Chapter 22, Problem 1Q

Interpretation Introduction

(a)

Interpretation:

The structure and chemical shifts of all the protons of ${\text{C}}_{\text{5}}{\text{H}}_{\text{11}}\text{Cl}$ should be estimated and the NMR signal should be assigned.

Concept introduction:

Nuclear magnetic resonance spectroscopy is applied for the identification of the structure of molecules. The energy in the radiofrequency region is suitable for NMR. Nuclear magnetic resonance results from the spin of the nucleus of an atom. The value of I is obtained using the atomic number and the mass number of an atom. The non-zero magnetic moment of an isotopic nucleus is detectable by the NMR technique.

Any nucleus with both an even atomic number and the mass number has 0 nuclear spins. There are a total of $\left(2\text{I}+1\right)$ energy levels that are allowed for a nuclear spin (I). In the absence of a magnetic field, energy levels are equal in energy that is degenerate.

However, energy levels become non-degenerate in the presence of a magnetic field.

Deuterated chloroform $\left({\text{CDCl}}_{\text{3}}\right)$ is a common NMR solvent generally used because it dissolutes a broad range of organic compounds and is inexpensive.

The total signal intensity of each set of proton is given by the height of each set of steps. The integration value defines the relative number of each kind of proton in the molecule.

In NMR spectrum, the intensity of signals is plotted against the magnetic field or frequency. Nuclei that are non-equivalent show only one peak in the NMR spectrum. However, protons absorb at different frequencies that are non-equivalent.

An increase in the electron density that surrounds the nucleus shields it from the applied field. This results in a net decrease in the field experienced by the nucleus. The value of the observed chemical shift of the signal therefore decreases, and, on a typical NMR spectrum, the signal moves to the right, which is called an upfield shift because, at a constant frequency, a slightly higher applied magnetic field is required for resonance to occur. De-shielding is the effect of a decline in the electron density around a nucleus which leads to shifting in the peaks of a chemical shift towards left in the NMR spectrum that results in an increase in delta values, hence downshift.

Explanation of Solution

Double bond equivalent(DBE) can be calculated as follows:

  $\text{DBE}=\text{C}+\frac{\text{N}-\text{H}-\text{X}}{2}+1$

Where,

• C is the number of Carbon.
• N is the number of nitrogen.
• H is the number of hydrogen.
• X is the number of halogen.

The DBE for the ${\text{C}}_{\text{5}}{\text{H}}_{\text{11}}\text{Cl}$ is calculated as follows:

  $\begin{array}{c}\text{DBE}=\text{C}+\frac{\text{N}-\text{H}-\text{X}}{2}+1\\ =5+\frac{0-11-1}{2}+1\\ =0\end{array}$

DBE zero implies the absence of any double bond or a ring.

Total there are 2 types of hydrogen. 9 hydrogen are singlet indicate the presence of methyl group on carbon with no hydrogen adjacent to it for splitting, 2 hydrogen singlet indicate the presence of $\text{CH}$ with no hydrogen adjacent to it for splitting and its higher chemical shift indicates it is attached to chlorine.

The signals and chemical shift values are indicated as follows:

  

The compound is $\text{1-Chloro-2,2-dimethylpropane}$ .

The chemical shifts of green hydrogen atoms can be estimated as follows:

  $\begin{array}{c}\text{base value of green protons}=\text{0}\text{.9 ppm}\\ \text{presence of γ chlorine group}=\text{0}\text{.2 ppm}\\ \text{estimated chemical shift}=\text{1}\text{.1 ppm}\\ \text{measured chemical shift}=\text{1}\text{.10 ppm}\end{array}$

The chemical shifts of red hydrogen atoms can be estimated as follows:

  $\begin{array}{c}\text{base value of red protons}=1.2\text{ ppm}\\ \text{presence of β chlorine}=2.1\text{ ppm}\\ \text{estimated chemical shift}=3.3\text{ ppm}\\ \text{measured chemical shift}=3.33\text{ ppm}\end{array}$

Interpretation Introduction

(b)

Interpretation:

The structure and chemical shifts of all the protons of ${\text{C}}_{\text{5}}{\text{H}}_{\text{10}}{\text{O}}_2$ should be estimated and the NMR signal should be assigned.

