📘 CodingMarble Learn

NMR Spectroscopy: Carbon-13 and Proton NMR

Nuclear magnetic resonance (NMR) uses a strong magnet and radio waves to see the different environments of ¹H or ¹³C atoms in a molecule. Each environment gives a peak at its own chemical shift δ (ppm), measured from TMS at δ = 0. In ¹³C NMR, the number of peaks = number of carbon environments. In ¹H NMR, the peak area (integration) gives the ratio of H atoms, and the n + 1 rule tells how many H are on the neighbouring carbon.

🎬 Step-by-step story

  1. Some nuclei, like ¹H and ¹³C, spin and act like tiny magnets. In a strong magnetic field B₀ they line up with it or against it. Radio waves of exactly the right energy flip them. The flip absorbs energy: that is the NMR signal.
  2. Every spectrum needs a zero. We add TMS, Si(CH₃)₄. All its 12 H atoms are the same, so it gives one sharp peak. We set it at δ = 0. Chemical shift δ is measured from there in ppm, and it grows to the left.
  3. Carbon-13 NMR of ethyl ethanoate: four different carbons, so four peaks. Each colour on the molecule matches one peak. The C=O carbon is far left, near 171 ppm, because oxygen pulls electrons away from it.
  4. Proton NMR of ethanol: three kinds of H, three peaks. The peak heights (areas) are in the ratio 3 : 2 : 1, the same as CH₃ : CH₂ : OH.
  5. Now the fine detail. A peak splits into n + 1 lines when there are n H on the next carbon. CH₃ next to CH₂ (2 H) is a triplet. CH₂ next to CH₃ (3 H) is a quartet. OH shows a singlet.
  6. Your turn. Pick a molecule. Switch between ¹³C and ¹H. Count peaks, read the areas and splitting, and work out the structure.

Tip: drag the 3D scene to turn it. Use two fingers to zoom.

🤔 Common doubts, cleared

Why do we need a magnet at all?

Without a field, the two spin states of a nucleus have the same energy, so nothing absorbs. The field B₀ splits them, and radio waves can then cause a flip. Step 1.

Why does δ increase to the left?

It is a convention: TMS (δ = 0) is drawn at the right edge, and more deshielded nuclei are plotted further left. Step 2.

Why does the C=O carbon have such a large shift?

Oxygen pulls electrons away from that carbon, so it is less shielded. Look at the far-left peak in step 3.

Why don't ¹³C peaks split like ¹H peaks?

¹³C is rare (about 1%), so two ¹³C atoms are almost never next to each other, and spectra are run in a way that removes H splitting. Each carbon shows as one line. Step 3.

Which H do I count for splitting?

Only the H on the carbon next door (not on the same carbon, and usually not O–H). CH₃ next to CH₂ → 2 + 1 = 3 lines. Step 5.

Why is the OH peak a singlet?

The O–H hydrogen swaps quickly between molecules, so it does not couple with neighbours. In ethanol it shows as a singlet. Step 5.

How NMR works: chemical shift and TMS

Nuclei with an odd number of protons or neutrons, such as 1H and 13C, have spin. In a strong magnetic field they can line up with the field (lower energy) or against it (higher energy). Radio-frequency radiation of the right energy flips them: this absorption is resonance.

Electrons around a nucleus shield it from the field. An electronegative atom nearby (O, N, Cl) pulls electrons away, deshielding the nucleus, so it absorbs at a higher chemical shift δ.

δ is measured in ppm relative to tetramethylsilane (TMS), Si(CH₃)₄, set at δ = 0. TMS is used because:

Samples are dissolved in solvents with no ¹H, such as CDCl₃ or CCl₄, so the solvent gives no proton peak.

Carbon-13 NMR

In a 13C spectrum, each peak is one carbon environment. Carbons in identical positions (by symmetry) give one peak. Peaks are single lines (no splitting is shown) and their heights are not used for counting.

Typical shift ranges (approximate, check your data sheet):

Example: propanone CH₃COCH₃ has only 2 peaks (the two CH₃ are identical): about 30 and 206 ppm.

Proton NMR and splitting

A 1H spectrum tells four things:

  1. Number of peaks = number of H environments.
  2. Chemical shift = type of H (e.g. R–CH₃ 0.7–1.2; H–C–C=O 2.1–2.6; H–C–O 3.3–4.3; R–OH 1–5, variable; aldehyde CHO 9–10; COOH 10–12).
  3. Integration (relative area, often shown as an integration trace) = ratio of H in each environment.
  4. Splitting (spin–spin coupling): a peak is split into n + 1 lines by n equivalent H on the adjacent carbon. 0 → singlet, 1 → doublet, 2 → triplet, 3 → quartet.

