Question

5.40 You are given a solution containing a pair of enantiomers (A and B). Careful measurements show that the solution contains 98% A and 2% B. What is the ee of this solution? 5.41 HO TOH Predict the value for the specific rotation of the following compound. Explain your answer. -OH HO. 5.42 The specific rotation of (S)-2-butanol is +13.5. If 1.00 g of its enantiomer is dissolved in 10.01 of ethanol and placed in a sample cell with a length of 1.00 dm, what observed rotation do you expect?

Answer 1

5.40) The solution contains 98%. A and 21. B. The enantiomeric excess (e.e) of the solution is the excess amount of peñaniomer.over. another. 2).ee of the solution = (/. amount of Henantiomer) : -(viamount of 'd'enantiomer) . Vie.e = 984.- 24. = 96%. 5.41) GOH . Can be written as following. - How etty ety-OH Hotel Xeherry Cheers. In the molecule, the marked carbon atom isholding two similar group (CH2OH). so, the carbon atom . is not chiral. As a result, the molecule is not Chiral. So, the molecule does not show optical activity. ... The specific rotation of the compound is oo (zerio degree).

5.42) specific rotation of the solution is given by a - [a] = e re . [2] 1 = Specific rotation of solution at temperature t' when 'a' wavelength light is irradiated: L = observed rotation a la path length in dm unit ... ea concentration of the solution in gmi" unit. specific rotation of (s)-2-butanol is (+)13.50 ... is specifie rotation of (R)-2-butanol is (-)13.50. Ag of (R)-2-butano! is dissolved in 10 mL of the ethanol solvent. . ca. 19 = Outg mil and sample cell path length 1 = Idm. 1. -13.5° = gun mor, d = (-) 13.5X0.1. L=(-) 1.350 The observed rotation of the solution containing (R)-2-butanol (ig) in 10 m2 ethand is td 1.35°... ... . 10 ml 1 X 0,1