Concept introduction:

Nuclear magnetic resonance spectroscopy is applied for the identification of the structure of molecules. The energy in the radiofrequency region is suitable for NMR. Nuclear magnetic resonance results from the spin of the nucleus of an atom. The value of I is obtained using the atomic number and the mass number of an atom. The non-zero magnetic moment of an isotopic nucleus is detectable by the NMR technique.

Any nucleus with both an even atomic number and the mass number has 0 nuclear spins. There are a total of $\left(2\text{I}+1\right)$ energy levels that are allowed for a nuclear spin (I). In the absence of a magnetic field, energy levels are equal in energy that is degenerate.

However, energy levels become non-degenerate in the presence of a magnetic field.

Deuterated chloroform $\left({\text{CDCl}}_{\text{3}}\right)$ is a common NMR solvent generally used because it dissolutes a broad range of organic compounds and is inexpensive.

The total signal intensity of each set of proton is given by the height of each set of steps. The integration value defines the relative number of each kind of proton in the molecule.

In NMR spectrum, the intensity of signals is plotted against the magnetic field or frequency. Nuclei that are non-equivalent show only one peak in the NMR spectrum. However, protons absorb at different frequencies that are non-equivalent.

An increase in the electron density that surrounds the nucleus shields it from the applied field. This results in a net decrease in the field experienced by the nucleus. The value of the observed chemical shift of the signal therefore decreases, and, on a typical NMR spectrum, the signal moves to the right, which is called an upfield shift because, at a constant frequency, a slightly higher applied magnetic field is required for resonance to occur. De-shielding is the effect of a decline in the electron density around a nucleus which leads to shifting in the peaks of a chemical shift towards left in the NMR spectrum that results in an increase in delta values, hence downshift.

Explanation of Solution

Double bond equivalent(DBE) can be calculated as follows:

  $\text{DBE}=\text{C}+\frac{\text{N}-\text{H}-\text{X}}{2}+1$

Where,

• C is the number of Carbon.
• N is the number of nitrogen.
• H is the number of hydrogen.
• X is the number of halogen.

The DBE for the compound ${\text{C}}_{\text{5}}{\text{H}}_{\text{10}}{\text{O}}_2$ is calculated as follows:

  $\begin{array}{c}\text{DBE}=\text{C}+\frac{\text{N}-\text{H}-\text{X}}{2}+1\\ =5+\frac{0-10-0}{2}+1\\ =1\end{array}$

DBE 1 implies the presence of any double bond or a ring. The presence of 2 oxygen implies either the compound is alcohol or carbonyl compound or both.

There are 3 types of hydrogen in total. 6 hydrogen are singlet indicate the presence of methyl group on a carbon with no hydrogen adjacent to it for splitting, 1 hydrogen singlet indicate the presence of $\text{CH}$ with no hydrogen adjacent to it for splitting and 3 hydrogen atoms are singlet indicate the presence of methyl group on a carbon with no hydrogen adjacent to it for splitting .

The signals and chemical shift values are indicated as follows:

  

The compound is $\text{3-hydroxy-3-methyl-3-butanone}$ .

