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NMR spectroscopy and chromatography (A-level)

Nuclear magnetic resonance (NMR) spectroscopy is the most powerful method for determining organic structure, and chromatography is the workhorse for separating and identifying mixtures. This A-level-only topic covers the principle of NMR and the chemical shift, carbon-13 and proton NMR (including integration and spin-spin splitting), and the thin-layer, column and gas chromatographic techniques.

4 sections~13 min reading time3 competenciesLevel Standard 1 · Advanced 3

T·181818 / 18

Exam profile

AO1 · Explain the basis of NMR and chromatography and the role of TMS and the two phasesAO2 · Deduce the number of environments, integration ratios and splitting patterns and calculate Rf valuesAO3 · Combine NMR, mass and infrared spectra to determine a complete structure

Operators:stateexplaindeduceinterpretcalculatedetermine

basic level

This entire topic is A-level only (A2); it is not assessed at AS.

higher level

The full A-Level treats the principle of NMR, carbon-13 and proton NMR with integration and spin-spin splitting, and thin-layer, column and gas chromatography, combining them with the other spectroscopic methods for structure determination.

Depth

Text

Contents · 4 sections

NMR spectroscopy and chromatography (A-level)

The principle of NMR and the chemical shift#

AdvancedAQA 7405 3.3.16.1DfE GCE Chemistry - NMR spectroscopy

Typical proton chemical-shift ranges

Fig. 1A data sheet of characteristic shifts lets each peak be assigned to a type of proton environment.

Nuclear magnetic resonance uses the fact that certain nuclei - notably

^1\text{H}

and

^{13}\text{C}

- behave like tiny magnets because of their nuclear spin. When placed in a strong external magnetic field, these nuclei can line up with or against the field, at two slightly different energy levels. Radio-frequency radiation of exactly the right energy makes a nucleus flip from the lower to the higher level (resonance), and the absorption is detected. The energy needed depends on the nucleus's electronic environment, which is what makes NMR so informative about structure.

The exact frequency at which a nucleus resonates depends on its chemical environment, because the electrons around it shield it from the external field to different extents. This difference is measured as the chemical shift,

\delta

, on a scale in parts per million (ppm). Nuclei in different environments absorb at different chemical shifts, so the number and positions of the peaks reveal how many different environments there are and what kind they are.

All chemical shifts are measured relative to a reference standard, tetramethylsilane (TMS,

\text{Si(CH}_3)_4

), which is assigned a chemical shift of exactly

0

. TMS is chosen because it is chemically inert, non-toxic and volatile (easily removed afterwards), it gives a single strong peak (all its hydrogens and carbons are equivalent), and its peak is well away from those of most organic compounds. A small amount is added to the sample as an internal standard.

Samples are often dissolved in a solvent with no

^1\text{H}

atoms, such as

\text{CDCl}_3

(deuterated chloroform) or

\text{CCl}_4

, so that the solvent does not produce peaks of its own that would obscure the sample's spectrum. Reading an NMR spectrum begins with counting the peaks (the number of environments) and noting their chemical shifts (which environment each represents), using a data sheet of characteristic shift ranges.

Worked example

Why TMS is the reference

Explain why tetramethylsilane is used as the standard in NMR, and what chemical shift it is given.

Assigned shift
TMS is assigned a chemical shift of exactly 0 ppm, the reference point of the scale.
Single peak
All twelve hydrogens (and all four carbons) are equivalent, so TMS gives one strong, sharp reference peak.
Inert and removable
It is chemically inert, non-toxic and volatile, so it does not react with the sample and is easily removed afterwards; its peak is well away from most organic signals.

Result: TMS (delta = 0) gives a single sharp peak, is inert and volatile, and absorbs away from other signals, making it an ideal reference.

Typical mistakes

Forgetting why TMS is chosen (inert, volatile, single peak, well away from other signals).
Using a solvent containing hydrogen, which would add its own peaks to a proton spectrum.

Active revision

State three reasons why tetramethylsilane (TMS) is used as the reference standard in NMR spectroscopy.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Chemistry 7405 specification (AQA)

Carbon-13 NMR and carbon environments#

AdvancedAQA 7405 3.3.16.1DfE GCE Chemistry - carbon-13 NMR

Carbon-13 NMR of ethanol

Fig. 2Ethanol has two carbon environments (CH3 and CH2), so its carbon-13 spectrum shows two peaks; TMS is at 0.

Carbon-13 NMR detects the

^{13}\text{C}

nuclei in a molecule (the small proportion of carbon that is the

^{13}\text{C}

isotope). The key rule is simple and powerful: the number of peaks in a

^{13}\text{C}

NMR spectrum equals the number of different carbon environments in the molecule. Two carbon atoms are in the same environment only if they are chemically equivalent - bonded to the same groups in the same way, often related by symmetry.

