10 100 3352 IR Spectrum (liquid film) 4000 3000 2000 1600 1200 800 V (cm³) 80 60 % of base peak 40-6 20 M+ 138 121 Mass Spectrum OH Problem 108 400 0 Singlet dublet OH UV Spectrum λmax 270 nm (log 10 € 3.1) λ max 282 nm (log 10 € 3.1) M solvent: hexane Wonol C8H1002 Note: UV spectrum not changed significantly on addition of base 12 16545)L 40 80 120 160 200 240 280 m/e 13C NMR,Spectrum (50.0 MHz, CDCI, solution) DEPT CH₂t CH₁t CH↑ proton decoupled aromatic aromatic solvent CHz ether 6 uneque C 200 160 120 80 40 'H NMR Spectrum (200 MHz. CDCI, solution) 6 alcohol exchanges with D₂O 0 $ (ppm) Sunique H TMS 8 7 6 5 4 3 2 1 Armani C $ (ppm)
10 100 3352 IR Spectrum (liquid film) 4000 3000 2000 1600 1200 800 V (cm³) 80 60 % of base peak 40-6 20 M+ 138 121 Mass Spectrum OH Problem 108 400 0 Singlet dublet OH UV Spectrum λmax 270 nm (log 10 € 3.1) λ max 282 nm (log 10 € 3.1) M solvent: hexane Wonol C8H1002 Note: UV spectrum not changed significantly on addition of base 12 16545)L 40 80 120 160 200 240 280 m/e 13C NMR,Spectrum (50.0 MHz, CDCI, solution) DEPT CH₂t CH₁t CH↑ proton decoupled aromatic aromatic solvent CHz ether 6 uneque C 200 160 120 80 40 'H NMR Spectrum (200 MHz. CDCI, solution) 6 alcohol exchanges with D₂O 0 $ (ppm) Sunique H TMS 8 7 6 5 4 3 2 1 Armani C $ (ppm)
Principles of Instrumental Analysis
7th Edition
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Chapter11: Atomic Mass Spectrometry
Section: Chapter Questions
Problem 11.2QAP
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Question
May you explain the H NMR and the C NMR please. If my structure is right please. Thanks
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