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Ch. 13 - Nuclear Magnetic Resonance Spectroscopy
Wade - Organic Chemistry 9th Edition
Wade9th EditionOrganic ChemistryISBN: 9780135213728Not the one you use?Change textbook
All textbooksWade 9th EditionCh. 13 - Nuclear Magnetic Resonance SpectroscopyProblem 8b
Chapter 13, Problem 8b

Draw the NMR spectra you expect for the following compounds.
(b)

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1
Step 1: Analyze the molecular structure of the compound. The given compound is CH3O-C=C-CH3-Cl. It contains a methoxy group (-OCH3), a methyl group (-CH3), a chlorine atom (-Cl), and a hydrogen atom attached to a double bond.
Step 2: Identify the distinct proton environments. The compound has four unique proton environments: (1) the protons in the methoxy group (-OCH3), (2) the protons in the methyl group (-CH3), (3) the proton attached to the double bond (C=CH), and (4) the proton on the carbon adjacent to the chlorine atom (C=CH-Cl).
Step 3: Predict the chemical shifts for each proton environment. (1) The methoxy protons (-OCH3) will appear in the range of 3.3-4.0 ppm due to the electronegative oxygen atom. (2) The methyl protons (-CH3) will appear in the range of 1.0-2.0 ppm. (3) The proton attached to the double bond (C=CH) will appear in the range of 5.0-6.5 ppm due to the deshielding effect of the double bond. (4) The proton adjacent to the chlorine atom (C=CH-Cl) will appear in the range of 4.5-5.5 ppm due to the electronegative chlorine atom.
Step 4: Determine the splitting patterns for each proton environment. (1) The methoxy protons (-OCH3) will appear as a singlet because there are no neighboring protons. (2) The methyl protons (-CH3) will appear as a doublet due to coupling with the adjacent proton on the double bond. (3) The proton attached to the double bond (C=CH) will appear as a quartet due to coupling with the three protons of the methyl group. (4) The proton adjacent to the chlorine atom (C=CH-Cl) will appear as a doublet due to coupling with the proton on the double bond.
Step 5: Summarize the expected NMR spectrum. The spectrum will include: (1) a singlet for the methoxy protons (-OCH3), (2) a doublet for the methyl protons (-CH3), (3) a quartet for the proton attached to the double bond (C=CH), and (4) a doublet for the proton adjacent to the chlorine atom (C=CH-Cl). The chemical shifts and splitting patterns will align with the predictions made in the previous steps.

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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, their environment, and connectivity in a molecule. The resulting spectra display peaks that correspond to different chemical environments, allowing chemists to deduce structural information.
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Chemical Shifts

Chemical shifts in NMR spectroscopy refer to the position of the peaks in the spectrum, which are influenced by the electronic environment surrounding the nuclei. Different functional groups and molecular structures cause variations in the magnetic field experienced by the nuclei, leading to shifts in resonance frequency. Understanding chemical shifts is crucial for interpreting NMR spectra and identifying the types of hydrogen or carbon present in a compound.
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Integration and Multiplicity

Integration in NMR refers to the area under the peaks, which correlates to the number of protons contributing to that signal. Multiplicity indicates the splitting of NMR signals due to neighboring hydrogen atoms, following the n+1 rule, where n is the number of adjacent protons. Together, integration and multiplicity provide insights into the number of hydrogen atoms and their arrangement, aiding in the elucidation of the compound's structure.
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Related Practice
Textbook Question

The NMR spectrum of toluene (methylbenzene) was shown in Figure 13-11.

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(a) How many different kinds of protons are there in toluene?

(b) Explain why the aromatic region around d7.2 is broad, with more than one sharp absorption.

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Textbook Question

Determine the number of different kinds of protons in each compound.

(a) 1-chloropropane

(b) 2-chloropropane

(c) 2,2-dimethylbutane

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Textbook Question

Propose a structure that corresponds to each spectrum.

(a) <IMAGE>

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Textbook Question

Draw a splitting tree, similar to Figures 13-32 and 13-33, for proton Hc in styrene. What is the chemical shift of proton Hc?

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Textbook Question

An unknown compound (C3H2NCl) shows moderately strong IR absorptions around 1650 cm–1 and 2200 cm–1. Its NMR spectrum consists of two doublets (J = 14 Hz) at δ5.9 and δ7.1. Propose a structure consistent with these data.

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Textbook Question

Draw the NMR spectra you expect for the following compounds.

(a)

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