Question

1) Predict the structure of the compounds for the IR and H- a) CH1002 s, 3H s, 3H d, 2H d. 2H 11 10 HSP-03-442 ppm TRANSMETTCa Hio o b) GIÁO S, 1H S, 3H s, 3H d, 1H d, 1H + S, IH 13 12 11 10 9 8 7 6 5 4 3 2 1 HSP-06-376 ppm 2000 1500 4000 1000 RAVENc) CH1002 s, 1H t, 3H 4,2H t, 17 , 1H d, IH d, 1H 12 10 1 2 0 HSP-01-142 ppm TRANSMETTANCE 4000 3000 2000 ISO 1000 NAVENUMBERe) CoH1202 t, 3H 9,2H s, 3H d, 2H d, 2H 11 10 e 8 7 6 4 3 1 0 5 ppm HSP-01-396 LOD TARNSMETTANCE1% D 4000 3000 2000 1500 1000# CoH120₂ 5, 3H s, 2H S, 3H 0,211 d, 2H 12 11 10 9 HSP-45-689 ppm TRANSMETTANCE 1000 4000 NAVENUMERI-1h) C1H1202 0,6H 0,2H 0,21 bs, IH heptet, IH 12 10 TTTTT 6 4 ² 8 HSP-01-312 ppm 12 13N LITHSH 0- 4000 2000 150g 1000 NAVENUMERi) CioH5N 0,2H 1,6H 9.4H L2H tIH 10 g 7 6 HSP-01-372 ppm LOG TRANSMETTRCEI 1000 2000 NAVENUMBERT 1)) CHAN s, 3H 0,2H d,2H 11 10 9 8 7 6 4 5 ppm HSP-03-128 TERNSHETTANCE: 4000 2000 1000 NAVENUMERI

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Answer #1

a) C9H10O2

a peak around 1700 cm-1 in IR implies presence of carbonyl group. A singlet around 4 ppm implies presence of O-R group. So structure is

b) C9H10O

singlet around 10.2 ppm implies presence of aldehyde group and singlets in between 2-3 ppm implies presence of aromatic-R group. As singlet around 7ppm is in shielded region so Ar-H must be in meta position with respect to aldehyde. So, structure is

CH3 CH3 07

c) C9H10O2

singlet around 10.5 ppm implies aldehyde group.

d)

ON

e)

f)

g)

ОН

h)

IO

i)

j)

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