Coupling Patterns of Common Groups

Introduction

Experienced spectroscopists rarely analyse every signal from scratch. Instead they look for familiar fingerprints of common groups : combinations of splitting and integration that immediately identify an ethyl, isopropyl or tert-butyl group. Recognising these patterns turns a busy spectrum into a small number of building blocks, which can then be joined together using chemical shift and the molecular formula.

Core explanation

The ethyl group, –CH₂CH₃. The CH₃ protons have two neighbours and appear as a triplet integrating to 3. The CH₂ protons have three neighbours and appear as a quartet integrating to 2. Because coupling is mutual, both multiplets have the same line spacing. Whenever you see a 2H quartet and a 3H triplet with matching spacing, you have an ethyl group — provided the CH₂ has no other neighbouring hydrogens.

The isopropyl group, –CH(CH₃)₂. The two methyl groups are equivalent: six protons with one neighbour, giving a doublet integrating to 6. The single CH proton has six neighbours and appears as a septet integrating to 1. The outer lines of a septet are very weak (intensities 1 : 6 : 15 : 20 : 15 : 6 : 1), so it can look like a quintet unless you inspect it closely.

The tert-butyl group, –C(CH₃)₃. Nine equivalent protons on three methyl groups attached to a quaternary carbon, which carries no hydrogen. There are no neighbouring protons, so the signal is a large singlet integrating to 9, usually near δ 0.9–1.3.

Where the pattern sits tells you what it is attached to. The chemical shift of the CH₂ or CH part of the group depends on its neighbour:

Fragment Typical δ of CH₂ or CH Typical δ of CH₃ --- --- --- CH₃CH₂–C (alkyl) 1.2–1.5 0.9 CH₃CH₂–C=O (ketone, ester acyl side) 2.2–2.5 1.0–1.1 CH₃CH₂–O (ether, ester, alcohol) 3.4–4.1 1.2–1.3 CH₃CH₂–Cl 3.5 1.5 (CH₃)₂CH–C=O 2.4–2.7 1.1 (CH₃)₂CH–O 3.9–5.0 1.2 (CH₃)₃C– none 0.9–1.3

Other useful patterns. A 3H singlet near δ 2.1 suggests CH₃–C=O; a 3H singlet near δ 3.7 suggests an OCH₃ group in an ester. A 5H multiplet near δ 7.2–7.4 signals a monosubstituted benzene ring. Two 2H triplets next to each other suggest an –X–CH₂CH₂–Y– unit with no further hydrogens on either side.

Putting fragments together. Once the fragments are identified, add up their atoms, subtract from the molecular formula and fit the remaining atoms (often C=O, O or a halogen) between them.

Step-by-step reasoning

To recognise groups in a spectrum:

1. List every signal with its δ, integration and multiplicity. 2. Look for a 2H quartet plus a 3H triplet (ethyl). 3. Look for a 1H septet plus a 6H doublet (isopropyl). 4. Look for a 9H singlet (tert-butyl) or 3H singlets (isolated methyl groups). 5. Use the δ values to decide what each fragment is attached to.

Visual explanation

The simulation places the spectra of ethyl ethanoate, 2-methylpropanal and 2,2-dimethylpropanoic acid side by side. Coloured brackets link the quartet and triplet of the ethyl group, the septet and doublet of the isopropyl group, and the lone tall singlet of the tert-butyl group, so each pattern can be seen as a matched pair or a single signature peak.

Real-world analogy

Recognising coupling patterns is like recognising familiar chords in music. A trained listener hears a major chord as a single unit instead of three separate notes. In the same way, a chemist sees a quartet and triplet together and immediately thinks "ethyl" without counting neighbours each time.

Real-world example

Many flavour esters contain ethyl or isopropyl groups. Ethyl butanoate, which smells of pineapple, shows the classic ethyl-ester quartet near δ 4.1 and triplet near δ 1.25, while isopropyl esters show a septet around δ 5.0. Flavour chemists use these patterns to confirm which ester has been made.

Why?

Why is the CH₃ triplet of an ethyl ester at about δ 1.25 while the CH₂ quartet is at δ 4.1? The CH₂ carbon is bonded directly to the electronegative oxygen, which withdraws electron density and deshields its protons strongly. The CH₃ is one carbon further away and is only slightly affected.

Common misconception

"A septet always has seven clearly visible lines." The outermost lines of a septet are only 1/20 of the height of the central line and are often lost in the baseline, so a septet may look like a quintet. Use the 1H integration and the partner 6H doublet to identify it correctly.

Worked example

Question: A compound C₄H₈O₂ shows a 3H triplet at δ 1.2, a 3H singlet at δ 2.0 and a 2H quartet at δ 4.1. Identify it.

Reasoning: The triplet and quartet form an ethyl group; the quartet at δ 4.1 means CH₂ is on oxygen. The 3H singlet at δ 2.0 is CH₃–C=O. The remaining atoms are C, O and O, which make an ester linkage.

Answer: Ethyl ethanoate, CH₃COOCH₂CH₃.

Quick check

1. Which group gives a 6H doublet and a 1H septet in a ¹H NMR spectrum? Answer: An isopropyl group, (CH₃)₂CH–, where six methyl protons couple to one CH proton.

Exam focus

Quote the evidence for each fragment in full: "a quartet of integration 2 and a triplet of integration 3 indicate CH₂CH₃". Examiners reward linking the multiplicity, the integration and the δ value to the same group rather than stating the group alone.

Advanced insight

In an ethyl group attached to a stereocentre, the two CH₂ protons are diastereotopic and can have different shifts, turning the simple quartet into a more complex pattern. A similar effect occurs in isopropyl groups next to a stereocentre, where the two methyl groups become non-equivalent and appear as two separate 3H doublets.

Summary

Common groups have signature patterns: ethyl gives a 2H quartet and 3H triplet, isopropyl gives a 1H septet and 6H doublet, and tert-butyl gives a 9H singlet. The chemical shift of the CH₂ or CH part shows what the group is bonded to. Recognising fragments makes it quick to assemble a complete structure.

Practice questions

1. Describe the ¹H NMR signals expected for the isopropyl group in propan-2-ol. Answer: A 6H doublet near δ 1.2 for the two methyl groups and a 1H septet near δ 4.0 for the CH on oxygen. 2. A spectrum contains a 9H singlet at δ 1.2. What group is present? Answer: A tert-butyl group, (CH₃)₃C–, whose nine equivalent protons have no neighbouring hydrogens. 3. The quartet of an ethyl group appears at δ 2.4. What is the CH₂ probably bonded to? Answer: A carbonyl carbon, as in a ketone such as butanone. 4. Why do the quartet and triplet of an ethyl group have the same line spacing? Answer: They arise from the same coupling interaction between CH₂ and CH₃ protons, so they share the same coupling constant.