Chapter 16, Problem 16.30P

Interpretation Introduction

Interpretation:

Based on the NMR spectrum in Figure 16-32, it is to be determined and explained whether the unknown 1 is ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{ or CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{OCH}}_{\text{2}}{\text{(C=O)C}}_{\text{6}}{\text{H}}_{\text{5}}$. Also explain why or why not.

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 the Number of signals, which tells 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, molecular fragments can be idenfiied. These can be used to build up the molecular structure. ${}^{13}\text{C}$ NMR provides valuable information about the carbon skeleton. Each signal in the ${}^{13}\text{C}$ NMR equals 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, 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 16.30P

Based on the NMR spectrum in Figure 16-32, the unknown 1 is

${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}$ because the compound will have a $\text{3H}$ split into a triplet, a $\text{2H}$ split into a multiplet, and a $\text{2H}$ split into a triplet. Thus, it will have two triplets and one multiplet, apart from the aromatic protons. This pattern matches with the proton spectrum given in Figure 16-32.

Explanation of Solution

The given proton NMR for the unknown 1 in Figure 16-32 is

The given molecular formula for unknown 1 is ${\text{C}}_{10}{\text{H}}_{\text{12}}{\text{O}}_2$. Presence of two oxygen atoms in the molecular formula indicates that the functional group may be an ester, a carboxylic acid, or a carbonyl and an ether.

The spectrum appears to have four signals. Therefore, there are at least four distinct types of protons in the unknown 1. The relative integrations tell the number of hydrogen atoms associated with that particular peak. The summation of intergrations given in the spectrum is $\text{1}\text{.5 H + 1 H + 1 H + 2}\text{.5 H = 6H}$. But the molecular formula shows that there are actually $12\text{H}$ atoms. Thus, each of the relative integration values given to us represents two hydrogen atoms. Thus, the signal $\text{δ}\text{1}\text{.1 ppm (1}\text{.5H)}$ actually corresponds to $3\text{H}$ atoms whereas the signal $\text{δ}\text{7}\text{.5 ppm (2}\text{.5H)}$ corresponds to $5\text{H}$ atoms.

The triplet at $\text{δ}\text{1}\text{.1 ppm (3H)}$ must be due to the methyl group, which is attached to the ${\text{CH}}_{\text{2}}$ group. Another triplet at $\text{δ}\text{4}\text{.3 ppm (2H)}$ must be due to the ${\text{CH}}_{\text{2}}$ group, which has two neighboring protons and is also attached to an electronegative element such as oxygen. The sextet at $\text{δ}\text{1}\text{.8 ppm (2H)}$ must be due to the ${\text{CH}}_{\text{2}}$ group, which has five neighboring protons.

The signal $\text{δ}\text{7}\text{.5 ppm (2}\text{.5H)}$ corresponding to $5\text{H}$ atoms is due to the aromatic ptotons; this also indicates that the compound is monosubstituted benzene.

The structures for ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{ and CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{OCH}}_{\text{2}}{\text{(C=O)C}}_{\text{6}}{\text{H}}_{\text{5}}$ are shown below:

Both compounds are monosubstituted benzenes. Therefore, the aromatic region of the proton NMR will be similar for both of them. The difference in the structures of the two compounds arises from the difference in connectivity of the methyl groups and two ${\text{CH}}_{\text{2}}$ groups.

In the first compound, which is on the left, there would be four signals. It will have a $\text{3H}$ split into a triplet, a $\text{2H}$ split into a multiplet, and a $\text{2H}$ split into a triplet. Thus, it will have two triplets and one multiplet, apart from the aromatic protons.

The second compound, which is on the right, would also have four signals. It will have a $\text{3H}$ split into a triplet, a $\text{2H}$ split into a quartet, and a $\text{2H}$ split into a singlet. Thus, it will have a triplet, a quartet, and a singlet, apart from the aromatic protons.

The spectrum shown in Figure 16-32 matches with the splitting patterns and integrations in ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}$ (which is the compound on the left). However, the chemical shift of the triplet signal for proton in the ${\text{CH}}_{\text{2}}$ group adjacent to the carbonyl group does not match very well. The proton in ${\text{CH}}_{\text{2}}$ group adjacent to the carbonyl group would be around $\text{δ 2}\text{.1 ppm}$ whereas the triplet in the spectrum appears at $\text{δ 4}\text{.3ppm}$, which is more consistent with the protons of the form $\text{H-C-O}$. These protons appear at $\text{δ 3}\text{.3 ppm}$ or higher.

Conclusion

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