The chemical shifts of green hydrogen atoms can be estimated as follows:

  $\begin{array}{c}\text{base value of green protons}=\text{0}\text{.9 ppm}\\ \text{presence of β-hydroxy group}=\text{0}\text{.3 ppm}\\ \text{estimated chemical shift}=\text{1}\text{.2 ppm}\\ \text{measured chemical shift}=\text{1}\text{.40 ppm}\end{array}$

The chemical shifts of red hydrogen atoms can be estimated as follows:

  $\begin{array}{c}\text{base value of red protons}=\text{0}\text{.9 ppm}\\ \text{presence of β-CO}\left(\text{R}\right)=\text{1}\text{.2 ppm}\\ \text{estimated chemical shift}=\text{2}\text{.1 ppm}\\ \text{measured chemical shift}=\text{2}\text{.25 ppm}\end{array}$

Interpretation Introduction

(c)

Interpretation:

The structure and chemical shifts of all the protons of ${\text{C}}_6{\text{H}}_{\text{12}}{\text{O}}_2$ should be estimated and the NMR signal should be assigned.

Concept introduction:

Nuclear magnetic resonance spectroscopy is applied for the identification of the structure of molecules. The energy in the radiofrequency region is suitable for NMR. Nuclear magnetic resonance results from the spin of the nucleus of an atom. The value of I is obtained using the atomic number and the mass number of an atom. The non-zero magnetic moment of an isotopic nucleus is detectable by the NMR technique.

Any nucleus with both an even atomic number and the mass number has 0 nuclear spins. There are a total of $\left(2\text{I}+1\right)$ energy levels that are allowed for a nuclear spin (I). In the absence of a magnetic field, energy levels are equal in energy that is degenerate.

However, energy levels become non-degenerate in the presence of a magnetic field.

Deuterated chloroform $\left({\text{CDCl}}_{\text{3}}\right)$ is a common NMR solvent generally used because it dissolutes a broad range of organic compounds and is inexpensive.

The total signal intensity of each set of proton is given by the height of each set of steps. The integration value defines the relative number of each kind of proton in the molecule.

In NMR spectrum, the intensity of signals is plotted against the magnetic field or frequency. Nuclei that are non-equivalent show only one peak in the NMR spectrum. However, protons absorb at different frequencies that are non-equivalent.

An increase in the electron density that surrounds the nucleus shields it from the applied field. This results in a net decrease in the field experienced by the nucleus. The value of the observed chemical shift of the signal therefore decreases, and, on a typical NMR spectrum, the signal moves to the right, which is called an upfield shift because, at a constant frequency, a slightly higher applied magnetic field is required for resonance to occur. De-shielding is the effect of a decline in the electron density around a nucleus which leads to shifting in the peaks of a chemical shift towards left in the NMR spectrum that results in an increase in delta values, hence downshift.

Explanation of Solution

Double bond equivalent(DBE) can be calculated as follows:

  $\text{DBE}=\text{C}+\frac{\text{N}-\text{H}-\text{X}}{2}+1$

Where,

• C is the number of Carbon.
• N is the number of nitrogen.
• H is the number of hydrogen.
• X is the number of halogen.

The DBE for the compound ${\text{C}}_6{\text{H}}_{\text{12}}{\text{O}}_2$ is calculated as follows:

  $\begin{array}{c}\text{DBE}=\text{C}+\frac{\text{N}-\text{H}-\text{X}}{2}+1\\ =6+\frac{0-12-0}{2}+1\\ =1\end{array}$

DBE 1 implies the presence of one double bond or a ring. The presence of 2 oxygen implies either the compound is alcohol or carbonyl compound or both.

There are 4 types of hydrogen in total. 6 hydrogen are singlet indicate the presence of methyl group on a carbon with no hydrogen adjacent to it for splitting, 1 hydrogen singlet indicate the presence of $\text{CH}$ with no hydrogen adjacent to it for splitting, 3 hydrogen are singlet indicate the presence of methyl group on a carbon with no hydrogen adjacent to it for splitting and 2 hydrogen singlet indicate the presence of ${\text{CH}}_2$ with no hydrogen adjacent to it for splitting.

