Answered: C11H11BrO2 No peaks exchange with D20

Science

Chemistry

C11H11BrO2 No peaks exchange with D20

Macroscale and Microscale Organic Experiments

7th Edition

ISBN:9781305577190

Author:Kenneth L. Williamson, Katherine M. Masters

Publisher:Kenneth L. Williamson, Katherine M. Masters

Chapter45: Synthesis And Bioassay Of Sulfanilamide And Derivatives

Section: Chapter Questions

Problem 5Q

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Concept explainers

Analyzing and Predicting Spectroscopic Data

The interaction of electromagnetic radiation with the structure of molecules in the sample and their study is referred spectroscopy. Several spectroscopic methods are used to acquire sample information, and these methods are proved to be accurate, fast, and non-destructive. The information obtained about the content and composition of molecules from these spectroscopic methods is called spectroscopic data.

Question

Determine the structure of the molecule using the chemical formula and NMR spectra. Can someone help me walk through this problem? Please show the structure and how you got it.

C11H11BrO2
No peaks exchange with D20
180
200
13C NMR:
160
140
120
100
80
60
40
20
1H NMR:
8 / || 7
10
2
1
integration: 1 22
1
2
3
(ppm)

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1Compound 1 has molecular formula C7H16. It shows three signals in the 1H-NMR spectrum, one at 0.85 ppm, one at 1.02 ppm, and one at 1.62 ppm. The relative integrals of these three signals are 6, 1, and 1, respectively. Compound 2 has molecular formula C7H14. It shows three signals in the 1H-NMR spectrum, one at 0.98 ppm, one at 1.36 ppm, and one at 1.55 ppm. The relative integrals of these three signals are 3, 2, and 2, respectively. Propose structures for compounds 1 and 2, explaining how you reach your conclusion.
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Compound D has the molecular formula C5H8O. Assign all peaks in 1H and 13C NMR spectrum of compound D.
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The question is attached..;
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There are four esters with molecular formula C4H8O2. How can they be distinguished by 1H NMR?
A compound has the molecular formula: C4H8O2 and gives the following C-13 NMR spectrum. Provide the most likely functional groups for each signal. nmrsim presentation 1 1 C:Bruken Topspin3.5pl7 examdata -170.7658 150 100 d 170.8 (O CH3) ; 60.4 (OCH2); 20.8, 14.1 (2 x C=0) 8170.8 (0 CH₂); 60.4 (OCH₂); 20.8, 14.1 (2 x CH3). 8 170.8 (CH3); 60.4 (C-0); 20.8, 14.1 (2 x OCH₂) 8170.8 (C=0); 60.4 (OCH₂); 20.8, 14.1 (2 x CH3). 8170.8 (C) ; 60.4 (O CH2); 20.8, 14.1 (2 x O CH3) -60.4293 50 -20.7902 -14.1385 [ppm] [+]