The chemical shift of each peak indicates the type of carbon environment, read from a data sheet. Carbons in different settings - an alkyl carbon, a carbon next to oxygen, a carbonyl carbon, an aromatic carbon - appear at characteristic shift ranges spread over roughly

0

220\ \text{ppm}

. So the spectrum tells you both how many carbon environments there are (the number of peaks) and what kind each is (the shift).

Counting carbon environments requires care with symmetry. In ethanol (

\text{CH}_3\text{CH}_2\text{OH}

) the two carbons are different, so there are two peaks; in propane (

\text{CH}_3\text{CH}_2\text{CH}_3

) the two end carbons are equivalent by symmetry, so there are only two environments (the two

\text{CH}_3

and the central

\text{CH}_2

), giving two peaks. Working out which carbons are equivalent, using the symmetry of the molecule, is the main skill.

Carbon-13 NMR is often the quickest way to narrow down a structure, because a simple peak count immediately tells you the number of carbon environments, which strongly constrains the possibilities. It is simpler to interpret than proton NMR because the peaks are single lines (there is no spin-spin splitting to consider in the routine spectra you meet). Combined with the molecular formula, the number and shifts of the peaks often identify the structure.

Worked example

Counting carbon environments

How many peaks would you expect in the carbon-13 NMR spectrum of (a) ethanol and (b) propanone?

Ethanol
CH3CH2OH has two different carbons (a CH3 and a CH2), so two environments and two peaks.
Propanone
CH3COCH3 has two equivalent CH3 carbons (by symmetry) and one C=O carbon, so two environments.
Conclude
Both molecules give two peaks, but for different reasons - ethanol's carbons are simply different, whereas propanone's two CH3 carbons are equivalent.

Result: Ethanol gives two peaks (two different carbons); propanone gives two peaks (the two CH3 carbons are equivalent, plus the C=O).

Typical mistakes

Counting the number of carbon atoms instead of the number of different carbon environments.
Missing symmetry-equivalent carbons (e.g. the two end carbons of propane are one environment).

Active revision

Predict the number of peaks in the carbon-13 NMR spectrum of propan-2-one ( $\text{CH}_3\text{COCH}_3$ ), explaining your answer.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

Proton NMR: integration and spin-spin splitting#

AdvancedAQA 7405 3.3.16.1DfE GCE Chemistry - proton NMR

Proton NMR of ethanol

Fig. 3Ethanol shows three peaks (CH3, CH2, OH) with integration 3:2:1; the CH3 is a triplet and the CH2 a quartet by the n+1 rule.

Proton (

^1\text{H}

) NMR detects the hydrogen atoms in a molecule and carries three kinds of information. The number of peaks (or groups of peaks) gives the number of different proton environments; the chemical shift of each gives the type of environment; and two further features - integration and spin-spin splitting - reveal how many protons are in each environment and how many are on the neighbouring carbon. This makes proton NMR the richest single technique for structure determination.

Integration measures the relative area under each peak, which is proportional to the number of protons in that environment. The spectrum shows the integration as a ratio (or a stepped integration trace), so a peak twice the area of another represents twice as many protons. For ethanol the three environments

\text{CH}_3

\text{CH}_2

and

\text{OH}

have integrations in the ratio

3:2:1

, matching the three, two and one hydrogens they contain.

Spin-spin splitting (coupling) splits a peak into a group of lines, because the magnetic field felt by a proton is slightly affected by the hydrogens on the adjacent carbon. The number of lines a peak is split into is given by the

n+1

rule: a proton with

n

equivalent protons on the neighbouring carbon is split into

n+1

lines. So a

\text{CH}_3

next to a

\text{CH}_2

(two neighbours) appears as a triplet (

2+1

), and a

\text{CH}_2

next to a

\text{CH}_3

(three neighbours) appears as a quartet (

3+1

The three pieces of information work together. From the spectrum of ethanol you deduce three environments (three peaks), integration

3:2:1

\text{CH}_3

, a

\text{CH}_2

and an

\text{OH}

), and a triplet and a quartet (from the

\text{CH}_3

and

\text{CH}_2

being adjacent) - which together confirm the ethyl group. Adding a few drops of

\text{D}_2\text{O}

makes any

\text{O-H}

\text{N-H}

proton peak disappear (the labile proton is exchanged for deuterium), which identifies those protons. Interpreting an unknown's proton NMR means combining the number of peaks, the shifts, the integration and the splitting.

\text{lines} = n + 1

The n+1 splitting rule

A proton with n equivalent protons on the neighbouring carbon is split into n+1 lines (singlet, doublet, triplet, quartet, ...).

Worked example

Interpreting the proton NMR of ethanol

Predict the proton NMR spectrum of ethanol (CH3CH2OH): the number of peaks, the integration ratio and the splitting of the CH3 and CH2 peaks.