The signals and chemical shift values are indicated as follows:

  

The chemical shifts of green hydrogen atoms can be estimated as follows:

  $\begin{array}{c}\text{base value of green protons}=\text{0}\text{.9 ppm}\\ \text{presence of β-hydroxy group}=\text{0}\text{.3 ppm}\\ \text{estimated chemical shift}=\text{1}\text{.2 ppm}\\ \text{measured chemical shift}=\text{1}\text{.26 ppm}\end{array}$

The chemical shifts of red hydrogen atoms can be estimated as follows:

  $\begin{array}{c}\text{base value of red protons}=\text{0}\text{.9 ppm}\\ \text{presence of β-CO}\left(\text{R}\right)=\text{1}\text{.2 ppm}\\ \text{estimated chemical shift}=\text{2}\text{.1 ppm}\\ \text{measured chemical shift}=\text{2}\text{.18 ppm}\end{array}$

The chemical shifts of pink hydrogen atoms can be estimated as follows:

  $\begin{array}{c}\text{base value of pink protons}=\text{1}\text{.2 ppm}\\ \text{presence of α-CO}\left({\text{CH}}_3\right)\text{ group}=\text{1}\text{.2 ppm}\\ \text{presence of β-hydroxy group}=\text{0}\text{.3 ppm}\\ \text{estimated chemical shift}=\text{2}\text{.7 ppm}\\ \text{measured chemical shift}=\text{2}\text{.63 ppm}\end{array}$

Interpretation Introduction

(d)

Interpretation:

The structure and chemical shifts of all the protons of ${\text{C}}_6{\text{H}}_{\text{12}}{\text{O}}_2$ should be estimated and the NMR signal should be assigned.

Concept introduction:

Nuclear magnetic resonance spectroscopy is applied for the identification of the structure of molecules. The energy in the radiofrequency region is suitable for NMR. Nuclear magnetic resonance results from the spin of the nucleus of an atom. The value of I is obtained using the atomic number and the mass number of an atom. The non-zero magnetic moment of an isotopic nucleus is detectable by the NMR technique.

Any nucleus with both an even atomic number and the mass number has 0 nuclear spins. There are a total of $\left(2\text{I}+1\right)$ energy levels that are allowed for a nuclear spin (I). In the absence of a magnetic field, energy levels are equal in energy that is degenerate.

However, energy levels become non-degenerate in the presence of a magnetic field.

Deuterated chloroform $\left({\text{CDCl}}_{\text{3}}\right)$ is a common NMR solvent generally used because it dissolutes a broad range of organic compounds and is inexpensive.

The total signal intensity of each set of proton is given by the height of each set of steps. The integration value defines the relative number of each kind of proton in the molecule.

In NMR spectrum, the intensity of signals is plotted against the magnetic field or frequency. Nuclei that are non-equivalent show only one peak in the NMR spectrum. However, protons absorb at different frequencies that are non-equivalent.

An increase in the electron density that surrounds the nucleus shields it from the applied field. This results in a net decrease in the field experienced by the nucleus. The value of the observed chemical shift of the signal therefore decreases, and, on a typical NMR spectrum, the signal moves to the right, which is called an upfield shift because, at a constant frequency, a slightly higher applied magnetic field is required for resonance to occur. De-shielding is the effect of a decline in the electron density around a nucleus which leads to shifting in the peaks of a chemical shift towards left in the NMR spectrum that results in an increase in delta values, hence downshift.

Explanation of Solution

Double bond equivalent(DBE) can be calculated as follows:

  $\text{DBE}=\text{C}+\frac{\text{N}-\text{H}-\text{X}}{2}+1$

Where,

• C is the number of Carbon.
• N is the number of nitrogen.
• H is the number of hydrogen.
• X is the number of halogen.

The DBE for the compound ${\text{C}}_5{\text{H}}_{10}\text{O}$ is calculated as follows:

  $\begin{array}{c}\text{DBE}=\text{C}+\frac{\text{N}-\text{H}-\text{X}}{2}+1\\ =5+\frac{0-10-0}{2}+1\\ =1\end{array}$

DBE 1implies the presence of one double bond or a ring. The presence of 1 oxygen implies either the compound is alcohol or carbonyl compound.

