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1) A H NMR doublet appears at chemical shifts of 2.585 and 2.597 ppm. 1.1) If the coupling constant J12 for this doublet is

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.J151-60) frequency of instrument (in HM2) So frequency of instrument (in H₂) = 8.4 Hz 2.597 - 2.585) ppm FooMHz. Bo 8 . gynoBo= Oxat 7 700 x 106 Hz X2 X3.14 267.83 x 100 radian / TX sect 16.437 = in da 18 x Bo 8 for c= 67.28 x 100 radion / Testa see

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