# Principles of ¹H NMR

> CIE A-Level Chemistry · 9701
> Source: https://www.owlsprep.com/study/cie-9701-u28-principles-of-1h-nmr/

This sub-topic covers the fundamental physical principles behind proton nuclear magnetic resonance (¹H NMR), a key analytical technique for determining organic molecule structure in CIE A-Level Chemistry.

**Prerequisites:** [Organic functional groups](https://www.owlsprep.com/study/cie-9701-organic-functional-groups/); [Infrared spectroscopy](https://www.owlsprep.com/study/cie-9701-u26-ir-spectroscopy/); [Mass spectrometry](https://www.owlsprep.com/study/cie-9701-u26-mass-spectrometry/)

## Learning objectives

- Explain the physical basis of ¹H NMR spectroscopy
- Identify NMR active nuclei and apply the activity rule
- Describe the origin of chemical shift and integration
- Explain spin-spin splitting and apply the n+1 rule

## NMR Active Nuclei and Basic Principle

**NMR active nucleus** — A nucleus with non-zero nuclear spin (odd mass number or odd atomic number) that interacts with an external magnetic field to produce a detectable NMR signal.

*Example:* ¹H and ¹³C are NMR active; ¹²C and ¹⁶O are inactive.

When placed in a strong external magnetic field, NMR active protons align either with (lower energy) or against (higher energy) the external field. The energy gap between these spin states corresponds to radio frequency radiation. When this frequency is applied, protons resonate (flip spin) to produce a measurable signal.

**Worked example:** State which of the following nuclei are NMR active: ¹²C, ¹H, ¹⁶O, ¹⁴N

1. Recall the rule for NMR activity: a nucleus is active if it has an odd mass number OR an odd atomic number.
2. Check each nucleus: ¹²C has mass number 12 (even) and atomic number 6 (even) → inactive
3. ¹H has mass number 1 (odd) → active
4. ¹⁶O has mass 16 (even) and atomic 8 (even) → inactive
5. ¹⁴N has mass 14 (even) but atomic number 7 (odd) → active

*Conclusion:* The NMR active nuclei are ¹H and ¹⁴N.

> **Exam tip:** CIE often asks to identify active nuclei from a list: always apply the mass/atomic number rule, don't just memorize common examples.

## Chemical Shift and TMS Reference

**Chemical shift** — The difference in resonance frequency of a proton relative to the reference standard TMS, measured in parts per million (ppm). It reflects the degree of shielding from the external magnetic field.

*Notation:* \delta

Electron density surrounding a proton shields it from the external magnetic field. Protons near electronegative groups (e.g. O, Cl, Br) have electron density withdrawn, leaving them deshielded. Deshielded protons resonate at higher chemical shift (higher $\delta$ value). Tetramethylsilane (TMS, $(CH_3)_4Si$) is used as the universal reference: all 12 of its protons are equivalent, it is inert, and its signal is set to $\delta = 0$ ppm, far from most organic proton signals.

**Worked example:** Explain why protons in $CH_3Cl$ have a higher chemical shift than protons in $CH_4$.

1. Chlorine is a highly electronegative atom, so it withdraws electron density from the C-H bonds in $CH_3Cl$.
2. Reduced electron density around the $CH_3$ protons means they are less shielded (deshielded) from the external magnetic field.
3. Deshielded protons require a higher resonance frequency, leading to a higher chemical shift than the fully shielded protons in $CH_4$.

> **Exam tip:** Always remember TMS is defined as $\delta = 0$ ppm, not 1 ppm, this is a common exam check.

## Equivalent Protons and Integration

**Chemically equivalent protons** — Protons that occupy identical chemical environments, so they have the same chemical shift and produce a single combined peak in the NMR spectrum.

The area under an NMR peak is directly proportional to the number of equivalent protons that produce the peak. This area is measured by an integration trace, which steps up after each peak: the height of each step equals the ratio of protons for that peak. This allows you to calculate the number of protons of each type in a molecule.

**Worked example:** A molecule with formula $C_2H_6O$ has two ¹H NMR peaks with integration step heights of 3 and 3. Deduce the identity of the molecule.

1. Total number of protons in the molecule is 6. Sum of the integration ratio is $3 + 3 = 6$, so the ratio equals the actual number of protons per peak.
2. Each peak corresponds to 3 equivalent protons, meaning there are two sets of 3 equivalent protons.
3. Ethanol ($CH_3CH_2OH$) has three sets of protons (3, 2, 1), so it cannot be the molecule. The only other isomer is methoxymethane ($CH_3OCH_3$), which has two identical $CH_3$ groups, giving two peaks of 3 protons each.

