🚀 START OF YEAR 12 SALE - OCT 7-21 🚀

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

  • mass spectrometry

  • 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

  • Chemical shift signals are different to those of carbon-13. This means both 1H and 13C nuclei of a sample can be analysed simultaneously

 

  • Splitting pattern (J-coupling) of a signal provides information on neighbouring hydrogen atoms and connectivity of the molecule

 

Number of splits = n + 1

 

Where n is the number of protons in the adjacent chemical environment

 

  • Area under the curve (AUC) of a signal provides information on the number of protons which share that particular 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

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)

 

Relative Intensity

  • 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.