Total there are 4 types of hydrogen. 1 hydrogen as multiplet indicates the presence of ${\text{CH}}_{\text{3}}{\text{-CH-CH}}_3$, 6 hydrogen atoms as doublet indicate the presence of ${\text{CH}}_{\text{3}}\text{-CH-}$, 2 hydrogen atoms are a doublet of doublet indicate the presence of ${\text{-CH-CH}}_{\text{2}}\text{-CH-}$ and 1 hydrogen multiplet indicate the presence of ${\text{-CH}}_{\text{2}}\text{-CH-}$ .

The signals and chemical shift values are indicated as follows:

  

The chemical shifts of green hydrogen atoms can be estimated as follows:

  $\begin{array}{c}\text{base value of green protons}=\text{0}\text{.9 ppm}\\ \text{presence of β}-\text{CO}\left(\text{H}\right)=\text{0}\text{.1 ppm}\\ \text{estimated chemical shift}=\text{1}\text{.0 ppm}\\ \text{measured chemical shift}=\text{0}\text{.98 ppm}\end{array}$

The chemical shifts of blue hydrogen atoms can be estimated as follows:

  $\begin{array}{c}\text{base value of blue protons}=\text{1}\text{.5 ppm}\\ \text{presence of β-CO}\left(\text{R}\right)=\text{0}\text{.0 ppm}\\ \text{estimated chemical shift}=\text{1}\text{.5 ppm}\\ \text{measured chemical shift}=\text{2}\text{.21 ppm}\end{array}$

The chemical shifts of pink hydrogen atoms can be estimated as follows:

  $\begin{array}{c}\text{base value of pink protons}=\text{1}\text{.2 ppm}\\ \text{presence of α-CO}\left({\text{CH}}_3\right)\text{ group}=\text{1}\text{.2 ppm}\\ \text{estimated chemical shift}=\text{2}\text{.4 ppm}\\ \text{measured chemical shift}=\text{2}\text{.31 ppm}\end{array}$

Interpretation Introduction

(e)

Interpretation:

The structure and chemical shifts of all the protons of ${\text{C}}_4{\text{H}}_8\text{O}$ should be estimated and the NMR signal should be assigned.

Concept introduction:

Nuclear magnetic resonance spectroscopy is applied for the identification of the structure of molecules. The energy in the radiofrequency region is suitable for NMR. Nuclear magnetic resonance results from the spin of the nucleus of an atom. The value of I is obtained using the atomic number and the mass number of an atom. The non-zero magnetic moment of an isotopic nucleus is detectable by the NMR technique.

Any nucleus with both an even atomic number and the mass number has 0 nuclear spins. There are a total of $\left(2\text{I}+1\right)$ energy levels that are allowed for a nuclear spin (I). In the absence of a magnetic field, energy levels are equal in energy that is degenerate.

However, energy levels become non-degenerate in the presence of a magnetic field.

Deuterated chloroform $\left({\text{CDCl}}_{\text{3}}\right)$ is a common NMR solvent generally used because it dissolutes a broad range of organic compounds and is inexpensive.

The total signal intensity of each set of proton is given by the height of each set of steps. The integration value defines the relative number of each kind of proton in the molecule.

In NMR spectrum, the intensity of signals is plotted against the magnetic field or frequency. Nuclei that are non-equivalent show only one peak in the NMR spectrum. However, protons absorb at different frequencies that are non-equivalent.

An increase in the electron density that surrounds the nucleus shields it from the applied field. This results in a net decrease in the field experienced by the nucleus. The value of the observed chemical shift of the signal therefore decreases, and, on a typical NMR spectrum, the signal moves to the right, which is called an upfield shift because, at a constant frequency, a slightly higher applied magnetic field is required for resonance to occur. De-shielding is the effect of a decline in the electron density around a nucleus which leads to shifting in the peaks of a chemical shift towards left in the NMR spectrum that results in an increase in delta values, hence downshift.

