Analysing Structure of Organic Compounds Using Proton NMR Spectroscopy
This is part of the HSC Chemistry course under Module 8 Section 2: Analysis of Organic Substances.
HSC Chemistry Syllabus
Investigate the processes used to analyse the structure of simple organic compounds addressed in the course, including but not limited to:
proton and carbon-13 NMR
infrared spectroscopy (ACSCH130)
How is Proton NMR Used to Analyse the Structure of Organic Compounds.
This video explains how to use proton NMR to analyse the structure of organic compounds. It shows how to interpret proton NMR spectra for organic molecules. Specifically, it discusses signal splitting (J-coupling) and using the relative intensity of each signal to gain additional information on the structure of an organic molecule.
How does Proton NMR Provide Different Information on the Structure of Organic Compounds?
1H (Proton) NMR
Number of splits = n + 1
Where n is the number of protons in the adjacent chemical environment
Example – Propanoic acid
- There are six protons in three distinct chemical environments, giving rise to three signals.
- –OH signal has the greatest chemical shift because the proton is the most de-shielded (directly bound to oxygen atom).
Splitting pattern in proton NMR provides information on the number of hydrogen atoms in an adjacent chemical environment.
- –CH2 hydrogens produce a quartet signal as there are three hydrogens bound to the adjacent carbon atom
- –CH3 hydrogens produce a triplet signal as there are two hydrogens bound to the adjacent carbon atom
- –OH hydrogen does not produce any splitting possibly due to hydrogen bonding with water (solvent)
- The relative intensity under the –CH3 signal is 3.00 because there are three protons in this environment.
- The relative intensity under the –CH2 signal is 2.00 because there are two protons in this environment.
- The relative intensity under the –OH signal is 1.00 because there is one proton in this environment.