Question 3 of 6: Infrared / Fourier Transform Infrared Spectroscopy (IR/FTIR)

Question 3: Infrared / Fourier Transform Infrared Spectroscopy (IR/FTIR) (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a) What FTIR Is Used For, and Why It Is an Absorption Technique

FTIR is used to identify chemical functional groups and molecular structure by measuring which infrared frequencies a sample absorbs. Every covalent bond has characteristic vibrational modes (stretching, bending, scissoring, wagging), and each mode has a natural vibrational frequency set by the bond's force constant and the masses of the atoms involved; when incident infrared light happens to match that frequency, the molecule absorbs a photon and is excited to a higher vibrational energy level. Scanning across the mid-infrared range (roughly 4000–400 cm⁻¹) and recording how much light is absorbed (or, equivalently, transmitted) at each wavenumber therefore produces a "fingerprint" that identifies the functional groups present — O–H, N–H and C–H stretches near 3000–3600 cm⁻¹, C=O stretches near 1650–1750 cm⁻¹, and so on. It is called an absorption technique precisely because the measured signal is the light absorbed by the sample at each frequency (rather than light emitted by the sample, as in fluorescence/emission spectroscopy, or scattered inelastically, as in Raman spectroscopy): the fundamental physical event being recorded is a molecule removing photons from the beam at its own resonant vibrational frequencies. In biochemistry, FTIR is widely used to probe protein secondary structure (via the amide I/II bands), to identify unknown organic/biomolecular samples, and for rapid, label-free quality-control identification of raw materials.

(b) Does N₂ Have an Infrared Spectrum?

No. N₂ is a homonuclear diatomic molecule (N≡N): both atoms are identical, so the molecule has no permanent dipole moment, and—critically for IR activity—stretching the N≡N bond does not create any change in dipole moment either, because the charge distribution remains symmetric about the bond's midpoint at every bond length. Infrared absorption requires a vibration to produce a net change in the molecule's dipole moment (only then can the oscillating electric field of the IR photon couple to and drive that vibration); since N₂'s stretch is dipole-silent, it is IR-inactive. (The same vibration is, however, Raman-active, because it does change the molecule's polarizability — Raman and IR selection rules are complementary for a centrosymmetric molecule like N₂.)

(c) What Is an Interferogram?

An interferogram is the raw, time-domain (or mirror-position-domain) signal recorded directly by an FTIR instrument's detector, before any Fourier transform is applied. Inside the instrument's Michelson interferometer, the IR beam is split into two paths, one reflected off a fixed mirror and one off a mirror that moves continuously; the two beams recombine and interfere, and because every wavelength present in the source beam interferes at its own characteristic pattern as the moving mirror's position (optical path difference) changes, the detector records a single complex signal — intensity versus mirror displacement — that is the superposition of the interference patterns of every wavelength simultaneously. This raw signal, the interferogram, is not directly interpretable as a chemical spectrum; a Fourier transform must be applied to convert it from the mirror-position (optical path difference) domain into the familiar intensity-versus-wavenumber spectrum that chemists read. This "measure everything at once, then transform" strategy (the Fellgett/multiplex advantage) is what makes FTIR far faster and more sensitive than a scanning dispersive IR instrument that measures one wavelength at a time.

(d) Interpreting the Given IR Spectrum

Reading the trace feature by feature: there is a very strong, broad absorption centred near 3300 cm⁻¹ that dips to almost 0% transmittance — a band this broad and this intense is the signature of an O–H stretch involved in extensive, cooperative hydrogen bonding (a molecule rich in hydroxyl groups), rather than the sharper, weaker N–H stretch of an amine/amide or the much weaker, narrower C–H stretches seen here as the small sharp features near 2900–3000 cm⁻¹. There is a broad, moderate-intensity band around 1650–1600 cm⁻¹ (consistent with O–H bending/associated water, or a weak C=O), but there is no distinct, separate band near 1550 cm⁻¹ of comparable intensity — the pairing of a strong amide I (~1650 cm⁻¹) band together with a clearly resolved amide II (~1550 cm⁻¹) band is the diagnostic protein/peptide-backbone signature, and it is not present here. Most diagnostic of all is the very strong, sharp, near-total-absorption doublet in the 1000–1150 cm⁻¹ fingerprint region: intense C–O and C–O–C ring/glycosidic-linkage stretching vibrations in exactly this range, combined with the dominant broad O–H stretch, are the classic fingerprint of a carbohydrate/polysaccharide (e.g. a sugar polymer such as starch, cellulose, or a similar polyol). The absence of a sharp, strong carbonyl band near 1740 cm⁻¹ argues against a lipid/ester, and the absence of a resolved amide I/amide II pair argues against a protein — so, on the evidence of (i) the very strong, broad hydroxyl stretch, (ii) weak C–H stretching, and (iii) the intense sharp C–O/C–O–C fingerprint doublet, this spectrum is best assigned to a carbohydrate-type molecule.