Chapter 13, Problem 13.40AP

Interpretation Introduction

(a)

Interpretation:

The structure of the compound having molecular formula, ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}$ and NMR data, $\delta 0.93$, s, is to be stated.

Concept introduction:

Many nuclei and electrons have spin. Due to this spin magnetic moment arises. The energy of this magnetic moment depends on the orientation of the applied magnetic field. In NMR spectroscopy, every nucleus has a spin. There is an angular momentum related to the spin. The difference between its resonance frequency and that of the reference standard is known as the chemical shift of a nucleus. Tetramethylsilane (TMS) is taken as reference.

Answer to Problem 13.40AP

The structure of the compound having molecular formula, ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}$ and NMR data, $\delta 0.93$, s, is shown below.

Explanation of Solution

There is only one signal in the compound that means all the protons are equivalent and all the protons are from methyl group as per chemical shift value of $\delta 0.93$.

Therefore, the structure of the compound corresponding to NMR data is shown below.

Figure 1

Conclusion

The structure of the compound having molecular formula, ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}$ and NMR data, $\delta 0.93$, s, is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation:

The structure of the compound having molecular formula ${\text{C}}_{\text{5}}{\text{H}}_{\text{10}}$ and NMR data, $\delta 1.5$, s, is to be stated.

Concept introduction:

Many nuclei and electrons have spin. Due to this spin magnetic moment arises. The energy of this magnetic moment depends on the orientation of the applied magnetic field. In NMR spectroscopy, every nucleus has a spin. There is an angular momentum related to the spin. The difference between its resonance frequency and that of the reference standard is known as the chemical shift of a nucleus. Tetramethylsilane (TMS) is taken as reference.

Answer to Problem 13.40AP

The structure of the compound having molecular formula ${\text{C}}_{\text{5}}{\text{H}}_{\text{10}}$ and NMR data, $\delta 1.5$, s is shown below.

Explanation of Solution

There is only one signal in the compound that means all the protons are equivalent and all the protons are from methylene group as per chemical shift value of $\delta 1.5$.

Therefore, the structure of the compound corresponding to NMR data is shown below.

Figure 2

Conclusion

The structure of the compound having molecular formula ${\text{C}}_{\text{5}}{\text{H}}_{\text{10}}$ and NMR data, $\delta 1.5$, s is shown in Figure 2.

Interpretation Introduction

(c)

Interpretation:

The structure of the compound having molecular formula ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}{\text{O}}_{\text{2}}$ and NMR data, $\delta 1.36$ ($\text{3H}$, d, $J=5.5\text{ Hz}$); $\delta 3.32$ ($6H$, s); $\delta 4.63$ ($1H$, q, $J=5.5\text{ Hz}$) is to be stated.

Concept introduction:

Many nuclei and electrons have spin. Due to this spin magnetic moment arises. The energy of this magnetic moment depends on the orientation of the applied magnetic field. In NMR spectroscopy, every nucleus has a spin. There is an angular momentum related to the spin. The difference between its resonance frequency and that of the reference standard is known as the chemical shift of a nucleus. Tetramethylsilane (TMS) is taken as reference.

Answer to Problem 13.40AP

The structure of the compound having molecular formula ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}{\text{O}}_{\text{2}}$ and NMR data, $\delta 1.36$ ($\text{3H}$, d, $J=5.5\text{ Hz}$); $\delta 3.32$ ($6H$, s); $\delta 4.63$ ($1H$, q, $J=5.5\text{ Hz}$) is shown below.

Explanation of Solution

There are three signals in the compound. One is at $\delta 4.63$ which corresponds to the methine proton. The signal at $\delta 3.32$ corresponds to the two methyl groups attached to the oxygen atom. The signal at $\delta 1.36$ corresponds to the methyl group which undergoes coupling with methine proton.

Therefore, the structure of the compound corresponding to NMR data is shown below.

Figure 3

Conclusion

The structure of the compound having molecular formula ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}{\text{O}}_{\text{2}}$ and NMR data, $\delta 1.36$ ($\text{3H}$, d, $J=5.5\text{ Hz}$); $\delta 3.32$ ($6H$, s); $\delta 4.63$ ($1H$, q, $J=5.5\text{ Hz}$) is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation:

The structure of the compound having molecular formula ${\text{C}}_{\text{7}}{\text{H}}_{\text{16}}\text{O}$ and the given NMR spectrum is to be stated.

