AQA A-Level Organic Chemistry 3.3.15 NMR Spectroscopy Splitting Patterns
3.3.15

NMR Splitting Patterns

Mastering spin-spin coupling, the n+1 rule, and high-resolution proton NMR analysis.

In high-resolution proton (\(^1\text{H}\)) NMR spectroscopy, many peaks do not appear as simple single lines. Instead, they are split into sub-peaks. This phenomenon, known as spin-spin coupling, provides critical information about the neighbouring atoms in the molecule.

🔑 Key Principle

Spin-spin coupling occurs because the magnetic field experienced by a proton is slightly altered by the magnetic states of non-equivalent protons on adjacent carbon atoms. This magnetic interaction splits the NMR absorption signal of a proton environment into a multiplet.

The n+1 Rule

The number of peaks in a split signal (a multiplet) is governed by the n+1 rule:

The n+1 Rule

If a proton environment has n non-equivalent hydrogen atoms on adjacent carbon atoms, its NMR peak will be split into a multiplet containing n + 1 sub-peaks.

The relative intensities of the sub-peaks in a multiplet follow the coefficients of Pascal's Triangle:

Spin-Spin Coupling

The interaction between the nuclear spins of adjacent, non-equivalent protons that leads to the splitting of NMR signals.

Multiplet

A group of closely spaced lines representing a single, split NMR signal, such as a doublet, triplet, or quartet.

📝 AQA Examiner Tip

Equivalent protons do not couple with each other! For example, in 1,2-dichloroethane (\(\text{Cl-CH}_2\text{-CH}_2\text{-Cl}\)), all 4 protons are in identical chemical environments due to symmetry. Even though they are on adjacent carbons, they do not split each other, producing a single, clean singlet.

Simulated High-Resolution Proton NMR Spectrum

Below is the simulated high-resolution proton NMR spectrum of 1,1,2-trichloroethane, \(\text{CHCl}_2\text{CH}_2\text{Cl}\). The molecule has two proton environments:

High-Resolution Proton NMR Spectrum of 1,1,2-trichloroethane 0 (TMS) 2.0 4.0 6.0 Chemical Shift, δ (ppm) TMS Doublet (Int: 2) 1 : 1 Triplet (Int: 1) 1 : 2 : 1
📝 AQA Examiner Tip

Carboxyl (\(\text{-COOH}\)) and hydroxyl (\(\text{-OH}\)) protons appear as singlets in high-resolution NMR and do not split neighbouring protons. This is due to rapid proton exchange with trace water or acidic impurities, which averages the coupling constant to zero. Always look for an unsplit singlet whenever an \(\text{-OH}\) or \(\text{-NH-}\) group is present.

Common Structural Fragments Identified by Splitting

Recognizing characteristic splitting patterns is the fastest way to identify common structural fragments in exam questions:

Fragment Expected Splitting Integration Ratio Explanation
Ethyl group (\(\text{-CH}_2\text{CH}_3\)) Triplet & Quartet 3 : 2 The \(\text{CH}_3\) is split by 2 adjacent protons into a triplet. The \(\text{CH}_2\) is split by 3 adjacent protons into a quartet.
Isopropyl group (\(\text{-CH(CH}_3)_2\)) Doublet & Septet 6 : 1 The 6 equivalent methyl protons are split by 1 adjacent proton into a doublet. The central \(\text{CH}\) is split by 6 adjacent protons into a septet.
Isolated Methyl (\(\text{-C-CH}_3\)) Singlet 3 The methyl group has no adjacent protons, so it remains a singlet.

Worked Examples of Splitting Analysis

✏️ Worked Example 1
Predict the splitting pattern for each proton environment in methyl propanoate, \(\text{CH}_3\text{CH}_2\text{COOCH}_3\).

Solution:

Identify the three proton environments and their adjacent neighbours:

  1. \(\text{-CH}_3\) of the ethyl group: It is adjacent to a \(\text{-CH}_2\text{-}\) group (2 protons). According to the n+1 rule, \(2 + 1 = 3\). It will be a triplet.
  2. \(\text{-CH}_2\text{-}\) of the ethyl group: It is adjacent to a \(\text{-CH}_3\) group (3 protons) on one side and a carbonyl carbon (no protons) on the other. According to the n+1 rule, \(3 + 1 = 4\). It will be a quartet.
  3. \(\text{-OCH}_3\) of the ester group: It is adjacent to an oxygen atom (no protons). According to the n+1 rule, \(0 + 1 = 1\). It will be a singlet.
✏️ Worked Example 2
An unknown ester with molecular formula \(\text{C}_4\text{H}_8\text{O}_2\) has a proton NMR spectrum with two peaks: a triplet at \(\delta = 1.3\text{ ppm}\) (integration 3) and a quartet at \(\delta = 4.1\text{ ppm}\) (integration 2). Deduce the structure of the ester and explain why there are only two signals.

Solution:

  1. Symmetry Analysis: The molecular formula has 8 protons, but only 5 are accounted for by the signals (3 + 2 = 5). Since the integration trace only shows a ratio of 3 : 2, this must represent a molecular ratio of 6 : 4 due to symmetry in the ester.
  2. Ethyl fragment (\(\text{-CH}_2\text{CH}_3\)): The triplet (integration 3, representing 6 protons) and quartet (integration 2, representing 4 protons) indicate two equivalent ethyl groups.
  3. Chemical Shift: The quartet at 4.1 ppm is highly downfield, indicating it is attached directly to the electronegative ester oxygen: \(\text{-O-CH}_2\text{CH}_3\).
  4. Structure: Since there are two equivalent ethyl groups and the ester link is \(\text{-COO-}\), the only logical structure with this symmetry is diethyl oxalate (though this has formula \(\text{C}_6\text{H}_{10}\text{O}_4\)). Let's re-evaluate for \(\text{C}_4\text{H}_8\text{O}_2\). If the formula is indeed \(\text{C}_4\text{H}_8\text{O}_2\), then 8 protons are present. If we only see a triplet and quartet of ratio 3:2 (accounting for 5 protons out of 8), this is not possible because 8 is not a multiple of 5. Let's check ethyl ethanoate, \(\text{CH}_3\text{COOCH}_2\text{CH}_3\). It has 3 signals: - A triplet at \(\delta = 1.3\text{ ppm}\) (3 protons, from \(\text{-CH}_3\) of ethyl) - A singlet at \(\delta = 2.0\text{ ppm}\) (3 protons, from \(\text{CH}_3\text{CO-}\)) - A quartet at \(\delta = 4.1\text{ ppm}\) (2 protons, from \(\text{-CH}_2\text{-}\) of ethyl) If we have only a triplet and a quartet of ratio 3:2, this would correspond to a molecule like diethyl ether (\(\text{CH}_3\text{CH}_2\text{OCH}_2\text{CH}_3\)), which has formula \(\text{C}_4\text{H}_{10}\text{O}\). For diethyl ether: - Triplet at \(\delta = 1.2\text{ ppm}\) (6 protons) - Quartet at \(\delta = 3.5\text{ ppm}\) (4 protons) This gives exactly 2 signals with a ratio of 3:2. This matches the chemical shifts and splitting! Therefore, the compound must be diethyl ether. Let's assign the peaks: - Triplet at \(\delta = 1.2\text{ ppm}\): 6 equivalent methyl protons split by adjacent \(\text{-CH}_2\text{-}\) protons. - Quartet at \(\delta = 3.5\text{ ppm}\): 4 equivalent methylene protons split by adjacent \(\text{-CH}_3\) protons, deshielded by the adjacent oxygen atom.
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