Question

the structure that fits the following data. Molecular formula: C6H12O H NMR spectral data: ppm 0.90 1.53 2.34 9.72 Integration6 4 1 1 Multiplicity triplet multiplet multiplet doublet C NMR spectrum:

Draw the structure that fits the following data. Molecular formula: C6H12O 1H NMR spectral data: 13C NMR spectrum:

PLEASE HELP

Answer 1

Concepts and reason

A compound is given with the molecular formula ${{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{\rm{O}}$ , $^1{\rm{H - NMR}}$ , and $^{13}{\rm{C - NMR}}$ having various chemical shift values. The structure of the compound can be deduced by calculating the degree of unsaturation (DU) in the compound and using the given NMR data.

The number of signals obtained in NMR is also depend on the different type of protons present in the compound. Thee splitting pattern is given by $\left( {{\rm{n}} + {\rm{1}}} \right)$ rule and intensity of the peak is shown by the pascals triangle.

Fundamentals

• NMR stands for nuclear magnetic resonance which is a useful spectroscopic technique for determining the type of protons and the number of protons in the compounds.

• A set of protons which are present in same chemical environment, that is, present between the same group of atoms are said to be chemically equivalent protons. All chemically equivalent proton generates 1 signal or 1 peak in $^1{\rm{H - NMR}}$ .

• A set of protons which are present in different chemical environment, that is, present between the different group of atoms are said to be chemically non-equivalent protons. All chemically non-equivalent proton generates distinct signal in $^1{\rm{H - NMR}}$ .

• Chemical shift generally depicts the resonance frequency of the nucleus under various electron distribution environment in a compound under magnetic field.

• Splitting of peak: It is also referred as spin-spin splitting. It follows $\left( {{\rm{n}} + {\rm{1}}} \right)$ rule, where n is the number of adjacent protons.

As proton NMR shows different chemical shifts for various protons in a compound similarly, $^{13}{\rm{C - NMR}}$ shows different chemical shifts value for chemically non-equivalent carbons.

Chemical shift range for $^{13}{\rm{C - NMR}}$ is higher than that of proton NMR and the range is from $0 - 250\,{\rm{ppm}}$ .

Degree of unsaturation (DU) can be calculated as follows:

${\rm{DU}} = {n_C} + 1 - \frac{{{n_H} + {n_N} - {n_X}}}{2}$

Here,

$\begin{array}{l}\\{n_C} = {\rm{number}}\,{\rm{of}}\,{\rm{carbon}}\,{\rm{atoms}}\\\\{n_H} = {\rm{number}}\,{\rm{of}}\,{\rm{carbon}}\,{\rm{atoms}}\\\\{n_N} = {\rm{number}}\,{\rm{of}}\,{\rm{carbon}}\,{\rm{atoms}}\\\\{n_X} = {\rm{number}}\,{\rm{of}}\,{\rm{carbon}}\,{\rm{atoms}}\\\end{array}$

The value of DU can be used to arrive at following predictions about the structure of the compound:

Similarly, for higher values such predictions can be made.

Calculate the degree of unsaturation present in the given compound having molecular formula ${{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{\rm{O}}$ as follows:

$\begin{array}{c}\\{\rm{DU}} = 6 + 1 - \frac{{12}}{2}\\\\ = 7 - 6\\\\ = 1\\\end{array}$

DU is calculated as one hence, either a double bond or a ring structure is present in the given compound.

Interpret the chemical shift values given for $^{13}{\rm{C - NMR}}$ . The chemical shift values in ppm from the given NMR are estimated as follows:

$205,\,52,\,21,\,11$

The chemical shift is given for four carbons, out of which for three the values $0 - 50\,{\rm{ppm}}$ implies the presence of carbon which is bonded to carbon itself, that is, ${\rm{C - C}}$ and the value around $205\,{\rm{ppm}}$ implies the presence of carbonyl carbon.

Interpret the chemical shift values given for $^1{\rm{H - NMR}}$ as follows:

$\begin{array}{l}\\0.90\,{\rm{ppm}}\left( {6{\rm{H, triplet}}} \right)\\\\1.53\,{\rm{ppm}}\left( {{\rm{4H, multiplet}}} \right)\\\\2.34\,{\rm{ppm}}\left( {{\rm{1H, multiplet}}} \right)\\\\9.72\,{\rm{ppm}}\left( {1{\rm{H, doublet}}} \right)\\\end{array}$

The chemical shift $0.90\,{\rm{ppm}}\left( {6{\rm{H, triplet}}} \right)$ indicates the presence of methyl protons, this implies two methyl groups are present in the structure with the carbon containing one two hydrogens.

The chemical shift $1.53\,{\rm{ppm}}\left( {{\rm{4H, multiplet}}} \right)$ indicates the presence of methylene protons, this implies two methylene groups are present in the structure with the carbons containing more than two chemically different protons.

$2.34\,{\rm{ppm}}\left( {{\rm{1H, multiplet}}} \right)$ indicates the presence of ${\rm{CH}}$ proton which is attached with the carbons having more than two chemically different protons.

$9.72\,{\rm{ppm}}\left( {1{\rm{H, doublet}}} \right)$ represents the peak of aldehydic proton.

Write the structure of the compound that best suits which the given data as shown below:

Ans:

Hence, the correct structure of the compound is as follows: