Investigate the processes used to analyse the structure of simple organic compounds, including proton and carbon-13 NMR

What this dot point is asking

NESA wants you to explain the principle of nuclear magnetic resonance, identify the four pieces of information in a proton NMR spectrum (number of signals, chemical shift, integration, multiplicity by the n+1 rule), use a carbon-13 NMR spectrum to count and classify carbon environments, and deduce the structure of a simple organic compound from $^1H$ and $^{13}C$ NMR together.

The answer

The physics in one paragraph

Nuclei with a non-zero spin (spin-half nuclei like $^1H$ and $^{13}C$) behave as tiny magnets. In a strong external magnetic field they take one of two orientations (aligned or opposed) with a small energy gap between them. Irradiating the sample with radiofrequency energy matched to that gap causes the lower-energy nuclei to flip to the higher state; this is resonance. The frequency at which resonance happens depends slightly on the local electron density around the nucleus, which differs for each chemical environment. The differences in resonance frequency are reported as a chemical shift $\delta$ in parts per million (ppm) from a reference compound (tetramethylsilane, TMS, $\delta = 0$).

The reference: TMS

Tetramethylsilane $(CH_3)_4Si$ has 12 equivalent protons in a single environment, all with very low resonance frequency (silicon is more electropositive than carbon, so the methyls are electron-rich and shielded). Chemical shifts are positive to the left (downfield, deshielded) and zero at TMS.

Proton NMR: four features per spectrum

1. Number of signals. Each unique proton environment gives one signal. Symmetry can make two formally different protons equivalent. For example, all three protons in $CH_3$ are equivalent. The methyl, methylene and OH of ethanol are three signals.

2. Chemical shift ($\delta$ in ppm). Tells you the electronic environment of the proton.

3. Integration. The area under each signal is proportional to the number of equivalent protons in that environment. The spectrometer reports areas as a step trace; the ratio of step heights gives the proton ratio. Integration is what distinguishes a methyl (3H) from a methylene (2H) at similar chemical shift.

4. Multiplicity (splitting) and the n+1 rule. Spin-spin coupling to neighbouring protons splits each signal into a multiplet. The rule:

where $n$ is the number of protons on the immediately adjacent carbon(s). So a $CH_3$ next to a $CH_2$ appears as a triplet (n = 2, peaks = 3) and the $CH_2$ next to the $CH_3$ appears as a quartet (n = 3, peaks = 4). The relative heights of the peaks follow Pascal's triangle: 1:1 (doublet), 1:2:1 (triplet), 1:3:3:1 (quartet), 1:4:6:4:1 (pentet).

Coupling is normally only seen between protons on adjacent carbons; protons on the same carbon are typically equivalent and do not split each other.

Carbon-13 NMR

$^{13}C$ NMR is run proton-decoupled as standard, which collapses all couplings and gives a singlet for each unique carbon environment. The spectrum tells you two things:

1. Number of signals = number of unique carbon environments.
2. Chemical shift classifies each carbon.

The shift range is much wider than for $^1H$, about 0 to 220 ppm:

Carbon-13 NMR does not have integration in the conventional sense (relaxation times differ between carbons, so peak heights are not reliable counts), but the number of peaks is a hard constraint on the structure.

Reading both spectra together: a workflow

1. From molecular formula (from mass spectrometry), compute the degree of unsaturation $= (2C + 2 - H + N - X)/2$.
2. Count $^{13}C$ peaks. That fixes the number of unique carbon environments.
3. Classify each $^{13}C$ peak by its chemical shift range.
4. Count $^1H$ peaks and their integrations. Confirm the total proton count.
5. Assign each $^1H$ peak by chemical shift to a type of environment.
6. Use multiplicity to determine which protons are adjacent.
7. Assemble the fragments into a structure consistent with all evidence.

Why we use NMR for structure

NMR is the most informative single technique for organic structure determination. Combined with mass spectrometry (molecular mass) and IR (functional groups), the three give an essentially complete picture.

Strengths and limits

Strengths. Non-destructive (sample is recovered after analysis), enormously information-rich, distinguishes isomers that IR and mass spectrometry cannot (e.g. propan-1-ol vs propan-2-ol from chemical shift and multiplicity patterns).

Limits. Needs tens of milligrams of dissolved sample (compared to nanograms for mass spec). $^{13}C$ has poor sensitivity due to 1.1% natural abundance. Solvent peaks (and water from $-OH$) can obscure regions of the spectrum. Magnetically equivalent groups give one peak even when their environments are subtly different.

Examples in context

Example 1. Structural identification at the Bragg Crystallography Facility, UNSW. Researchers at UNSW use a 600 MHz NMR spectrometer to confirm the structure of newly synthesised pharmaceutical candidates. A typical screen records the proton NMR (chemical shift, integration, multiplicity) and the carbon-13 NMR (count of carbon environments), then combines them with mass-spec data to confirm the molecule. For a candidate with the formula $C_5H_{10}O$, three proton signals plus three carbon signals suggests pentan-3-one (symmetric ketone). The HSC framework of "count environments, read shifts, apply $n+1$" is exactly the analysis a PhD student runs to confirm the synthesis worked before sending the compound for biological testing.

Example 2. Ethanol identification in NSW HSC depth study. A common Stage 6 NMR task gives students the proton NMR of ethanol and asks them to assign every signal. The spectrum shows three signals: a triplet at 1.2 ppm (3 H, $CH_3$, coupled to 2 H of $CH_2$, multiplicity $2+1$), a quartet at 3.7 ppm (2 H, $CH_2$, coupled to 3 H of $CH_3$, multiplicity $3+1$), and a broad singlet at 2.6 ppm (1 H, OH, exchanges with water). The carbon-13 spectrum shows two signals at 18 and 57 ppm. NESA marking rewards correct assignment of every feature plus the $n+1$ rule logic.

Try this

Q1. State the four pieces of structural information that a proton NMR spectrum provides. [4 marks]

• Cue. Number of signals (number of H environments); chemical shift (electronic environment); integration (relative number of H); multiplicity (number of adjacent H via $n+1$ rule).

Q2. A compound with molecular formula $C_3H_6O$ shows two proton signals: a singlet (6 H) at 2.1 ppm. The carbon-13 NMR shows two signals at 30 and 207 ppm. Identify the compound and justify. [3 marks]

• Cue. $C_3H_6O$ with one degree of unsaturation; only 2 H environments with 6 H singlet plus carbonyl carbon at 207 ppm; the compound is propanone ($CH_3COCH_3$).

Q3. A compound $C_2H_6O$ shows two proton signals: a triplet (3 H) at 1.2 ppm and a quartet (2 H) at 3.7 ppm, plus a broad singlet (1 H) at 2.6 ppm that exchanges with $D_2O$. (a) Identify the compound. (b) Justify the multiplicity of the triplet and quartet. (c) Explain the chemical shift of the OH signal. [2+2+1 marks]

• Cue. (a) Ethanol. (b) Triplet from $CH_3$ adjacent to 2 H of $CH_2$ ($2+1$); quartet from $CH_2$ adjacent to 3 H of $CH_3$ ($3+1$). (c) OH is deshielded by electronegative oxygen; broad and exchangeable in $D_2O$.