Question

68 The (a] of pure quinine, an antimalarial drug, is -165 a. Calculate the ee of a solution with the following (a] values: -50, -83, and -120. b. For each ee, calculate the percent of each enantionmer present. c. What is [a) for the enantiomer of quinine? но. d. If a solution contains 80 % quinine and 20 % of its enantiomer, what is the ee of H" the solution? CH30. e. What is [a)] for the solution described in part (d)? quinine (antimalarial drug)

Answer 1

Answer :- lee = [x] mixture x 200 [2] when pure mavar (a) [x] values with ave sign -50,- 83 and -120 indicates given quinine is maron [x]=-50 = %.ee = OOOOUUUUUUUUUUUUUUU +50 x 100 = 30-3.1. +165 [a]=-83 olce= 50.3.1 783 x 100 = +165 {x}=-no ol.ee = 1120 x 600= 72.72-1 7165 minos = 100 - .l.ee 2 mawar = 100-minal fee=30-30% 7 minal = 100-30.3 = 34-85 - minar maior = 200-34-85 = 65-15 - main 1.ce= 50-3.l. =) menar = 400-50-3 = 24. 85 - minos madal = 100-24.85= 75.15 - mavoy minou lec=72.72.8 = renal = 100-72.72 = 1.64 maray = 100-13.64 = 86-36-madar

(7 +165 same but = enantiomers have [L] value with opporsite sign Quinine (x2=-165 enantiomer (x)= +165 olce= mawar.l. - minarl. = 80-20 1.ee = 6001 [ l.ee = 60% 60 = (27 salion & Loo - 165 Elsad = 60 x-165 100 = -9900 = -99 loo Say solution manage