11.1 Spectroscopy
Key Takeaways
- Infrared (IR) spectroscopy probes bond vibrations and identifies functional groups via characteristic absorption bands (e.g., O–H ~3200–3600 cm⁻¹, C=O ~1700 cm⁻¹).
- 1H NMR reveals the chemical environment and number of non-equivalent protons; integration gives proton ratios and splitting follows the n+1 rule.
- 13C NMR distinguishes the number of unique carbon environments; DEPT experiments classify carbons as CH3, CH2, CH, or quaternary.
- Mass spectrometry gives the molecular weight (molecular ion M+) and structural clues from fragmentation patterns and the M+1/M+2 isotope ratios.
- UV-Vis spectroscopy reports π→π* and n→π* transitions in conjugated systems; longer conjugation shifts λmax to longer wavelengths (red shift).
Spectroscopy: The Molecular Lens
Quick Answer: Spectroscopy interrogates how molecules absorb, emit, or scatter electromagnetic radiation and ions. For the PA-CAT (Bulletin of Information, rev. 20240815), the organic chemistry Table 5 leaf "Spectroscopy" expects you to know what each technique reveals about molecular structure and how to read the key signals.
The PA-CAT Chemistry domain is 16% of the exam—about 38 scored items drawn from general and organic chemistry. Within the organic subtree, Spectroscopy is a named leaf, so you should be able to (1) match a functional group to its IR absorption, (2) predict the number of NMR signals and splitting patterns, (3) read a mass spectrum for the molecular ion and isotope pattern, and (4) recognize conjugation effects in UV-Vis.
Electromagnetic Spectrum Map
| Region | Wavelength | Energy transition | Technique |
|---|---|---|---|
| UV-Vis | 200–800 nm | Electronic (π→π*, n→π*) | UV-Vis spectroscopy |
| Infrared | 2.5–25 µm (4000–400 cm⁻¹) | Molecular vibrations | IR spectroscopy |
| Radio | ~0.5–10 m (1–500 MHz) | Nuclear spin flip | NMR spectroscopy |
| (ionization) | — | Electron removal | Mass spectrometry |
IR Spectroscopy — Bond Vibrations
A bond absorbs IR when its vibration changes the dipole moment. The wavenumber (cm⁻¹) depends on bond strength and reduced mass: stronger bonds and lighter atoms vibrate at higher frequency. Fingerprint region (1500–400 cm⁻¹) is unique per molecule; functional-group region (4000–1500 cm⁻¹) is diagnostic.
| Absorption (cm⁻¹) | Bond | Notes |
|---|---|---|
| 3200–3600 (broad) | O–H | Alcohols, carboxylic acids (very broad 2500–3300) |
| 3300 (sharp) | N–H | Amines, amides |
| 3000–3100 | =C–H | Alkenes, aromatics |
| 2850–2960 | C–H | Alkanes |
| 2100–2260 | C≡C, C≡N | Alkynes, nitriles |
| 1700–1750 | C=O | Ketones, aldehydes, acids, esters |
| 1620–1680 | C=C | Alkenes |
NMR Spectroscopy — Nuclear Spin Environments
1H NMR (proton NMR) is the most structurally rich. Key readouts:
- Number of signals = number of chemically non-equivalent proton sets.
- Chemical shift (δ, ppm) = electronic environment (downfield = deshielded by electronegative atoms or π systems). Typical ranges: alkyl ~0.9–1.5 ppm; O–H/N–H 1–5 ppm (variable); alkenyl ~4.5–6.5; aromatic ~6.5–8.5; aldehyde ~9–10; carboxylic acid ~10–12.
- Integration = relative proton count per signal.
- Splitting (multiplicity) = n+1, where n = neighboring nonequivalent protons (singlet, doublet, triplet, quartet).
13C NMR gives one peak per unique carbon; chemical shifts span ~0–220 ppm, with carbonyls far downfield (160–220 ppm). DEPT (Distortionless Enhancement by Polarization Transfer) classifies each carbon as CH3, CH2, CH, or C(quaternary).
Mass Spectrometry — Mass and Fragmentation
Electron-impact ionization removes one electron to give the molecular ion M+ (m/z = molecular weight). Fragment peaks reveal substructure: e.g., m/z 43 (acetyl CH3CO+), m/z 91 (tropylium, benzyl fragment), m/z 29 (CHO+ or ethyl). M+1 grows with carbon count (1.1% per carbon from 13C); M+2 is large for Cl (≈1:3 M:M+2) and Br (≈1:1), diagnostic for halogens.
UV-Vis — Conjugation Depth
UV-Vis measures electronic transitions. Only π→π* and n→π* fall in the accessible 200–800 nm window, so UV-Vis is sensitive to conjugated π systems. The Woodward–Fieser rules estimate λmax: each additional conjugated double bond adds ~30 nm; an auxochrome (–OH, –OR) can add 10–30 nm. A molecule with one isolated C=C absorbs ~170 nm (vacuum UV, not observed); butadiene λmax ≈ 217 nm; β-carotene with 11 conjugated double bonds absorbs ~450 nm (visible orange).
Application to the PA-CAT sample item. The Bulletin highlights a carboxylic acid whose ionization forms a salt that increases water solubility. Spectroscopically, the deprotonated carboxylate (COO⁻) shows a strong, broad IR band at ~1550–1610 cm⁻¹ (asymmetric stretch) and loses the broad O–H of the acid; in 13C NMR the carbonyl carbon shifts slightly upfield from ~175 to ~175–180 ppm. Recognizing these signatures confirms that salt formation occurred.
Combined-Spectroscopy Structure Elongation
PA-CAT items often present data from two or more techniques and ask for a structure. A reliable workflow:
- Molecular formula (from mass spectrometry) → calculate the degree of unsaturation (DU) = (2C + 2 + N − H − X) / 2; DU ≥ 4 suggests an aromatic ring.
- IR → identify functional groups (O–H, N–H, C=O, C≡N).
- 1H NMR → count proton environments, integration ratios, splitting by the n+1 rule, and chemical shift.
- 13C NMR / DEPT → count unique carbons and the number of hydrogens on each.
Worked example: A compound C4H8O2 (DU = 1) shows a strong IR band at 1730 cm⁻¹ (a C=O, consistent with an ester or carboxylic acid), a 1H NMR singlet near 2.0 ppm integrating for 3H, and a quartet near 4.1 ppm (2H) paired with a triplet near 1.2 ppm (3H). The 2H quartet plus 3H triplet signals a –OCH2CH3 ethyl group (the CH2 and CH3 split each other); the 3H singlet at 2.0 ppm is a –COCH3; the C=O together with a C–O indicates an ester. The structure is ethyl acetate, CH3COOCH2CH3. The PA-CAT rewards assembling all four data types rather than over-relying on a single spectrum.
A compound shows a strong IR peak near 1710 cm⁻¹ and a broad absorption spanning 2500–3300 cm⁻¹. Which functional group is most consistent with these data?
In a 1H NMR spectrum, a proton signal appears as a quartet. How many chemically non-equivalent neighboring protons cause this splitting, assuming first-order coupling?