Unit 13: Infrared (IR) Spectroscopy
Using molecular vibrations to identify the functional groups within a compound.
13.4 Basic Principles (Vibrational Modes, IR Active/Inactive)
Infrared (IR) spectroscopy works by irradiating a sample with infrared light. Covalent bonds within a molecule are not rigid; they are constantly vibrating, behaving much like springs. They can stretch, bend, and twist in various ways called vibrational modes.
Each vibrational mode has a specific, quantized frequency. If a bond is irradiated with IR light of the exact same frequency as its natural vibration, the bond will absorb that energy and vibrate with a greater amplitude. An IR spectrometer measures which frequencies of IR radiation are absorbed by the sample.
IR Active vs. Inactive
For a bond to absorb IR radiation, its vibration must cause a change in the molecule's dipole moment. Such a vibration is said to be IR active.
- IR Active: Most bonds in organic molecules, especially polar bonds like C=O, O-H, and C-Cl, have a significant dipole moment that changes as the bond stretches. These produce strong signals in an IR spectrum.
- IR Inactive: Symmetrical, non-polar bonds, such as those in diatomic molecules like $O_2$ or $N_2$, do not have a dipole moment. Stretching these bonds does not change the dipole moment (it remains zero), so they do not absorb IR radiation and are IR inactive. The C=C bond in a symmetrical alkene like ethene is also IR inactive or very weak.
Solved Examples:
- Will the C=O bond in propanone ($CH_3COCH_3$) be IR active?
Explain.
Solution: Yes. The C=O bond is highly polar, creating a strong dipole moment. When this bond stretches, the dipole moment changes, so it will strongly absorb IR radiation and be IR active. - Will the N≡N bond in a nitrogen molecule be IR active?
Explain.
Solution: No. The N≡N bond is perfectly symmetrical and non-polar. Its dipole moment is zero. When the bond vibrates, the dipole moment remains zero. Since there is no change in dipole moment, it is IR inactive. - What are the two main types of molecular
vibrations?
Solution: Stretching (the distance between two atoms changes) and bending (the angle between bonds changes). - Why is IR spectroscopy useful for organic
chemistry?
Solution: Because organic molecules are rich in covalent bonds (C-H, C-C, C=O, O-H, etc.) that have characteristic vibrational frequencies, making them easy to identify. - What property must a vibration have to be IR
active?
Solution: The vibration must cause a change in the overall dipole moment of the molecule. - Is the C-H bond in methane ($CH_4$) IR
active?
Solution: Yes. Although the overall molecule is non-polar due to its symmetry, the individual C-H bonds are polar. When they stretch or bend, the symmetry is temporarily broken, causing a change in the net dipole moment, making the vibrations IR active. - Would you expect the C=C stretch in ethene ($H_2C=CH_2$) to be IR
active?
Solution: No. Because the molecule is symmetrical, stretching the C=C bond does not change the overall dipole moment (which is zero). It is IR inactive. - Would you expect the C=C stretch in propene ($CH_3CH=CH_2$) to be IR
active?
Solution: Yes. The molecule is asymmetrical. The C=C bond has a small dipole moment, and stretching it will cause a change, resulting in a weak but observable IR absorption. - What is a dipole moment?
Solution: A measure of the separation of positive and negative charges in a molecule, resulting from differences in electronegativity in a polar covalent bond. - What does it mean for energy to be
"quantized"?
Solution: It means that energy can only be absorbed or emitted in discrete, specific amounts, not in a continuous range. A bond can only absorb a photon of IR radiation if its energy exactly matches the energy gap between vibrational levels.
13.5 Interpretation of IR Spectra (Functional Groups)
An IR spectrum is a plot of percent transmittance on the y-axis versus wavenumber ($cm^{-1}$) on the x-axis. Wavenumber is proportional to frequency and energy. The peaks (or absorption bands) point downwards, indicating that radiation at that frequency has been absorbed.
The spectrum is typically divided into two regions:
- Functional Group Region (4000 - 1500 cm⁻¹): This region contains clear, well-defined peaks that are characteristic of specific bonds and functional groups.
