1. Organic Synthesis
Organic synthesis now has its own page: Organic Synthesis.
2. Carbon-13 (¹³C) NMR Spectroscopy
Carbon-13 is a stable isotope of carbon with nuclear spin (unlike the abundant carbon-12, which has zero spin). ¹³C NMR spectroscopy provides direct structural information about the carbon skeleton:
¹³C spectra are simpler than ¹H spectra. At this level there is no splitting in a ¹³C spectrum. Each carbon environment gives one peak.
- Number of peaks = Number of non-equivalent carbon environments. If a molecule has 6 carbons but exhibits only 3 peaks, molecular symmetry is present.
- Chemical shift (δ in ppm): Indicates the electronic environment of the carbon atom. These are typical values. In the exam, always use the data sheet you are given:
- Aliphatic alkyl carbons (C-C): 0 to 50 ppm.
- Carbons bonded to oxygen (C-O in alcohols, ethers and esters): 50 to 90 ppm.
- Alkene and aromatic carbons (C=C, aromatic ring): 100 to 160 ppm.
- Carbonyl carbons (C=O in esters, acids, amides): 160 to 185 ppm.
- Carbonyl carbons (C=O in aldehydes, ketones): 190 to 220 ppm.
3. High-Resolution ¹H NMR Spectrum Diagram
The spectrum below displays the high-resolution ¹H NMR spectrum of ethyl ethanoate (CH₃COOCH₂CH₃), illustrating chemical shift positions, integration values, and splitting patterns:
Relative peak areas (integration) are 3 : 2 : 3; peak heights here are not to scale.
Ethyl ethanoate, CH₃COOCH₂CH₃: the ¹H NMR peaks (typical values)
| Group | δ / ppm | Splitting | Integration |
|---|---|---|---|
| CH₃ next to C=O | about 2.0 | singlet | 3 |
| OCH₂ | about 4.1 | quartet | 2 |
| CH₃ of the ethyl | about 1.3 | triplet | 3 |
- The CH₃ next to C=O has no H on the next carbon, so it is a singlet.
- The OCH₂ is next to a CH₃ (3 H), so it is a quartet (3 + 1 = 4). It has a high shift because it is bonded to O.
- The ethyl CH₃ is next to a CH₂ (2 H), so it is a triplet (2 + 1 = 3).
4. Proton (¹H) NMR Spectroscopy Principles
High-resolution proton NMR is the most powerful spectroscopic tool for organic structure determination. Every spectrum provides four crucial pieces of information:
| Spectral Feature | Physical Information Revealed | Example Deduction |
|---|---|---|
| 1. Number of Signals | Number of non-equivalent proton (hydrogen) environments. | Propan-1-ol has 4 signals (-CH₃, -CH₂-, -CH₂-O, -OH). Propan-2-ol has 3 signals (two equivalent -CH₃ groups). |
| 2. Chemical Shift (δ / ppm) | The chemical environment and shielding/deshielding by nearby electronegative atoms relative to TMS at 0 ppm. | Protons on -CH₃ appear at ~0.9 ppm; protons on -CH₂-O- appear deshielded at ~4.0 ppm. |
| 3. Integration Trace | The relative ratio of protons responsible for each signal. | Integration ratio of 3 : 2 : 1 indicates 3 protons in one environment, 2 in another, and 1 in the third. |
| 4. Splitting Pattern | The number of non-equivalent protons on immediately adjacent carbon atoms. | A triplet indicates 2 adjacent protons (2 + 1 = 3). A quartet indicates 3 adjacent protons (3 + 1 = 4). |
Typical ¹H chemical shifts
These are typical values, matching the examples on this page. In the exam, always use the data sheet you are given.
| Type of proton | Typical δ / ppm |
|---|---|
| R-CH₃ in an alkyl chain | about 0.9 to 1.3 |
| CH₃ or CH₂ next to C=O | about 2.0 to 2.6 |
| H on a carbon bonded to O (alcohol, ether or ester O-CH) | about 3.3 to 4.1 |
| H on a benzene ring | about 7 to 8 |
| Aldehyde H (CHO) | about 9 to 10 |
Why TMS and Deuterated Solvents (CDCl₃) are Used
- Solvents: use a deuterated solvent such as CDCl₃, or CCl₄. Neither contains ¹H atoms, so the solvent gives no peak in a ¹H NMR spectrum. The solvent dissolves the sample.
- Standard: tetramethylsilane (TMS), Si(CH₃)₄. Its peak is set at δ = 0.
- All 12 H atoms are equivalent, so TMS gives a single peak.
- It is inert and volatile (b.p. 27 °C), so it is easy to remove from the sample afterwards.
- Its peak lies away from most other peaks.
