Question
determine the structure of the compound and label the peaks on spectra

left to right compound 3: C.H. H NMR integration, right to left: 1:6 11 10 9 ppm 140120 1000


left to right sound 3: CsH14. H NMR integration, right to left: 1:6 Compound 3: CH. CC 11 7 10 9 8 Cửa - Chº ppm RECH/ /R₂ C
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Answer #1

Hi the unknown is 2,5-dimethyl-2,4-hexadiene

Given M.F: C8H14

IHD = [(2 x 8) + 2 - 14] / 2 = 2, means either 2 double bonds, 2 rings, or combination

HNMR shows 2 signals, 1 from alkene region & 1 from aliphatic region

CNMR shows 2 signals from alkene region & 2 signals from aliphatic region.

IR:

3024 cm-1, medium, sp2 C-H stretch

2967 to 2857 cm-1, strong, due to sp3 C-H stretch

1621 & 1650 cm-1, medium, due to sp2 C=C stretch

1388 & 1466 cm-1, strong, due to sp3 C-H bending

841 cm-1, strong, due to sp2 =C-H bending

(0) (A) H. (B) A ppm 5.972 1.788 1.732 (B) H (A) 11 10 9 8 7 6 4 3 2 1 0 5 ppm HSP-00-467

200 180 160 140 120 100 80 60 40 20 0 CDS-04-135 ppm I- ppm 132.01 121.53 26.28 18.03 WN

sp2 C=C stretch EM sp2 =C-H str sp3 C-H bend sp2 =C-H bend DH 4000 sp3 C-H stretch 9000 2000 1000 500 Cmy 15.60 H I-

Hope this helped you!

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