Chapter 16, Problem 16.31P

Interpretation Introduction

Interpretation:

The structure for the compound having molecular formula ${\text{C}}_9{\text{H}}_{\text{10}}{\text{O}}_2$ is to be proposed based on the distinct signals in its ${}^1\text{H}$ NMR and ${}^{13}\text{C}$ NMR spectra.

Concept introduction:

In ^{${}^1\text{H}$} NMR spectroscopy, protons in different environments within a molecule have different chemical shifts, that is, they experience different degrees of shielding.

In addition to chemical shift, a ${}^1\text{H}$ NMR spectrum provides structural information based on: Number of signals, which tells us how many different kinds of protons are there; Integrated areas, which tells the ratios of the various kinds of protons; and, Splitting pattern, which gives information about the number of protons that are within two or three bonds of the one giving the signal. Spin–spin splitting of NMR signals results from the coupling of the nuclear spins that are separated by two bonds (geminal coupling) or three bonds (vicinal coupling). In these cases, the number of peaks into which a signal is split is equal to $\text{(n+1)}$, where n is the number of protons to which the proton in question is coupled. Protons that have the same chemical shift do not split each other’s signals.

Complicated splitting patterns can result when a proton is unequally coupled to two or more protons that are different from one another.

The ideal range for alkane protons is $\text{δ}0.9\text{- δ}1.4$; for alcohol it is $\text{δ}2\text{- δ}5.0$; and, for aromatic protons it is $\text{δ}6.5\text{- δ}8.5$. The chemical shift of a signal prompts about the aromatic rings, double bonds, or nearby electronegative atoms. The integration of each signal suggests the number of protons responsible for that signal. The splitting pattern of a signal indicates the number of neighboring protons that are distinct from the protons responsible for that signal. To deduce the structure of an unknown compound, the first step is to find the index of hydrogen deficiency if the molecular formula is given. Based on the data given in the ${}^1\text{H}$ NMR build molecular fragments with multiple carbon atoms.

${}^{13}\text{C}$ NMR provides valuable information about the carbon skeleton. Each signal in the ${}^{13}\text{C}$ NMR equals to the number of distinct carbon atoms in the given unknown. In most of the ${}^{13}\text{C}$ NMR spectra almost every time all the signals would appear as singlets. The saturated carbon atoms appear in the range $\text{δ}\text{0-35 ppm}$ if it is a simple alkane fragment. If it is attached to any electronegative element such as halogens, or nitrogen then the range is $\text{δ}\text{25-70 ppm}$. Triple bonded carbon atoms range from $\text{δ}\text{65-85 ppm}$. Alkene carbons range from $\text{δ}\text{105-150 ppm}$. Carbonyl carbon atoms in acids, esters, amides and anhydrides range from $\text{δ}\text{120-185 ppm}$. Carbonyl carbons in aldehydes and ketones range from $\text{δ}\text{190-220 ppm}$.

Answer to Problem 16.31P

The proposed structure for the compound having molecular formula ${\text{C}}_9{\text{H}}_{\text{10}}{\text{O}}_2$ based on the distinct signals in its ${}^1\text{H}$ NMR and ${}^{13}\text{C}$ NMR spectra is:

Explanation of Solution

The molecular formula for the given unknown compound is ${\text{C}}_9{\text{H}}_{\text{10}}{\text{O}}_2$.

The molecular formula shows an index of hydrogen deficiency of five.

The ${}^1\text{H}$ NMR of the compound has four distinct signals, therefore there must be four distinct protons in the structure, and the data is as follows:

$\begin{array}{l}{\text{C}}_9{\text{H}}_{\text{10}}{\text{O}}_2\text{: δ 7}\text{.0 ppm (2H), doublet}\\ \text{ δ 2}\text{.3 ppm (6H), singlet}\\ \text{ δ 7}\text{.2 ppm (1H), triplet}\\ \text{ δ 12}\text{.9 ppm (1H) very broad singlet}\end{array}$

IHD of five and the doublet at $\text{δ 7}\text{.0 ppm (2H)}$ and a triplet at $\text{δ 7}\text{.2 ppm (1H)}$ indicates trisubstituted-substituted benzene.

The broad singlet at $\text{δ 12}\text{.9 ppm (1H)}$ is because of the highly desheilded proton and the signal belongs to the proton of the carboxylic acid group. Thus, one of the substituents in the trisubstituted benzene is the carboxylic acid group.

The singlet at $\text{δ 2}\text{.3 ppm (6H)}$ represents two ${\text{-CH}}_3$ groups with no adjacent proton attached. One signal for two ${\text{-CH}}_3$ groups indicates that they must be present in an identical environment and might be attached to the benzene ring.

For the aromatic protons, the signal at $\text{δ 7}\text{.0 ppm (2H)}$ which is a doublet indicates that these two protons that are attached to the benzene ring have identical chemical environment. There are two possibilities for the symmetrically trisubstituted benzene:

Both the trisubstituted benzenes have only two types of aromatic protons, ${\text{H}}_{\text{a}}{\text{ and H}}_{\text{b}}$.

In the structure on left, the protons highlighted in blue will appear as a doublet as both of them have identical chemical environment, while the proton in red will split into a triplet as it has two neighboring protons attached.

In the structure on right, the protons in pink will appear as singlets, as they do not have any neighboring protons while the proton in orange will also appear as a singlet.

Thus, the second structure doesn’t match with the data given in the ${}^1\text{H}$ NMR as we do not have a singlet consisting of $\text{2H}$ atoms.

The ${}^{13}\text{C}$ NMR spectrum shows six signals; this suggests that there must be six distinct carbon atoms in the structure.

The ${}^{13}\text{C}$ NMR spectrum data is:

$\begin{array}{l}{\text{C}}_9{\text{H}}_{\text{10}}{\text{O}}_2\text{: δ 19}\text{.4 ppm }\\ \text{ δ 127}\text{.2 ppm}\\ \text{ δ 128}\text{.5 ppm}\\ \text{ δ 133}\text{.6 ppm}\\ \text{ δ 135}\text{.3 ppm}\\ \text{ δ 170}\text{.9 ppm}\end{array}$

The signal $\text{δ 170}\text{.9 ppm}$ is the carbonyl carbon in the carboxylic acid group. The four signals $\text{δ 127}\text{.2, 128}\text{.5, 133}\text{.6, 135}\text{.3 ppm}$ are the carbon atoms in the benzene ring. The signal $\text{δ 19}\text{.4 ppm}$ is the signal for the saturataed carbon in alkanes. This signal might be because of the methyl groups attached to the benzene ring.

Considering both the ${}^1\text{H}$ and ${}^{13}\text{C}$ NMR, the proposed structure for the unknown compound having molecular formula ${\text{C}}_9{\text{H}}_{\text{10}}{\text{O}}_2$ is:

Conclusion

The structure of the unknown compound can be proposed based on its ${}^1\text{H}$ and ${}^{13}\text{C}$ NMR sanalysis.