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) \(\mathrm{C}_{10} \mathrm{H}_{10} \mathrm{O}_{2}\) $$ \begin{array}{cc} \hline{ }^{1} \mathbf{H}-\mathrm{NMR} & { }^{13} \mathrm{C}-\mathrm{NMR} \\ \hline 2.49(\mathrm{t}, 2 \mathrm{H}) & 173.89 \\ 2.80(\mathrm{t}, 2 \mathrm{H}) & 157.57 \\ 3.72(\mathrm{~s}, 3 \mathrm{H}) & 132.62 \\ 6.78(\mathrm{~d}, 2 \mathrm{H}) & 128.99 \\ 7.11(\mathrm{~d}, 2 \mathrm{H}) & 113.55 \\ 12.4(\mathrm{~s}, 1 \mathrm{H}) & 54.84 \\ & 35.75 \\ & 29.20 \\ \hline \end{array} $$ $$ \begin{array}{cc} \hline{ }^{1} \text { H-NMR } & { }^{13} \text { C-NMR } \\ \hline 2.34(\mathrm{~s}, 3 \mathrm{H}) & 167.82 \\ 6.38(\mathrm{~d}, 1 \mathrm{H}) & 143.82 \\ 7.18(\mathrm{~d}, 1 \mathrm{H}) & 139.96 \\ 7.44(\mathrm{~d}, 2 \mathrm{H}) & 131.45 \\ 7.56(\mathrm{~d}, 2 \mathrm{H}) & 129.37 \\ 12.0(\mathrm{~s}, 1 \mathrm{H}) & 127.83 \\ & 111.89 \\ & 21.13 \\ \hline \end{array} $$

Step by step solution

01

Underline important points from the given data

First, let's underline the important aspects of the exercise. The given data indicates strong and sharp absorption between 1700 and \(1720 \mathrm{~cm}^{-1}\) and broad absorption over the region \(2500-3000 \mathrm{~cm}^{-1}\). This suggests the presence of carbonyl groups (C=O) and O-H or N-H bonds in both compounds.

02

Analyze the NMR data for the first compound: \({\mathrm{C}}_{10}{\mathrm{H}}_{12}{\mathrm{O}}_{3}\)

In the given \({ }^{1}\mathrm{H}\) NMR spectrum, we can see that it has 5 peaks. The peaks at 2.49 ppm (triplet, 2H) and 2.80 ppm (triplet, 2H) indicate two CH2 groups that are adjacent to each other. The peak at 3.72 ppm (singlet, 3H) suggests the presence of a methoxy (-OCH3) group. The two peaks at 6.78 ppm (doublet, 2H) and 7.11 ppm (doublet, 2H) indicate two pairs of protons in an aromatic system (each pair 1H) that are not adjacent to each other. The peak at 12.4 ppm (singlet, 1H) indicates an acidic hydrogen, most likely on a carboxylic acid group (COOH). Now, the \({ }^{13}\mathrm{C}\) NMR spectrum shows 7 peaks. The values of carbon shifts indicate the presence of carbonyl (C=O) carbon atoms, two types of aromatic carbons, and one methoxy (OCH3) group. Taking all this information into account, we can now propose a structure for this compound.

03

Propose a structural formula for the first compound#\({\mathrm{C}}_{10}{\mathrm{H}}_{12}{\mathrm{O}}_{3}\)

Based on the analysis we have done in Step 2, we can propose the following structure for the first compound: $$ \begin{array}{c} \includegraphics{images/HNMR_exercise_solution_1.png} \end{array} $$ The structure of the first compound is composed of an aromatic ring with a carboxylic acid group (COOH) and a methoxy group (OCH3) at para positions. Moreover, this compound has an ethylene group (-CH2-CH2-) attached to the carboxylic acid group.

04

Analyze the NMR data for the second compound: \({\mathrm{C}}_{10}{\mathrm{H}}_{10}{\mathrm{O}}_{2}\)

Now, let's analyze the given \({ }^{1}\mathrm{H}\) NMR spectrum for the second compound. It also has 5 peaks. The peak at 2.34 ppm (singlet, 3H) suggests the presence of a methyl group (CH3) in the compound. The peaks at 6.38 ppm (doublet, 1H) and 7.18 ppm (doublet, 1H) indicate one set of aromatic protons (each 1H) that are not adjacent to each other. The remaining two peaks at 7.44 ppm (doublet, 2H) and 7.56 ppm (doublet, 2H) also indicate two pairs of protons in an aromatic system (each pair 1H) that are not adjacent to each other. The peak at 12.0 ppm (singlet, 1H) once again suggests the presence of an acidic hydrogen, most likely on a carboxylic acid group (COOH). Now, the \({ }^{13}\mathrm{C}\) NMR spectrum shows 7 peaks. The values of carbon shifts indicate the same type of groups as in the first compound - carbonyl (C=O) group, aromatic carbons, and an aliphatic group. Taking all this information into account, we can now propose a structure for the second compound.

05

Propose a structural formula for the second compound#\({\mathrm{C}}_{10}{\mathrm{H}}_{10}{\mathrm{O}}_{2}\)

Based on our analysis in Step 4, we can propose the following structure for the second compound: $$ \begin{array}{c} \includegraphics{images/HNMR_exercise_solution_2.png} \end{array} $$ The structure of the second compound also involves an aromatic ring with a carboxylic acid group (COOH) at one position and a methyl group (CH3) at the ortho position.

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