Interactive revision notes for ¹H NMR, ¹³C NMR, IR, mass spec, TLC and GLC
What is spectroscopy?
Spectroscopy and chromatography are analytical techniques used to identify unknown compounds and confirm molecular structure.
The 6 techniques you need to know
¹H NMR
Hydrogen environments
¹³C NMR
Carbon skeleton
IR spectroscopy
Functional groups
Mass spec
Molecular mass + fragments
TLC
Separate + identify mixtures
GLC
Separate volatile mixtures
Exam tip: These techniques are often combined. You might be given an IR spectrum and a mass spectrum for the same compound and asked to identify it. Practice using them together!
¹H NMR — proton NMR
Tells you about the hydrogen atoms in a molecule
What it doesIdentifies the number of different hydrogen environments and how many H atoms are in each group
X-axis (δ / ppm)Chemical shift — how far a proton is shifted from TMS (0 ppm). Higher δ = more deshielded.
Number of peaks= number of different H environments
IntegrationRelative ratio of H atoms in each environment
Splitting (n+1)n neighbouring H atoms → n+1 peaks. Singlet (no neighbours), doublet (1), triplet (2), quartet (3)
Diagram — ¹H NMR spectrum of ethanol (CH₃CH₂OH)
CH₃—CH₂—OH
← Click a group on the molecule to highlight its peak
Click each group (CH₃, CH₂, OH) to highlight its peak. Use the buttons to switch between spectrum, integration, and splitting views.
Common chemical shifts
Group
δ (ppm)
Memory trick
TMS reference
0
Starting point
–CH₃, –CH₂ (alkyl)
0.5–2.0
Low shift = shielded
–O–CH₃ (ether)
3.1–3.9
O pulls electrons away
–C=C–H (vinyl)
4.5–6.5
—
Aromatic (ArH)
6.5–8.5
Ring current deshields
–CHO (aldehyde)
9.5–10.0
Very deshielded
–COOH (carboxylic acid)
10–12
Highest shift
Splitting shortcut: Doublet = 1 neighbour, triplet = 2, quartet = 3. Singlet = no H neighbours at all.
¹³C NMR — carbon NMR
Tells you about the carbon skeleton of a molecule
What it doesShows the number of different carbon environments in a molecule
Number of peaks= number of different carbon environments (symmetry reduces this!)
Peak heightNOT proportional to number of carbons — unlike ¹H NMR, peak area cannot count carbons
No splittingPeaks are always single lines — much simpler than ¹H NMR
ReferenceTMS at 0 ppm, same as ¹H NMR
Diagram — ¹³C NMR of propanone (CH₃COCH₃)
Propanone has 3 carbons but gives only 2 peaks — the two CH₃ groups are chemically equivalent by symmetry.
Common carbon chemical shifts
Carbon type
δ (ppm)
Alkyl C–C
5–50
C–O (ether, alcohol)
50–90
C=C (alkene)
100–150
Aromatic C
110–160
C=O (ester, amide)
155–185
C=O (aldehyde, ketone)
185–220
Key difference from ¹H NMR: Peak height means nothing. Only the number of peaks matters — it tells you how many unique carbon environments exist.
Infrared (IR) spectroscopy
Identifies functional groups by how bonds absorb infrared radiation
PrincipleMolecules absorb IR radiation at specific frequencies, causing bonds to vibrate (stretch/bend)
X-axisWavenumber (cm⁻¹), read right to left from ~4000 to 500 cm⁻¹
Y-axis% Transmittance — dips (troughs) downward mean absorption = that bond is present
Fingerprint regionBelow 1500 cm⁻¹ — unique to each molecule. Used to confirm identity against a database.
Diagram — IR spectrum of ethanol with key absorptions labelled
Troughs (dips) = absorption. The broad O–H trough is the key identifier for alcohols. The fingerprint region (shaded) is unique to each compound.
Key absorptions to memorise
Bond / Group
Wavenumber (cm⁻¹)
Notes
O–H (alcohol)
3200–3550
Broad, strong
O–H (carboxylic acid)
2500–3300
Very broad, overlaps C–H
N–H (amine/amide)
3100–3500
Medium, broad
C–H (alkane)
2850–3000
Medium
C∹N (nitrile)
2200–2260
Sharp, medium
C=O (carbonyl)
1630–1820
Strong, sharp — very important!
C=C (alkene)
1620–1680
Medium
C–O (ether/ester/alcohol)
1000–1300
Strong
Top tip: The C=O peak position tells you the type: ketone/aldehyde (~1715), ester (~1735), carboxylic acid (~1710). Broad O–H = alcohol or acid.
