Question

CHE-353-DyeExpmt-MeOH 2.3 0-OH-3-56 PP- (s) Ar-Hn 6.75 p ) -2.2 -2.1 wi e ct doen AH-7-S2 Pm )2.0 AT-H-7-32 Pptri) -1.8 O-Ar-H-7.60 Ppm (d) 17 -1.9 CXAF-H 32 p--(6 1.5 Ar-Han742 Pp-d) -1.4 S-2xAr-H7 92 PPmld) 13 OH -Ar -H-N? -46 PPm (du) -1.2 H NMR(0 Mta coCs: 1.1 1.0 F0.9 F0.8 F0.7 F0.6 F0.5 0.4 F0.3 F0.2 0.1 -0.0 -0.1 -0.2 7.0 6.9 6.8 6.7 6.6 6.5 6.4 6.3 7.6 7.5 f1 (ppm) 8.8 8.7 8.6 8.5 8.4 8.3 8.2 8.1 8.0 7.9 7.8 7.7 7.4 7.3 7.2 71 Bengene

how I can write this data in mhz and find the coupling constant for this data. basically I wanna know how I can write this data in data summary. thank you so much for your help.

I don't have those values

can you please describe it with out those values

Answer 1

ANSWER:-

for data summary

1. first you have to write how many signals are observed in this spectrum
2. second you have to see that how signals are split(splitting pattern)
3. and then you should notice the chemical shift
4. and last one you have to notice relative intensity (number of proton present or you can say integration

so all things you have done correctly , you assign chemical shift correctly

now you want to know how to convert ppm into mhz so easily you can convert this by formula

frequency (MHz) = frequency (ppm) x RF carrier frequency )MHZ

to determine the coupling constant you have to know the splitting pattern

for example you have doublet so you should first minus the chemical shift of 1 peak from another peak then you have to multiply the result by spectrometer frequency in MHZ than you get the desired coupling constant value in hZ.