Combining NMR with Other Techniques

Introduction

Mass spectrometry and IR tell you how big a molecule is and which functional groups it carries, but they rarely show how the atoms are joined. Nuclear magnetic resonance fills that gap. ¹H NMR reveals how many kinds of hydrogen there are, how many of each, and what their neighbours are; ¹³C NMR counts carbon environments. Combined with a molecular formula and IR evidence, NMR usually leads to a single, confident structure.

Core explanation

A logical order of analysis.

1. Molecular formula from high-resolution mass spectrometry (or Mᵣ plus elemental composition). 2. Degree of unsaturation from the formula: how many rings and π bonds. 3. IR : which functional groups account for the unsaturation and the heteroatoms. 4. ¹³C NMR : number of carbon environments and their types — carbonyl carbons at about 160–220 ppm, aromatic and alkene carbons at 100–150 ppm, carbons next to oxygen at 50–90 ppm, alkyl carbons below about 50 ppm. 5. ¹H NMR : chemical shift, integration and splitting of each signal.

Extracting fragments from ¹H NMR. Each signal is interpreted in three ways:

- Chemical shift shows the environment. Protons on carbon next to an ester oxygen appear near 3.7–4.1 ppm; protons next to a C=O near 2.0–2.5 ppm; aldehyde protons near 9.5–10 ppm. - Integration gives the relative number of protons. A 3:2 ratio with a total matching the formula suggests CH₃ and CH₂. - Splitting gives the number of neighbouring protons by the n+1 rule. A triplet (3H) with a quartet (2H) is the classic ethyl group, CH₃CH₂–. A singlet 3H is a methyl group with no neighbouring hydrogens, such as CH₃CO– or CH₃O–.

Assembling the skeleton. Write down each fragment, add up its atoms, and compare with the molecular formula. Any atoms left over (often O atoms or a carbonyl carbon with no hydrogens) must link the fragments. The chemical shifts then tell you which way round they are joined, which is crucial for isomers such as esters.

Final check. Count the signals you would predict for your structure in both ¹H and ¹³C NMR, predict the key IR bands and the main mass spectrum fragments, and compare them with the data.

Step-by-step reasoning

1. Calculate the DBE from the formula. 2. Use IR to assign the unsaturation to specific groups. 3. Count ¹³C signals and note carbonyl or aromatic carbons. 4. Turn each ¹H signal into a fragment using shift, integration and splitting. 5. Join the fragments so that the formula is satisfied and the shifts make sense.

Visual explanation

Think of the fragments as jigsaw pieces laid out on a table: a CH₃CH₂– piece with a "next to oxygen" label, a CH₃– piece with a "next to C=O" label, and a lone –COO– piece from IR. There is only one way to fit them together so that every label is satisfied.

Real-world analogy

Building a structure from spectra is like assembling flat-pack furniture. The parts list (molecular formula) tells you what is in the box, the labelled bags (IR functional groups) show the special fittings, and the instructions showing which piece joins which (NMR splitting) reveal how they connect.

Real-world example

Chemists who isolate new compounds from plants or marine organisms rely on exactly this workflow. High-resolution MS gives the formula, IR indicates carbonyl or hydroxyl groups, and a suite of one- and two-dimensional NMR experiments pieces together skeletons with dozens of carbon atoms, often with only milligrams of sample available.

Why?

Why is NMR the central technique for connectivity? Spin–spin coupling occurs mainly through bonds between protons on adjacent carbons, so a splitting pattern directly reports which groups are neighbours. No other routine technique gives such direct information about how atoms are joined.

Common misconception

"Any quartet at 4.1 ppm and triplet at 1.2 ppm means ethanol." This ethyl pattern appears in many compounds, including ethyl esters. The shift of the CH₂ tells you it is next to oxygen, but IR and the formula must decide whether that oxygen belongs to an alcohol or an ester.

Worked example

Question: Compound Q, C₄H₈O₂, shows an IR band at 1740 cm⁻¹ and no O–H band. Its ¹H NMR spectrum shows 4.1 ppm (quartet, 2H), 2.0 ppm (singlet, 3H) and 1.2 ppm (triplet, 3H). Identify Q.

Reasoning: DBE = (8 + 2 − 8) ÷ 2 = 1, accounted for by the C=O. The 1740 cm⁻¹ band and two O atoms without O–H indicate an ester. The quartet/triplet pair is CH₃CH₂–, and its CH₂ at 4.1 ppm is attached to oxygen. The singlet 3H at 2.0 ppm is CH₃ attached to C=O. The isomer methyl propanoate would instead show a singlet near 3.7 ppm (OCH₃) and a quartet near 2.3 ppm.

Answer: Ethyl ethanoate, CH₃COOCH₂CH₃.

Quick check

1. In a ¹H NMR spectrum, a 3H singlet appears at 3.7 ppm. What fragment does this suggest? Answer: A methyl group attached to oxygen with no neighbouring hydrogens, such as the OCH₃ group of a methyl ester.

Exam focus

For each ¹H signal, examiners expect three statements: the environment (from shift), the number of protons (from integration) and the number of neighbours (from splitting). Link every deduction to a specific piece of data, and finish by confirming the structure against the formula.

Advanced insight

Two-dimensional NMR experiments extend the same logic. COSY maps which protons couple to each other, HSQC links each proton to the carbon it is attached to, and HMBC shows protons coupling to carbons two or three bonds away. Together they can connect fragments across quaternary carbons and heteroatoms where simple splitting gives no information.

Summary

A full structure determination combines a molecular formula, the degree of unsaturation, IR functional groups and NMR connectivity. ¹³C NMR counts carbon environments; ¹H NMR supplies environments, proton counts and neighbours. Fragments are joined to satisfy the formula, and chemical shifts decide how they are connected, distinguishing isomers such as ethyl ethanoate and methyl propanoate.

Practice questions

1. How many ¹³C NMR signals would ethyl ethanoate give? Answer: Four, because all four carbon atoms are in different environments. 2. Predict the ¹H NMR spectrum of methyl propanoate, CH₃CH₂COOCH₃. Answer: A singlet (3H) near 3.7 ppm for OCH₃, a quartet (2H) near 2.3 ppm for CH₂ next to C=O, and a triplet (3H) near 1.1 ppm for CH₃. 3. What does a DBE of 0 and an IR band at 3350 cm⁻¹ tell you about a compound C₃H₈O? Answer: It has no rings or π bonds and contains an O–H group, so it is a saturated alcohol, propan-1-ol or propan-2-ol. 4. Why is IR alone unable to distinguish ethyl ethanoate from methyl propanoate? Answer: Both are esters with similar C=O and C–O stretches; only NMR shows how the alkyl groups are arranged about the ester linkage.