Question

B. a. Cholesterol isolated from natural resources is enantiopure. The observed rotation of a 0.3 g or sample of cholesterol in 15 mL of chloroform solution contained in a 10 cm polarimeter tube is -0.78 °. Calculate the specific rotation of cholesterol. b. A sample of synthetic cholesterol consisting entirely of (+) cholesterol was mixed with some natural (-)-cholesterol. The specific rotation of the mixture was -13. What fraction of the mixture was (+)-cholesterol? C. A sample of the chiral molecule limonene is 94% enantiopure. What percent of each enantiomer is present?

Please help with these three questions

Answer 1

B.a) observed rotation = -0.78° mass of sample=0.39 in 15ml 1 ml contain = 0.39 0.029 mL 3 in 15mL solution 15 ML - length of tube = 10cm = 1 dm (..1dm=locm) [2)T: Specific rotation

L: obsermed rotation l length of folarimeter tube in dm C: concentration in g/m2 Putting values; [2] T = -0.78 1 dm x 0.02g m2 = -390 specific rotation of cholesterol was found to be 1391 (+) cholesterol + (-) chlester J Mix mixture 7 -13° (specific rotation) oftical rotatues of fese enantiomes (-) foud in part a -39° / optical furity = loox Specific rotation of specific sotection of pure enantional = 100 x (13) (-39°) 33.331. mixture LD

= We have 33.33% optical hure enantiones of (-) cholesterol and . 66.67% racemic mixture I having 33.33%. E) Cholesterol 33.33.1. (+)-cholesterol (-) Cholesterol total = 33.33 +33.33 = 66.67a1. (t) cholesterol = 33.33al. 33.33% of I mixture is (1) Cholesterol. To A sample of limonene is 947enantiohuse 6. racunic mixture 341 of 34. of another One enantioner enantiomer

u One enantiomer = 94./.+3.1. 97, Another enantioner = Balo are present Thus 2 enantioners is 97 3.