Spectroscopy and Structure Determination
Spectroscopy and Structure Determination
Lesduur is: 50 minOnderdelen in deze les
Spectroscopy and Structure Determination
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Learning objective
I can use mass spectrometry, infrared spectroscopy, NMR and chromatography evidence to work out the structure of an organic compound.
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What clues can help identify an unknown organic compound?
Use this as prior-knowledge activation after the learning objective. Likely responses: Mr, peaks, functional group, IR, NMR, fragments, chromatography, formula, isomers. Invite students to notice which methods they already know and which are less familiar.
Introduction
This KS5 chemistry lesson explores Spectroscopy and Structure Determination. Students combine mass spectrometry, infrared spectroscopy, NMR and chromatography evidence to identify organic compounds. It is suitable for A level classroom teaching, independent study or exam revision.
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Why structure determination matters
Chemists rarely identify an unknown organic compound from one test alone.
Instead, they combine several pieces of evidence:
mass spectrometry for relative molecular mass and fragmentation
infrared spectroscopy for bond vibrations and functional groups
NMR spectroscopy for different hydrogen or carbon environments
chromatography for separating mixtures before analysis
The key skill is to build one structure that fits all the evidence.
Students often treat each method in isolation. Emphasise that A level questions reward combining evidence rather than naming a result from one spectrum.
A logical order for analysis
A useful exam approach is:
start with molecular formula if given
use mass spectrometry to check Mr
use IR to spot likely functional groups
use NMR to count environments and splitting
use chromatography when purity or separation matters
Each method narrows the possibilities.
If students know index of hydrogen deficiency, you may mention it orally, but it is not essential unless already part of their course specification.
Mass spectrometry: what it shows
In mass spectrometry, molecules are ionised and detected according to their mass-to-charge ratio, m/z.
At A level, the most useful ideas are:
the molecular ion peak gives the mass of the molecule as an ion
its m/z value usually shows the relative molecular mass, Mr
smaller peaks often come from fragment ions
the pattern of fragments can support a proposed structure
Keep the focus on interpretation rather than instrument design. Students do not need a full account of time-of-flight unless required by their board.
Mass spectrometry: molecular ion and fragments
A mass spectrum usually contains:
a molecular ion peak, M⁺
fragment peaks from bonds breaking
a base peak, the tallest peak, showing the most abundant ion
Example idea:
an alcohol may show a molecular ion peak for the whole molecule
it may also show fragment peaks formed when C–C or C–O bonds break
Fragments do not identify the whole structure on their own, but they give useful clues.
Students may confuse the base peak with the molecular ion peak. Point out that the tallest peak is not always the whole molecule.
Using Mr to narrow options
If the molecular ion peak is m/z 60, the compound has Mr 60.
That immediately rules out many possible structures.
If the molecular formula is also known, chemists can compare fragment peaks with likely bond breakages and ask which structure best matches the spectrum.
You can briefly model how two isomers with the same Mr need extra evidence from IR or NMR to be distinguished.
Which peak is usually used to find Mr?
base peak
molecular ion peak
fragment peak
integration peak
Checks the common confusion flagged on the mass spectrometry slides: the tallest peak is not necessarily the molecular ion peak.
Mass spectrometry recap
In mass spectrometry, the molecular ion peak usually gives the relative molecular mass. Smaller peaks are often fragment ions formed when bonds break. The base peak is the tallest peak and shows the most abundant ion.
All three blanks are load-bearing terms from slides 5 to 7.
Infrared spectroscopy: bond vibrations
Infrared spectroscopy works because different bonds absorb infrared radiation at different frequencies.
When a bond absorbs energy, it vibrates more strongly. The absorption appears as a peak on an IR spectrum.
Chemists use the positions of peaks to identify functional groups rather than the full carbon skeleton.
Stress particle-level reasoning: absorption matches bond vibration frequencies. Students should not think IR gives a complete structure by itself.
