What is the IR and HNMRAndCNMR and mass in this example IR Spectrum(liquid film)4000300059100806040204013C NMR Spectrum(100 MHz CDCI, solution)% of base peak8780DEPT CH CH. CHproton decoupled200¹H NMR Spectrum(400 MHz, CDCI, solution)109817391600v (cm¹)2000115M 146-1%120 160m/emm1200800Mass SpectrumC6H10042801607200 24012065solvent804Problem 44No significant UVabsorption above 220 nm0403218 (ppm)TMS08 (ppm)13

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IR Spectrum (liquid film) 4000 100 80 % of base peak 3000 60 40 20 40 13C NMR Spectrum (100 MHz CDCI, solution) 59 80 DEPT CH CH. CH proton decoupled www. 200 ¹H NMR Spectrum (400 MHz, CDCI, solution) 1739 2000 v (cm') 115 120 1600 M 145-1% 160 m/e 160 MY 800 Mass Spectrum C6H1004 280 solvent 80 1200 200 240 120 Problem 44 No significant UV absorption above 220 nm 0 40 8 (ppm)

IR Spectrum (liquid film) 4000 100 80 % of base peak 3000 60 40 20 40 13C NMR Spectrum (100 MHz CDCI, solution) 59 80 DEPT CH CH. CH proton decoupled www. 200 ¹H NMR Spectrum (400 MHz, CDCI, solution) 1739 2000 v (cm') 115 120 1600 M 145-1% 160 m/e 160 MY 800 Mass Spectrum C6H1004 280 solvent 80 1200 200 240 120 Problem 44 No significant UV absorption above 220 nm 0 40 8 (ppm)

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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I have an NMR spectrum. There are 5 peaks. There is a doublet at 9.6 ppm (integrates to 2H) where aldehydes usually arise, don't know why there is a doublet. There is another doublet at 6.8 ppm, again integrating to 2H. There is a singlet at 4.5 ppm, integrating to 1H. There is a quartet at 3.4 ppm integrating to 2H, and lastly there is triplet at 1.2 ppm integrating to 3H. i recofnize the ethyl group but don't understand the peak at 9.6 unless it is not an aldehyde.
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From the proton NMR spectrum in Figure 19-44, deduce the structure of this hydrocarbon
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Answer Q40, 41, 42 showing detailly all explanations