Question

Construct a simulated 1H NMR spectrum for the given structural formula. Drag the appropriate splitting patterns to the approximate chemical shift positions; place the integration values in the small bins above the associated chemical shift. Splitting patterns and integrations may be used more than once, or not at all, as needed. Likewise, some bins might remain blank. Note that peak heights are arbitrary and do not indicate proton integrations.

Answer 1

Concepts and reason

The concept used in this problem is based on Nuclear magnetic resonance $\left( {^{\rm{1}}{\rm{H NMR}}} \right)$ spectroscopy. The peaks in the ${\rm{NMR}}$ spectrum are used to identify the structure of the compound by splitting pattern, integration value, equivalent, non-equivalent protons.

First, identify the number of signals for the given compound. After that, assign the values of the chemical shift to identified signals in the spectrum.

Fundamentals

Nuclear magnetic resonance $\left( {^{\bf{1}}{\bf{H NMR}}} \right)$ spectroscopy:

Nuclear magnetic resonance spectroscopy is an analytical technique used in quality control and research for determining the molecular structure of any compound.

Chemical shift:

It is the difference between the frequency of observed proton and that of protons in solvent/reference. The commonly used solvent in ${\rm{NMR}}$ spectroscopy is tetramethylsilane $\left( {{\rm{TMS}}} \right)$ . The chemical shift of ${\rm{TMS}}$ is ${\rm{0 ppm}}$ .

Spin multiplicity:

The formula to calculate spin multiplicity is $n + 1$ , here $n$ is the number of hydrogens on neighboring carbon.

Interpretation of the $^{\bf{1}}{\bf{H NMR}}$ spectrum:

• Firstly, identify the different type of signals (protons).

• Look for the spin multiplicity on the basis of a number of protons on the neighboring carbon.

• Then, according to the position of the peaks, identify the chemical shift.

Some of the values of the chemical shift of $^{\rm{1}}{\rm{H NMR}}$ are given as follows.

The structure of the given compound showing the number of signals and spin multiplicity as follows.

The spin multiplicity of all different protons are calculated as follows using the.

For the carbon ${{\rm{C}}_{\rm{1}}}$ : $n = 0$ ; Spin multiplicity: $\left( {0 + 1 = 1} \right)$ .

For the carbon ${{\rm{C}}_2}$ : $n = 3$ ; Spin multiplicity: $\left( {3 + 1 = 4} \right)$ .

For the carbon ${{\rm{C}}_2}$ : $n = 2$ ; Spin multiplicity: $\left( {2 + 1 = 3} \right)$ .

The $^{\rm{1}}{\rm{H NMR}}$ peaks corresponding to different protons of the given compound are shown as follows.

The proton of the alcohol group is highly deshielded. The value of the chemical shift of protons is upfield approx to ${\rm{6}}{\rm{.2 ppm}}$ .

The carbon $\left( {{{\rm{C}}_{\rm{1}}}} \right)$ is deshielded due to the oxygen present in the furan ring and also due to the vinylic position. Therefore the value of the chemical shift of protons is upfield approx to ${\rm{4}}{\rm{.7 ppm}}$ .

The carbon $\left( {{{\rm{C}}_2}} \right)$ is deshielded due to the vinylic position. Therefore the value of the chemical shift of protons is upfield approx to ${\rm{2}}{\rm{.3 ppm}}$ .

The carbon $\left( {{{\rm{C}}_3}} \right)$ is attached to the methylene $\left( {{\rm{C}}{{\rm{H}}_{\rm{2}}}} \right)$ which has little effect on the chemical shift. Therefore the value of the chemical shift of protons is upfield approx to ${\rm{1}}{\rm{.3 ppm}}$ .

Ans:

The peaks of $^{\bf{1}}{\bf{H NMR}}$ corresponding to different protons of a given compound are shown as follows.