Question

If someone explain how to find these unknowns..And what they are and how you got the answer. I am having a hard time really understanding NMR and IR.. Thanks in advance

Unknowns for 1H NMR Final Report 13) Chemical Formula: CHO IR: strong peak 1720cm, strong broad 3300cm, Зн 1H 1H1H,1H THI TH PPM 14) Chemical Formula: CHO IR: strong peak 1720cm, strong broad 3300cm 2H 2H2H THI TH PPM 15) Chemical Formula: CHO IR: strong peak 1720cm, 1650cm (hard to see but, all peaks that integrate to 1H are doublet of doublets) I 3H 1H 1H Page 5 of 6

Answer 1

Molecular formula 1 H1403 S.O.0. = 05. IR data 1720 carbonyl group is present 3600 Indicate Oh stretching freq. o 'H NMR 1.5 doublet, 34) ch-ch group 13.2 (singler, 34) > o-chz group 4.7 (t, 14) presence of CH-ch a group 4.8 (d, 2H) - cu group. 4.9 Cbs, 1H) - 0-4 stretching (broad singlet

CIASSACe page o 74 to 8.0 ppm Indicate presence of benzone ring protons By considering all above data structure is 4.7(E) 3.2 (6) cho CH-CH₂ 4 1.5(d) <H264 ,8 (4) 3300 cmt c=6 1720 cont 4.9 (S) o 88.0 (4) . 7. Lpp ñ 35 7.8 CV) In this spectrum all pattern of benzene ring here two equivalent protons are present 6.8 doubler, 2H) 7 7.3 ( double data is proton sets of similar only is different benzene ring so stucture of given organic compound is- 13.2 (5) hoch - Ch1.5 (d. - Lo crv = 4.8 G) F-3 (d) Du 47.3 (d) 6.8 (d) x L 16.8 (d) 3300 cmt

Page Molecular formula : CG H 802 site of unsaturation - 02 IR: 1720 carbonyl group is present I 1650 c c is present I'H NMR- 2.2 (s, 34 presence of cbr c=1 group 2.4 (s, 3H) presence of chc=0 group 4.3 4.6 By (d, IH) (d, 14) considering → all presence of presence of above data H-c=c group H-c=c group structure is 403 (4) 0 720 cm 2.4 (5) 1650 cmt