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Following are \({ }^{1} \mathrm{H}\)-NMR and \({ }^{13} \mathrm{C} N \mathrm{NMR}\) spectral data for compound \(\mathrm{G}\left(\mathrm{C}_{10} \mathrm{H}_{10} \mathrm{O}\right)\). From this information, deduce the structure of compound \(G\). $$ \begin{array}{lrr} { }^{1} \text { H-NMR } & \multicolumn{2}{c}{{ }^{13} \text { C-NMR }} \\ \hline 2.50(\mathrm{t}, 2 \mathrm{H}) & 210.19 & 126.82 \\ 3.05(\mathrm{t}, 2 \mathrm{H}) & 136.64 & 126.75 \\ 3.58(\mathrm{~s}, 2 \mathrm{H}) & 133.25 & 45.02 \\ 7.1-7.3(\mathrm{~m}, 4 \mathrm{H}) & 128.14 & 38.11 \\ 127.75 & & 28.34 \\ \hline \end{array} $$

Short Answer

Expert verified
Answer: The structure of compound G is para-methoxyethylbenzoate.

Step by step solution

01

Analyzing the given \({ }^{1}\mathrm{H}\)-NMR spectrum

From the given \({ }^{1}\mathrm{H}\)-NMR spectrum, we have four sets of signals: 1. A triplet at 2.50 ppm integrating to 2 H 2. A triplet at 3.05 ppm integrating to 2 H 3. A singlet at 3.58 ppm integrating to 2 H 4. A multiplet between 7.1 and 7.3 ppm integrating to 4 H
02

Analyzing the given \({ }^{13}\mathrm{C}\)-NMR spectrum

From the given \({ }^{13}\mathrm{C}\)-NMR spectrum, we have eight different signals: 1. 210.19 ppm 2. 136.64 ppm 3. 133.25 ppm 4. 128.14 ppm 5. 127.75 ppm 6. 126.82 ppm 7. 126.75 ppm 8. 45.02 ppm 9. 38.11 ppm 10. 28.34 ppm
03

Assigning hydrogen environments

Now, let's assign the environments for the hydrogens using the \({ }^{1}\mathrm{H}\)-NMR spectrum: 1. A triplet at 2.50 ppm integrating to 2 H: This is characteristic for a methylene (-CH2-) group next to an unsaturated system (alkenes, carbonyls). 2. A triplet at 3.05 ppm integrating to 2 H: This is also characteristic for a methylene (-CH2-) group but next to an electronegative atom (oxygen or nitrogen). 3. A singlet at 3.58 ppm integrating to 2 H: A singlet indicates no coupling, indicating these hydrogens are not near another hydrogen. This suggests these hydrogens are attached to an oxygen in a methoxy group (-O-CH3). 4. A multiplet between 7.1 and 7.3 ppm integrating to 4 H: This corresponds to a group of four hydrogens in an aromatic environment.
04

Assigning carbon environments

Using the \({ }^{13}\mathrm{C}\)-NMR spectrum, we can assign the environments for the carbons: 1. 210.19 ppm: Carbonyl carbon (C=O) 2. 136.64 ppm, 133.25 ppm, 128.14 ppm, 127.75 ppm, 126.82 ppm, and 126.75 ppm: These signals are characteristic of aromatic carbons. 3. 45.02 ppm: Methoxy carbon attached to an oxygen (-O-CH3) 4. 38.11 ppm: Methylene (-CH2-) carbon next to a carbonyl group or double bond 5. 28.34 ppm: Methylene (-CH2-) carbon next to an electronegative atom (oxygen or nitrogen)
05

Deduce the structure of compound G

Combining the information obtained from the \({ }^{1}\mathrm{H}\)-NMR and \({ }^{13}\mathrm{C} \mathrm{NMR}\) spectra, we can propose a structure for compound G: 1. The presence of four aromatic hydrogens with six aromatic carbons suggests a disubstituted benzene ring. 2. The presence of a carbonyl group and a methoxy group suggests an ester functional group attached to the benzene ring. 3. Two different methylenes with triplets suggest two ethyl groups with different chemical environments. One ethyl group is attached to the benzene ring, and the other is linked to the carbonyl carbon of the ester. Therefore, the structure of compound G is ethyl benzoate with a methoxy group para to the ester, or para-methoxyethylbenzoate with the molecular formula \(\mathrm{C}_{10}\mathrm{H}_{10}\mathrm{O}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

1H-NMR
Nuclear Magnetic Resonance (NMR) spectroscopy is a powerful technique for determining the molecular structure of organic compounds. The \(^{1} \mathrm{H}\)-NMR, or proton NMR, is particularly useful as it provides information about the hydrogen atoms in a molecule. Hydrogens in different chemical environments absorb energy differently, resulting in distinct peaks in the \(^{1} \mathrm{H}\)-NMR spectrum.

