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## Organic chemistry

### Course: Organic chemistry>Unit 4

Lesson 2: 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? • 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.
• 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? • 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? • The units certainly aren't consistent with SI. The units of measurement are:
α = °
c = g/mL
l = dm
[α] = α/(cl), so the units are °·mL·g⁻¹dm⁻¹, but this is usually shortened to °, with the rest of the unit being understood.
So, you must express the length of the tube in decimetres if you want to get the same values as in the literature.
• will using SI units yield the same result? • I still don't get why that leftover 14% from the last problem is racemic mixture. There was literally nothing that indicates that whatever is left from 14% is racemic mixture • At , why is it that optical purity is equal to the percentage of one enantiomer minus the percentage of the other enantiomer? Where does the basis for that equation come from?
(1 vote) • It's logical. Take the last of his three examples where you have a solution containing 75% of one enantiomer and 25% of the other enantiomer. Let's call these enantiomers A and B and imagine there are 75 molecules of A and 25 molecules of B in the solution.

The optical rotation caused by the 25 molecules of B will be cancelled out by the opposite optical rotation of 25 molecules of A - in other words, the 25 molecules of B and 25 of the 75 molecules of A form a racemic mixture.

This leaves 50 molecules of A whose optical rotation is not cancelled by B molecules. These 50 are reason for the observed optical activity. Hence, the optical purity is defined as 50% which, you will notice, is also the enantiomeric excess (= 75 minus 25).