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

(a) C₅H₁₂S (An –SH proton absorbs near 1.6$\mathbf{\delta }$)

(b)C₉H₁₁BRO

(c)C₅H₁₂O₂

NMR spectroscopy is used for molecules with atoms having odd atomic numbers or mass numbers. The most commonly used NMR spectroscopy is H-1 NMR spectroscopy and C-13 NMR spectroscopy.

The nuclei of these atoms spin. Therefore, they can absorb radiofrequency when kept in a strong electromagnetic field.The nucleus excites to a higher energy state which can cause a change in its spin. The frequency absorbed by the nucleus is recorded to study the molecule's structure.

In NMR spectroscopy, all the hydrogen atoms in the same chemical environment absorb the same frequency of radiation. These are called equivalent hydrogen atoms. The reference used for NMR is tetramethylsilane (TMS). The chemical shift of TMS is taken as 0, and the distance between the corresponding peak from TMS is called the chemical shift of the peak.

The number of peaks in NMR gives the number of types of hydrogen in the compound. The peak area gives the number of hydrogen in each type, and the multiplicity of the peak gives the number of neighbouring hydrogen.

(a) The given spectrum contains three peaks, and therefore, there are three types of distinguishable hydrogen.

Type a has a chemical shift of 0.99. It is split into three peaks and has two hydrogenatoms on the neighbouring carbon.

Type b has a chemical shift of 1.34. It is not split into any peaks, and therefore, it has no hydrogen atoms in neighbouring carbon.

Type c has a chemical shift of 1.61. It is split into four peaks and has three hydrogenatoms on the neighbouring carbon.

Therefore, the structure of the compound is

Compound (a)

(b) The given spectrum contains four peaks, and therefore, there are three types of distinguishable hydrogen.

Type a has a chemical shift of 2.31. It is split into five peaks and has four hydrogen atoms on the neighbouring carbon.

Type b has a chemical shift of 3.58. It is not split into three peaks, and therefore, it has two hydrogen atoms in neighbouring carbon.

Type c has a chemical shift of 4.08. It is split into three peaks and has four hydrogen atoms on the neighbouring carbon.

Type d has a chemical shift of 6.90-7.25. These indicate protons of a benzene ring.

Therefore, the structure of the compound is

Compound (b)

(c) The given NMR spectrum contains two peaks, peak a having an integral height of 3.97 and peak b having a height of 4.09.

The integral height per hydrogen can be calculated by dividing the total integral height of all peaks by the total number of hydrogen as:

$$\begin{gathered} \mathrm{IntegraltperH}=\frac{4.09+3.97}{12} \\ =0.6717 \end{gathered}$$

Then, the number of hydrogen per peak can be calculated as:

$\mathrm{No}.\mathrm{ofH}=\frac{\mathrm{integralheightofthepeak}}{\mathrm{integralheightperH}}$

The number of hydrogen atoms corresponding to peaks can be calculated as:

$$\begin{gathered} {\mathrm{N}}_{\mathrm{a}}=\frac{3.97}{0.6717} \\ \approx 6 \\ {\mathrm{N}}_{\mathrm{b}}=\frac{4.09}{0.6717} \\ \approx 6 \end{gathered}$$

Therefore, both peaks in NMR correspond to 6 hydrogen atoms.

The chemical shift for a peak a is between 1-1.5 and represents methyl protons. The peak at 3-3.5 corresponds to methoxy protons.

Therefore, the structure of the compound is:

Compound (c)