the principles and interpretation of proton (^1H) and carbon-13 (^13C) NMR spectroscopy (chemical shift, integration, n+1 splitting and number of carbon environments) and high performance liquid chromatography (HPLC, retention time and calibration curves) for the identification and quantification of organic compounds

What this dot point is asking

VCAA wants you to interpret ^1H NMR (number of environments, chemical shift, integration, n+1 splitting) and ^13C NMR (number of carbon environments), and to describe how HPLC identifies components by retention time and quantifies them by using calibration curves.

The answer

Proton (^1H) NMR: what it tells you

A proton NMR spectrum gives four pieces of information for each set of equivalent hydrogens (each "environment") in the molecule:

1. The number of signals = the number of unique hydrogen environments. Hydrogens are equivalent (one environment) if you can interchange them by symmetry of the molecule. CH3CH2OH has 3 environments (the 3 CH3 protons, the 2 CH2 protons, the 1 OH proton).

2. The chemical shift (delta, in ppm) reflects the local electronic environment. More electronegative neighbours (O, N, halogens, C=O) shift the signal downfield (higher ppm). Reference compound is TMS (tetramethylsilane), set at 0 ppm.

Approximate chemical shift ranges to know:

3. Integration (peak area) is proportional to the number of hydrogens in that environment. A 3:2 integration ratio in an ethyl group corresponds to CH3 : CH2.

4. Splitting (multiplicity) follows the n+1 rule: a hydrogen environment with n equivalent neighbouring hydrogens on adjacent carbons appears as an (n+1)-multiplet. So:

• 0 neighbours: singlet.
• 1 neighbour: doublet.
• 2 neighbours: triplet.
• 3 neighbours: quartet.
• 4 neighbours: pentet, etc.

Hydrogens on the same carbon do not split each other if they are equivalent. Hydrogens on -OH and -NH typically do not split (exchange happens quickly) and appear as broad singlets.

Carbon-13 (^13C) NMR

Detects the natural ^13C isotope (1.1% abundance) instead of ^1H. The output is simpler:

• The number of signals = number of carbon environments. CH3CH2OH has 2 carbon environments.
• Chemical shifts span 0 to 220 ppm. Useful ranges: alkane C around 10 to 40, alcohol C-O around 50 to 80, aromatic C around 110 to 150, carbonyl C around 165 to 215 (ester/acid 165 to 180, ketone/aldehyde 190 to 215).
• Splitting is not used in routine ^13C NMR (protons are decoupled out; each carbon appears as a singlet).
• Integration is unreliable and usually not used.

The main use of ^13C NMR in VCE is counting carbon environments to distinguish isomers.

How to interpret a ^1H NMR spectrum (workflow)

1. Count the number of signals = number of H environments.
2. Read each chemical shift and look it up in your table to suggest the type of H.
3. Read the integration to get the ratio of Hs in each environment. Scale to match the molecular formula.
4. Read the splitting and use n+1 to count neighbours.
5. Combine all of the above to propose a structure consistent with the data and with the molecular formula.

A useful starting trick: the molecular formula plus the count of environments narrows the candidates fast.

High performance liquid chromatography (HPLC)

HPLC separates components of a liquid mixture for both identification and quantification.

Principle. A liquid (the mobile phase) is pumped at high pressure (5 to 400 bar) through a column packed with a solid (the stationary phase). The sample is injected, and components travel through the column at rates that depend on their partition between the two phases. More polar species in a non-polar (reversed-phase) column move quickly; less polar species spend more time on the stationary phase and move slowly.

A detector (UV-vis absorbance, fluorescence, or mass spectrometry) at the column exit gives a chromatogram: detector response (y) versus time (x). Each component appears as a peak at its characteristic retention time Rt.

Use 1: identification

Compare the retention time of an unknown peak to that of a known standard run under the same conditions. Matching Rt is consistent with the same compound (but not proof; many compounds may share an Rt).

Use 2: quantification

The area under the peak is proportional to the amount of that component injected. A calibration curve of peak area against known concentrations of pure standard converts the unknown's peak area into a concentration.

Calibration curve procedure

1. Prepare a series of standards of pure analyte at known concentrations.
2. Run each through HPLC under identical conditions and record the peak area at the analyte's Rt.
3. Plot peak area (y) versus concentration (x); expect a linear relationship at low concentration.
4. Run the unknown sample; measure the peak area at the same Rt.
5. Read the concentration from the calibration line.

