Question

Need to find what is this unknown compound with this NMRs. Explain how did you know what it was using all the graphs.

13C NMR 46,7 → 11.78 200 180 160 140 TTTTTTTTTTT 120 100 80 60 40 20 0 ppm 'H NMR €0.970 triplet 2.425 (6) quartet 10 9 8 7 6 4 3 2 1 ppm

Answer 1

Two signal indicate Iwo nonequivalent Carbon atomy ave present plump quartet at 2.1ppm, triplet at 1.0 PIM. indicates presonu al cu Citq- group al wilt electronegative atom, tached 6 IR 18. felds no carbonyl grauz at 17 20 chil no alkene at 3030 cal Singnal at 1150 om is due to con band Mudd - m13 2.101 mdicates odd way having odd nost N'atan CH-HCN -CH.CH mil a 101