The pure (+) enantiomer of a chiral compound has a specific rotation of +40.0 under the conditions used. A sample of the compound, 0.250 g dissolved to make 10.0 mL of solution, rotates plane-polarized light by +0.600° in a 1.00 dm tube under the same conditions. Find the sample’s specific rotation, its enantiomeric excess and the percent of each enantiomer.
[α]=lcα
1
Find the concentration
Specific rotation uses grams per milliliter. The sample holds 0.250 g in 10.0 mL, so c=0.250÷10.0=0.0250 g/mL.
c=10.00.250=0.0250
2
Substitute into the specific rotation
The observed angle is +0.600°, the path length 1.00 dm and the concentration 0.0250 g/mL. Each goes where the definition puts it, the angle on top and the path length times the concentration below.
[α]=(1.00)(0.0250)0.600
3
Divide
The product below is 0.0250, and 0.600÷0.0250=24.0. The sample’s specific rotation is +24.0, positive like the pure (+) enantiomer’s but smaller.
[α]=24.0
4
Compare with the pure enantiomer
Only the excess of (+) over (−) rotates the light, so the sample’s rotation is that fraction of the pure value. Both values were measured the same way, which is what makes the comparison fair. The enantiomeric excess is 60.0%.
40.024.0×100%=60.0%
5
Split the mixture
The rest of the sample, 40.0%, is racemic and holds equal amounts of each enantiomer, 20.0% apiece. The (+) form has its 60.0% excess plus 20.0%, which is half of 100% plus the excess.
2100%+60.0%=80.0%
6
Find the minor enantiomer
The (−) form is what remains. The difference 80.0%−20.0%=60.0% gives back the excess, which checks the split.
100%−80.0%=20.0%
Result
The sample’s specific rotation is +24.0, its enantiomeric excess 60.0%, and it holds 80.0% of the (+) enantiomer and 20.0% of the (−).
Your turn
1. One like the example
The pure (−) enantiomer of a compound has a specific rotation of −52.0. A sample, 0.400 g in 20.0 mL of solution, rotates the light by −1.56° in a 2.00 dm tube under the same conditions. Find its enantiomeric excess and the percent of the (−) enantiomer.
Show the answer and explanation
An excess of 75.0%, with 87.5% of the (−) enantiomer.
c=0.400÷20.0=0.0200 g/mL.[α]=−1.56÷(2.00×0.0200)=−39.0. The excess is 39.0÷52.0×100%=75.0%, and the (−) form is (100%+75.0%)÷2=87.5%.
A sample has an enantiomeric excess of 90.0% of the (+) form, whose pure specific rotation is +30.0. What angle should a 0.0500 g/mL solution of it show in a 1.00 dm tube?
Show the answer and explanation
About +1.35°.
Work backward. The sample’s specific rotation is 90.0% of the pure value, 0.900×30.0=27.0. The observed angle is [α]×l×c=27.0×1.00×0.0500=1.35°, positive like the major form.
100%90.0%×30.0=27.0α=(27.0)(1.00)(0.0500)α=1.35
3. Somewhere new
A sample of 2,3-dibromobutane shows no optical rotation. Is it the meso compound or a racemic mixture of (2R,3R) and (2S,3S)? How could you tell?
Show the answer and explanation
The rotation alone cannot say. A melting point or a chiral chromatography column can.
Both show zero rotation, the meso compound because it is achiral and the racemic mixture because its two enantiomers cancel. The meso compound is a diastereomer of the enantiomers, so it has its own melting point. A chiral column would split a racemic mixture into two peaks and give one peak for the meso compound.
Keep exploring
Check the rotation in Math, then change the observed angle in row 3 from 0.600 to 0.300. Row 4 reads ✗ until you write 12.0, and the excess falls to 30.0%, which is 65.0% and 35.0%.
