Chapter 15, Problem 20P

Name each compound and state how many lines are observed in its ${}^{13}\text{C}$ NMR spectrum.

a. b.

Interpretation Introduction

(a)

Interpretation: The name of the given compound is to be determined. The number of lines observed in its ${}^{\text{13}}\text{C}$ NMR spectrum is to be stated.

Concept introduction: ${}^{\text{13}}\text{C}$ NMR spectroscopy is used to determine the number of equivalent and non equivalent carbon atoms and to identify the type of carbon atoms present in the compound.

Answer to Problem 20P

The IUPAC name of the given compound is $\text{1-isopropyl-4-methylbenzene}$. Seven lines are present in ${}^{\text{13}}\text{C}$ NMR spectrum.

Explanation of Solution

The given molecule is

Figure 1

Black coloured atoms have four bonds. So, these are the carbon atoms. The grey coloured balls have one bond. So, these are the hydrogen atoms. The molecular structure of the compound is,

Figure 2

The parent name of the compound is benzene. The numbering will be in alphabetical order. Isopropyl group is present on first carbon atom and methyl group is present on fourth carbon atom. The IUPAC name of the given compound is $\text{1-isopropyl-4-methylbenzene}$. In the given compound there are seven types of carbons are present which are labeled as a, b, c, d, e, f and g.

Thus, seven lines are present in ${}^{\text{13}}\text{C}$ NMR spectrum.

Conclusion

The IUPAC name of the given compound is $\text{1-isopropyl-4-methylbenzene}$. Seven lines are present in ${}^{\text{13}}\text{C}$ NMR spectrum.

Interpretation Introduction

(b)

Interpretation: The name of the given compound is to be determined. The number of lines observed in its ${}^{\text{13}}\text{C}$ NMR spectrum is to be stated.

Concept introduction: ${}^{\text{13}}\text{C}$ NMR spectroscopy is used to determine the number of equivalent and non equivalent carbon atoms and to identify the type of carbon atoms present in the compound.

Answer to Problem 20P

The IUPAC name of the given compound is $\text{3-bromo-5-nitrophenol}$. Six lines are present in ${}^{\text{13}}\text{C}$ NMR spectrum.

Explanation of Solution

The given molecule is

Figure 3

The red coloured balls have two bonds. So, these are the oxygen atoms. The blue coloured ball has three bonds. So, this is the nitrogen atom. The maroon coloured ball has one bond. So, this is the halogen atom. The black coloured atoms have four bonds. So, these are the carbon atoms. The grey coloured balls have one bond. So, these are the hydrogen atoms. The molecular structure of the compound is,

Figure 4

The parent name of the compound is phenol. The numbering will be in alphabetical order. Bromine atom is present on third carbon atom and nitro group is present on fifth carbon atom. The IUPAC name of the given compound is $\text{3-Bromo-5-nitrophenol}$. In the given compound there are six types of carbons are present which are labeled as a, b, c, d, e and f.

Thus, six lines are present in ${}^{\text{13}}\text{C}$ NMR spectrum.

Conclusion

The IUPAC name of the given compound is $\text{3-bromo-5-nitrophenol}$. Six lines are present in ${}^{\text{13}}\text{C}$ NMR spectrum.