Question

II 8.6 8.4 8.2 8.0 7.8 7.6 7.4 7.2 ppm 8 6 5 4 3 2 1 ppm 8.540 8.540 8.528 8.528 7.828 00L 000 T 7.819 CO 90' 90 86S L 7.586 7.476 7.463 1.08 092L 7.182 7.172 7.160 7.828 1.061 7.819 86S 7.586 1.056 7.476 7.463 1.072 7.260 7.182 8LO T 7.172 7.160 -1.246 990 0

please provide the chemical shift, splitting, how many protons in each peak, and calculate the J coupling constant for above spectrum.

The structure is

the 1H NMR was 400 MHz and solvent CDCl3.

Answer 1

chemical shift value for doublet peak is the average value of the doublet peaks.

(14, 7.17, t, 24 of 2 tr = 3.6min) 07.59 (1H, d, do 4.800 ) 1 1 r 8.53(d, 1H, J = 4.8 Hz) 7.47 d. = 5.2mm) do doublet t- triplet de coupling Const) J (8.54 – 8.528) x 400 MHz och -8.528) x Coupling Coast alculation Calculation H A = 4. 82