Question

17.       What mass of (S)-(–)-mandelic acid is present in a 10 g sample which has an enantiomeric excess of 20% of the S-enantiomer?

18. The specific rotation of the S-enantiomer of a compound is -120º. What is the enantiomeric excess of a sample of the compound that has a measured specific rotation of -30º.

19. The specific rotation of the S-enantiomer of a compound is -120º. A student determines that a sample of the compound has a specific rotation of +60º. What mass of the S-enantiomer is present in 10 g of the sample

Answer 1

17) The sample has enantiomeric excess of 20% S- enantiomer.

That is in 100, 20 is pure S- isomer and the remaining is racemized.

In the 80% racemic mixture, 40 is s and 40 i R isomer.

Thus total of S - isomer in the sample is (40+20) = 60% .

Thus in 10g sample the mass of S -enantiomer = 10g x 60/100 = 6g

18) specific rotation of pure S- enantiomer = -120

rotatio n of sample = -30

enantiomeric excess = rotation of sample x 100/ specific rotation of pure sample

= -30 x 100/-120

= 25%

19)

Now the rotation of sample = +60

so enenatiomeric excess = +60x100 /+120

= 50%

That is 50% is excess R- enantiomer (rotation is positive)and the remaining is racemized.

In the racemized 50% half is S and half is R isomer.

That is percentage of S enantiomer in the sample is 25% and R is (50+25) = 75%

Thus in 10g sample mass of S- enantiomer = 10g x 25/100

= 2.5 g