Concept introduction:

Many nuclei and electrons have spin. Due to this spin magnetic moment arises. The energy of this magnetic moment depends on the orientation of the applied magnetic field. In NMR spectroscopy, every nucleus has a spin. There is an angular momentum related to the spin. The difference between its resonance frequency and that of the reference standard is known as the chemical shift of a nucleus. Tetramethylsilane (TMS) is taken as reference.

Answer to Problem 13.40AP

The structure of the compound having molecular formula ${\text{C}}_{\text{7}}{\text{H}}_{\text{16}}\text{O}$ and the given NMR spectrum is shown below.

Explanation of Solution

The given spectrum is shown below.

Figure 4

There are three signals in the compound. One is at $\delta 0.9$ for nine hydrogens corresponds to the three methyl group. The signal at $\delta 1.2$ corresponds to the hydrogen of the hydroxide group as this peak disappears after ${\text{D}}_{\text{2}}\text{O}$ shake. The signal at $\delta 1.36$ is for six hydrogens of the three methylene groups which undergoes coupling with methyl group protons.

Therefore, the structure of the compound corresponding to NMR data is shown below.

Figure 5

Conclusion

The structure of the compound having molecular formula ${\text{C}}_{\text{7}}{\text{H}}_{\text{16}}\text{O}$ and the given NMR spectrum is shown in Figure 5.

Interpretation Introduction

(e)

Interpretation:

The structure of the compound having molecular formula ${\text{C}}_{\text{8}}{\text{H}}_{\text{16}}$ and given NMR spectrum is to be stated.

Concept introduction:

Many nuclei and electrons have spin. Due to this spin magnetic moment arises. The energy of this magnetic moment depends on the orientation of the applied magnetic field. In NMR spectroscopy, every nucleus has a spin. There is an angular momentum related to the spin. The difference between its resonance frequency and that of the reference standard is known as the chemical shift of a nucleus. Tetramethylsilane (TMS) is taken as reference.

Answer to Problem 13.40AP

The structure of the compound having molecular formula ${\text{C}}_{\text{8}}{\text{H}}_{\text{16}}$ and given NMR spectrum is shown below.

Explanation of Solution

The given spectrum is shown below.

Figure 6

It is given that the ${\text{C}}_{\text{8}}{\text{H}}_{\text{16}}$ on catalytic hydrogenation gives $\text{2, 2, 4-trimethylpentane}$ which means the compound is an alkene.

The alkenes those produce $\text{2, 2, 4-trimethylpentane}$ on catalytic hydrogenation are shown below.

Figure 7

There are five signals in the compound. The two signal at $\delta 4.8-4.9$ corresponds to the two vinylic protons and the two vinylic protons are only present in one alkene that is $\text{2, 4, 4-trimethylpent-1-ene}$.

Therefore, the structure of the compound corresponding to NMR data is shown below.

Figure 8

Conclusion

The structure of the compound having molecular formula ${\text{C}}_{\text{8}}{\text{H}}_{\text{16}}$ and given NMR spectrum is shown in Figure 8.

Interpretation Introduction

(f)

Interpretation:

The structure of the compound having molecular formula ${\text{C}}_{\text{7}}{\text{H}}_{\text{12}}{\text{Cl}}_{\text{2}}$ and NMR data, $\delta 1.07$ ($9H$, s); $\delta 2.28$ ($2H$, d, $J=6\text{ Hz}$); $\delta 5.77$ ($1H$, t, $J=6\text{ Hz}$) is to be stated.

Concept introduction:

Many nuclei and electrons have spin. Due to this spin magnetic moment arises. The energy of this magnetic moment depends on the orientation of the applied magnetic field. In NMR spectroscopy, every nucleus has a spin. There is an angular momentum related to the spin. The difference between its resonance frequency and that of the reference standard is known as the chemical shift of a nucleus. Tetramethylsilane (TMS) is taken as reference.

Answer to Problem 13.40AP

The structure of the compound having molecular formula ${\text{C}}_{\text{7}}{\text{H}}_{\text{12}}{\text{Cl}}_{\text{2}}$ and NMR data, $\delta 1.07$ ($9H$, s); $\delta 2.28$ ($2H$, d, $J=6\text{ Hz}$); $\delta 5.77$ ($1H$, t, $J=6\text{ Hz}$) is shown below.

