Reference
Structure elucidation with IR, NMR & MS: a practical guide
A practical introduction, meant to be read alongside the exercises, to how MS, IR and NMR are combined to determine the structure of an unknown. The three methods have different strengths; alone each often lacks a decisive answer, but combined they cover each other's weaknesses. This is a summary, not a substitute for the textbooks themselves.
1. The combined workflow
Typically you build up information in this order. First use MS for the molecular weight and, where possible, composition (isotope pattern, nitrogen rule), then compute the degree of unsaturation. Next use IR to narrow down functional groups (O–H, N–H, C=O, C≡N — present or absent). Finally use NMR to assemble the carbon skeleton, the environment of each H/C and their connectivity, and check that everything is consistent with the MS fragmentation. The key is not to decide from any one method but to verify that all of them point to the same structure.
2. Degree of unsaturation (DBE)
Given a molecular formula, you can count the total number of rings and multiple bonds. For CcHhNnOoXx (X = halogen):
DBE = c − (h + x)/2 + n/2 + 1
Oxygen and sulfur do not affect the formula. DBE ≥ 4 often signals an aromatic ring (a benzene ring is 1 ring + 3 double bonds = 4). C=O, C=C and C≡N contribute 1, 1 and 2; C≡C contributes 2. Checking that the unsaturation found by IR/NMR matches the DBE total helps catch what you missed.
3. What MS tells you
The m/z of the molecular ion M⁺ is the nominal molecular weight. Its intensity varies with structure — strong for stable molecules such as aromatics, weak (sometimes absent) for branched alcohols. Nitrogen rule: with only C, H, O the molecular weight is even; an odd value means an odd number of nitrogens. Isotopes: chlorine M:M+2 ≈ 3:1, bromine ≈ 1:1, sulfur ~4% at M+2. The M+1 intensity (≈ 1.1% × number of carbons) estimates carbon count. Read part-structures from neutral losses (mass difference from M) and fragment ions (−15=CH₃, −18=H₂O, −28=CO, −31=OCH₃, −45=COOH / m/z 43=CH₃CO⁺, 31=CH₂OH⁺, 30=CH₂NH₂⁺, 77=C₆H₅⁺, 91=tropylium C₇H₇⁺). The McLafferty rearrangement (a characteristic even-mass ion from carbonyls with a γ-hydrogen) and tropylium from a benzylic position are diagnostic.
4. What IR tells you
IR quickly judges the presence or absence of functional groups. 4000–1500 is the functional-group region, 1500–400 the fingerprint region. Scan the high-wavenumber side first for O–H/N–H (3200–3600), triple bonds (2100–2260) and the carbonyl (1650–1780). The "present/absent" call is especially powerful: the absence of a broad O–H, for instance, rules out an alcohol or acid. Whether there is a C=O, and if so what its position implies, is IR's main job (next section).
5. What NMR tells you
¹H-NMR gives part-structures and their connectivity from chemical shift (electronic environment), integration (H-number ratios) and multiplicity (number of neighbouring H, the n+1 rule). ¹³C-NMR (proton-decoupled) shows symmetry directly — the number of signals equals the number of inequivalent carbons — and reveals quaternary carbons such as carbonyls. Learning ¹H patterns like "ethyl (t+q)", "isopropyl (d+septet)" and "monosubstituted benzene (5H near δ7)" ties a signal straight to a substructure.
6. Why the same group's peak moves
Reference values like "methyl at δ0.9" or "C=O at 1715" are only a starting point. The same functional group and the same bond can shift clearly with its environment. Understanding why they move turns the deviation itself into a structural clue.
