37.4 Proton NMR spectroscopy

Syllabus
9701–2028–2029
Topic
37.4
Level
A2

Learning objectives

Combine shift, integration and splitting to deduce a proton NMR structure

For a simple 1H NMR spectrum: 1) count sample signals for distinct proton environments; 2) use chemical shifts to assign likely local groups; 3) reduce relative peak areas to the simplest proton ratio; 4) use each splitting pattern and n+1 to count equivalent protons on adjacent carbon atoms; 5) assemble candidate fragments; 6) reject any structure that fails the formula, total integral, shift or splitting evidence.

Spectrum feature Structural information
number of signals number of chemically distinct proton environments
chemical shift, delta / ppm electronic environment and nearby functional groups/pi systems
relative integrated area relative number of protons in each environment
singlet/doublet/triplet/quartet/multiplet adjacent non-equivalent proton count through n+1 within the simple model

A 3H triplet paired with a 2H quartet supports CH3-CH2-: CH3 is split by two adjacent protons, while CH2 is split by three. Its shifts then show what group is attached to the CH2 end.

O-H and N-H signals are often broad and exchangeable, and they do not reliably participate in simple n+1 splitting. Confirm them by D2O exchange rather than forcing them into the neighbouring-proton pattern.

Multiplicity is not proton count: a quartet means splitting into four sub-peaks, commonly by three adjacent protons. No single clue proves the structure; all four evidence types must agree.

Predict each proton environment's shift and simple splitting pattern

Label chemically equivalent proton sets. For each set, use its nearest functional/electronic environment to estimate chemical shift, then count non-equivalent protons on directly adjacent carbon atom(s) and apply n+1 within the syllabus's simple splitting model.

Approximate delta / ppm Common proton environment
0.9-1.7 saturated alkyl C-H
2.0-3.0 C-H next to C=O or aryl/alkene system
3.2-4.0 C-H next to O or another electronegative atom
4.5-6.0 alkene H
6.0-9.0 arene H
9.3-10.6 aldehyde H
about 9-13 carboxylic-acid O-H; broad/variable
Adjacent equivalent protons, n n+1 result
0 singlet
1 doublet
2 triplet
3 quartet
more/overlapping simple neighbours multiplet as appropriate

Equivalent protons do not split each other. Do not count all hydrogens in the molecule, and do not treat exchangeable O-H/N-H protons as reliable fixed splitting partners.

TMS defines zero on the chemical-shift scale

Tetramethylsilane, Si(CH3)4, is added as the NMR reference and assigned delta = 0 ppm. Every sample signal is reported by its shift relative to this standard, allowing spectra recorded under different field strengths to use a common ppm scale.

Property of TMS Why it helps
all 12 protons are equivalent gives one sharp, strong signal
highly shielded protons signal lies at 0 ppm, away from most organic sample peaks
chemically inert is unlikely to react with the sample
volatile can be removed readily after measurement

A sample signal at 2.1 ppm is 2.1 ppm downfield from the TMS reference. The TMS peak defines the axis and is not part of the unknown compound.

Do not assign the TMS peak as a sample environment or impurity. Its job is to provide the zero reference, not structural information about the analyte.

Deuterated solvents avoid a dominant solvent proton signal

An NMR sample must be dissolved, but an ordinary proton-containing solvent would produce a large 1H signal that could mask the sample. A deuterated solvent such as CDCl3 replaces most 1H with 2H, which resonates outside the ordinary proton NMR observation used here.

The solvent must dissolve the sample and not react with it. Deuterated solvent also supports the instrument lock, while small residual protonated-solvent peaks can remain.

Using CDCl3 instead of CHCl3 greatly suppresses the chloroform proton signal so the analyte's proton environments can be observed.

Deuterated does not mean absolutely proton-free. Recognise labelled residual-solvent peaks rather than assigning them to the unknown structure.

D2O exchange identifies O-H and N-H proton signals

Add D2O and compare the proton NMR spectrum before and after. Exchangeable O-H or N-H protons are replaced by deuterium, which is not observed at the same position in ordinary 1H NMR, so their signal disappears or becomes much weaker.

ROH+DX2OROD+HODRNHX2+DX2ORNHD+HOD\ce{R-OH + D2O <=> R-OD + HOD}\qquad\ce{R-NH2 + D2O <=> R-NHD + HOD}

A disappearing broad signal supports assignment to O-H or N-H. Combine this with chemical structure or other evidence because disappearance alone does not distinguish oxygen from nitrogen.

Ordinary C-H signals should remain: their protons do not exchange rapidly with D2O under the test conditions. Do not treat every changed baseline feature as an exchangeable proton without a before/after peak comparison.