Question

1H-NMR (500 MHz]: Combined Class Sample -3.904 Насо - ŏ 0 --1.873 TTTTTTTTTTTTTT water -2.399 --1.581 Normalized Intensity --Based on the given structure, please label the peaks with letters A-J.

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Answer #1

Base value (8) of group- 6= 8 = = 0.9 1.2 is ppm ppm ppm for for for – CH₂. YCH₂ –CH group. Increment - For CH₂ -X if -6=0 acb OH ŷ = j attached to aromatic ring means - ph so increment = 1.4 Base value of j ECHg) group = 0.9 So overall s= 0.g+ 14 =Base value of – CH₂1 group = 0.9 ppm. = 9.5 Increment Due to in - CH3 group bare valm o CH = 3.5ppm = 3.5+ 0.g= 4.4 ppm. Tota- All aromatic proton give &- valu blw (6.S- 81 ppm. Hear are So not not any aromatic given spectra of (6.5 - 8.0) ppm. HydroIntensity = 3:45) Intensity = . tensity = 2:38) (6= 3-9, 3H) 118= 2.93H) . G=(8= 1187, 34) (1) (0=4.794 ppm 12H Intensity = 3

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Based on the given structure, please label the peaks with letters A-J. 1H-NMR (500 MHz]: Combined...
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