Chapter 14, Problem 14.17P

Describe the ${}^{\text{1}}\text{H}$ NMR spectrum of each compound. State how many NMR signals are present,

the splitting pattern for each signal, and the approximate chemical shift?

Interpretation Introduction

(a)

Interpretation:

The ${}^{\text{1}}\text{H}$ NMR spectrum of the given compound is to be described. The number of NMR signals, the splitting pattern for each signal and the approximate chemical shift values are to be stated.

Concept introduction:

The number of NMR signal in a compound is equal to the number of chemically non-equivalent protons present in that compound. In ${}^1\text{H}-\text{NMR}$ all chemically equivalent protons generates one signal or one peak, whereas non-equivalent proton generates different signals. On the horizontal axis, the position of absorption is generally referred to as chemical shift.

Answer to Problem 14.17P

The ${}^{\text{1}}\text{H}$ NMR spectrum of the given compound is,

The number of NMR signals, the splitting pattern for each signal and the approximate chemical shift values are rightfully stated.

Explanation of Solution

The given compound is ethyl methyl ether $\left({\text{CH}}_{\text{3}}-\text{O}-{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}\right)$. It contains three types of proton ${\text{H}}_{\text{a}}{\text{, H}}_{\text{b}}$ and ${\text{H}}_{\text{c}}$. Therefore, it gives three signals in ${}^{\text{1}}\text{H}$ NMR spectrum. The ${}^{\text{1}}\text{H}$ NMR spectrum of the given compound is shown below.

Figure 1

The proton ${\text{H}}_{\text{a}}$ gets split into a singlet because it has no adjacent proton, the proton ${\text{H}}_{\text{b}}$ gets split into a quartet by $3$${\text{H}}_{\text{c}}$ protons and the proton ${\text{H}}_{\text{c}}$ gets split into a triplet by $2$${\text{H}}_{\text{b}}$ protons. The chemical shift of ${\text{H}}_{\text{a}}$ is approximately $3.3\text{ ppm}$. The chemical shift of ${\text{H}}_{\text{b}}$ is approximately $3.5\text{ ppm}$ and the chemical shift of ${\text{H}}_{\text{c}}$ is approximately $\text{1}\text{.1 ppm}$.

Conclusion

The ${}^{\text{1}}\text{H}$ NMR spectrum of the given compound is shown in Figure 1. It gives three signals with one, four and three peaks.

Interpretation Introduction

(b)

Interpretation:

The ${}^{\text{1}}\text{H}$ NMR spectrum of the given compound is to be described. The number of NMR signals, the splitting pattern for each signal and the approximate chemical shift values are to be stated.

Concept introduction:

The number of NMR signal in a compound is equal to the number of chemically non-equivalent protons present in that compound. In ${}^1\text{H}-\text{NMR}$ all chemically equivalent protons generates one signal or one peak, whereas non-equivalent proton generates different signals. On the horizontal axis, the position of absorption is generally referred to as chemical shift.

Answer to Problem 14.17P

The ${}^{\text{1}}\text{H}$ NMR spectrum of the given compound is,

The number of NMR signals, the splitting pattern for each signal and the approximate chemical shift values are rightfully stated.

Explanation of Solution

The given compound is isopropylpropionate $\left({\text{CH}}_{\text{3}}{\text{CH}}_2\text{COOCH}{\left({\text{CH}}_{\text{3}}\right)}_2\right)$. It contains four types of proton ${\text{H}}_{\text{a}}{\text{, H}}_{\text{b}}{\text{, H}}_{\text{c}}$ and ${\text{H}}_{\text{d}}$. Therefore, it gives four signals in ${}^{\text{1}}\text{H}$ NMR spectrum. The ${}^{\text{1}}\text{H}$ NMR spectrum of the given compound is shown below.

Figure 2

The proton ${\text{H}}_{\text{a}}$ gets split into a triplet because it has two adjacent proton, the proton ${\text{H}}_{\text{b}}$ gets split into a quartet by $3$${\text{H}}_{\text{a}}$ protons, the proton ${\text{H}}_{\text{c}}$ gets split into a septet by $6$${\text{H}}_{\text{d}}$ protons and the proton ${\text{H}}_{\text{d}}$ gets split into a doublet by $1$${\text{H}}_{\text{c}}$ proton. The chemical shift of ${\text{H}}_{\text{a}}$ is approximately $\text{1}\text{.2 ppm}$. The chemical shift of ${\text{H}}_{\text{b}}$ is approximately $\text{2}\text{.4 ppm}$. The chemical shift of ${\text{H}}_{\text{c}}$ is approximately $\text{5}\text{.0 ppm}$ and the chemical shift of ${\text{H}}_{\text{d}}$ is approximately $\text{1}\text{.8 ppm}$.

