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While ¹H NMR focuses on hydrogen environments, ¹³C NMR provides complementary information about the carbon skeleton. Together, ¹H and ¹³C NMR give a comprehensive picture of a molecule's framework. ¹³C NMR is particularly useful for determining how many distinct carbon environments exist and, through chemical shift values, what types of carbons are present.
The most common carbon isotope, ¹²C, has zero nuclear spin and is therefore NMR-inactive. Only ¹³C (spin = ½) produces an NMR signal. Since ¹³C has a natural abundance of only 1.1%, ¹³C NMR is much less sensitive than ¹H NMR — signals are weaker and require longer acquisition times or more concentrated samples.
Key differences from ¹H NMR:
| Feature | ¹H NMR | ¹³C NMR |
|---|---|---|
| Natural abundance | 99.98% (¹H) | 1.1% (¹³C) |
| Sensitivity | High | Low (~1/6000 of ¹H) |
| Chemical shift range | 0–12 ppm | 0–220 ppm |
| Splitting in routine spectra | Yes (n+1 rule) | No (proton-decoupled) |
| Integration | Quantitative (proportional to number of H) | Not reliably quantitative in routine spectra |
In routine ¹³C NMR spectra, a technique called broadband proton decoupling is used. This involves irradiating the sample at all ¹H frequencies simultaneously, which eliminates ¹³C–¹H coupling. The result is that each carbon environment appears as a single line (a singlet), regardless of how many hydrogens are attached.
Without decoupling, ¹³C spectra would show complex splitting patterns (a CH₃ carbon would be split into a quartet by its three attached hydrogens, for example). Decoupling greatly simplifies the spectrum and makes it much easier to count the number of distinct carbon environments.
The downside is that you lose information about how many hydrogens are attached to each carbon. This information can be recovered using DEPT spectra (discussed below).
The chemical shift scale in ¹³C NMR is much wider than in ¹H NMR, spanning from about 0 to 220 ppm (referenced to TMS at δ = 0). This wider range means there is less overlap between peaks, making it easier to distinguish different carbon environments.
| Type of carbon | δ / ppm | Example compound | Typical δ |
|---|---|---|---|
| R–CH₃ (primary alkyl) | 5–30 | Ethane CH₃CH₃ | 6 |
| R–CH₂–R (secondary alkyl) | 15–55 | Propane CH₃CH₂CH₃ | 16 |
| R₃CH (tertiary alkyl) | 20–60 | 2-Methylpropane | 25 |
| R₄C (quaternary alkyl) | 30–45 | Neopentane | 32 |
| C–Cl | 25–50 | Chloromethane | 25 |
| C–Br | 25–65 | Bromomethane | 10 |
| C–N (amine) | 30–65 | Methylamine | 27 |
| C–O (alcohol, ether) | 50–90 | Methanol CH₃OH | 50 |
| C≡C (alkyne) | 65–90 | Ethyne | 72 |
| C=C (alkene) | 100–150 | Ethene CH₂=CH₂ | 123 |
| Aromatic C–H | 110–145 | Benzene | 128 |
| Aromatic C (substituted) | 120–160 | Toluene ipso-C | 137 |
| C=O (acid, ester, amide) | 160–185 | Ethanoic acid COOH | 178 |
| C=O (aldehyde) | 190–205 | Ethanal CHO | 200 |
| C=O (ketone) | 195–220 | Propanone C=O | 206 |
Notice the diagnostic separation:
Just as in ¹H NMR, the number of peaks in a ¹³C NMR spectrum equals the number of distinct carbon environments. Equivalent carbons (those related by symmetry) give a single peak.
Ethanol (CH₃CH₂OH): Two carbon environments — the CH₃ carbon and the CH₂ carbon. Two peaks.
Propanone (CH₃COCH₃): Two carbon environments — the two equivalent CH₃ carbons (one environment) and the C=O carbon (one environment). Two peaks.
Ethanoic acid (CH₃COOH): Two carbon environments — CH₃ and COOH. Two peaks.
Benzene (C₆H₆): All six carbons are equivalent. One peak (at about δ = 128 ppm).
Methylbenzene (toluene, C₆H₅CH₃): Five carbon environments — the CH₃ carbon, and four different ring carbon types (C-1 attached to CH₃, C-2/C-6 which are equivalent, C-3/C-5 which are equivalent, and C-4). Five peaks.
Para-dimethylbenzene (1,4-C₆H₄(CH₃)₂): Three carbon environments — the two equivalent CH₃ carbons, the two equivalent C–CH₃ ring carbons, and the four equivalent unsubstituted ring carbons. Three peaks.
Three isomers of dimethylbenzene (xylene) can be distinguished by ¹³C NMR peak count:
| Isomer | Structure | Number of ¹³C peaks | Carbon environments |
|---|---|---|---|
| Ortho-xylene | 1,2-C₆H₄(CH₃)₂ | 4 | CH₃, C-1/C-2, C-3/C-6, C-4/C-5 |
| Meta-xylene | 1,3-C₆H₄(CH₃)₂ | 5 | CH₃, C-1/C-3, C-2, C-4/C-6, C-5 |
| Para-xylene | 1,4-C₆H₄(CH₃)₂ | 3 | CH₃, C-1/C-4, C-2/C-3/C-5/C-6 |
Simply counting the number of ¹³C peaks distinguishes all three isomers. This is a classic exam question.