*Conclusion:* The molecule is methoxymethane.

## Spin-Spin Splitting and the n+1 Rule

**Spin-spin splitting (coupling)** — The splitting of an NMR peak into multiple smaller peaks caused by magnetic interaction with the spin of adjacent non-equivalent protons.

For CIE A-Level, splitting follows the simple n+1 rule: if a set of equivalent protons has $n$ adjacent non-equivalent protons, the peak will split into $n+1$ peaks. Coupling only occurs between non-equivalent protons on adjacent carbon atoms; equivalent protons do not split each other.

> **mnemonic**
>
> n adjacent non-equivalent protons: add 1 to get the number of peaks. That's all you need for CIE exams.

**Worked example:** Predict the splitting pattern for the $CH_3$ and $CH_2$ protons in $CH_3CH_2Br$.

1. For the $CH_3$ protons: they are adjacent to the $CH_2$ group, which has 2 non-equivalent protons, so $n=2$.
2. Apply n+1 rule: $2 + 1 = 3$, so the $CH_3$ peak splits into a triplet.
3. For the $CH_2$ protons: they are adjacent to the $CH_3$ group, which has 3 non-equivalent protons, so $n=3$.
4. Apply n+1 rule: $3 + 1 = 4$, so the $CH_2$ peak splits into a quartet.

*Conclusion:* The spectrum will have a triplet (3H, low δ) and a quartet (2H, higher δ).

> **Exam tip:** Never count equivalent adjacent protons when calculating n for the n+1 rule.

## Common pitfalls

- **Wrong:** Claiming ¹²C is NMR active because it has 6 protons.
  - Why it fails: NMR activity depends on nuclear spin (linked to mass/atomic number), not number of protons.
  - Correct: A nucleus is NMR active if it has an odd mass number OR odd atomic number; ¹²C is even-even so inactive.
- **Wrong:** Assuming all peaks must be split by adjacent protons.
  - Why it fails: If there are no adjacent non-equivalent protons, no splitting occurs.
  - Correct: If n=0, n+1 = 1, so the peak is a singlet (unsplit).
- **Wrong:** Using chemical shift values to count the number of protons in a peak.
  - Why it fails: Chemical shift only indicates the chemical environment of the proton, not the number of protons.
  - Correct: Use the integration trace step height to find the ratio/number of protons per peak.
- **Wrong:** Counting equivalent adjacent protons when applying the n+1 rule.
  - Why it fails: Only non-equivalent adjacent protons cause coupling and splitting. Equivalent protons do not interact.
  - Correct: Only count adjacent protons that are in a different chemical environment when calculating n.
- **Wrong:** Setting the TMS reference signal to δ = 1 ppm.
  - Why it fails: TMS is the universal reference standard defined as 0 ppm.
  - Correct: TMS always has a chemical shift of δ = 0 ppm in ¹H NMR.

## Cheatsheet

| Concept | Key Fact | Exam Note |
| --- | --- | --- |
| NMR Activity | Active if odd mass/odd atomic number | Check all nuclei in the question |
| Chemical Shift | Higher δ = more deshielded | Electronegative groups increase δ |
| TMS Reference | δ = 0 ppm, all protons equivalent | Always the reference standard |
| Integration | Step height = ratio of proton count | Sum steps to get total protons |
| n+1 Splitting Rule | n = adjacent non-equivalent protons | n+1 = number of split peaks |
| Equivalent Protons | Same environment = one peak | Do not split each other |

## What's next

Now you have mastered the core principles of ¹H NMR, the next step is applying these rules to interpret full ¹H NMR spectra of unknown organic molecules, matching chemical shift values to specific proton environments, and combining NMR data with other analytical techniques to deduce complete molecular structures. ¹H NMR is one of the most heavily weighted topics in CIE A-Level Chemistry theory papers, so mastering interpretation is critical for a high grade. These principles also form the foundation for understanding ¹³C NMR, which gives additional information about a molecule's carbon skeleton.

- [Interpretation of NMR spectra](https://www.owlsprep.com/study/cie-9701-u28-interpretation-of-nmr-spectra/)
- [Advanced analytical techniques](https://www.owlsprep.com/study/cie-9701-u29-overview/)

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