Question

Find the structure using NMR

C7H1402 TTTTTTTTTTTTTT 1.25 1.20 1.15 1.10 PPM.05 5. Proton NMR 4.5 4.5 4.0 1.0 PPM 43.5 12.9 Integral Values 77. 2 28.6 27.6 -- 71.6 1'61 - 9.3 173.7 68.8 31.7 - 22.5 TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT 180 160 120 100 Carbon 13 NMR TTTTTTTTTT 20 PPM

Answer 1

Simulated 1H and 13C spectra is also attached

we have to find structure step by step # No peaks above 5 indicates that structure is non aromatic # Now we have to find out the probabilty of having double bond So Double bond equivalent (DBE)= C + 1 - H/2 - X/2 + N2 = 7 + 1 -14/2 DBE= 1 #Now 13C spectrum give clear picture about functional group in 13C spectrum peaks at 173 ppm indicates that either it is acid or ester but in 1H NMR no peaks at 10-11 ppm so we can dischard the probability of Acid Thus 1DBE and 13C Spectrum confirms the functional group is ester # if we will look at integration ratio it will come as 1:2:2:6:3 # 1 proton at 4.7 ppm confirms that it is next to O atom 0 Η # Now 2 proton quartet at 2.3 ppm confirms that it is next to Carbonyl group and 3 proton triplet around 1 ppm confirms that it is ethyl group connected next to carbonyl group Ć I # 2 proton multiplet around 1.5 ppm shows that R1 is CH2 group, if CH2 group is there so there must be CH3 attached to it around 1 ppm which is not clear visible in given Spectrum. Now R must be CH3 group with doublet around 1.2 or 1.5 ppm which is also merged in the spectrum So based upon the given data the best possible structure is

ChemNMR 13C Estimation o 19.5 9.4 9.7 173.1074.0 27.9 29.5 173.1074.005 Estimation quality is indicated by color: good, medium, rough 180.160.140.120100 80 60 40 20 PPM

ChemNMR 1H Estimation 4.67 01.30 4.67 1.21 0.89 2.29 1.55 Estimation quality is indicated by color: good, medium, rough PPM Protocol of the H-1 NMR Prediction (Lib=SU Solvent=DMSO 300 MHz): Node Shift Base + Inc. Comment (ppm rel. to TMS)