- Fingerprint Region (below 1500 cm⁻¹): This region contains a complex pattern of many overlapping peaks, caused by the vibrations of the molecule as a whole. While difficult to interpret fully, the fingerprint region is unique to each molecule.
Key Characteristic Absorptions:
| Bond | Functional Group | Wavenumber (cm⁻¹) | Appearance |
|---|---|---|---|
| O-H | Alcohols, Phenols | 3200 - 3600 | Very broad, strong |
| O-H | Carboxylic Acids | 2500 - 3300 | Extremely broad, strong |
| C-H | Alkanes/Alkyls | 2850 - 3000 | Strong, sharp |
| C=O | Aldehydes, Ketones, Acids, Esters | 1680 - 1750 | Very strong, sharp |
| C-O | Alcohols, Ethers, Esters, Acids | 1000 - 1300 | Strong, sharp |
Solved Examples:
- An IR spectrum shows a very strong, sharp peak at 1710 cm⁻¹. What functional
group is almost certainly present?
Solution: A very strong, sharp peak in the 1680-1750 cm⁻¹ range is the classic signal for a carbonyl group (C=O). - A spectrum shows an extremely broad absorption band centered around 3000
cm⁻¹. What functional group does this
indicate?
Solution: An extremely broad peak spanning from 2500-3300 cm⁻¹ is characteristic of the O-H bond in a carboxylic acid. - How could you use IR spectroscopy to distinguish between propan-1-ol and
propanone?
Solution: The spectrum of propan-1-ol would show a very broad O-H peak around 3200-3600 cm⁻¹. The spectrum of propanone would show a very strong, sharp C=O peak around 1715 cm⁻¹ and would have no broad O-H peak. - What is the fingerprint region used
for?
Solution: While the functional group region identifies the parts of a molecule, the fingerprint region is unique to the entire molecule. It is used to confirm the identity of a compound by matching its fingerprint region exactly to that of a known sample. - A compound's IR spectrum has a strong, broad peak at 3400 cm⁻¹ but no peak
around 1700 cm⁻¹. What type of compound is it likely to
be?
Solution: The peak at 3400 cm⁻¹ indicates an O-H group. The absence of a C=O peak at 1700 cm⁻¹ rules out a carboxylic acid. Therefore, the compound is likely an alcohol. - What does a low percent transmittance value on an IR spectrum
signify?
Solution: A low transmittance value means that a large amount of IR radiation at that specific wavenumber has been absorbed by the sample. This corresponds to a strong peak. - What is a "wavenumber"?
Solution: Wavenumber is the unit used on the x-axis of an IR spectrum. It is the reciprocal of wavelength ($1/\lambda$) and is directly proportional to frequency and energy. - Why is the O-H peak in an alcohol so
broad?
Solution: The O-H bond can participate in hydrogen bonding. Since the strength of hydrogen bonds varies within the sample, the O-H bonds vibrate at a wide range of slightly different frequencies, causing the single peak to broaden. - A spectrum shows sharp peaks just below 3000 cm⁻¹. What bonds do these
represent?
Solution: These are characteristic of C-H bonds in alkanes/alkyl groups. - Two isomers have the same functional groups. Will their functional group
regions look similar or different? What about their fingerprint
regions?
Solution: Their functional group regions will look very similar because they contain the same types of bonds. However, because they are different molecules with different overall structures, their fingerprint regions will be different.
13.6 Applications of IR Spectroscopy
IR spectroscopy is a fast, reliable, and non-destructive technique with many applications:
- Functional Group Identification: Its primary use is to determine which functional groups are present in a molecule, which is a crucial first step in structure elucidation.
- Monitoring Reaction Progress: By taking IR spectra of a reaction mixture over time, a chemist can monitor the disappearance of a reactant's characteristic peak (e.g., the broad O-H of an alcohol) and the appearance of a product's peak (e.g., the sharp C=O of a ketone) to see if a reaction is complete.
- Quality Control: In industry, IR spectroscopy can be used to check the purity of a sample. The spectrum of a manufactured batch can be compared to the spectrum of a pure standard. The presence of unexpected peaks would indicate impurities.
