Principles of ¹H NMR

CIE A-Level Chemistry· Unit 28: Nuclear magnetic resonance spectroscopy· 45 min read

1. NMR Active Nuclei and Basic Principle★★☆☆☆⏱ 15 min

📘 Definition

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

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.

2. Chemical Shift and TMS Reference★★☆☆☆⏱ 20 min

📘 Definition

Chemical shift

\delta

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.

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 Protons near electronegative groups (e.g. O, Cl, Br) have electron density withdrawn, leaving them deshielded.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 $δ$ value). Tetramethylsilane (TMS, $(C H_{3})_{4} S i$ ) is used as the universal reference: all 12 of its protons are equivalent, it is inert, and its signal is set to $δ = 0$ ppm, far from most organic proton signals.

📐 Worked Example

Explain why protons in $C H_{3} C l$ have a higher chemical shift than protons in $C H_{4}$ .

1
Chlorine is a highly electronegative atom, so it withdraws electron density from the C-H bonds in $C H_{3} C l$ .
2
Reduced electron density around the $C H_{3}$ 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 $C H_{4}$ .

Exam tip:

Always remember TMS is defined as $δ = 0$ ppm, not 1 ppm, this is a common exam check.

3. Equivalent Protons and Integration★★★☆☆⏱ 20 min

📘 Definition

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 $C_{2} H_{6} O$ 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 ( $C H_{3} C H_{2} O H$ ) has three sets of protons (3, 2, 1), so it cannot be the molecule. The only other isomer is methoxymethane ( $C H_{3} O C H_{3}$ ), which has two identical $C H_{3}$ groups, giving two peaks of 3 protons each.

4. Spin-Spin Splitting and the n+1 Rule★★★☆☆⏱ 20 min

📘 Definition

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.

📐 Worked Example

Predict the splitting pattern for the $C H_{3}$ and $C H_{2}$ protons in $C H_{3} C H_{2} B r$ .

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

Exam tip:

Never count equivalent adjacent protons when calculating n for the n+1 rule.

5. Common Pitfalls

Wrong move:

Claiming ¹²C is NMR active because it has 6 protons.

Why:

NMR activity depends on nuclear spin (linked to mass/atomic number), not number of protons.

Correct move:

A nucleus is NMR active if it has an odd mass number OR odd atomic number; ¹²C is even-even so inactive.

Wrong move:

Assuming all peaks must be split by adjacent protons.

Why:

If there are no adjacent non-equivalent protons, no splitting occurs.

Correct move:

If n=0, n+1 = 1, so the peak is a singlet (unsplit).

Wrong move:

Using chemical shift values to count the number of protons in a peak.

Why:

Chemical shift only indicates the chemical environment of the proton, not the number of protons.

Correct move:

Use the integration trace step height to find the ratio/number of protons per peak.

Wrong move:

Counting equivalent adjacent protons when applying the n+1 rule.

Why:

Only non-equivalent adjacent protons cause coupling and splitting. Equivalent protons do not interact.

Correct move:

Only count adjacent protons that are in a different chemical environment when calculating n.

Wrong move:

Setting the TMS reference signal to δ = 1 ppm.

Why:

TMS is the universal reference standard defined as 0 ppm.

Correct move:

TMS always has a chemical shift of δ = 0 ppm in ¹H NMR.

6. Quick Reference 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

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 2
Identify NMR active nuclei
2023 · 4
Explain origin of chemical shift
2021 · 4
Apply n+1 splitting rule

Going deeper

revision videoCIE 9701 ¹H NMR Principles SummaryVisual walkthrough of core concepts

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.