Question

Propose structures for compounds that have the following ¹H NMR spectra:

(a) ${\mathbf{\text{C}}}_{\mathbf{\text{4}}}{\mathbf{\text{H}}}_{\mathbf{\text{10}}}{\mathbf{\text{O}}}_{\mathbf{\text{2}}}$

(b) ${\mathbf{\text{C}}}_{\mathbf{\text{9}}}{\mathbf{\text{H}}}_{\mathbf{\text{10}}}\mathbf{\text{O}}$

Short answer

The answer is

(a)

2,3-Butane diol

(b)

1-benzyl-3-hydroxy propanone

Step-by-step solution

Description

The NMR spectroscopy deals with the interaction of the magnetic field with the nucleus. It can be used to characterise the compounds based on the resonant frequencies with which the nucleus vibrates.

Subpart (a)

The molecular formula along with the NMR spectrum shows the structure of the compound as shown below.

2,3-Butane diol

The 2,3-Butane diol have peaks corresponding to the methyl groups and hydroxy groups. Since it is a symetric compound, it gives three peaks in the proton NMR spectrum. The carbon atom with three hydrogen atom will give triplet on coupling with neighbouring hydrogen atoms. Similarly carbon atom with two hydrogen atom will give a quartet and the hydroxyl hydrogen gives a singlet.

Subpart (b)

The molecular formula along with the NMR spectrum shows the structure of the compound as shown below.

1-benzyl-3-hydroxy propanone

The compound gives a six peaks with different type of protons. This compound is symmetrical and hence six peaks only. The benzene protons gives a triplet and two doublets. The methylene protons gives a singlet, hydroxyl group gives a triplet and the nearby carbon shows a doublet.