In the \(^{1} \mathrm{H}\)-NMR spectrum, each set of peaks or signals corresponds to hydrogen atoms in similar environments. The position of these signals, known as the chemical shift, is measured in parts per million (ppm) and reflects the environment's relative shielding or deshielding effect on the hydrogen nuclei.
  • The triplet at 2.50 ppm in the exercise is a result of hydrogens in a methylene group next to an unsaturated system, such as a double bond or carbonyl group.
  • Another triplet at 3.05 ppm indicates a methylene group adjacent to an electronegative atom, like oxygen.
  • The singlet at 3.58 ppm suggests hydrogens are connected to an oxygen, possibly in a methoxy group.
  • A multiplet signal between 7.1 and 7.3 ppm reveals the presence of hydrogens in an aromatic environment, typical of a benzene ring.
Each pattern within the \(^{1} \mathrm{H}\)-NMR spectrum provides clues about the number of hydrogens and their surrounding atoms, making it an essential tool for structure determination.
13C-NMR
Carbon-13 NMR spectroscopy, or \(^{13} \mathrm{C}\)-NMR, complements \(^{1} \mathrm{H}\)-NMR by revealing information about the carbon skeleton of a compound. While proton NMR focuses on hydrogen atoms, \(^{13} \mathrm{C}\)-NMR provides insights into carbon environments, helping to pinpoint carbon connections in a molecule.

The chemical shifts in \(^{13} \mathrm{C}\)-NMR are usually in a higher range than those in \(^{1} \mathrm{H}\)-NMR. Each peak corresponds to a unique carbon environment, with multiplicity details often omitted for clarity. In the exercise, ten signals were examined:
  • The peak at 210.19 ppm is unmistakably indicative of a carbonyl group (\(\text{C=O}\)).
  • The signals at around 126-136 ppm point to carbons part of an aromatic system, like a benzene ring.
  • The signal at 45.02 ppm corresponds to a methoxy carbon.
  • Signals at lower ppm values such as 38.11 and 28.34 ppm indicate methylene carbons, one next to a carbonyl, another near a more electronegative atom.
\(^{13} \mathrm{C}\)-NMR, although less sensitive than \(^{1} \mathrm{H}\)-NMR, plays a crucial role in complementing proton data to provide a more complete molecular picture.
Chemical Shifts
Chemical shifts form the backbone of interpreting NMR spectra as they convey the unique environment each nucleus experiences. The values are expressed in parts per million (ppm) and are measured relative to a reference compound, typically tetramethylsilane (TMS).

In both \(^{1} \mathrm{H}\)-NMR and \(^{13} \mathrm{C}\)-NMR, the chemical shift indicates how shielding or deshielding affects the nuclei, providing insights about their electronic environment.
  • Protons in electron-rich environments are more shielded, absorbing at lower ppm values, hence appearing upfield.
  • Protons or carbons adjacent to electronegative atoms or within unsaturated systems are deshielded, experiencing downfield shifts (higher ppm values).
  • In aromatic systems, like benzene, unique delocalized electrons cause characteristic shifts between 7-8 ppm for hydrogens and 120-150 ppm for carbons.
Understanding chemical shifts is crucial in deciphering NMR data, as it leads to identifying functional groups and piecing together a molecule's framework.
Molecular Structure Deduction
Piecing together a molecular structure from NMR data involves combining both \(^{1} \mathrm{H}\)-NMR and \(^{13} \mathrm{C}\)-NMR spectra to identify potential environments for each atom and simulate the real structure.

The deduction process involves:
  • Identifying functional groups from characteristic shifts, such as carbonyl, methoxy, or aromatic systems.
  • Analyzing the integration and pattern of \(^{1} \mathrm{H}\)-NMR signals to suggest the number of neighboring hydrogens and their arrangements.
  • Assigning carbon peaks in \(^{13} \mathrm{C}\)-NMR to specific environments, confirming the presence of functional groups and the backbone structure.
In the given problem, the combination of aromatic and carbonyl chemical shifts, along with specific triplet and singlet patterns, directed us to propose ethyl benzoate as the likely structure. This methodology is an essential part of spectroscopic analysis in organic chemistry, enabling scientists to unravel even complex molecular structures with confidence.