Number of environments
There are three proton environments: CH3, CH2 and OH, so three peaks.
Integration
The integration ratio is 3 : 2 : 1, matching three, two and one hydrogens.
Splitting
The CH3 has two neighbouring protons (the CH2), so it is a triplet (2+1); the CH2 has three neighbouring protons (the CH3), so it is a quartet (3+1); the OH is usually a singlet.

Result: Three peaks with integration 3:2:1; the CH3 is a triplet and the CH2 a quartet, confirming the ethyl group.

Typical mistakes

Applying the n+1 rule to the protons within the same group rather than to those on the neighbouring carbon.
Confusing integration (number of protons in an environment) with the number of lines from splitting.

Active revision

Predict the number of peaks, the integration ratio and the splitting pattern in the proton NMR spectrum of ethanal, $\text{CH}_3\text{CHO}$ .

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

Chromatography: TLC, column and gas chromatography#

StandardAQA 7405 3.3.16.2DfE GCE Chemistry - chromatography

Thin-layer chromatography and the Rf value

Fig. 4Rf is the distance moved by the spot divided by the distance moved by the solvent front, measured from the baseline.

Chromatography separates the components of a mixture by their different distributions between two phases: a stationary phase (fixed in place) and a mobile phase (which moves through or over it). Each component spends a different proportion of its time in each phase according to how strongly it is attracted to the stationary phase, so the components travel at different speeds and separate. The more strongly a component is held by the stationary phase, the more slowly it moves.

In thin-layer chromatography (TLC) the stationary phase is a thin layer of silica (or alumina) on a plate, and the mobile phase is a liquid solvent that rises up the plate by capillary action. A spot of the mixture is placed on a pencil baseline; as the solvent rises, the components separate into spots. Each component is described by its retention factor,

R_f = \dfrac{\text{distance moved by the spot}}{\text{distance moved by the solvent front}}

, a value between

0

and

1

that is characteristic of the substance under those conditions and can be compared with known standards.

Column chromatography uses a vertical column packed with the stationary phase, through which the mobile-phase solvent flows under gravity; the components emerge (elute) from the bottom at different times and can be collected separately, so it can be used to purify as well as to identify. Gas chromatography (GC) uses an inert carrier gas as the mobile phase and a long column coated with a stationary phase; each component has a characteristic retention time (the time to pass through), and the area under each peak is proportional to the amount of that component.

Chromatography is widely used to separate, identify and quantify the components of mixtures - in forensic analysis, drug testing, and checking the purity of a product. It is often coupled to another technique: gas chromatography-mass spectrometry (GC-MS) first separates a mixture by GC and then identifies each component by its mass spectrum, combining the separating power of chromatography with the identifying power of mass spectrometry. When you use a chromatogram, measure distances (for

R_f

) or read retention times, and compare with standards to identify the components.

R_f = \dfrac{\text{distance moved by the spot}}{\text{distance moved by the solvent front}}

Retention factor

Both distances are measured from the baseline; Rf lies between 0 and 1 and is characteristic of the substance under set conditions.

Worked example

Calculating an Rf value

On a TLC plate the solvent front travels 6.0 cm from the baseline and a spot travels 4.5 cm. Calculate the Rf value of that component.

Identify the distances
Distance moved by the spot = 4.5 cm; distance moved by the solvent front = 6.0 cm (both from the baseline).
Apply the formula
$R_f = \dfrac{4.5}{6.0}$ .
$R_f = \dfrac{4.5}{6.0} = 0.75$
Check the range
0.75 lies between 0 and 1, as an Rf must.

Result: The Rf value is 0.75, which can be compared with reference values to identify the component.

Typical mistakes

Measuring the spot distance from the bottom of the plate instead of from the baseline.
Confusing the roles of the mobile and stationary phases, or forgetting that Rf must lie between 0 and 1.

Active revision

On a TLC plate the solvent front moves $8.0\ \text{cm}$ from the baseline and a component's spot moves $5.6\ \text{cm}$ . Calculate the $R_f$ value of the component.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

References & sources

Sources

Department for Education

GCE AS and A level subject content for the sciences

AQA

AQA A-level Chemistry 7405 specification

NMR spectroscopy and chromatography (A-level)

4 sections~13 min reading time3 competenciesLevel Standard 1 · Advanced 3

NMR spectroscopy and chromatography (A-level)

Key points

Why TMS is the reference

Key points

Counting carbon environments

Key points

Interpreting the proton NMR of ethanol

Key points

Calculating an Rf value

Sources

NMR spectroscopy and chromatography (A-level)

Key points

Why TMS is the reference

Key points

Counting carbon environments

Key points

Interpreting the proton NMR of ethanol

Key points

Calculating an Rf value

Sources