Mass spectrometry
Determines molecular mass and structural fragments of a molecule
PrincipleMolecules are vaporised and bombarded with electrons, forming positive ions separated by mass-to-charge ratio (m/z)
Molecular ion M⁺Peak at highest m/z = molecular mass of the compound (also called the parent ion)
Base peakThe tallest peak — the most stable fragment produced
FragmentationM⁺ breaks into smaller pieces. The pattern of fragments helps identify structure.
m/z axisEffectively the mass of each fragment (charge is usually +1)
Diagram — mass spectrum of pentan-2-one (Mr = 86)
Pentan-2-one (Mr = 86). Loss of CH₃CO (mass 43) from M⁺ gives the base peak. The M⁺ peak at m/z 86 confirms the molecular mass.
Common fragment masses
m/z lost
Fragment lost
Suggests…
15
CH₃
Methyl group present
17
OH
Alcohol or acid
29
CHO or C₂H₅
Aldehyde or ethyl group
43
CH₃CO
Acetyl / methyl ketone
45
OC₂H₅ or COOH
Ethoxy or carboxyl
77
C₆H₅
Benzene ring
M+2 peaks: A large M+2 peak (similar height to M⁺) strongly suggests chlorine. Two similar M+2 peaks suggest bromine — both isotopes are roughly equally abundant.
M+1 peak — calculating the number of carbon atoms
The M+1 method
Using natural ¹³C abundance to count carbon atoms
Why M+1 existsCarbon-13 (¹³C) makes up 1.1% of all naturally occurring carbon. In a molecule with n carbon atoms, the probability that at least one of them is ¹³C is roughly 1.1% × n. This creates a small peak one mass unit above M⁺, called the M+1 peak.
The formulan (carbon atoms) = (M+1 intensity %) ÷ 1.1 where M+1 intensity % = (height of M+1 peak ÷ height of M⁺ peak) × 100
Worked exampleAn unknown compound has M⁺ at m/z = 72. The M+1 peak has intensity 4.4% relative to M⁺.
n = 4.4 ÷ 1.1 = 4 carbon atoms
This tells us the molecular formula contains 4 C. With Mr = 72 and 4 C (mass 48), the remaining mass = 24, which could be H₂₄ (impossible) so we consider other elements. With 4C + 8H + 1O = 48+8+16 = 72 ✓ → possible formula C₄H₈O.
LimitationsThe M+1 peak is small and can overlap with noise. Results should always be rounded to the nearest whole number. Other elements also contribute small M+1 peaks (e.g. ³³S, ²H, ¹⁵N) but ¹³C dominates for most organic molecules. This method works best alongside high-resolution MS data.
M+1 carbon calculator
Enter peak intensities to find the number of carbon atoms
Step-by-step method for exam questions
Step 1Measure the height of M⁺ (the molecular ion peak) and the M+1 peak from the spectrum. Use mm or any consistent unit.
Step 3Divide by 1.1 to get the number of carbon atoms: n = M+1% ÷ 1.1. Round to the nearest whole number.
Step 4Use n along with the molecular mass (from M⁺) to work out possible molecular formulae. Subtract the mass of n carbons from Mr, then account for the remaining mass using H, O, N etc.
Watch out: If the molecule contains Cl, Br, or S, these elements also contribute to M+1/M+2 peaks. The 1.1% × n formula is most reliable for purely C/H/O/N compounds.
Thin layer chromatography (TLC)
Separates mixtures and identifies compounds by comparing Rf values
Stationary phaseSilica gel (polar) coated on an aluminium or glass plate
Mobile phaseA solvent that travels up the plate by capillary action
PrincipleComponents travel different distances depending on how strongly they adsorb to the stationary phase vs. dissolve in the mobile phase
Rf value= distance moved by spot ÷ distance moved by solvent front. Always between 0 and 1.
VisualisationUV lamp (fluorescent plate), or iodine vapour if spots are colourless
Diagram — TLC plate showing a mixture vs reference compounds
Two spots in the mixture lane match the Rf values of references A and B — the mixture contains both compounds.
Step-by-step procedure
1.Draw a pencil baseline 1 cm from the bottom of the plate
2.Spot the sample(s) using a capillary tube onto the baseline
3.Place in a beaker with solvent below the baseline (don't submerge the spots!)
4.Allow solvent to rise nearly to the top, remove and mark the solvent front
5.Visualise spots, measure distances and calculate Rf values
Rf interpretation: A pure compound always gives the same Rf in the same solvent. Same Rf = likely same compound. Multiple spots = mixture.
Gas-liquid chromatography (GLC)
Separates volatile mixtures and can quantify each component
Stationary phaseA high-boiling liquid coated on an inert solid inside a long coiled column in an oven
Mobile phaseAn inert carrier gas — usually nitrogen or helium
PrincipleComponents partition between carrier gas and liquid stationary phase. Those that dissolve more in the liquid take longer to travel through the column.
