IB ChemistryStructure 33.23.2.12

Combined Structural Analysis

Synthesising MS + IR + NMR data to deduce an unknown structure: the ultimate analytical skill.

Structure 3.2 HL Extension ⏱️ ~5 min revision
Analytical Workflow

3-Step Spectroscopic Structure Deduction

  1. Mass Spectrometry (MS) → Determine \(M_{\text{r}}\) & Fragments: The molecular ion peak gives the molecular mass. Fragment losses indicate attached functional units (e.g. loss of 15 = \(-\text{CH}_3\), loss of 17 = \(-\text{OH}\), loss of 29 = \(-\text{C}_2\text{H}_5\) or \(-\text{CHO}\)).
  2. Infrared (IR) → Identify Functional Groups: Look for key diagnostic bands: broad O–H acid (\(2500\text{--}3300\text{ cm}^{-1}\)), broad O–H alcohol (\(3200\text{--}3600\text{ cm}^{-1}\)), sharp C=O carbonyl (\(1700\text{--}1750\text{ cm}^{-1}\)), C=C alkene (\(1620\text{--}1680\text{ cm}^{-1}\)).
  3. NMR Spectroscopy → Assemble the Molecular Skeleton: Count environments (number of peaks), integration ratios (number of H atoms), and splitting multiplicity (\(n+1\) rule to connect adjacent carbons).

Worked Example

Worked Example

Combined Spectral Deduction: Butanone

Data: \(M^+ = 74\) | IR: strong C=O at \(1715\text{ cm}^{-1}\), no broad O–H | ¹H NMR: 3 peaks (singlet \(2.1\delta\) : quartet \(2.5\delta\) : triplet \(1.1\delta\)) with ratio 3:2:3.

Analysis:

  • \(M_{\text{r}} = 74\), C=O present, no O–H → ketone
  • Singlet (3H) at \(2.1\delta\) → \(-\text{CH}_3\) directly adjacent to C=O (0 adjacent protons)
  • Quartet (2H) + Triplet (3H) → \(-\text{CH}_2\text{CH}_3\) ethyl group
  • Deduction: Butanone (CH₃COCH₂CH₃)

¹H NMR Spectrum of Butanone

3.2.12 Combined Structural Analysis (HL) - IB | ChemEasy 5 4 3 2 1 0 TMS Chemical Shift, δ (ppm) 3H 3H 2H Triplet -CH₃ (next to CH₂) Singlet -CH₃ (next to C=O) Quartet -CH₂- (next to CH₃)
AQA GCSE & IB Chemistry

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