11. Provide a structure for Compound A, (C&H100), based on its IR and NMR spectra. Showfull analysis of both the IR and NMR spectra.882.12.343.77.2B40204000NMR:environ.3304Ratio1H,2H,2H,5 H,M3015 2981#H spl pat.broad singlettriplettripletmultiplet2000#nei.16101900Pieces

The molecular formula of the organic compound given is C8H10O. The degree of unsaturation =…

11. Provide a structure for Compound A, (CsH₁0O), based on its IR and NMR spectra. Show full analysis of both the IR and NMR spectra. 8 82.1 2.34 3.7 7.2 NO MI 4000 NMR: 3304 environ. Ratio 1H, 2H, 2H, 5 H, 3015 2981 #H spl pat. broad singlet triplet triplet multiplet. #nei. 1610 Pieces

11. Provide a structure for Compound A, (CsH₁0O), based on its IR and NMR spectra. Show full analysis of both the IR and NMR spectra. 8 82.1 2.34 3.7 7.2 NO MI 4000 NMR: 3304 environ. Ratio 1H, 2H, 2H, 5 H, 3015 2981 #H spl pat. broad singlet triplet triplet multiplet. #nei. 1610 Pieces

Macroscale and Microscale Organic Experiments

7th Edition

ISBN:9781305577190

Author:Kenneth L. Williamson, Katherine M. Masters

Publisher:Kenneth L. Williamson, Katherine M. Masters

Chapter13: Mass Spectrometry

Section: Chapter Questions

Problem 3Q

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Concept explainers

NMR Spectroscopy

NMR stands for Nuclear Magnetic Resonance in which the study of different molecules is done based on the interaction of radiofrequency electromagnetic radiations with the nuclei of molecules placed in a strong magnetic field. A precise study of spectra of visible light and all bands of electromagnetic radiation is called spectroscopy.

Question

11. Provide a structure for Compound A, (C&H100), based on its IR and NMR spectra. Show
full analysis of both the IR and NMR spectra.
8
82.1
2.34
3.7
7.2
B
40
20
4000
NMR:
environ.
3304
Ratio
1H,
2H,
2H,
5 H,
M
3015 2981
#H spl pat.
broad singlet
triplet
triplet
multiplet
2000
#nei.
1610
1900
Pieces

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