H on O–H usually show a singlet and do not split neighbours, because they exchange quickly. A quick test: shake with D₂O; the O–H peak disappears.

A classic pattern: an ethyl group CH₃CH₂– gives a triplet (3H) and a quartet (2H).

Solving a structure, step by step (Try it)

  1. Use the molecular formula to count H and C.
  2. Count ¹³C peaks → carbon environments; check for C=O above 160.
  3. In ¹H: count peaks, read integration ratios, convert to actual H numbers.
  4. Use n + 1 splitting to find neighbours: triplet + quartet → CH₃CH₂–; singlet 3H near 2.1 → CH₃C=O.
  5. Join the pieces and check every peak fits.

Try it: in free play, pick ethyl ethanoate, read the ¹H peaks (2.0 singlet, 4.1 quartet, 1.3 triplet) and draw the structure on paper before you look at the molecule above the spectrum.

Key formulas and definitions

Worked examples

1. How many peaks are in the ¹³C NMR spectrum of propan-1-ol, CH₃CH₂CH₂OH, and of propan-2-ol, CH₃CH(OH)CH₃?

Step 1: Propan-1-ol: three carbons, all in different positions → 3 peaks. Step 2: Propan-2-ol: the two CH₃ are identical by symmetry; the CH(OH) is different → 2 peaks. Answer: 3 and 2.

2. Why does the C=O carbon of propanone appear near 206 ppm while the CH₃ carbons appear near 30 ppm?

Step 1: Oxygen is very electronegative and pulls electron density from the C=O carbon. Step 2: Less shielding means the nucleus feels more of the field and resonates at higher δ. Answer: deshielded C=O carbon → large δ; CH₃ carbons are shielded → small δ.

3. Predict the ¹H NMR spectrum of chloroethane, CH₃CH₂Cl.

Step 1: Two H environments: CH₃ (3H) and CH₂ (2H). Step 2: CH₃ is next to CH₂ (2 H) → triplet; CH₂ is next to CH₃ (3 H) → quartet. Step 3: CH₂ is next to Cl → larger δ (about 3.6); CH₃ about 1.5. Answer: triplet 3H at δ ≈ 1.5 and quartet 2H at δ ≈ 3.6, ratio 3 : 2.

4. A compound C₃H₆O shows one singlet in its ¹H NMR spectrum at δ 2.1. Identify it.

Step 1: One peak → all 6 H are equivalent. Step 2: δ 2.1 fits H–C–C=O. Step 3: Singlet → no H on the neighbouring carbons (the neighbour is C=O). Answer: propanone, CH₃COCH₃.

5. Integration values for three peaks are 1.5 cm, 1.0 cm and 0.5 cm. The compound has 6 H. How many H in each environment?

Step 1: Ratio 1.5 : 1.0 : 0.5 = 3 : 2 : 1. Step 2: 3 + 2 + 1 = 6, matching the formula. Answer: 3H, 2H and 1H (as in ethanol).

6. An ester C₄H₈O₂ gives ¹H peaks: δ 1.3 (triplet, 3H), δ 2.0 (singlet, 3H), δ 4.1 (quartet, 2H). Find the structure.

Step 1: Triplet 3H + quartet 2H → an ethyl group, CH₃CH₂–. Step 2: The CH₂ is at 4.1 → bonded to O (H–C–O). Step 3: Singlet 3H at 2.0 → CH₃ next to C=O, no H neighbours. Step 4: Join: CH₃–C(=O)–O–CH₂CH₃. Answer: ethyl ethanoate.

Common mistakes

Practice quiz

1. TMS is used in NMR because:
2. How many ¹³C peaks does propanone give?
3. A CH₃ next to a CH₂ in ¹H NMR appears as a:
4. The relative area under ¹H peaks gives:
5. A carbon next to oxygen appears at:

Practice: answer these yourself

Type or choose your answer, then press Check. Use a hint if you are stuck; the full solution appears after you answer.

Frequently asked questions

What is chemical shift in NMR?

The position of a peak, δ in ppm, relative to TMS at 0. It depends on how shielded the nucleus is by electrons; electronegative neighbours increase δ.

What is the n + 1 rule?

A ¹H peak is split into n + 1 lines by n equivalent hydrogens on the adjacent carbon(s).

What is the difference between ¹³C and ¹H NMR?

¹³C NMR counts carbon environments (one line each). ¹H NMR shows hydrogen environments, their relative numbers (integration) and their neighbours (splitting).

Where this is taught

England (GCSE, A level)Year 133.3 Organic chemistry

Learn first

Learn next

Related lessons

All Chemistry lessons