Conclusion

The ${}^{\text{1}}\text{H}$ NMR spectrum of the given compound is shown in Figure 2. It gives four signals with three, four, seven and two peaks.

Interpretation Introduction

(c)

Interpretation:

The ${}^{\text{1}}\text{H}$ NMR spectrum of the given compound is to be described. The number of NMR signals, the splitting pattern for each signal and the approximate chemical shift values are to be stated.

Concept introduction:

The number of NMR signal in a compound is equal to the number of chemically non-equivalent protons present in that compound. In ${}^1\text{H}-\text{NMR}$ all chemically equivalent protons generates one signal or one peak, whereas non-equivalent proton generates different signals. On the horizontal axis, the position of absorption is generally referred to as chemical shift.

Answer to Problem 14.17P

The ${}^{\text{1}}\text{H}$ NMR spectrum of the given compound is,

The number of NMR signals, the splitting pattern for each signal and the approximate chemical shift values are rightfully stated.

Explanation of Solution

The given compound is $1,3-\text{dimethoxypropane}$$\left({\text{CH}}_{\text{3}}-\text{O}-{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}-\text{O}-{\text{CH}}_{\text{3}}\right)$. It contains three types of proton ${\text{H}}_{\text{a}}{\text{, H}}_{\text{b}}$ and ${\text{H}}_{\text{c}}$. Therefore, it gives three signals in ${}^{\text{1}}\text{H}$ NMR spectrum. The ${}^{\text{1}}\text{H}$ NMR spectrum of the given compound is shown below.

Figure 3

The proton ${\text{H}}_{\text{a}}$ gets split into a singlet because it has no adjacent proton, the proton ${\text{H}}_{\text{b}}$ gets split into a triplet by $2$${\text{H}}_{\text{c}}$ protons and the proton ${\text{H}}_{\text{c}}$ gets split into a quintet, five peaks by $2+2=4$${\text{H}}_{\text{b}}$ protons. The chemical shift of ${\text{H}}_{\text{a}}$ is approximately $3.3\text{ ppm}$. The chemical shift of ${\text{H}}_{\text{b}}$ is approximately $3.4\text{ ppm}$ and the chemical shift of ${\text{H}}_{\text{c}}$ is approximately $\text{1}\text{.7 ppm}$.

Conclusion

The ${}^{\text{1}}\text{H}$ NMR spectrum of the given compound is shown in Figure 3. It gives three signals with one, three and five peaks.

Interpretation Introduction

(d)

Interpretation:

The ${}^{\text{1}}\text{H}$ NMR spectrum of the given compound is to be described. The number of NMR signals, the splitting pattern for each signal and the approximate chemical shift values are to be stated.

Concept introduction:

The number of NMR signal in a compound is equal to the number of chemically non-equivalent protons present in that compound. In ${}^1\text{H}-\text{NMR}$ all chemically equivalent protons generates one signal or one peak, whereas non-equivalent proton generates different signals. On the horizontal axis, the position of absorption is generally referred to as chemical shift.

Answer to Problem 14.17P

The ${}^{\text{1}}\text{H}$ NMR spectrum of the given compound is,

The number of NMR signals, the splitting pattern for each signal and the approximate chemical shift values are rightfully stated.

Explanation of Solution

The given compound is $cis-3-\text{hexene}$$\left({\text{CH}}_{\text{3}}{\text{CH}}_2\text{CH}={\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{3}}\right)$. It contains three types of proton ${\text{H}}_{\text{a}}{\text{, H}}_{\text{b}}$ and ${\text{H}}_{\text{c}}$. Therefore, it gives three signals in ${}^{\text{1}}\text{H}$ NMR spectrum. The ${}^{\text{1}}\text{H}$ NMR spectrum of the given compound is shown below.

Figure 4

The proton ${\text{H}}_{\text{a}}$ gets split into a triplet because it has two adjacent proton, the proton ${\text{H}}_{\text{b}}$ gets split into a quintet, five peaks by $3+1=4$${\text{H}}_{\text{c}}{\text{ and H}}_{\text{b}}$ protons and the proton ${\text{H}}_{\text{c}}$ gets split into a triplet by two ${\text{H}}_{\text{b}}$ protons. The chemical shift of ${\text{H}}_{\text{a}}$ is approximately $\text{5}\text{.4 ppm}$. The chemical shift of ${\text{H}}_{\text{b}}$ is approximately $\text{2}\text{.0 ppm}$ and the chemical shift of ${\text{H}}_{\text{c}}$ is approximately $\text{1}\text{.1 ppm}$.

Conclusion

The ${}^{\text{1}}\text{H}$ NMR spectrum of the given compound is shown in Figure 4. It gives three signals with three, five and three peaks.