Textbook Question
Show how you would accomplish the following multistep syntheses. You may use any additional reagents and solvents you need.
(d)
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Ch. 20 - Carboxylic Acids
Wade9th EditionOrganic ChemistryISBN: 9780135213728
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Table of contents
- Ch.1 - Structure and Bonding107
- Ch. 2 - Acids and Bases; Functional Groups115
- Ch.3 - Structure and Stereochemistry of Alkanes100
- Ch.4 - The Study of Chemical Reactions99
- Ch.5 - Stereochemistry93
- Ch.6 - Alkyl Halides; Nucleophilic Substitution118
- Ch. 7 - Structure and Synthesis of Alkenes; Elimination158
- Ch.8 - Reactions of Alkenes171
- Ch.9 - Alkynes91
- Ch.10 - Structure and Synthesis of Alcohols143
- Ch.11 - Reactions of Alcohols104
- Ch. 12 - Infrared Spectroscopy and Mass Spectrometry22
- Ch. 13 - Nuclear Magnetic Resonance Spectroscopy45
- Ch. 14 - Ethers, Epoxides, and Thioethers72
- Ch. 15 - Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy89
- Ch. 16 - Aromatic Compounds74
- Ch. 17 - Reactions of Aromatic Compounds126
- Ch. 18 - Ketones and Aldehydes193
- Ch. 19 - Amines129
- Ch. 20 - Carboxylic Acids77
- Ch. 21 - Carboxylic Acid Derivatives141
- Ch. 22 - Condensations and Alpha Substitutions of Carbonyl Compounds175
- Ch. 23 - Carbohydrates and Nucleic Acids93
- Ch. 24 - Amino Acids, Peptides, and Proteins54
Chapter 20, Problem 40a
The following NMR spectra correspond to compound of formula (A) C9H10O2. Propose structure, and show how it is consistent with the observed absorptions.
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Verified step by step guidance1
Step 1: Analyze the molecular formula C9H10O2. This indicates the presence of 9 carbons, 10 hydrogens, and 2 oxygens. The degree of unsaturation can be calculated using the formula: (2C + 2 - H)/2. For C9H10O2, the degree of unsaturation is (2(9) + 2 - 10)/2 = 5. This suggests the presence of rings and/or double bonds.
Step 2: Examine the NMR spectrum. The peak at approximately 12 ppm corresponds to a highly deshielded proton, likely from a carboxylic acid (-COOH) group. This is consistent with the presence of one hydrogen in this region.
Step 3: The multiplet at approximately 7-8 ppm corresponds to aromatic protons. The integration of 5H suggests a monosubstituted benzene ring, as the benzene ring typically has 5 protons when one substituent is attached.
Step 4: The singlet at approximately 4 ppm corresponds to 2 hydrogens. This is likely from a methylene group (-CH2-) attached to an electronegative group, such as oxygen. This is consistent with the presence of an ester or alcohol functional group.
Step 5: Combine the observations. The molecular formula, degree of unsaturation, and NMR data suggest the structure is a benzene ring with a carboxylic acid group (-COOH) and a methoxy group (-OCH3) attached. Verify this structure by ensuring the integration and chemical shifts match the observed spectrum.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Nuclear Magnetic Resonance (NMR) Spectroscopy
NMR spectroscopy is a powerful analytical technique used to determine the structure of organic compounds. It relies on the magnetic properties of certain nuclei, primarily hydrogen (1H) and carbon (13C), to provide information about the number of hydrogen atoms in different environments within a molecule. The chemical shifts observed in the spectrum indicate the electronic environment surrounding the nuclei, allowing chemists to infer structural details.
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General NMR Features
Integration of NMR Peaks
In NMR spectroscopy, the area under each peak corresponds to the number of hydrogen atoms contributing to that signal, known as integration. For example, a peak labeled '5 H' indicates that five hydrogen atoms are in a similar environment, while '2 H' and '1 H' indicate two and one hydrogen atoms, respectively. This information is crucial for deducing the molecular structure and understanding how different hydrogen environments relate to the overall compound.
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Chemical Shifts and Splitting Patterns
Chemical shifts in an NMR spectrum are measured in parts per million (PPM) and provide insight into the electronic environment of hydrogen atoms. Peaks appearing at lower PPM values typically indicate protons attached to carbons that are more electron-rich, while higher PPM values suggest protons on less electron-rich carbons. Additionally, splitting patterns (multiplicity) reveal how many neighboring hydrogen atoms are present, which helps in determining the connectivity and arrangement of atoms in the molecule.
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Related Practice
Textbook Question
Peroxyacetic acid (bp = 105 °C) has a lower boiling point than acetic acid (bp = 118 °C), even though peroxyacetic acid has a higher molecular weight. Explain why peroxyacetic acid is more volatile than acetic acid.
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Textbook Question
In the presence of a trace of acid, δ-hydroxyvaleric acid forms a cyclic ester (lactone).
(a) Give the structure of the lactone, called δ-valerolactone.
(b) Propose a mechanism for the formation of δ-valerolactone.
2
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Textbook Question
Show how you would accomplish the following multistep syntheses. You may use any additional reagents and solvents you need.
(e)
1
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Textbook Question
The following NMR spectra correspond to compound of formula (C) C6H10O2. Propose structure, and show how it is consistent with the observed absorptions.
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Textbook Question
Show how you would accomplish the following multistep syntheses. You may use any additional reagents and solvents you need.
(f)
3
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