact numbers
On exams, expect qualitative and comparative questions: for example, predict whether a CH stretch would appear above 3,000 cm⁻¹ given a bond environment, or describe whether a peak will be sharp or broad and what that implies
When comparing two spectra, analyze: (1) a large, broad OH peak vs (2) sharp CH peaks, and (3) the pattern in the fingerprint region to identify the molecule
Quick recap of the main ideas
Each IR peak has three important descriptors: wave number, intensity, and shape
The wave number is primarily dictated by bond strength and reduced mass: higher bond strength and lighter masses push the wave number higher; heavier masses push it lower
The spectrum ranges from 4000 cm⁻¹ (left) to 400 cm⁻¹ (right); the left side is diagnostic, the right side is the fingerprint region
OH bonds tend to produce broad peaks; CH bonds tend to produce sharper peaks (with the alkene CH usually above 3000 cm⁻¹ and alkyl CH around 2900 cm⁻¹)
To identify a molecule, combine interpretation of the diagnostic region with database matching of the fingerprint region; don’t rely on a single peak
Useful formulas and quick references (from the lecture context)
Wave number relation (approximate): ν~=2πc1μf
Reduced mass for two masses: μ=m<em>1+m</em>2m<em>1m</em>2
Typical spectrum range: left ≈ 4000 cm⁻¹, right ≈ 400 cm⁻¹
Intensity concept (transmittance vs absorbance, qualitative):
If transmittance T ≈ 0.15, absorbance A ≈ -\log_{10}(0.15) ≈ 0.82–0.85 (approximate in lecture context)