Infrared spectroscopy: key absorptions
Useful A level absorptions include:
O–H in alcohols: broad peak around 3230–3550 cm⁻¹
C=O: strong peak around 1680–1750 cm⁻¹
C–H: absorptions around 2850–3300 cm⁻¹ depending on bond type
These ranges help identify which functional groups are present or absent.
Keep values in ranges rather than pretending every spectrum gives an exact single number.
What IR can and cannot tell you
IR is strong at identifying functional groups.
It is weaker at telling you:
the exact arrangement of every carbon atom
which structural isomer is present when isomers share the same functional group
So IR is most useful when combined with mass spectrometry and NMR.
A useful contrast is ethanol vs dimethyl ether: different functional groups, so IR helps. But propanone vs propanal both contain C=O, so extra evidence is needed.
What is IR most useful for identifying?
carbon skeleton
exact isomer
functional groups
Mr value
Distractors match likely overclaims students make after seeing a spectrum. IR identifies functional groups, not the full structure.
Infrared spectroscopy recap
Infrared spectroscopy works because bonds absorb infrared radiation and vibrate more strongly. Chemists use peak positions to identify functional groups rather than the full carbon skeleton. A broad O–H absorption appears around 3230–3550 cm⁻¹.
The final blank uses the taught O–H range rather than an untaught exact value.
NMR spectroscopy: chemical environments
NMR spectroscopy shows how nuclei in a molecule experience different magnetic environments.
For A level structure determination, students mainly use:
the number of peaks
chemical shift values
peak area or integration
splitting patterns
Together, these features reveal how atoms are arranged.
If useful, remind students that 1H NMR and 13C NMR give different information, but both are based on chemical environment.
1H NMR: number of peaks and integration
In ¹H NMR:
each peak represents a different proton environment
the integration trace shows the relative number of protons in each environment
Example:
a ratio of 3:2:1 means three sets of protons are present
their actual numbers could be 3, 2 and 1, or any whole-number multiple with the same ratio
Equivalent protons produce the same signal.
Students often think integration gives the total number of hydrogens directly. Clarify that it gives a ratio first.
1H NMR: splitting patterns
Splitting happens because neighbouring non-equivalent protons affect each other.
The A level rule is:
n neighbouring protons → n + 1 peaks
Common patterns:
singlet: 0 neighbours
doublet: 1 neighbour
triplet: 2 neighbours
quartet: 3 neighbours
Splitting helps connect proton environments within the molecule.
You may mention that exchangeable protons such as O–H can behave differently, but keep the main slide on the standard rule.
13C NMR: carbon environments
In ¹³C NMR:
each peak represents a different carbon environment
the number of peaks helps show symmetry in the molecule
fewer peaks than carbons means some carbons are equivalent
¹³C NMR is especially useful for comparing possible structural isomers.
Students do not usually use splitting in 13C NMR at this level. Keep the focus on counting environments.
In 1H NMR, a triplet shows how many neighbouring protons?
1
2
3
4
Applies the n + 1 rule from the splitting slide.
NMR recap
In ¹H NMR, each peak represents a different proton environment. The integration trace shows the relative number of protons in each environment as a ratio. In ¹³C NMR, fewer peaks than carbons suggests some carbons are equivalent.
Targets three common ideas students mix up: environment, ratio, and equivalence.
What does a triplet–quartet pair suggest in 1H NMR?
Expected elements: a triplet means 2 neighbouring protons; a quartet means 3 neighbouring protons; together this often suggests adjacent CH3 and CH2 groups, such as an ethyl pattern. Accept concise explanations using the n + 1 rule.
Chromatography: separation before identification
Chromatography separates substances because components move at different speeds between a mobile phase and a stationary phase.
In structure determination, chromatography is useful when:
a sample contains more than one substance
a pure compound must be isolated before spectra are interpreted
retention data can support identification
Link to practical chemistry: many real samples are mixtures, so separation may come before spectral analysis.
Paper and thin-layer chromatography
At A level, chromatography evidence may include Rf values.