DEPT (Distortionless Enhancement by Polarisation Transfer) is an experiment that reveals how many hydrogen atoms are attached to each carbon:
By comparing the DEPT-135 spectrum with the standard broadband-decoupled ¹³C spectrum, you can classify each carbon as CH₃, CH₂, CH, or quaternary C.
Some spectrometers provide edited DEPT or DEPT-90 spectra:
| Peak in standard ¹³C? | Peak in DEPT-135? | Direction in DEPT-135? | Carbon type |
|---|---|---|---|
| Yes | Yes | Positive (up) | CH₃ or CH |
| Yes | Yes | Negative (down) | CH₂ |
| Yes | No | — | Quaternary C (no H attached) |
To distinguish CH₃ from CH, compare DEPT-135 with DEPT-90: CH appears in both (positive), while CH₃ appears in DEPT-135 (positive) but not in DEPT-90.
Because of proton decoupling, you will not observe splitting patterns in routine ¹³C NMR spectra. Each carbon environment appears as a single vertical line. Do not apply the n+1 rule to ¹³C spectra.
Carbon-carbon coupling (¹³C–¹³C) is not observed in routine spectra because the probability of two adjacent ¹³C atoms is extremely low (1.1% × 1.1% ≈ 0.012%).
Unlike ¹H NMR, the peak heights (or areas) in routine proton-decoupled ¹³C NMR spectra are not directly proportional to the number of carbons. This is because the Nuclear Overhauser Effect (NOE) enhancement from proton decoupling varies for different types of carbons, and different carbons relax at different rates.
Therefore, you should not use peak heights to determine the number of carbons in each environment. Instead, count the number of peaks and use other information (molecular formula, ¹H NMR) to determine the structure.
A compound with molecular formula C₃H₆O shows three peaks in its ¹³C NMR spectrum:
Analysis:
If it were propanone (CH₃COCH₃), there would be only two ¹³C peaks (the two CH₃ groups are equivalent), not three. The three peaks confirm propanal.
Mistake 1: Trying to use peak heights to count carbons. Unlike ¹H NMR, peak heights in ¹³C NMR are NOT proportional to the number of carbons. A quaternary carbon often gives a shorter peak than a CH₃ carbon, even though each represents one carbon environment.
Mistake 2: Applying the n+1 rule to ¹³C spectra. Routine ¹³C spectra are proton-decoupled — there is no splitting. Each carbon appears as a singlet.
Mistake 3: Forgetting that equivalent carbons give one peak. If a molecule has 6 carbons but only 3 ¹³C peaks, it has significant symmetry. Use this information to narrow down the structure.
Mistake 4: Confusing the ¹³C chemical shift regions. Remember: alkyl C < C–O/C–N < aromatic/alkene C < acid/ester C=O < aldehyde/ketone C=O.
¹³C NMR reveals the number of distinct carbon environments and, through chemical shift values, the types of carbons present. Routine spectra are proton-decoupled (no splitting), and DEPT experiments reveal how many H atoms are attached to each carbon. Combined with ¹H NMR and mass spectrometry, ¹³C NMR provides a powerful tool for complete structure determination.
Edexcel 9CH0 specification, Topic 19 — Modern Analytical Techniques II, sub-strand 19.1 introduces ¹³C nuclear magnetic resonance (NMR) spectroscopy: the principle that ¹³C nuclei in different chemical environments resonate at different frequencies; the use of TMS as the chemical-shift reference at δ = 0 ppm; characteristic chemical-shift ranges for carbon environments (data booklet: 5–55 ppm sp³ alkyl C; 60–90 sp³ C–O; 100–150 alkene/aromatic C; 110–160 C in C–N; 170–185 ester/acid carbonyl; 190–220 ketone/aldehyde carbonyl); and the convention that proton-decoupled ¹³C spectra show a single peak per chemically distinct carbon environment, with integration not normally used for quantitative interpretation due to relaxation differences (refer to the official specification document for exact wording). Examined on Paper 2 and Paper 3.
Question (7 marks):
(a) Predict the number of peaks in the ¹³C NMR spectrum of each compound: propan-1-ol, propan-2-one (acetone), methylbenzene (toluene). (3)
(b) For each compound, suggest the approximate chemical shift of each peak. (3)
(c) State why ¹³C NMR is usually run in proton-decoupled mode. (1)
Solution with mark scheme:
(a)
(b)
(c) Without proton decoupling, every ¹³C signal would split via ¹³C–¹H coupling (^1J(C–H) ~125 Hz, large), producing complicated multiplets. Proton decoupling collapses each ¹³C peak to a singlet, simplifying interpretation. A1.
Total: 7 marks (M6 A1).
Question (6 marks): An unknown organic compound E has molecular formula C₄H₈O₂ and shows three peaks in its ¹³C NMR spectrum at approximately δ = 173, 60, and 21 ppm.
(a) Calculate the degree of unsaturation. (1)
(b) Identify the carbon environment corresponding to each peak. (3)
(c) Suggest a structure for E and explain how the ¹³C peak count supports your assignment. (2)
Mark scheme decomposition by AO:
(a)
(b)
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