The compound whose ¹H NMR spectrum is shown has the molecular

formula C₃H₆Br₂. Propose a structure.

The proton NMR spectroscopy can be used to identify the structure of different molecules. The different kinds of electronically non equivalent hydrogens can be identified from the ¹H NMR spectroscopy.

The degree of unsaturation for the formulaC₃H₆Br₂ can be given as:

The compound does not contain rings or double bonds as the degree of unsaturation is 0.

The relative integration value may or may not be the exact value. The molecular compound has six hydrogen atoms, so the relative values are the exact values. The chemical shift values are 2.33 ppm and 3.56 ppm. The multiplicity of each signal is triplet and multiplets with the chemical shift values , respectively.

The signal at 2.33 ppmrepresents two protons.

The signal at 3.56 ppm represent one proton.

Fragments obtained from the molecule

The signal at 3.54 ppm and 2.33 ppm resulting in a triplet and multiplet indicate the presence of methylene groups fragment.

The fragments can be combined to generate the resulting structure as:

Structure of the compound