Retention timeTime for a component to pass through the column — used to identify by comparison to standards
Peak areaProportional to the amount of each component — used for quantitative analysis
Diagram — GLC apparatus and chromatogram output
Component B has a longer retention time (more soluble in stationary phase) and a larger peak area (greater quantity in sample).
Factors affecting separation
Factor
Effect
Higher oven temperature
Shorter retention times — faster but less separation
Longer column
Better separation but longer analysis time
Polar stationary phase
Separates polar compounds better
Faster carrier gas flow
Shorter time but less resolution
GLC vs TLC: GLC only works for volatile substances but gives precise quantitative results. TLC is simpler and works for non-volatile compounds too.
Technique
What it tells you
What you measure
Key output
Limitations
¹H NMR
H environments, ratio, neighbours
δ (ppm), integration, splitting
Number & ratio of H environments
Complex spectra; TMS needed; expensive
¹³C NMR
Number of different C environments
δ (ppm); number of peaks
Carbon skeleton; symmetry
Peak height not proportional to C count
IR spec
Functional groups present
Wavenumber (cm⁻¹) of troughs
Functional group ID; fingerprint
Can't determine molecular mass alone
Mass spec
Molecular mass and fragments
m/z of M⁺ and fragments
Exact Mr; fragmentation pattern
Fragmentation complex to interpret
TLC
Components in mixture; purity
Rf = spot dist ÷ solvent dist
Identity by Rf; 1 spot = pure
Qualitative only; non-volatile only
GLC
Components and amounts in volatile mixtures
Retention time; peak area
Identity + quantitative by peak area
Volatile substances only; expensive
Combined analysis: Work systematically — mass spec → molecular mass; IR → functional group; NMR → H/C environments. Cross-check everything!
¹H NMR — advanced points
Deuterated solvents & TMS
Why we use special solvents in NMR
Deuterated solventCDCl₃ (deuterated chloroform) or D₂O are used because ¹H NMR only detects hydrogen-1. Deuterium (²H) is invisible, so the solvent doesn't interfere with the spectrum.
TMS purposeTetramethylsilane Si(CH₃)₄ is the reference standard at 0 ppm. It is chemically inert, volatile, and all 12 H atoms are equivalent — giving one sharp peak well away from most sample peaks.
D₂O shake testAdding D₂O causes exchangeable protons (–OH, –NH) to disappear from the spectrum, as H is replaced by D. This confirms which peak is O–H or N–H.
Deshielding explained
Why electronegative groups increase chemical shift
ShieldingElectron density around a proton opposes the applied magnetic field, reducing the effective field felt — so the proton resonates at lower δ (more upfield).
DeshieldingElectronegative atoms (O, N, Cl) withdraw electron density, reducing shielding — the proton resonates at higher δ (downfield). E.g. –OCH₃ at ~3.5 ppm vs alkyl –CH₃ at ~0.9 ppm.
Aromatic ring currentThe delocalised π electrons in benzene create a ring current that greatly deshields protons on the ring, pushing ArH peaks to 6.5–8.5 ppm — higher than you'd expect from electronegativity alone.
Interpreting splitting patterns — worked rules
Step-by-step approach for exam questions
Step 1Count the number of peaks in a given signal → subtract 1 → this is the number of neighbouring H atoms (n+1 rule).
Step 2Integration ratios tell you how many H atoms are in each environment. Always express as the simplest whole-number ratio.
Step 3OH and NH protons often appear as broad singlets and do not split adjacent CH protons (they exchange too rapidly in solution).
Step 4Cross-check: the splitting of peak A should be consistent with the number of H atoms shown by integration of peak B (the neighbouring environment).
Common trap: A singlet does NOT mean only one proton — it means no neighbouring H atoms. E.g. the –OCH₃ group in methyl ethanoate gives a singlet integrating for 3H.
¹³C NMR — advanced points
DEPT NMR (context — not always examined)
Tells you whether a carbon is CH₃, CH₂, CH or quaternary C
What it doesDEPT (Distortionless Enhancement by Polarisation Transfer) separates carbon signals by type. CH₃ and CH point up; CH₂ points down; quaternary C (e.g. C=O) is absent — it has no attached H.
Exam useIf the question gives you a DEPT spectrum alongside a ¹³C, use it to distinguish C=O (missing in DEPT) from other carbons in the same δ region.
Symmetry and ¹³C peak counting
RuleChemically equivalent carbons give ONE peak. Always check for symmetry before predicting the number of peaks.
ExampleBenzene (C₆H₆): all 6 carbons equivalent → 1 peak at ~128 ppm. Ethanoic acid (CH₃COOH): 2 different carbons → 2 peaks (~20 ppm and ~178 ppm).
Exam tipIf a molecule has a plane of symmetry, the number of ¹³C peaks ≤ half the number of carbons. E.g. propanone (3C, 2 peaks).