Explanation of Solution

There are three signals in the compound. The signal at $\delta 1.07$ is for nine hydrogens that corresponds to the three methyl groups. The triplet signal at $\delta 5.77$ corresponds to the hydrogen directly attached to the double bond. The doublet signal at $\delta 2.28$ for two hydrogens corresponds to the methylene group and the protons coupled with hydrogen directly attached to double bond that means these two are close to each other.

Therefore, the structure of the compound corresponds to NMR data as shown below.

Figure 9

Conclusion

The structure of the compound having molecular formula ${\text{C}}_{\text{7}}{\text{H}}_{\text{12}}{\text{Cl}}_{\text{2}}$ and NMR data, $\delta 1.07$ ($9H$, s); $\delta 2.28$ ($2H$, d, $J=6\text{ Hz}$); $\delta 5.77$ ($1H$, t, $J=6\text{ Hz}$) is shown in Figure 9.

Interpretation Introduction

(g)

Interpretation:

The structure of the compound having molecular formula ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{Br}}_{\text{2}}{\text{F}}_{\text{2}}$ and NMR data, $\delta 4.02$ (t, $J=16\text{ Hz}$) is to be stated.

Concept introduction:

Many nuclei and electrons have spin. Due to this spin magnetic moment arises. The energy of this magnetic moment depends on the orientation of the applied magnetic field. In NMR spectroscopy, every nucleus has a spin. There is an angular momentum related to the spin. The difference between its resonance frequency and that of the reference standard is known as the chemical shift of a nucleus. Tetramethylsilane (TMS) is taken as reference.

Answer to Problem 13.40AP

The structure of the compound having molecular formula ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{Br}}_{\text{2}}{\text{F}}_{\text{2}}$ and NMR data, $\delta 4.02$ (t, $J=16\text{ Hz}$) is shown below.

Explanation of Solution

There is only one signal in the compound for two hydrogens that means both are equivalent. The triplet signal is given by the two protons that means the splitting of the signal is done by two equivalent fluorine groups.

Therefore, the structure of the compound corresponding to NMR data is shown below.

Figure 10

Conclusion

The structure of the compound having molecular formula ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{Br}}_{\text{2}}{\text{F}}_{\text{2}}$ and NMR data, $\delta 4.02$ (t, $J=16\text{ Hz}$) is shown in Figure 10.

Interpretation Introduction

(h)

Interpretation:

The structure of the compound having molecular formula ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{F}}_{\text{3}}\text{I}$ and NMR data, $\delta 3.56$ (q, $J=10\text{ Hz}$) is to be stated.

Concept introduction:

Many nuclei and electrons have spin. Due to this spin magnetic moment arises. The energy of this magnetic moment depends on the orientation of the applied magnetic field. In NMR spectroscopy, every nucleus has a spin. There is an angular momentum related to the spin. The difference between its resonance frequency and that of the reference standard is known as the chemical shift of a nucleus. Tetramethylsilane (TMS) is taken as reference.

Answer to Problem 13.40AP

The structure of the compound having molecular formula ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{F}}_{\text{3}}\text{I}$ and NMR data, $\delta 3.56$ (q, $J=10\text{ Hz}$) is shown below.

Explanation of Solution

There is only one signal in the compound for two hydrogens that means both are equivalent. The quartet signal is given by the two protons that means the splitting of the signal is done by three equivalent fluorine groups.

Therefore, the structure of the compound corresponding to NMR data is shown below.

Figure 11

Conclusion

The structure of the compound having molecular formula ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{F}}_{\text{3}}\text{I}$ and NMR data, $\delta 3.56$ (q, $J=10\text{ Hz}$) is shown in Figure 11.

Interpretation Introduction

(i)

Interpretation:

The structure of the compound having molecular formula ${\text{C}}_{\text{7}}{\text{H}}_{\text{16}}{\text{O}}_{\text{4}}$ and NMR data, $\delta 1.93$ (t, $J=6\text{ Hz}$); $\delta 3.35$ (s); $\delta 4.49$ (t, $J=6\text{ Hz}$); relative integral $1:6:1$ is to be stated.

Concept introduction:

Many nuclei and electrons have spin. Due to this spin magnetic moment arises. The energy of this magnetic moment depends on the orientation of the applied magnetic field. In NMR spectroscopy, every nucleus has a spin. There is an angular momentum related to the spin. The difference between its resonance frequency and that of the reference standard is known as the chemical shift of a nucleus. Tetramethylsilane (TMS) is taken as reference.

