Chapter 12C, Problem 36P

(a) How many ${}^{\text{1}}\text{H}$ NMR signals does each of the following compounds exhibit? (b) How many ${}^{\text{13}}\text{C}$ NMR

signals does each of the following compounds exhibit?

Interpretation Introduction

(a)

Interpretation: The number of ${}^{\text{1}}\text{H}$ NMR signals present in each of the given compounds is to be identified.

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.

Answer to Problem 36P

The given compound A gives four signals and compound B gives five signals in ${}^{\text{1}}\text{H}$ NMR spectrum.

Explanation of Solution

The given compound A is the ball and stick model of $2-\left(\text{chloromethoxy}\right)-1-\text{methoxy}-2-\text{methylpropane}$. The bond-line representation of the compound A with all chemically non-equivalent protons is shown below.

Figure 1

There are four non-equivalent protons $\left({\text{H}}_{\text{a}}{\text{, H}}_{\text{b}}{\text{, H}}_{\text{c}}{\text{ and H}}_{\text{d}}\right)$ present in compound A. The number of signals in each compound is equal to the number of hydrogen atoms present in a different chemical environment. Therefore, the compound A gives four signals in ${}^{\text{1}}\text{H}$ NMR spectrum.

The given compound B is the ball and stick model of propyl butyrate. The bond-line representation of the compound B with all chemically non-equivalent protons is shown below.

Figure 2

There are five non-equivalent protons $\left({\text{H}}_{\text{a}}{\text{, H}}_{\text{b}}{\text{, H}}_{\text{c}}{\text{, H}}_{\text{d}}{\text{ and H}}_{\text{e}}\right)$ present in compound B. The number of signals in each compound is equal to the number of hydrogen atoms present in a different chemical environment. Therefore, the compound B gives five signals in ${}^{\text{1}}\text{H}$ NMR spectrum.

Conclusion

The given compound A gives four signals and compound B gives five signals in ${}^{\text{1}}\text{H}$ NMR spectrum.

Interpretation Introduction

(b)

Interpretation: The number of ${}^{\text{13}}\text{C}$ NMR signals present in each of the given compounds is to be identified.

Concept introduction: The number of NMR signal in a compound is equal to the number of carbon atoms present in that compound. In ${}^{\text{13}}\text{C}$ NMR all chemically equivalent carbon atoms generates one signal or one peak, whereas non-equivalent carbon atoms generates different signals.

Answer to Problem 36P

The given compound A gives five signals and compound B gives six signals in ${}^{\text{13}}\text{C}$ NMR spectrum.

Explanation of Solution

The given compound A is the ball and stick model of $2-\left(\text{chloromethoxy}\right)-1-\text{methoxy}-2-\text{methylpropane}$. The bond-line representation of the compound A with all chemically non-equivalent carbon atoms is shown below.

Figure 3

There are five non-equivalent carbon atoms $\left({\text{C}}_{\text{a}}{\text{, C}}_{\text{b}}{\text{, C}}_{\text{c}}{\text{, C}}_{\text{c}}{\text{ and C}}_{\text{e}}\right)$ present in compound A. The number of signals in each compound is equal to the number of carbon atoms present in a different chemical environment. Therefore, the compound A gives five signals in ${}^{\text{13}}\text{C}$ NMR spectrum.

The given compound B is the ball and stick model of propyl butyrate. The bond-line representation of the compound B with all chemically non-equivalent carbon atoms is shown below.

Figure 4

There are six non-equivalent carbon atoms $\left({\text{C}}_{\text{a}}{\text{, C}}_{\text{b}}{\text{, C}}_{\text{c}}{\text{, C}}_{\text{c}}{\text{, C}}_{\text{e}}{\text{ and C}}_{\text{f}}\right)$ present in compound B. The number of signals in each compound is equal to the number of carbon atoms present in a different chemical environment. Therefore, the compound B gives six signals in ${}^{\text{13}}\text{C}$ NMR spectrum.

Conclusion

The given compound A gives five signals and compound B gives six signals in ${}^{\text{13}}\text{C}$ NMR spectrum.