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Most popular questions from this chapter

Given here are \({ }^{1}\) H-NMR and \({ }^{13}\) C-NMR spectral data for two compounds. Each shows strong, sharp absorption between 1700 and \(1720 \mathrm{~cm}^{-1}\), and strong, broad absorption over the region \(2500-3000 \mathrm{~cm}^{-1}\). Propose a structural formula for each compound. (a) \(\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_{3}\) (b) \(\mathrm{C}_{10} \mathrm{H}_{10} \mathrm{O}_{2}\) $$ \begin{array}{cc} \hline{ }^{1} \mathbf{H}-\mathrm{NMR} & { }^{13} \mathrm{C}-\mathrm{NMR} \\ \hline 2.49(\mathrm{t}, 2 \mathrm{H}) & 173.89 \\ 2.80(\mathrm{t}, 2 \mathrm{H}) & 157.57 \\ 3.72(\mathrm{~s}, 3 \mathrm{H}) & 132.62 \\ 6.78(\mathrm{~d}, 2 \mathrm{H}) & 128.99 \\ 7.11(\mathrm{~d}, 2 \mathrm{H}) & 113.55 \\ 12.4(\mathrm{~s}, 1 \mathrm{H}) & 54.84 \\ & 35.75 \\ & 29.20 \\ \hline \end{array} $$ $$ \begin{array}{cc} \hline{ }^{1} \text { H-NMR } & { }^{13} \text { C-NMR } \\ \hline 2.34(\mathrm{~s}, 3 \mathrm{H}) & 167.82 \\ 6.38(\mathrm{~d}, 1 \mathrm{H}) & 143.82 \\ 7.18(\mathrm{~d}, 1 \mathrm{H}) & 139.96 \\ 7.44(\mathrm{~d}, 2 \mathrm{H}) & 131.45 \\ 7.56(\mathrm{~d}, 2 \mathrm{H}) & 129.37 \\ 12.0(\mathrm{~s}, 1 \mathrm{H}) & 127.83 \\ & 111.89 \\ & 21.13 \\ \hline \end{array} $$

Account for the fact that water-insoluble carboxylic acids ( \(\mathrm{p} K_{\mathrm{a}} 4-5\) ) dissolve in \(10 \%\) aqueous sodium bicarbonate \((\mathrm{pH} 8.5)\) with the evolution of a gas but that waterinsoluble phenols ( \(\mathrm{p} K_{\mathrm{a}} 9.5-10.5\) ) do not dissolve in \(10 \%\) sodium bicarbonate.

Molecules of \(6,6^{\prime}\)-dinitrobiphenyl-2,2'-dicarboxylic acid have no tetrahedral chiral center, and yet they can be resolved to a pair of enantiomers. Account for this chirality.

Following is a synthesis for toremifene, a nonsteroidal estrogen antagonist whose structure is closely related to that of tamoxifen. (a) This synthesis makes use of two blocking groups, the benzyl (Bn) group and the tetrahydropyranyl (THP) group. Draw a structural formula of each group, and describe the experimental conditions under which it is attached and removed. (b) Discuss the chemical logic behind the use of each blocking group in this synthesis. (c) Propose a mechanism for the conversion of D to \(E\). (d) Propose a mechanism for the conversion of \(F\) to toremifene. (e) Is toremifene chiral? If so, which of the possible stereoisomers are formed in this synthesis?

In certain clinical situations, there is need for an injectable \(\beta\)-blocker with a short biological half-life. The clue to development of such a drug was taken from the structure of atenolol, whose corresponding carboxylic acid (the product of hydrolysis of its amide) has no \(\beta\)-blocking activity. Substitution of an ester for the amide group and lengthening the carbon side chain by one methylene group resulted in esmolol. Its ester group is hydrolyzed quite rapidly to a carboxyl group by serum esterases under physiological conditions. This hydrolysis product has no \(\beta\)-blocking activity. Propose a synthesis for esmolol from 4-hydroxycinnamic acid, epichlorohydrin, and isopropylamine. (a) Propose a synthesis for esmolol from 4-hydroxycinnamic acid, epichlorohydrin, and isopropylamine. (b) Is esmolol chiral? If so, which of the possible stereoisomers are formed in this synthesis?

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