Question

provide the structure for the compound while using as much of thr NMR/IR as possible. show work.

Answer 1

The degree of unsaturation for the compound CisH33N can be calculated using following formula, Dou - (20)+2+N-X- Here, c= number of carbons (15), N= number of nitrogens (1), X = number of halogens (0) and H= number of hydrogens (33) DoU = (2x (2 X 15) +2+1-0-33 - = 0 Therefore, degree of unsaturation in the given compound is 0. 1H NMR analysis. There are 4 set of proton in the NMR Signal at 2.4 ppm (2H, t) corresponds to CH2 attached to CH2 Signal at 1.4 (2H, m) may correspond CH2 attached to two CH2 groups Signal at 1.3 ppm (4H, m) correspond to CH2 attached to two CH2 Signal at 0.9 ppm (3H, t) correspond to CH3 attached to CH2 13C NMR analysis. There are 5 sets of carbon in 13C NMR Based on the above information, the structure of the compound is, www 1.3 (4H, m) - 2.4 (2H ) 1 1 .4 (2H, m) 0.9 (3H,t)