Chapter 13, Problem 13.44AP

Interpretation Introduction

(a)

Interpretation:

The difference in the proton NMR spectra of the given compound is to be stated.

Concept introduction:

The NMR stands for nuclear magnetic resonance. NMR spectroscopy deals with the interaction between electromagnetic radiation and the nucleus of an atom. NMR spectroscopy is used to determine the structural information about compounds.

Answer to Problem 13.44AP

The difference in the proton NMR spectra of the compound A and compound B is that compound A contains a doublet and septet and compound B contains only singlet.

Explanation of Solution

The given compounds are shown below.

$\begin{array}{l}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CH-Cl}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CD-Cl}\\ \text{A}\text{B}\end{array}$

The formula used to determine NMR splitting is shown below.

$\text{Number of}\text{peaks}=\text{n}+\text{1}$ … (1)

Where

• n is the number of hydrogen atom on the adjacent atom.

The structure of compound A is shown below.

Figure 1

The compound A shown in Figure 1 contains six methyl protons and one methyne proton. The number of hydrogen atoms on adjacent atom of methyl protons $\left(\text{n}\right)$ is $1$. Substitute the value of n in equation (1) as shown below.

$\begin{array}{rll}\text{Number of}\text{peaks}& =& 1+\text{1}\\ & =& \text{2}\end{array}$

Therefore, the NMR spectra of methyl protons splits into a doublet as the number of peaks is $2$. On the other hand, the number of hydrogen atoms on adjacent atom of the methyne proton $\left(\text{n}\right)$ is $6$. Substitute the value of n in equation (1) as shown below.

$\begin{array}{rll}\text{No}\text{.}\text{of}\text{peaks}& =& 6+\text{1}\\ & =& 7\end{array}$

NMR spectra of methyne protons splits into septet as the number of peaks is $6$.

The structure of the compound $\text{B}$ is shown below.

Figure 2

The compound B shown in Figure 2 contains six methyl protons. The number of hydrogen atoms on adjacent atom of methyl protons $\left(\text{n}\right)$ is $0$. Substitute the value of n in equation (1) as shown below.

$\begin{array}{rll}\text{Number of}\text{peaks}& =& 0+\text{1}\\ & =& 1\end{array}$

The NMR spectra of methyl protons splits into singlet as the number of peaks is 0. As a result, there is no coupling. Therefore, the methyl protons give a singlet peak.

Conclusion

The compound A gives septet and doublet whereas compound B gives a singlet in proton NMR spectra.

Interpretation Introduction

(b)

Interpretation:

The difference in the proton NMR spectra of the given compound is to be stated.

Concept introduction:

The NMR stands for nuclear magnetic resonance. NMR spectroscopy deals with the interaction between electromagnetic radiation and the nucleus of an atom. NMR spectroscopy is used to determine the structural information about compounds.

Answer to Problem 13.44AP

The difference in the proton NMR spectra of the compound A and compound B is that compound A contains two triplets and compound B contains two triplets and one quintet (multiplet).

Explanation of Solution

The given compounds are shown below.

$\begin{array}{l}{\text{Cl-CD}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{-Cl}{\text{Cl-CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{-Cl}\\ \text{A}\text{B}\end{array}$

The formula used to determine NMR splitting is shown below.

$\text{Number of}\text{peaks}=\text{n}+\text{1}$ …(1)

Where

• n is the number of hydrogen atom on the adjacent atom.

The structure of compound A is shown below.

Figure 3

The compound A shown in Figure 3 contains four methylene protons. The methylene protons are adjacent to each other. Therefore, both of them couple with each other. The value of hydrogen atoms $\left(\text{n}\right)$ on adjacent atom of $\text{C}-1$ is $2$. Substitute the value of $\text{n}$ in equation (1) as shown below.

$\begin{array}{rll}\text{Number of}\text{peaks}& =& 2+\text{1}\\ & =& 3\end{array}$

Therefore, the NMR spectra of methyl protons of $\text{C}-1$ splits into a triplet as the number of peaks is $3$. Similarly, the value of a number of hydrogen atoms $\left(\text{n}\right)$ on adjacent atom of $\text{C}-2$ is $2$. Substitute the value of $\text{n}$ in equation (1) a shown below.

$\begin{array}{rll}\text{Number of}\text{peaks}& =& 2+\text{1}\\ & =& 3\end{array}$

Therefore, the NMR spectra of methyl protons of $\text{C}-2$ splits into a triplet as the number of peaks is $3$.

The structure of compound B is shown below.

Figure 4

The compound B shown in Figure 4 contains six methylene protons. The methylene protons are adjacent to each other. The value of hydrogen atoms $\left(\text{n}\right)$ on adjacent atom of $\text{C}-1$ is $2$. Substitute the value of $\text{n}$ in equation (1) as shown below.

