# Important Ochem calculation problem

Discussion in 'DAT Discussions' started by harrygt, Jun 21, 2008.

1. ### harrygt 2+ Year Member

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As you know, there are not many choices for the test designers to make a problem in ochem that needs calculations.
Here is one of them that I think is important, and everyone should know: enantiomerics excess or optical purity! [Two different names for the same concept]
In order to find the enantiomeric excess, we divide the specific rotation of the mixture by the specific rotation of one of the pure enantiomers, and multiply it by 100 to get the % [Remember: the specific rotation of the two enantiomers are the same numbers with different signs]. Let's say we get 30% for optical purity [for example for the (+) enantiomer].This means 30% of the mixture is (+), and the other 70% is racemic [meaning you have another 35% (+) and finally 35% (-) ]. Add the 30 and 35 to get the percentage of (+). You will get 65% (+) and 35% (-).

Another Example:
The specific rotation of 2-butanol is +13.52 for the (d) enantiomer, and -13.52 for the (l) enantiomer. Suppose we have a sample of 2-butanol which gives [alpha]D = -5.41 . Find the percentage of each of the two enantiomers in the mixture.

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2. ### Glycogen 2+ Year Member

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harry,I don't think they will cover this on DAT.This is way too much to know for DAT.I `might be wrong though.I just have not seen it anywhere on practice tests.By the way I have so much problem to get the concept of optical activity and all that would you mind explaining it!
I just memorized that enantiomers have optical activity and diastremers do not!
wait,maybe vice versa!!!!!!!!!!!!
Such crazy and useless topic!

3. ### harrygt 2+ Year Member

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Hey bud, this is not complicated. It is covered in every OCHEM class.
What is it that you don't understand? let me know.
Basically, every pure enantiomer is able to rotate plane polarized light. if the one enantiomer rotates it +30 degrees [30 to right], the other will rotate it 30 to left [-30]. If you mix equal amounts of both enantiomer, you will have a racemic mixture which is optically INACTIVE! It still has enantiomers, but it is inactive. Get it so far?

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4. ### Orgodox 7+ Year Member

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Hey yea its an interesting point. I think the more relavent part is what you said about having "30% for optical purity [for example for the (+) enantiomer].This means 30% of the mixture is (+), and the other 70% is racemic" itself is a nice point an is one that is actually in the Destroyer already. I think %EE is more of a lab type concept and usually seams to be stuff on silica or something like that if you get an OChem lab question. Like was said I haven't seen this in any practice material and although you are right this is a common concept covered in OChem I think its not up the DATs alley but hey you never know so thanks for pointing it out.

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6. ### harrygt 2+ Year Member

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Another Example:
The specific rotation of 2-butanol is +13.52 for the (d) enantiomer, and -13.52 for the (l) enantiomer. Suppose we have a sample of 2-butanol which gives [alpha]D = -5.41 . Find the percentage of each of the two enantiomers in the mixture.[/quote]

We should divide -5.41 by -13.52 and multiply it by hundred to get the optical purity. the result is 40%.
A racemic mixute would have 0 degree rotaion. Our mixture has a negative rotation, which means there is more of the (l) enantiomer [40% more].
So we have 40% (l), and 60% racemic [30% (l) and 30% (d)]. It means there is 70% (l) and 30% (d).

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