- Forensic Science: Used to identify substances like drugs, fibers, and paint chips by matching their IR spectra to known databases.
Solved Examples:
- How could IR spectroscopy be used to monitor the oxidation of ethanol to
ethanoic acid?
Solution: You would monitor the disappearance of the broad O-H peak of the alcohol (around 3300 cm⁻¹) and the appearance of two new peaks: the extremely broad O-H peak of the carboxylic acid (2500-3300 cm⁻¹) and the strong, sharp C=O peak (around 1710 cm⁻¹). - A pharmaceutical company produces aspirin. How can they use IR to ensure
each batch is pure?
Solution: They can run an IR spectrum of a sample from the batch and compare it to the standard spectrum of pure aspirin. The "fingerprint" regions must match exactly, and there should be no significant extra peaks that would indicate an impurity. - Why is IR spectroscopy considered a "non-destructive"
technique?
Solution: Because the sample can be recovered unchanged after the analysis is complete. The energy from IR radiation only causes bonds to vibrate; it is not strong enough to break them. - A police lab receives a paint chip from a hit-and-run accident. How can IR
help?
Solution: The lab can take an IR spectrum of the paint chip. This spectrum can be compared to a database of spectra from different car manufacturers and models to help identify the suspect's vehicle. - Can IR spectroscopy be used to distinguish between enantiomers
(non-superimposable mirror images)?
Solution: No. Enantiomers have the exact same bonds and functional groups, so their IR spectra are identical. Other techniques are needed to distinguish them. - How might IR spectroscopy be used in the food
industry?
Solution: It can be used for quality control, for example, to measure the trans fat content in oils or to verify the authenticity of products like olive oil by checking its spectral fingerprint. - A chemist is trying to reduce a ketone (which has a C=O bond) to a secondary
alcohol (which has an O-H bond). How will the IR spectrum change if the reaction
is successful?
Solution: The strong, sharp C=O peak around 1715 cm⁻¹ will disappear, and a new, strong, broad O-H peak will appear around 3300 cm⁻¹. - Can IR spectroscopy determine the molecular weight of a
compound?
Solution: No. IR spectroscopy only provides information about the types of covalent bonds (functional groups) present in a molecule. Mass spectrometry is used to determine molecular weight. - Why is the C=O absorption band typically one of the strongest in an IR
spectrum?
Solution: Because the carbon-oxygen double bond is highly polar, meaning it has a large dipole moment. The stretching of this bond causes a large change in the dipole moment, leading to a very strong absorption of IR radiation. - A compound has the formula C₃H₆O. Its IR spectrum shows a strong, sharp peak
at 1715 cm⁻¹. What is the likely structure of the
compound?
Solution: The peak at 1715 cm⁻¹ indicates a C=O group. The formula C₃H₆O fits with the structure of propanone ($CH_3COCH_3$), a ketone.
🧠 Quiz
Answer: Molecular vibrations
(stretching and bending).
Answer: It must cause a change in the
molecule's dipole moment.
Answer: A carbonyl group (C=O).
Answer: The fingerprint region.
Answer: By observing the disappearance
of a reactant's peak and/or the appearance of a product's peak over time.
Answer: No, because it is symmetrical
and has no dipole moment, so it is IR inactive.
Answer: A strong and very broad peak
around 3200-3600 cm⁻¹.
Answer: Wavenumber (cm⁻¹).
Answer: Due to very strong and
extensive hydrogen bonding (dimerization).
Answer: A carboxylic acid.
Answer: To confirm the identity of a
compound by matching it to a known spectrum.
Answer: Non-destructive.
Answer: C-H bonds of alkanes/alkyl
groups.
Answer: Down, because the y-axis is
percent transmittance.
Answer: Strong absorption.
Answer: Yes, though it's less common
than other techniques, the intensity of an absorption is related to
concentration.
Answer: It naturally vibrates, but it
absorbs IR energy to vibrate with a greater amplitude.
Answer: An alkane (or a compound with
only C-H and C-C single bonds).
Answer: Quality control to check for
purity and product identity.
Answer: The C=O bond. Double bonds are
stronger and stiffer than single bonds, so they vibrate at a higher frequency
(wavenumber).