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IR Spectrum 4000 hahahahahad D-W) 3000 100 BO 60 40 20 40 13C NMR Spectrum (100 0 MHz CDC), solution) 80 103 DEPT CH CH. CH proton decoupled 200 ¹H NMR Spectrum (200 MHz, CDCI, solution) 10 9 8 2000 v (cm) Mote blanch M-118 (15) 120 160 m/e 160 7 1200 120 200 240 6 800 Mass Spectrum + C6H140₂ 280 5 solvent 80 Problem 9 20+1=3 2x -(12+27-12 + 2/1/2 =14-14 4 =2016 часо 40 Exchanges with D₂0 s 3 2 O 1 8 (ppm) TMS 0 8 (ppm

IR Spectrum 4000 hahahahahad D-W) 3000 100 BO 60 40 20 40 13C NMR Spectrum (100 0 MHz CDC), solution) 80 103 DEPT CH CH. CH proton decoupled 200 ¹H NMR Spectrum (200 MHz, CDCI, solution) 10 9 8 2000 v (cm) Mote blanch M-118 (15) 120 160 m/e 160 7 1200 120 200 240 6 800 Mass Spectrum + C6H140₂ 280 5 solvent 80 Problem 9 20+1=3 2x -(12+27-12 + 2/1/2 =14-14 4 =2016 часо 40 Exchanges with D₂0 s 3 2 O 1 8 (ppm) TMS 0 8 (ppm

Chemistry
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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What is the IR and HNMRAndCNMR and mass
3458 IR Spectrum (CC), solut 98 4000 D-H) S of base peak 3000 100 BO 60 40 20 40 13C NMR Spectrum (100 0 MHz CDC, solution) 59 80 103 DEPT CH CH. CH proton decoupled 200 ¹H NMR Spectrum (200 MHz, CDCI, solution) 10 9 8 2000 1600 v (cm) Mote banch M-118 (1%) 120 160 200 m/e 160 7 120 6 1200 Mass Spectrum + C6H1402 280 240 5 solvent L 80 Problem 9 2n+1=3 No significant UV absorption above 220 m 2X (12)+(X2) - (12 +27-√2+2/1/1/3 = ₁212-12 +2 =14/-14 =2016 40 Exchanges with D₂0 4 3 2 0 1 8 (ppm) TMS 0 8 (ppm)
Transcribed Image Text:3458 IR Spectrum (CC), solut 98 4000 D-H) S of base peak 3000 100 BO 60 40 20 40 13C NMR Spectrum (100 0 MHz CDC, solution) 59 80 103 DEPT CH CH. CH proton decoupled 200 ¹H NMR Spectrum (200 MHz, CDCI, solution) 10 9 8 2000 1600 v (cm) Mote banch M-118 (1%) 120 160 200 m/e 160 7 120 6 1200 Mass Spectrum + C6H1402 280 240 5 solvent L 80 Problem 9 2n+1=3 No significant UV absorption above 220 m 2X (12)+(X2) - (12 +27-√2+2/1/1/3 = ₁212-12 +2 =14/-14 =2016 40 Exchanges with D₂0 4 3 2 0 1 8 (ppm) TMS 0 8 (ppm)
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