Textbook Question
The 1H NMR spectra of three isomers with molecular formula C4H9Br are shown here. Which isomer produces which spectrum?
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Ch. 14 - NMR Spectroscopy
Bruice8th EditionOrganic ChemistryISBN: 9780135213711
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Table of contents
- Ch. 1 - Remembering General Chemistry: Electronic Structure and Bonding (Part 1)31
- Ch. 1 - Remembering General Chemistry: Electronic Structure and Bonding (Part 2)65
- Ch. 2 - Acids and Bases: Central to Understanding Organic Chemistry84
- Ch. 3 - An Introduction to Organic Compounds:Nomenclature, Physical Properties, and Structure183
- Ch. 4 - Isomers: The Arrangement of Atoms in Space121
- Ch. 5 - Alkenes: Structure, Nomenclature, and an Introduction to Reactivity • Thermodynamics and Kinetics83
- Ch. 6 - The Reactions of Alkenes • The Stereochemistry of Addition Reactions175
- Ch. 7 - The Reactions of Alkynes • An Introduction to Multistep Synthesis119
- Ch. 8 - Delocalized Electrons: Their Effect on Stability, pKa, and the Products of a Reaction • Aromaticity and Electronic Effects: An Introduction to the Reactions of Benzene148
- Ch. 9 - Substitution and Elimination Reactions of Alkyl Halides212
- Ch. 10 - Reactions of Alcohols, Ethers, Epoxides, Amines, and Sulfur-Containing Compounds131
- Ch. 11 - Organometallic Compounds60
- Ch. 12 - Radicals90
- Ch. 13 - Mass Spectrometry; Infrared Spectroscopy; UV/Vis Spectroscopy62
- Ch. 14 - NMR Spectroscopy95
- Ch. 15 - Reactions of Carboxylic Acids and Carboxylic Acid Derivatives123
- Ch. 16 - Reactions of Aldehydes and Ketones • More Reactions of Carboxylic Acid Derivatives141
- Ch. 17 - Reactions at the Alpha-Carbon101
- Ch. 18 - Reactions of Benzene and Substituted Benzenes144
- Ch. 19 - More About Amines • Reactions of Heterocyclic Compounds49
- Ch. 20 - The Organic Chemistry of Carbohydrates80
- Ch. 21 - Amino Acids, Peptides, and Proteins57
- Ch. 22 - Catalysis in Organic Reactions and in Enzymatic Reactions35
- Ch. 23 - The Organic Chemistry of the Coenzymes—Compounds Derived from Vitamins1
- Ch. 28 - Pericyclic Reactions58
Chapter 15, Problem 56
Identify the compound with molecular formula C7H14O that gives the following proton-coupled 13C NMR spectrum:
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Verified step by step guidance1
Analyze the molecular formula C7H14O to determine the degree of unsaturation. Use the formula: Degree of Unsaturation = (2C + 2 - H + N - X)/2, where C = number of carbons, H = number of hydrogens, N = number of nitrogens, and X = number of halogens. For C7H14O, calculate the degree of unsaturation to identify the presence of rings or double bonds.
Examine the proton-coupled 13C NMR spectrum. Look for the number of distinct carbon signals, which corresponds to the number of unique carbon environments in the molecule. This will help determine the symmetry and structure of the compound.
Identify the splitting patterns in the proton-coupled 13C NMR spectrum. The splitting of each carbon signal provides information about the number of hydrogens directly attached to that carbon. For example, a triplet indicates two hydrogens attached, a doublet indicates one hydrogen, and a singlet indicates no hydrogens.
Consider the functional group indicated by the molecular formula and the NMR data. The presence of oxygen (O) suggests a functional group such as an alcohol, ether, or carbonyl. Use the chemical shifts in the 13C NMR spectrum to identify the specific functional group (e.g., carbonyl carbons typically appear in the range of 190-220 ppm).
Propose a structure for the compound that matches the molecular formula, degree of unsaturation, and the NMR data. Verify that the structure accounts for all observed signals and splitting patterns in the spectrum, ensuring consistency with the molecular formula C7H14O.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Molecular Formula Interpretation
The molecular formula C7H14O indicates a compound containing seven carbon atoms, fourteen hydrogen atoms, and one oxygen atom. Understanding the molecular formula is crucial as it provides insights into the compound's structure, functional groups, and potential isomers. The ratio of hydrogen to carbon suggests that the compound may be an alkane, alkene, or alcohol, influencing its reactivity and properties.
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Proton-Coupled 13C NMR Spectroscopy
Proton-coupled 13C NMR spectroscopy is a technique used to determine the carbon environment in organic compounds. In this method, the interaction between protons and carbon nuclei provides information about the number of unique carbon environments and their connectivity. Analyzing the spectrum helps identify functional groups and structural features, which are essential for deducing the compound's identity based on its molecular formula.
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Isomerism in Organic Compounds
Isomerism refers to the existence of compounds with the same molecular formula but different structural or spatial arrangements. For C7H14O, various isomers can exist, including alcohols, ethers, and ketones. Recognizing the potential isomers is vital for accurately interpreting NMR data and determining the correct compound, as different isomers can exhibit distinct chemical behaviors and spectral characteristics.
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Related Practice
Textbook Question
An alkyl halide reacts with an alkoxide ion to form a compound whose 1H NMR spectrum is shown here. Identify the alkyl halide and the alkoxide ion. (Hint: See Section 9.15.)
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Textbook Question
Compound A, with molecular formula C4H9Cl, shows two signals in its 13C NMR spectrum. Compound B, an isomer of compound A, shows four signals, and in the proton-coupled mode, the signal farthest downfield is a doublet. Identify compounds A and B.
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Textbook Question
Identify each of the following compounds from the 1H NMR data and molecular formula:
c. C5H10O2: a 3H triplet at 1.15 ppm
a 3H triplet at 1.25 ppm
a 2H quartet at 2.33 ppm
a 2H quartet at 4.13 ppm
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Textbook Question
Would it be better to use 1H NMR or 13C NMR spectroscopy to distinguish 1-butene, cis-2-butene, and 2-methylpropene? Explain your answer.
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Textbook Question
Identify each of the following compounds from the 1H NMR data and molecular formula:
b. C8H9Br: a 3H doublet at 2.01 ppm a 1H quartet at 5.14 ppm
a 5H broad singlet at 7.35 ppm
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