$\begin{array}{rll}\text{Number of}\text{peaks}& =& 2+\text{1}\\ & =& 3\end{array}$

Therefore, the NMR spectra of methyl protons of $\text{C}-1$ splits into a triplet as the number of peaks is $3$. Similarly, the value of a number of hydrogen atoms $\left(\text{n}\right)$ on adjacent atom of $\text{C}-3$ is $2$. Substitute the value of $\text{n}$ in equation (1) a shown below.

$\begin{array}{rll}\text{Number of}\text{peaks}& =& 2+\text{1}\\ & =& 3\end{array}$

Therefore, the NMR spectra of methyl protons of $\text{C}-3$ splits into a triplet as the number of peaks is $3$.

The protons on the adjacent of $\text{C}-2$ are chemically equivalent. The value of hydrogen atoms $\left(\text{n}\right)$ on adjacent atom of $\text{C}-2$ is $4$. Substitute the value of $\text{n}$ in equation (1) as shown below.

$\begin{array}{rll}\text{Number of}\text{peaks}& =& 4+\text{1}\\ & =& 5\end{array}$

Therefore, the NMR spectra of methyl protons of $\text{C}-2$ splits into a quintet (multiplet) as the number of peaks is $5$.

Conclusion

The compound A gives triplet at both the carbons whereas compound B gives triplet on $\text{C}-1$ and $\text{C}-3$ but proton NMR spectra.

Interpretation Introduction

(c)

Interpretation:

The difference in the proton NMR spectra of the given compound is to be stated.

Concept introduction:

The NMR stands for nuclear magnetic resonance. NMR spectroscopy deals with the interaction between electromagnetic radiation and the nucleus of an atom. NMR spectroscopy is used to determine the structural information about compounds.

Answer to Problem 13.44AP

The difference in the proton NMR spectra of the compound A and compound B is that compound A contains a doublet and septet and compound B contains only singlet.

Explanation of Solution

The formula used to determine NMR splitting is shown below.

$\text{Number of}\text{peaks}=\text{n}+\text{1}$ …(1)

Where

• n is the number of hydrogen atom on the adjacent atom.

The structure of compound A and B are shown below.

Figure 5

The compound A and B shown in Figure 5 both have similar spectrums. In the compound A, the value of number of methyl protons on adjacent to $\text{C}-1$ is $0$. Substitute the value of $\text{n}$ in equation (1) as shown below.

$\begin{array}{rll}\text{Number of}\text{peaks}& =& 0+\text{1}\\ & =& 1\end{array}$

Therefore, the NMR spectra of methyl protons of $\text{C}-1$ splits into a singlet as the number of peaks is $1$. Similarly the value of a number of hydrogen atoms $\left(\text{n}\right)$ on adjacent atom of $\text{C}-2$ is $0$. Substitute the value of $\text{n}$ in equation (1) a shown below.

$\begin{array}{rll}\text{Number of}\text{peaks}& =& 0+\text{1}\\ & =& 1\end{array}$

Therefore, the NMR spectra of methyl protons of $\text{C}-2$ splits into a singlet as the number of peaks is $1$.

Compound A and compound B both are diastereotopic. The chemical shift value of both the compounds is slightly different. The spectrum of both compounds is the same as there are diastereoisomers. Therefore, compound B also forms singlet on both the carbons.

The structure of compound C is shown below.

Figure 6

The compound C shown in Figure 6 contains five equivalent protons. The value of hydrogen atoms $\left(\text{n}\right)$ on adjacent atom of $\text{C}-3$ is $0$. Substitute the value of $\text{n}$ in equation (1) as shown below.

$\begin{array}{rll}\text{Number of}\text{peaks}& =& 0+\text{1}\\ & =& 1\end{array}$

Therefore, the NMR spectra of methyl protons of $\text{C}-3$ splits into a singlet as the number of peaks is $1$ as the neighboring protons are equivalent. Similarly the value of a number of hydrogen atoms $\left(\text{n}\right)$ on adjacent atom of $\text{C}-2$ is $1$. Substitute the value of $\text{n}$ in equation (1) is shown below.

$\begin{array}{rll}\text{No}\text{.}\text{of}\text{peaks}& =& 1+\text{1}\\ & =& 2\end{array}$

Therefore, the NMR spectra of methyl protons of $\text{C}-2$ splits into a doublet as the number of peaks is $2$.

The number of peaks of both $\text{C}-1$ and $\text{C}-2$ is same as both the protons are surrounded by same neighboring protons. The number of protons $\left(\text{n}\right)$ adjacent to $\text{C}-1$ is $1$. Substitute the value of $\text{n}$ in equation (1) is shown below.

$\begin{array}{rll}\text{Number of}\text{peaks}& =& 1+\text{1}\\ & =& 2\end{array}$

Therefore, the NMR spectra of methyl protons of $\text{C}-1$ splits into a doublet as the number of peaks is $2$.

Conclusion

The compound A gives triplet at both the carbons whereas compound B gives triplet on $\text{C}-1$ and $\text{C}-3$ but proton NMR spectra.