The formula is:
Rf = distance moved by substance / distance moved by solvent
Important points:
Rf values are compared under the same conditions
matching Rf values can support identification
chromatography shows whether a sample is likely to be pure or mixed
A matching Rf does not prove identity on its own; students should combine it with other evidence.
Why one method is not enough
A compound may share one piece of evidence with several others.
Examples:
the same Mr can belong to different isomers
the same functional group can appear in different molecules
a similar NMR pattern may still fit more than one skeleton
Chemists reach stronger conclusions by combining independent lines of evidence.
This is the key transition slide before the worked example. Emphasise convergence of evidence.
When can Rf values be compared fairly?
same conditions
same colour
same mass
same formula
Checks the key condition stated on the chromatography slide.
Chromatography and combined evidence
Chromatography separates substances between a mobile phase and a stationary phase. It can show whether a sample is pure or mixed, but identification becomes stronger when evidence from several methods is combined.
This bridges chromatography with the transition slide about combining evidence.
Worked example: combining the evidence
Suppose an unknown compound has:
molecular ion peak at m/z 60
strong IR absorption near 1700 cm⁻¹
¹H NMR showing one peak only
Reasoning:
m/z 60 suggests Mr 60
IR near 1700 cm⁻¹ suggests a C=O group
one ¹H NMR peak means all protons are in the same environment
A structure consistent with all three clues is propanone.
Use this to model the exam habit of ruling in and ruling out. Propanone fits because the two methyl groups are equivalent.
A step-by-step exam method
A reliable structure-determination routine is:
Identify Mr from the molecular ion peak.
Check IR for key functional groups.
Count NMR environments.
Use integration and splitting to connect groups.
Check whether the final structure matches every piece of evidence.
If one clue does not fit, the structure must be revised.
Encourage students to annotate spectra and eliminate impossible isomers as they go.
Common pitfalls
Watch for these errors:
choosing the base peak instead of the molecular ion peak for Mr
assuming IR gives the full structure
forgetting that integration gives a ratio
misapplying the n + 1 rule
accepting a structure that matches most, but not all, of the data
Good structure determination is about consistency.
This slide anticipates typical A level mistakes and can be tied to exam technique.
Which structure best fits Mr 60, C=O and one 1H NMR peak?
propanal
propanone
propan-1-ol
ethanoic acid
Applied task based directly on the worked example. The distractors are plausible Mr/functional-group confusions, but only propanone fits all three clues together.
Which clue do you find most helpful when starting a structure question?
Mr
IR peaks
NMR peaks
chromatography
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What is one thing you'll remember from this lesson?
Reflection only. Invite students to name one durable takeaway such as using the molecular ion peak for Mr, IR for functional groups, or checking that every clue fits.
Which part of structure determination still feels hardest?
Reflection only. Encourage honest responses about difficulty, for example NMR splitting, integration ratios, or combining evidence from different methods.
Glossary
mass spectrometry - an analytical method that forms ions and detects them by mass-to-charge ratio
molecular ion peak - the peak produced by the whole molecule as an ion, usually used to find Mr
fragment ion - a smaller ion formed when the molecular ion breaks apart
base peak - the tallest peak in a mass spectrum, showing the most abundant ion
infrared spectroscopy - a method that identifies bonds and functional groups by their absorption of infrared radiation
functional group - a group of atoms that gives a molecule characteristic chemical properties
chemical shift - the position of an NMR signal, showing the magnetic environment of a nucleus
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Glossary
integration - the relative area under an NMR signal, showing the ratio of protons in each environment
splitting pattern - the way an NMR peak is divided by neighbouring non-equivalent protons
proton environment - a set of equivalent hydrogen atoms in the same chemical environment
carbon environment - a carbon atom in a distinct chemical environment within a molecule
stationary phase - the phase in chromatography that stays in place while substances move past it
mobile phase - the phase in chromatography that carries substances through or over the stationary phase
Rf value - the distance moved by a substance divided by the distance moved by the solvent
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