Strengths: separation and quantification of complex mixtures, can detect small amounts, works for non-volatile species that gas chromatography cannot handle.

Limitations: similar polarity compounds may co-elute, sample preparation can be involved, expensive instrument.

Examples in context

Example 1. Caffeine quantification in V Energy drinks at NMI Sydney. Analysts at the National Measurement Institute, Lindfield, quantify caffeine in V Energy ($80 \, \text{mg}$ per $250 \, \text{mL}$ can) using HPLC with UV detection at $273 \, \text{nm}$. Caffeine elutes at $R_t = 4.8 \, \text{min}$ on a C18 reverse-phase column with $\text{methanol-water}$ ($20:80$) mobile phase. A four-point calibration curve from $10$ to $200 \, \text{mg/L}$ shows peak area $= 18.5 \times [\text{caffeine}]$ (mg/L). A diluted sample shows area $7400$, giving $[\text{caffeine}] = 400 \, \text{mg/L}$ in the cuvette and (with $\times 1.25$ dilution) $320 \, \text{mg/L}$ in the drink; total per can $= 320 \times 0.250 = 80 \, \text{mg}$. The result matches the label and meets FSANZ Standard 2.6.4.

Example 2. NMR confirmation of synthesised aspirin at WEHI medicinal chemistry. WEHI chemists confirm a drug-development batch of aspirin using $^1$H NMR. The product gives four peaks: singlet $\delta = 11.2$ (1H, COOH), multiplet $\delta = 7.5$ to $8.1$ (4H, aromatic), singlet $\delta = 2.35$ (3H, $\text{O-COCH}_3$). Integration ratio $1:4:3$ matches the eight protons of $\text{C}_9 \text{H}_8 \text{O}_4$. The $^{13}$C NMR shows seven carbon environments: $\delta = 170.5$ (acid C=O), $169.4$ (ester C=O), $151$ (aromatic C-O), $133$, $126$, $124$, $123$ (other aromatic C), $20.8$ ($\text{CH}_3$). Match to NIST library confirms identity to $99.5\%$ purity, supporting clinical-trial release.

Try this

Q1. A compound shows two singlets in its $^1$H NMR at $\delta = 2.1$ (3H) and $\delta = 3.7$ (3H). It also has $M^+ = 74$. (a) Suggest a structure. (b) Justify the absence of coupling. [3 marks]

• Cue. (a) Methyl ethanoate, $\text{CH}_3 \text{COOCH}_3$, $M_r = 74$. (b) The two $\text{CH}_3$ groups are on different sides of the C=O and have no adjacent H atoms; n+1 = 1 = singlet for each.

Q2. Ethanol gives three peaks in $^1$H NMR. (a) State the chemical shift, multiplicity and integration ratio for each. (b) Explain the splitting using n+1. [4 marks]

• Cue. (a) $\delta \sim 1.2$ triplet (3H, $\text{CH}_3$); $\delta \sim 3.6$ quartet (2H, $\text{CH}_2$); $\delta \sim 2.5$ singlet (1H, OH, broad). (b) $\text{CH}_3$ has 2 neighbours ($n+1 = 3$, triplet); $\text{CH}_2$ has 3 ($n+1 = 4$, quartet); OH exchangeable so singlet.

Q3. HPLC of a mixture of acetaminophen and caffeine shows peaks at $R_t = 3.2 \, \text{min}$ (area $4200$) and $R_t = 5.8 \, \text{min}$ (area $8800$). Standards give calibration slopes $42.0$ and $22.0$ ($\text{area/(mg/L)}$). (a) Identify which $R_t$ is which. (b) Calculate concentration of each. (c) State one limitation of identifying by $R_t$ alone. [2+2+2 marks]

• Cue. (a) Earlier $R_t$ more polar; acetaminophen more polar than caffeine, so $3.2$ is acetaminophen. (b) Acetaminophen $= 4200 / 42.0 = 100 \, \text{mg/L}$; caffeine $= 8800 / 22.0 = 400 \, \text{mg/L}$. (c) Other compounds may share $R_t$; confirm with MS or NMR.