Problem 11117IR Spectrumquid e4000S730300020001001600v (cm')1200B00B0Mass Spectrum60No significant Uvabsorption above 220 nmCHO4080120160200240280m/e13C NMR Spectrum(100 O MHZ. CDCI, solution)DEPT CH CHt CHtICHproton decoupledsolvent408 (ppm)20016012080'H NMR Spectrumexpansion(400 MHz, CDCI, solution)triplerTMS221.8 PPM4328 (ppm)7.6.9.8100iof base peak10

Based on the given spectrum data, the structure of C5H8O will be deduced as shown in the following…

Problem 11117 IR Spectrum uit em 4000 1738 3000 2000 1600 v (em') 100 1200 B00 COO 80 Mass Spectrum M 84 No significant uv absorption above 220 nm Cuclo Penanon CHO 40 80 120 160 200 240 280 m/e 13C NMR Spectrum (100 O MHZ. CDC, sokution) DEPT CH CH,t CHt ICH CH proton decoupled solvent 40 8 (ppm) 200 160 120 80 'H NMR Spectrum (400 MHz, CDCI, solution) expansion tripler TMS 2.2 1.8 PPM 3 2 4 8 (ppm 6. 9. 8. 10

Problem 11117 IR Spectrum uit em 4000 1738 3000 2000 1600 v (em') 100 1200 B00 COO 80 Mass Spectrum M 84 No significant uv absorption above 220 nm Cuclo Penanon CHO 40 80 120 160 200 240 280 m/e 13C NMR Spectrum (100 O MHZ. CDC, sokution) DEPT CH CH,t CHt ICH CH proton decoupled solvent 40 8 (ppm) 200 160 120 80 'H NMR Spectrum (400 MHz, CDCI, solution) expansion tripler TMS 2.2 1.8 PPM 3 2 4 8 (ppm 6. 9. 8. 10

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question 11. Answer the following questions towards the identification of an unknown compound given the MS, IR, ¹H and 13C NMR spectra. MS IR LOD 100 80 Relative Intensity IKANSOLITAMLETZT 60 10 20 20 0 10 -20 30 40 -8 50 M+* = 116 = 60 70 80 ge 90 100 110 120 m/z D 4000 3000 2000 1500 1000 500 HAVENUMBERI -11
How do I find %T for the following? Thank you
Problem 8 114 IR Spectrum (gud fm 4000 3000 2000 1000 1200 v (cm') 800 100 More branch Coo 80 Mass Spectrum 60 40 No significant UN M 114 (1N)) absorption above 220 nm 20 99 C&H18 40 80 120 160 m/e 200 240 280 16+ 18-(6) 13C NMR Spectrum (50 0 M. CDC, solution) DEPT CH CH, CHf KAly Asymi proton decoupled 1500 solvent CAl3 Sym s(ppm) 390 200 160 120 80 40 1H NMR Spectrum (200 MHz. CDCI3 solution) TMS 10 9. 8 6. 4 1. 0. 8 (ppm) wd eps
Problem 7 2249 IR Spectrum (KBr disc) 4000 3000 2000 1600 1200 800 v (cm') 100 Mass Spectrum 80 53 60 No significant UV 40 M+' = 80 absorption above 220 nm 40 20 C4HẠN2 160 m/e 40 80 120 200 240 280 T 13C NMR Spectrum (100.0 MHz, CDCI, solution) DEPT CHat CH, сн+ proton decoupled solvent 200 160 120 80 40 8 (ppm) H NMR Spectrum (400 MHz, CDCI, solution) TMS 10 1 8 (ppm) 9 7 6. 4 % of base peak 5
Draw the structure of molecule
Identify each peak (peak splitting and ppm) and identify the unknown compound. Please draw the structure, and explain why its that structure. Attached are the spectra for the unknown compound.
A student performed a series of reactions and obtained the spectral data provided. The reaction sequence is shown below.. What are the structures for A, B, C, and D? Compound B NMR (relative integrated areas below peaks) CH;CH,O°N 1. BHTHF N-bromosuccnimide A CH;CH,OH B 2. H,O,/OH- D hv reflux Compound A NMR (relative integrated areas below peaks) PPM 5 Compound B IR INFRARED SPECTRUM PPM 5 1 6 0.8 Compound A IR INFRARED SPECTRUM 0.6 0.8 0.4 0.4 0.2- 3000 2000 Wavenumber (cm-1) 1000 The MS of compound B shows an M+ peak at 198 and an M+2 peak at 200 with a relative abundance of 1:0.97 2000 Wavenumber (cm-1) 3000 1000 Transmitance Relative Transmittance
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опанит 1715 1600 IR Spectrum 100 4000 (liquid film) 43 80 75 60 M 40 20 % of base peak 10 3000 40 C 13C NMR Spectrum (20.0 MHz, CDCI solution) proton decoupled 9 80 200 ¹H NMR Spectrum (100 MHz, CDCI, solution) 101 8 2000 v (cm¹) M** 132 120 160 m/e 160 7 6 1200 120 Mass Spectrum 200 240 280 800 C-H 5 C6H1203 solvent L 80 4 -CH3 No significant UV absorption above 220 nm 3 H-C-H u -CH3 40 2 59 TMS 0 1 8 (ppm) TMS 0 8 (ppm)
Problem 8 114 IR Spectrum (quid fim) 4000 3000 2000 1600 1200 V (cm') 800 100 80 Mass Spectrum 57 More branch 60 E Co M 114 ( 1%) No significant UV 40 absorption above 220 nm 20 99 CBH18 2メ8k8-み8) 16 18-(6) 40 80 120 160 200 240 280 m/e 13C NMR Spectrum (50 O MHz. CDCI, solution) DEPT CH, CH,t CHt KAS ASym. 1500 solvent proton decoupled CAl3 Sym o 8(ppm) 340 200 160 120 80 40 1H NMR Spectrum (200 MHz. CDCIg solution) TMS 2 8. 7 6. 5 4 8 (ppm) 10 Sof base peak
is the product supposed to be 2- methyl-3-pentanol?