Question

In (R)-limonene isolation lab (from oranges), we observed the rotation of (R)-limonene.

1) If a little amount of water is mixed with limonene in the sample, how would the presence of water affect %ee(enantiomeric excess) determined for the isolated limonene?

2) What is the effect of a small amount of (S)-limonene to the experimental enantiomeric excess determined for the isolated (R)-limonene?

3) Would the same mass of water vs. (S)-limonene impurity have the same magnitude effect of the experimental enantiomeric excess?

Answer 1

1. (R)-Limonene is a purely non polar compound and when water comes to the compound, there is actually no reaction observed. So the optical activity and enantiomeric excess remains as it is and doesn't change.

2. But when (S)-Limonene is used then there will be reduction on the value of ee since ee= (R-S)/(R+S) × 100 %. Thus if S increase then the denominator increases and the numerator decrease and as a result ee will decrease.

3. No since from the above discussion we can conclude that water impurity has more ee value than (S)-Limonene, since water doesn't effect the ee.