Electronic effects (inductive and resonance)
The nearer an electronegative atom or electron-withdrawing group, the further downfield (larger δ) the ¹H and ¹³C. Even a single methyl changes a lot with what it is bonded to.
| environment | ¹H δ (approx.) | why |
|---|
| CH₃–C (alkane) | ≈0.9 | near baseline |
| CH₃–C=C / CH₃–Ar | ≈1.6 / 2.3 | π anisotropy, mild deshielding |
| CH₃–C=O (acetyl) | ≈2.1 | carbonyl induction/anisotropy |
| CH₃–N | ≈2.2–3.0 | nitrogen electronegativity |
| CH₃–O (methoxy) | ≈3.3–3.9 | strong oxygen electronegativity |
| CH₃–NO₂ | ≈4.3 | very strong withdrawal |
The effect is additive: more adjacent electronegative groups shift it further downfield (CH₂Cl₂ > CH₃Cl; CHCl₃ further still).
Conjugation lowers a carbonyl
An IR C=O with conjugation (an adjacent C=C or ring) delocalises electrons, lowers the bond order and moves 15–40 cm⁻¹ lower. Against a saturated ketone ≈1715, aryl ketones and α,β-unsaturated ketones are ≈1680–1690. This is why benzaldehyde's C=O (≈1700) is lower than an aliphatic aldehyde's (≈1725).
Ring strain raises a carbonyl
A ring C=O moves higher as the ring shrinks: cyclohexanone ≈1715, cyclopentanone ≈1745, cyclobutanone ≈1780. Ring lactones and anhydrides follow the same trend.
| carbonyl type | C=O / cm⁻¹ |
|---|
| acyl halide | ≈1800 |
| anhydride | ≈1820 and 1760 (2 bands) |
| ester | ≈1735–1750 |
| aldehyde | ≈1725 |
| ketone (saturated) | ≈1715 |
| carboxylic acid (dimer) | ≈1710 (with broad O–H) |
| amide | ≈1630–1690 |
| conjugated / aromatic C=O | −15 to 40 from above |
Hydrogen bonding moves O–H and N–H (concentration/solvent dependent)
A free O–H is sharp and high (≈3600 in dilute solution), but hydrogen bonding makes it broad and lower (≈3200–3550). A carboxylic acid forms dimers and gives a very broad band spanning 2500–3300. In ¹H, O–H, N–H and COOH are exchangeable and their position and width change a lot with concentration, solvent, temperature and D₂O exchange. Intramolecular hydrogen bonding (e.g. the enol of a β-diketone) gives extreme downfield shifts (δ11–16).
Magnetic anisotropy
The ring current of an aromatic ring deshields the hydrogens on its outside, pushing aromatic H to δ7–8 and aldehyde H to δ9–10. Conversely, the π electrons of an alkyne shield the terminal H, keeping it at δ2–3. ¹³C behaves similarly; shifts cannot be explained by electronegativity alone.
7. Common pitfalls
- The highest m/z is not always M⁺ — don't mistake an isotope peak (M+1, M+2) or an M+1 adduct for M⁺.
- The same m/z can have different origins — e.g. m/z 29 is CHO⁺ or C₂H₅⁺; m/z 43 is CH₃CO⁺ or C₃H₇⁺. Judge with the neighbouring peaks.
- Weak bands at 2200–2260 — separate C≡N (medium) from C≡C (weak; symmetric alkynes vanish) by position and intensity.
- Exchangeable protons are variable — don't treat O–H/N–H δ as fixed; confirm with integration and D₂O exchange.
- Overlapping aromatic shifts — the 5H of a monosubstituted benzene often overlap near δ7.2, limiting how well multiplicity reveals the substitution.
8. A quick checklist
- Fix M⁺ (molecular weight) in the MS; get composition clues from the nitrogen rule and isotopes.
- Once the formula is narrowed, compute the degree of unsaturation (DBE) to anticipate rings and multiple bonds.
- Sort functional groups by "present/absent" in the IR (O–H, N–H, C=O, C≡N). Use the C=O position to narrow the type.
- Read symmetry from the ¹³C count, and part-structures and connectivity from ¹H integration and multiplicity.
- Check that any deviation from reference values is explained by conjugation, ring strain, hydrogen bonding or anisotropy.
- Re-read to confirm the MS fragmentation (−15, −18, 43, 91, …) is consistent with the structure you built.
You can practise these judgements in Drill and Solve. The Learn tab summarises each method with spectrum figures.