Enantiomeric Excess

[G1SG2]

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Enantiomeric excess is the ratio of the difference between compared enantiomers over the sum of both enantiomers. If the enantiomeric excess is 80%, what is the percentage of the target enantiomer in the mixture?

The answer is 90%. I don't really understand how TPR solved this problem.

x-(1-x)/1=80%, 2x-1=180%, x=90%. Can someone please explain this to me? Thanks.

 

[amine2086]

Enantiomeric excess is the ratio of the difference between compared enantiomers over the sum of both enantiomers. If the enantiomeric excess is 80%, what is the percentage of the target enantiomer in the mixture?

The answer is 90%. I don't really understand how TPR solved this problem.

x-(1-x)/1=80%, 2x-1=180%, x=90%. Can someone please explain this to me? Thanks.

Enantiomers are mirror images, so they always exist in pairs. Let,

Enantiomer 1 = x
Enantiomer 2 = 1-x (considering the two enantiomers must add up to 100%)

Enantiomeric excess = (enantiomer 1 - enantiomer 2)/(1) [as defined in the question stem]

So, x-(1-x)/(1) = 0.8 [it's given that the enantiomeric ratio is 80%]

x-(1-x)/(1) = 0.8
x-1+x = 0.8
2x = 0.8+1
x = 1.8/2
x = 0.9 or 90%

 

[G1SG2]

Enantiomers are mirror images, so they always exist in pairs. Let,

Enantiomer 1 = x
Enantiomer 2 = 1-x (considering the two enantiomers must add up to 100%)

Enantiomeric excess = (enantiomer 1 - enantiomer 2)/(1) [as defined in the question stem]

So, x-(1-x)/(1) = 0.8 [it's given that the enantiomeric ratio is 80%]

x-(1-x)/(1) = 0.8
x-1+x = 0.8
2x = 0.8+1
x = 1.8/2
x = 0.9 or 90%

Thank you!!!

 

[G1SG2]

And, this may be a stupid question but-rotating plane polarized light is a physical property, right?

 

[Hemichordate]

And, this may be a stupid question but-rotating plane polarized light is a physical property, right?

Yes because it maintains the molecular structure.

 

[SuperSaiyan3]

Enantiomers are mirror images, so they always exist in pairs. Let,

Enantiomer 1 = x
Enantiomer 2 = 1-x (considering the two enantiomers must add up to 100%)

Enantiomeric excess = (enantiomer 1 - enantiomer 2)/(1) [as defined in the question stem]

So, x-(1-x)/(1) = 0.8 [it's given that the enantiomeric ratio is 80%]

x-(1-x)/(1) = 0.8
x-1+x = 0.8
2x = 0.8+1
x = 1.8/2
x = 0.9 or 90%

😱

you have to know this stuff for the MCAT??! I've never once had a question like this that I've seen on a practice test nor have I ever seen it go this in depth in Kaplan! Should you know this sort of stuff for discrete questions??

 

[G1SG2]

😱

you have to know this stuff for the MCAT??! I've never once had a question like this that I've seen on a practice test nor have I ever seen it go this in depth in Kaplan! Should you know this sort of stuff for discrete questions??

I would DEFINITELY KNOW IT! Read my signature!!

 

[SuperSaiyan3]

I would DEFINITELY KNOW IT! Read my signature!!

😕

you don't have a signature.

 

[G1SG2]

😕

you don't have a signature.

I do, but sometimes it doesn't show up in my posts for some reason 😕 Look at the opening post, it's there.

 

[neil100]

Enantiomers are mirror images, so they always exist in pairs. Let,

Enantiomer 1 = x
Enantiomer 2 = 1-x (considering the two enantiomers must add up to 100%)

Enantiomeric excess = (enantiomer 1 - enantiomer 2)/(1) [as defined in the question stem]

So, x-(1-x)/(1) = 0.8 [it's given that the enantiomeric ratio is 80%]

x-(1-x)/(1) = 0.8
x-1+x = 0.8
2x = 0.8+1
x = 1.8/2
x = 0.9 or 90%

👍

You can get that equation by translating what they give you for the ratio.
the ratio of the difference between compared enantiomers over the sum of both enantiomers.

Enantiomers: R or S so you are dealing with 2 things.
x1= enantiomer 1
x2= enantiomer 2

Now use what they give you.

x1- x2/x1+x2 = en. excess

so we know x1 + x2 = 100% or 1 you have both of course.

substitute x1 for x2 in the numerator.
x1+x2 = 1 so, x2 = 1-x1
so you get ,

x1- (1-x1)/1 = .8 or 80%
solve for the amount you want, which is x1
You should get 0.9 when you do the algebra.
or 90%.