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### Course: Class 12 Chemistry (India)>Unit 7

Lesson 5: Enantiomers

# Optical activity calculations

How to calculate specific rotation and % enantiomeric excess.

## Want to join the conversation?

• How do we know that the 86% figure calculated refers to natural cholesterol and not its enantiomer? I guess what I'm asking is why do we assume cholesterol is in the higher quantity among the two entantiomers?
• At , the problem tells you that the sign of the rotation from natural cholesterol is -.
The sign of the enantiomer must be +.
If a mixture of the two enantiomers is -. the natural cholesterol must be in excess.
• is %ee always be the natutal compound?
• Think of this:
We have : x the dominant Enan, Y the other enan
Excess = Optical PURITY
100 = x + y (1)
Excess = x - y (2)
<-----------------------------> (1) - (2): 100 - Excess = 2y => y = (100 - excess) / 2
Hope this help!
• Please correct me if I'm wrong , at do we take 14% as the value of racemic mixture because we consider that some amount of the enantiomer of cholesterol would have formed a racemic mixture with a conc. of natural cholesterol equal to its own concentration and this part would have had zero optical activity?
thanx
• That's correct. You have 86 % natural cholesterol and 14 % racemic mixture.
The racemic mixture is 7 % (+)-cholesterol and 7 % (-)-cholesterol. It will have zero optically activity, so all of the observed rotation comes from the 86 % natural cholesterol.
The natural cholesterol then accounts for 86 % + 7 % = 93 % of the mixture.
• How do we know that the remaining 14% MUST be a racemic mixture?
• Enantiomers are referred to as optically active because they interact with polarized light. One enantiomer will cause a rotation with a positive angle, which is known as dextrorotatory. The other will cause a rotation with a negative angle of the same magnitude, known as levorotatory. Since they rotate the polarized light with the same magnitude, if you put equal amounts together they cancel each other out and no rotation will be observed, a racemic mixture.

If it were enantiomerically pure cholesterol, we would expect it to interact with polarized light and produce a certain rotation of -31.5 at that temperature and wavelength of light. But with the enantiomerically impure sample we only observe a rotation of -27. This tell us two things, that the dextrorotatory enantiomer is present and that the levorotatory enantiomer is the dominant enantiomer. The enantiomeric excess tells us how much more of one enantiomer we have after equal amounts of the two enantiomers have canceled out each other's effect on the polarized light. Essentially we expect a larger rotation of the polarized light but we observe less than that because a minor amount of the enantiomers are cancelling each other out in a racemic mixture with is optically inactive. Hope that helps.
• Jay says that %ee = [observed[a]] / [[a] of pure enantiomer] * 100.
So my question is, what is the difference between the numerator & denominator? Jay said that the specific rotation was a constant in the last video...so how could there be "observed specific rotation" and "specific rotation of a pure enantiomer?" And between those two and "normal" observed rotation? (not specific).
• The specific rotation of a particular enantiomer is a constant. Here, "observed specific rotation" is referring to the specific rotation induced by a mixture of two enantiomers, and it depends on the percentage composition of each enantiomer in the mixture. Nonspecific observed rotation is the rotation measured by the polarimeter; the conditions of the experiment (sample concentration, temperature, light wavelength, and container length) are then factored in to yield the specific rotation of the sample. :)
• At , why is it said that the remaining 14% MUST be a racemic mixture? Why isn't the remaining 14% the enantiomer?
• He's calculated that the enantiomeric excess is 86% - this is the percentage that is responsible for giving the observed specific rotation of -27. It therefore follows that the remaining 14% is not responsible for rotating plane polarised light. Therefore, this 14% must be the racemic mixture of the two enantiomers.
• Are we just assuming in this case that the enantiomer in excess is nat. cholesterol? How do we know it's not 86% excess of the other enantiomer?
• Enantiomers are either dextrorotatory or levorotatory, referring to how they rotate the polarized light and the sign of the angle. Dextrorotary refers to a clockwise rotation and a positive angle, while levorotary refers to a counterclockwise rotation and a negative angle. If we have a mixture, then we can know which one of the enantiomers is in excess based on the sign of the rotation, since the rotation is dominated by the one with a greater concentration.

For cholesterol, the natural enantiomer, known as nat-cholesterol, is levorotary with a negative angle. We know that from question #1 which has solely natural cholesterol and had a negative observed angle. The other enantiomer is known as ent-cholesterol and is dextrorotary and has a positive angle. Since the mixture produces a negative angle, nat-cholesterol is the excess enantiomer.

As a side note, technically there are 256 stereoisomers (2^(8)) because there are eight chiral carbons in cholesterol, but only nat-cholesterol and ent-cholesterol have an biological significance.

Hope that helps.
• Why is the D line of sodium used? Why that wavelength specifically?
• Why would we convert the length of the polarimeter tube 10.0cm into 1dm? How is this consistent with the rest of the units?