3.2.12Additional lessons
Structural Analysis of Molecules (HL) — IB Chemistry HL Revision Notes & Topical Questions
5 min read 5 parts Lesson 69 of 172
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Section 1 of 5Building the Case
Empirical formula
- Empirical formula Percentage composition leads to the simplest ratio of atoms, giving a first estimate of the formula.
Molecular formula
- Molecular formula The molecular ion peak in mass spectrometry gives the molar mass, which turns the empirical formula into the molecular formula.
Unsaturation
- Unsaturation The index of hydrogen deficiency, IHD = (2C + 2 + N − H − X) ÷ 2, shows the number of rings plus double bonds.
Functional groups
- Functional groups IR absorptions identify groups such as O–H, C=O and C≡N.
Connectivity
- Connectivity ¹H NMR chemical shifts, areas and splitting show how the groups are joined.
Section 2 of 5Strategy
1
Calculate the molecular formula and the IHD to see how many double bonds or rings are present.
2
Use IR to list the functional groups, ruling out any group whose characteristic absorption is missing.
3
Use the number of NMR signals and the ratio of areas to count hydrogen environments.
4
Use splitting and shifts to place groups next to each other and propose a structure.
5
Check the proposal against the mass spectrum fragments and make sure every piece of data is explained.
| Technique | Information |
|---|---|
| Elemental analysis | Empirical formula |
| MS | Molar mass and fragments |
| IR | Functional groups |
| ¹H NMR | Environments and connectivity |
Section 3 of 5Common Pitfalls
Common Pitfalls
Single technique
- Single technique No single technique is usually enough, so evidence must be cross-checked.
Isomers
- Isomers Isomers can have the same MS molecular ion, so fragments and NMR are needed to tell them apart.
Symmetry
- Symmetry Fewer NMR signals than expected points to a symmetric molecule.
Exchangeable H
- Exchangeable H OH and NH signals may be broad or absent in some samples, so use the formula to account for them.
Section 4 of 5Worked Example
Question
- A compound has formula C₄H₈O₂, M = 88.
- IR shows a strong peak at 1735 cm⁻¹ and no broad O–H absorption. ¹H NMR shows a singlet at δ 2.0 (3H), a quartet at δ 4.1 (2H) and a triplet at δ 1.2 (3H).
- Deduce the structure.
- IHD = (2 × 4 + 2 − 8) ÷ 2 = 1, so there is one C=O or ring.
- The peak at 1735 cm⁻¹ shows C=O and the lack of O–H rules out a carboxylic acid or alcohol, suggesting an ester.
- The quartet and triplet show an ethyl group attached to oxygen, and the singlet shows CH₃ next to C=O.
- Joining the pieces gives CH₃COOCH₂CH₃.
Answer Ethyl ethanoate
Section 5 of 5Exam Checklist & Tips
3.2.12 – Key Points for the Exam
- Index of hydrogen deficiency The number of rings plus double bonds (counting a triple bond as two).
- Molecular ion The ion with m/z equal to the molar mass of the compound.
- Functional group The part of a molecule responsible for its characteristic reactions.
- Fragmentation Breakdown of the molecular ion into smaller ions in the mass spectrometer.
Common Mistakes
- Forgetting to subtract halogens when calculating IHD.
- Choosing a structure that fits only the IR but not the NMR.
- Using the highest peak instead of the molecular ion for the molar mass.
- Ignoring the area ratios, which give the number of hydrogens in each environment.
Last updated:
\n