Question

7. What is the structure of the following compound based on the spectroscopic data you are provided with? (M"= 86) 60 10 20 3040 50 60 70 80 90 100 10 120 130 140 150 CH- CH. 2500 Wane 2000 1500 4000 3500 1000 500 3000 200 120 Clemsal sit & p I50 160 140 100 40 30 0 2.30 2.40 120 100 1.10 (ppm) (ppm) 10.0 9.0 80 70 6,0 50 4,0 3,0 2.0 1,0 0,0 Chemical shift (&, pem)

Answer 1

The important information obtained from the spectral data are as follows: MASS SPECTRUM: M at 86 indicates that the molecular mass of the compound is 86 I 13C NMR The compound has 3 types of carbon : CH2, CH3, and C. No CH carbon is present. IR SPECTRUM: Broad band at 2890-2975 range : Aliphatic C-H bond Strong band at approx 1700 : C-O stretching. IH NMR Only two signals One triplet at a chemical shift of approx lppm which corresponds to 6 protons. One quartet at a chemical shift of approx 2.5ppm which corresponds to 4 protons. observed are ANALYSIS Since the 1H NMR spectrum shows signals corresponding to 4 and 6H , it means that the compound is symmetrical. The H which gives triplet signal has two chemically equivalent H beside it, and the H which gives quartet, has three chemically equivalent H beside it Hence we conclude that the molecule has two (since it is symmetrical) of the following group more downfield° Н2 Molecular weight of this fragment = 29 Н,с- Also, from IR know that there is a carbonyl group present we Molecular weight = 28 mass of (2x29)+28 = 86, which is the molecular weight of the compound We know have a total We also have all the required carbon types from the 13C NMR. Thus, remembering that the compound required is symmetrical, and that the CH2 protons are more downfield than the CH3 protons, we join the fragments to get H2 H2 Н,С CH3