5. Spin-Spin Splitting & The (n + 1) Rule
The magnetic field experienced by a proton is perturbed by the nuclear spin magnetic fields of non-equivalent protons on adjacent carbon atoms. This spin-spin coupling causes peaks to split according to the (n + 1) rule, where n is the number of non-equivalent protons on adjacent carbons:
| Adjacent Protons (n) | Splitting Pattern (n + 1) | Relative Peak Area Ratios | Typical Structural Fragment |
|---|---|---|---|
| 0 | Singlet | 1 | Isolated methyl group (e.g. -CO-CH₃, -O-CH₃) or quaternary carbon. |
| 1 | Doublet | 1 : 1 | Proton adjacent to a -CH- group. |
| 2 | Triplet | 1 : 2 : 1 | Proton adjacent to a -CH₂- group (e.g. the -CH₃ of an ethyl group: -CH₂-CH₃). |
| 3 | Quartet | 1 : 3 : 3 : 1 | Proton adjacent to a -CH₃ group (e.g. the -CH₂- of an ethyl group: -CH₂-CH₃). |
9620 uses the n + 1 rule to explain doublets, triplets and quartets only, in aliphatic compounds. A singlet means there are no H atoms on the neighbouring carbons. The integration ratio gives the relative number of H atoms in each environment.
Whenever you see a triplet (integrating to 3H) paired with a quartet (integrating to 2H) in a ¹H NMR spectrum, you have identified an ethyl group (-CH₂-CH₃)!
6. D₂O Exchange for Identifying -OH and -NH Protons (Beyond the OxfordAQA specification)
Beyond the OxfordAQA specification: 9620 does not require D₂O exchange. This section is background only.
Protons bonded to electronegative oxygen (-OH in alcohols and carboxylic acids) or nitrogen (-NH in amines and amides) undergo rapid proton exchange in solution. Consequently, they usually appear as broad singlets that do NOT participate in spin-spin splitting.
- Run the initial ¹H NMR spectrum and record the peak position.
- Add a few drops of deuterium oxide (D₂O) and shake the NMR tube.
- The acidic labile proton rapidly exchanges with deuterium:
R-OH + D₂O ⇌ R-OD + HOD - Because deuterium produces no signal in ¹H NMR, the original -OH (or -NH) peak completely disappears from the spectrum!
7. Worked Spectral Deductions
IR: Sharp strong absorption at 1740 cm⁻¹.
¹H NMR:
- δ 1.25 ppm: Triplet, integration = 3H
- δ 2.05 ppm: Singlet, integration = 3H
- δ 4.10 ppm: Quartet, integration = 2H
Deduce the structure of the ester and justify your answer.
Step 1: Functional group identification:
IR absorption at 1740 cm⁻¹ corresponds to a carbonyl group (C=O) of an ester.
Step 2: Analyse ¹H NMR signals:
- Triplet at 1.25 ppm (3H) and Quartet at 4.10 ppm (2H): Proves an ethyl group (-CH₂-CH₃). Because the -CH₂- quartet is deshielded at 4.10 ppm, it must be bonded directly to the ester oxygen: -O-CH₂-CH₃.
- Singlet at 2.05 ppm (3H): Represents a methyl group with zero adjacent protons, bonded to the carbonyl carbon: CH₃-C=O.
Step 3: Assemble the molecule:
CH₃-CO-O-CH₂-CH₃ (Ethyl ethanoate)
(a) Pentan-2-one: CH₃-CO-CH₂-CH₂-CH₃
(b) Pentan-3-one: CH₃-CH₂-CO-CH₂-CH₃
Solution:
- (a) Pentan-2-one: Asymmetrical molecule. All 5 carbons are in different electronic environments (C₁ methyl, C₂ carbonyl, C₃ methylene, C₄ methylene, C₅ methyl). Expected: 5 peaks.
- (b) Pentan-3-one: Symmetrical molecule with a plane of symmetry through the C₃ carbonyl group. Carbons 1 and 5 are identical; Carbons 2 and 4 are identical; Carbon 3 is unique. Expected: 3 peaks.
Go deeper
These UK AQA pages cover the same ideas in more detail. They may include content that is not in 9620.
8. Practice Questions
Show Answer & Mark Scheme
Answer:
- All 12 H atoms are equivalent, so it gives a single peak (1).
- It is inert and volatile, so it is easy to remove from the sample (1).
- Its peak (δ = 0) lies away from most other peaks (1).
Show Answer & Mark Scheme
Answer: CDCl₃ contains no ¹H atoms, so it gives no peak in the ¹H NMR spectrum (1).
Show Answer & Mark Scheme
Answer:
- CH₃: triplet, integration 3 (1).
- CH₂: quartet, integration 2 (1).
- The CH₂ peak has the higher chemical shift, because the CH₂ is bonded to the electronegative Cl atom (1).
Show Answer & Mark Scheme
Answer:
- (a) 3 peaks: all three carbons are in different environments (1).
- (b) 2 peaks: the two CH₃ carbons are equivalent (1).