The ee of a mixture is 99%. The percent of the enantiomer in excess in the mixtu

Question

The ee of a mixture is 99%. The percent of the enantiomer in excess in the mixture is: 5% 99% 99.5% 48.5% 2.5% The ee of a mixture is 30%. The percent of the enantiomer not in excess is: 35.5% 70% 34.5% 17.5% 65.5% 1.00 g of an enantiomer was dissolved in 50.0 mL of acetone and analyzed by a polarimeter. (1 = 1.00cm). The observed rotation was +16.1degree. The enantiomer is optically inactive an / isomer a d isomer a meso a (-) isomer Calculate the specific rotation of the compound in Question 31. +8.03 Times103 8.03 Times 103 +8.03 Times 103 +8.03 Times 102 -8.03 Times 102

Explanation / Answer

29) ee = percent of enantiomer in excess - percent of enantiomer not in excess

ee = E1 - E2

so

99 = E1 - E2

but we know that

percent of enantiomer in excess + percent of enantiomer not in excess = 100

so

E1 + E2 = 100

and we got

E1- E2 = 99

solving both we get

E1 = 99.5 and E2 = 0.5

so

the percent of enantiomer in excess is 99.5

so the answer is option C

30 )

ee = percent of enantiomer in excess - percent of enantiomer not in excess

ee = E1 - E2

so

30= E1 - E2

but we know that

percent of enantiomer in excess + percent of enantiomer not in excess = 100

so

E1 + E2 = 100

and we got

E1- E2 = 30

solving both we get

E1 = 65 and E2 = 35

so

the percent of enantiomer not in excess is 35

31)

+ indicates d isomer and - indicates l isomer

so the answer is C) a d isomer

32)

for a solution conc c should be in g/100 ml

conc c = mass / volume

c = 1 / 50

c = 0.02 g/ ml

c= 2 g/100 ml

path lenght l should in deci meters

specific rotation = 100a / cl

specific rotation =   +16.1 x 100/ 2 x 0.1

specific rotation= +8.05 x 103

so the answe is option A

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