IR spectroscopy — advanced points
Distinguishing C=O environments precisely
The exact wavenumber of the carbonyl peak is diagnostic
Compound type
C=O position (cm⁻¹)
Other clue
Aldehyde
~1720–1740
C–H stretch at 2720 & 2820 cm⁻¹ (two weak peaks)
Ketone
~1705–1725
No O–H, no N–H peaks
Carboxylic acid
~1700–1725
Very broad O–H (2500–3300 cm⁻¹)
Ester
~1730–1750
C–O stretch at 1000–1300; no O–H peak
Amide
~1630–1690
N–H peak at 3100–3500 cm⁻¹
Acyl chloride
~1770–1815
Highest C=O wavenumber; no O–H
Memory tip: More electron-withdrawing groups attached to the C=O raise the wavenumber. Acyl chloride > ester > ketone/aldehyde > amide.
Hydrogen bonding and broad peaks
Why broad?O–H groups in alcohols and carboxylic acids form hydrogen bonds. This means O–H bonds exist in a range of slightly different environments, so absorption is spread over a wide wavenumber range rather than a sharp peak.
Carboxylic acid vs alcoholCarboxylic acid O–H is even broader (2500–3300 cm⁻¹) and often overlaps with the C–H region. Alcohol O–H (3200–3550 cm⁻¹) is broad but narrower and sits above the C–H peaks.
Mass spectrometry — advanced points
Isotope patterns in detail
Using M⁺, M+1 and M+2 to identify elements
Chlorine isotopes³⁵Cl : ³⁷Cl ≈ 3 : 1. A molecule with one Cl gives M⁺ and M+2 peaks in roughly 3:1 ratio. Two Cl atoms give M, M+2, M+4 in ~9:6:1.
Bromine isotopes⁷⁹Br : ⁸¹Br ≈ 1 : 1. Two peaks of almost equal height separated by 2 mass units — the classic "bromine doublet".
M+1 peakDue to ¹³C (1.1% natural abundance). Roughly 1.1% × number of C atoms gives the M+1 percentage — useful for counting carbons in high-resolution MS.
Quick rule: Equal-height M and M+2 → bromine. M+2 is 1/3 of M → chlorine.
High-resolution mass spectrometry (HRMS)
What it doesMeasures molecular mass to 4+ decimal places. This allows the molecular formula to be determined unambiguously — different formulae (e.g. C₂H₆O vs CH₂O₂) have slightly different exact masses.
Exact masses to knowH = 1.00794, C = 12.00000, N = 14.0031, O = 15.9949, Cl = 34.9689. These are not the same as Ar values — HRMS uses exact nuclear masses.
Chromatography — advanced points
Why Rf values can vary
Solvent polarityA more polar solvent increases Rf values for all spots (components spend more time in mobile phase). Changing solvent changes all Rf values — always state which solvent was used.
TemperatureHigher temperature slightly increases Rf by increasing solubility in the mobile phase.
Stationary phaseSilica (polar) is standard. Reversed-phase TLC uses non-polar stationary phase — polar compounds then have higher Rf.
GLC: quantitative analysis in practice
Peak areaArea under each GLC peak is proportional to the mole fraction of that component (assuming similar detector response). Used in reaction monitoring and purity analysis.
Internal standardA known amount of a reference compound is added to the sample. Comparing peak areas relative to the internal standard corrects for injection volume variations.
Coupled GC-MSGLC separates components; each component then passes directly into a mass spectrometer. This gives both retention time (identification) and a mass spectrum (confirmation of structure) — very powerful for unknowns.
GLC advantage over TLC: GLC is quantitative (peak area → amount) and separates very similar compounds with different boiling points. TLC is quicker and works for non-volatile compounds.
Combined structure determination — strategy
How to tackle a "identify the unknown compound" exam question
Step 1 — Mass specFind M⁺ → this gives Mr. Check for M+2 pattern (Cl or Br?). Note the base peak and any fragment losses (e.g. loss of 15 = CH₃, loss of 17 = OH).
Step 2 — IRLook for O–H (broad), N–H, C=O, C≡N. This narrows down functional group(s). Eliminate impossible structures immediately.
Step 3 — ¹³C NMRCount distinct carbon environments. Check the 150–220 ppm region for C=O. Note symmetry (fewer peaks than carbon count → equivalent carbons).
Step 4 — ¹H NMRCount H environments. Use integration for H ratios. Use splitting to find neighbours. Use δ values to assign environments to groups.
Step 5 — AssembleDraw possible structures consistent with all data. Cross-check every piece of evidence. One structure should fit all spectra — if not, re-examine your assignments.
Exam strategy: Write the molecular formula first (from Mr + degree of unsaturation if given). Then eliminate using IR. Then confirm with NMR. Show your reasoning step by step — marks are awarded for method.
20-question quiz
Test yourself across all six techniques. Click an answer to reveal the explanation.