Explanation of Solution

Double bond equivalent(DBE) can be calculated as follows:

  $\text{DBE}=\text{C}+\frac{\text{N}-\text{H}-\text{X}}{2}+1$

Where,

• C is the number of Carbon.
• N is the number of nitrogen.
• H is the number of hydrogen.
• X is the number of halogen.

The DBE for the compound ${\text{C}}_4{\text{H}}_8\text{O}$ is calculated as follows:

  $\begin{array}{c}\text{DBE}=\text{C}+\frac{\text{N}-\text{H}-\text{X}}{2}+1\\ =4+\frac{0-8-0}{2}+1\\ =1\end{array}$

DBE 1 implies the presence of one double bond or a ring. The presence of 1 oxygen implies either the compound is alcohol or carbonyl compound.

Total there are 6 types of hydrogen and 3 hydrogen atoms are vinyl hydrogens per their chemical shift values.

The signals and chemical shift values are indicated as follows:

  

The chemical shifts of red hydrogen atoms can be estimated as follows:

  $\begin{array}{c}\text{base value of red protons}=\text{1}\text{.5 ppm}\\ \text{presence of β-}\left(\text{C=C}\right)=\text{0}\text{.2 ppm}\\ \text{presence of α-}\left(\text{OH}\right)=-2.2\text{ppm}\\ \text{estimated chemical shift}=-\text{1}\text{.5 ppm}\\ \text{measured chemical shift}=\text{1}\text{.27 ppm}\end{array}$

The chemical shifts of pink hydrogen atoms can be estimated as follows:

  $\begin{array}{c}\text{base value of pink protons}=\text{1}\text{.5 ppm}\\ \text{presence of β-}\left(\text{C=C}\right)=\text{0}\text{.2 ppm}\\ \text{presence of α-}\left(\text{OH}\right)=2.2\text{ppm}\\ \text{estimated chemical shift}=\text{3}\text{.9 ppm}\\ \text{measured chemical shift}=\text{4}\text{.30 ppm}\end{array}$

The chemical shifts of ${\text{H}}_{\text{c}}\text{vinyl}$ hydrogen atoms can be estimated as follows:

  $\begin{array}{c}{\text{base value of H}}_{\text{c}}\text{vinyl protons}=\text{5}\text{.28 ppm}\\ {\text{presence of geminal CH}}_2\text{OH}=\text{0}\text{.64 ppm}\\ \text{estimated chemical shift}=\text{5}\text{.92 ppm}\\ \text{measured chemical shift}=\text{5}\text{.90 ppm}\end{array}$

The chemical shifts of ${\text{H}}_{\text{a}}\text{vinyl}$ hydrogen atoms can be estimated as follows:

  $\begin{array}{c}{\text{base value of H}}_{\text{a}}\text{vinyl protons}=\text{5}\text{.28 ppm}\\ {\text{presence of cis CH}}_2\text{OHgroup}=-\text{0}\text{.01 ppm}\\ \text{estimated chemical shift}=\text{5}\text{.27 ppm}\\ \text{measured chemical shift}=\text{5}\text{.19 ppm}\end{array}$

The chemical shifts of ${\text{H}}_{\text{b}}\text{vinyl}$ hydrogen atoms can be estimated as follows:

  $\begin{array}{c}{\text{base value of H}}_{\text{b}}\text{vinyl protons}=\text{5}\text{.28 ppm}\\ {\text{presence of trans CH}}_2\text{OHgroup}=-\text{0}\text{.02 ppm}\\ \text{estimated chemical shift}=\text{5}\text{.26 ppm}\\ \text{measured chemical shift}=\text{5}\text{.06 ppm}\end{array}$