Structure elucidation from formula, IR and NMR
In the exam all the data come together: formula, IR, ¹³C and ¹H. With a fixed method you find the structure instead of guessing.
Rules and values
- 1. DBE from the formula: rings, double bonds, a benzene ring?
- 2. IR: which functional group (O–H, N–H, C=O, C≡N, C–O)?
- 3. ¹³C: how many signals, so how much symmetry?
- 4. ¹H: build fragments from signals, integrals and splitting (ethyl: q + t, isopropyl: sept + d, tert-butyl: s 9 H)
- 5. Join the fragments using the positions (next to O, C=O or a ring) and check against all data
Your turn
Three problems on this topic. Type whole numbers, choose splitting and structures.
Task
A compound with the formula C₄H₉Cl shows this ¹H NMR spectrum. Which structure fits?
Formula: C₄H₉Cl
¹H NMR spectrum
| δ (ppm, approx.) | Splitting | Integral |
|---|---|---|
| 3.5 | doublet (d) | 2 H |
| 2 | multiplet (m) | 1 H |
| 0.9 | doublet (d) | 6 H |
Structure
Worked example
Structure elucidation: formula C₄H₈O₂, 4 signals in the ¹³C NMR, plus IR and ¹H NMR data. Identify the compound.
Formula: C₄H₈O₂
IR spectrum
- 2500–3300 cm⁻¹ (very broad)
- 2850–2960 cm⁻¹
- 1700–1725 cm⁻¹ (strong)
- 1000–1300 cm⁻¹ (strong)
¹H NMR spectrum
| δ (ppm, approx.) | Splitting | Integral |
|---|---|---|
| 10.5–12.5 | broad singlet (br s) | 1 H |
| 2.3 | triplet (t) | 2 H |
| 1.6 | sextet (sext) | 2 H |
| 0.9 | triplet (t) | 3 H |
Solution
- 1.DBE = C − (H + X)/2 + N/2 + 1 = 4 − (8 + 0)/2 + 0/2 + 1 = 1
- 2.DBE 1: exactly one double bond (usually C=O here) or one ring.
- 3.IR: carboxylic acid (2500–3300 cm⁻¹).
- 4.¹³C: 4 signals for 4 C atoms.
- 5.δ ≈ 10.5–12.5 ppm, br s, 1 H: COOH acid H, broad and exchangeable
- 6.δ ≈ 2.3 ppm, t, 2 H: CH₂ next to the COO group, 2 neighbouring H
- 7.δ ≈ 1.6 ppm, sext, 2 H: CH₂ with COO one bond further (shifted a little), 5 neighbouring H
- 8.δ ≈ 0.9 ppm, t, 3 H: CH₃ in the alkyl region, 2 neighbouring H
- 9.→ butanoic acid (butyric acid)
- 10.ethyl acetate: 3 signals instead of 4.
- 11.methyl propanoate: 3 signals instead of 4.
- 12.2-methylpropanoic acid (isobutyric acid): 3 signals instead of 4.
Typical mistake: Looking at each piece of data on its own. DBE and IR narrow down the class first; then the ¹H NMR (integral, splitting, position) decides between the isomers.
A compound with the formula C₅H₁₂O shows this ¹H NMR spectrum. Which structure fits?
Formula: C₅H₁₂O
¹H NMR spectrum
| δ (ppm, approx.) | Splitting | Integral |
|---|---|---|
| 3.3 | singlet (s) | 3 H |
| 1.2 | singlet (s) | 9 H |
Solution
- 1.δ ≈ 3.3 ppm, s, 3 H: CH₃ on the ether oxygen, 0 neighbouring H
- 2.δ ≈ 1.2 ppm, s, 9 H: CH₃ (3 equivalent) with O one bond further (shifted a little), 0 neighbouring H
- 3.→ tert-butyl methyl ether (MTBE)
- 4.3-methylbutan-1-ol (isoamyl alcohol): 5 signals instead of 2.
- 5.ethyl propyl ether (1-ethoxypropane): 5 signals instead of 2.
- 6.2,2-dimethylpropan-1-ol (neopentyl alcohol): 3 signals instead of 2.
Typical mistake: Guessing when the pattern is the same. Ethyl acetate and methyl propanoate both show s, q, t: what decides is whether the CH₂ (q) sits at ≈ 4.1 ppm on oxygen or at ≈ 2.3 ppm next to C=O.
Common questions
Ethyl acetate or methyl propanoate: how do I tell them apart?
Both show s (3 H), q (2 H) and t (3 H). In ethyl acetate the CH₂ sits on oxygen (≈ 4.1 ppm) and the singlet is the acetyl CH₃ (≈ 2.0 ppm). In methyl propanoate the CH₂ is next to C=O (≈ 2.3 ppm) and the singlet is the OCH₃ (≈ 3.7 ppm).