Answer to Problem 13.40AP

The structure of the compound having molecular formula ${\text{C}}_{\text{7}}{\text{H}}_{\text{16}}{\text{O}}_{\text{4}}$ and NMR data, $\delta 1.93$ (t, $J=6\text{ Hz}$); $\delta 3.35$ (s); $\delta 4.49$ (t, $J=6\text{ Hz}$); relative integral $1:6:1$ is shown below.

Explanation of Solution

There are three signals in the compound and total hydrogens are sixteen. The relative intergral tells about the ratio of number of hydrogen in each signal. The relative integral is $1:6:1$ that means there are twelve hydrogens at $\delta 3.35$ which corresponds to four equivalent methyl groups directly attached to the oxygen. The signal at $\delta 1.93$ is for two hydrogens which corresponds to the methylene group coupling with hydrogens at $\delta 4.49$ which corresponds to two methine group in the compound directly linked to the oxygen.

Therefore, the structure of the compound corresponding to NMR data is shown below.

Figure 12

Conclusion

The structure of the compound having molecular formula ${\text{C}}_{\text{7}}{\text{H}}_{\text{16}}{\text{O}}_{\text{4}}$ and NMR data, $\delta 1.93$ (t, $J=6\text{ Hz}$); $\delta 3.35$ (s); $\delta 4.49$ (t, $J=6\text{ Hz}$); relative integral $1:6:1$ is shown in Figure 12.

Interpretation Introduction

(j)

Interpretation:

The structure of the compound having molecular formula ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}\text{O}$ and NMR data, $\delta 0.91$ ($6H$, d, $J=7\text{ Hz}$); $\delta 1.17$ ($6H$, s); $\delta 1.48$ ($1H$, s; disappears following ${\text{D}}_{\text{2}}\text{O}$ shake); $\delta 1.65$ ($1H$, septet, $J=7\text{ Hz}$); ${}^{13}\text{C}$ NMR: $\delta 17.6,\delta 26.5,\delta 38.7,\delta 73.2$ is to be stated.

Concept introduction:

Many nuclei and electrons have spin. Due to this spin magnetic moment arises. The energy of this magnetic moment depends on the orientation of the applied magnetic field. In NMR spectroscopy, every nucleus has a spin. There is an angular momentum related to the spin. The difference between its resonance frequency and that of the reference standard is known as the chemical shift of a nucleus. Tetramethylsilane (TMS) is taken as reference.

Answer to Problem 13.40AP

The structure of the compound having molecular formula ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}\text{O}$ and NMR data, $\delta 0.91$ ($6H$, d, $J=7\text{ Hz}$); $\delta 1.17$ ($6H$, s); $\delta 1.48$ ($1H$, s; disappears following ${\text{D}}_{\text{2}}\text{O}$ shake); $\delta 1.65$ ($1H$, septet, $J=7\text{ Hz}$); ${}^{13}\text{C}$ NMR: $\delta 17.6,\delta 26.5,\delta 38.7,\delta 73.2$ is shown below.

Explanation of Solution

There are three signals in the compound and total hydrogens are fourteen. The doublet signal at $\delta 0.91$ for six hydrogens correspond to the two methyl groups and these protons are coupling with the one proton at $\delta 1.65$ corresponding to the methine proton.

The signal at $\delta 1.48$ for one hydrogen which disappears following ${\text{D}}_{\text{2}}\text{O}$ shake is for hydroxide group. The signal at $\delta 1.17$ for six hydrogens corresponds to the two methyl groups close the hydroxide group.

Therefore, the structure of the compound corresponding to NMR data is shown below.

Figure 13

Conclusion

The structure of the compound having molecular formula ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}\text{O}$ and NMR data, $\delta 0.91$ ($6H$, d, $J=7\text{ Hz}$); $\delta 1.17$ ($6H$, s); $\delta 1.48$ ($1H$, s; disappears following ${\text{D}}_{\text{2}}\text{O}$ shake); $\delta 1.65$ ($1H$, septet, $J=7\text{ Hz}$); ${}^{13}\text{C}$ NMR: $\delta 17.6,\delta 26.5,\delta 38.7,\delta 73.2$ is shown in Figure 13.