Question

G: CH, NO Hz ppm 650.31 7.262 858 645.88 7.212 46 392.56 4.384 230 386.81 4.320 218 174.25 1.946 1000 m, 5H s, 3H d, 2H s, 1H 11 10 9 8 7 6 5 4 3 2 1 0 HSP-01-452 ppm

Answer 1

Molewlar formula is CGH , NO From na 'H NMR spectra peak at 9.0 8 Prom (5,34) indicates --CH3 458H 6.5 spim (5,11) indicates -NH group 7.3 pppcos (m, 5H) indicates aromatic proting From H NMR conclude that compound contain one benzene ring, one -NH group and one - e-CH2 Group 6 - NH , -8-CH3 one double at 4.3 8ppm (, 2H) indicates - CH2 group adjusant to _NH group & benzene ring. .so the Structure of Compound is D C- 4N9 – – CH3 u (5) (5) Yoh (s) m, 5H

BC NMR: geak at no indicates amide group _{-NH2 peaks of 138,128,129,127.5 ,127.12 indicates aromatic carbon. pealcs at 43.38, 22.15 indicates methyl Carbon. (43.38) Chy-NH-C- CH3 1138.53 (70) (22.15) 128. 42 7 12843 1276521 1 27.52 127.12 So, the structure of compound i8 CH-NH-C-CH2 Coupling valves (3): J = line spacing x HZ Coupling constant for m,5H = 7.262–7.212

coupling constant (I) in H NMR: For m.& doublet we can calculate I value from H2. I= difference in HZ for dooblet (2H) = One peak at 4.384- peak 24:320 (H2) (Hz) = 392.556 - 3,86.81 = 5.75 HZ Got I value for droblet is 5.15 HZ For multiplet J = 650-31- 645-88 = 4:43 HZ