Calculations with Enantiomeric Percentages - Online Tutor, Practice Problems & Exam Prep

Optical activity involves understanding enantiomeric excess (ee) and its calculation using specific and observed rotations. The equation for observed rotation is observed=specific×ee, rearranged to find ee as ee=observed/specific. The total of the higher and lower enantiomers equals 100%. Knowing the higher enantiomer allows calculation of the lower one. This framework is essential for analyzing mixtures in polarimetry.

How to solve for the percentage of each enantiomer.

So we're going to keep talking about optical activity, but it turns out that there's another way that your professor could ask these questions. That's not based on the specific and observed rotations that we talked about already. And it has to do with enantiomeric percentages. So, another way that your professor could ask this type of question is that they give you both the specific and the observed rotation, and you have to solve for the percentages of each enantiomer. I know that sounds like you should be able to figure it out, and I'm sure many of you could, but it can get surprisingly tricky. So, I want to go over this now. It turns out that there are three equations that we want to use for this; two of them you already know.

This one right here is just the shuffled-around version of the equation that we were using earlier, which was that observed equals specific times enantiomeric excess. So that's the same equation, I just rearranged it so that we're solving for enantiomeric excess now. The observed over the specific is your enantiomeric excess. Cool. Then another one that you know because we've already used it, is that the higher enantiomer plus the lower enantiomer always has to equal 100%. That's just kind of common sense.

But then we've got this one in the middle that's kind of just a derivation of the definition of enantiomer. The way that it works is that if you're given the enantiomeric excess, let's say that I say that my enantiomeric excess is 50%, and you're asked to figure out how much each enantiomer represents in that, then you'd use this equation. What this equation says is that you take your percent highest enantiomer and that would be \( (50\% + 50\%)/2 \) and that's going to give you your percentage higher enantiomer. Then we can use the last equation to figure out what the lower enantiomer percentage is. If we know the higher one, we can easily figure out the lower one. So let's go ahead and get started with this question.

I'm going to go ahead and read it for you, give you some hints, and then I'm going to have you solve it on your own. It says that the specific rotation of pure S epinephrine is \( +50 \). So I've been given the specific rotation, that would be right here. I'm going to say it's \( +50 \). Calculate the enantiomeric excess (ee) of the solution for the following observed value. The following observed value is going to be this guy right here. So that means I'm going to put here from my observed what I'm actually getting out of it, even though I have a 50% theoretical, I'm only getting a 25%. So it tells you that I have some mixture of enantiomers in here. It's definitely not 100% of the S.

Then we have to calculate the percentage of each enantiomer from the enantiomeric excess and then we have to sketch the approximate mixture in our polarimeter tube. What that means is that I want you to actually sketch out like we were doing before. Say like this part would be the S and this part would be the R or whatever based on percentages. So if this is 50/50, then I would want you to draw that. Now these numbers are wrong, but I just want you to draw something like that so that you can understand why we're only getting half of the efficacy from this 50. Instead of rotating at 50, it's only 25. So go ahead and try to work with the equations above, see if you can get the right answer, fill in those three blanks below right here, and then I'll go ahead and solve it for you.

Calculating EE, percent of each enantiomer, and sketching mixture

Alright, so this is a pretty fun problem. First, I'm going to take my equation that relates specific rotation and observed and solve for ee. So what it says is that basically 25 for my observed over 50 for my specific is going to give me my enantiomeric excess. When you divide those together, what you're going to wind up getting is equal to 0.5 or if you convert that into a percentage, you would just multiply by 100 and that would give you 50%. So my enantiomeric excess is 50%. Is that cool so far? So I'm going to write here 50. Okay. And that actually already makes sense to me because notice that I'm only getting half of the rotation that I would expect. I would expect a 50-degree rotation but I'm getting 25, so that means that it must be half as effective. So it means that only half of it is in excess, is the enantiomer in excess. Okay?

Now we just have to use that kind of weird equation to figure out what each percentage enantiomer is. So the way we do this is it says that the percentage highest enantiomer is equal to 10050+50. When I do this math, what I'm gonna get is 25 plus 50 equals 75%. So what this is telling me is that I must have one of my enantiomers at 75%. Okay. Then if I want to figure out what the other enantiomer is, I just use the equation that 100 minus 75 is equal to 25. So that means that my other enantiomer is 25%. Cool. So now we have our percentages.

Now the only thing we need to do is figure out which letter is which. Okay? The way we do this is by looking at the sign of the rotation. So notice that my specific rotation is positive. And it's positive for which letter? S or R? S. Okay? That's something to be very careful about. So I have a positive rotation for the S enantiomer. When the S enantiomer is in excess, then we're going to have a positive. Now what is my observed? It's also positive. So what that means is that my higher enantiomer must be the S because that's the one that would give it a positive rotation. So that means that my 75% must belong to my S enantiomer and my 25% must belong to my R enantiomer. If I then wanted to backtrack and calculate to see if I did everything right, I could subtract 25 from 75 and I would get back to the 50 percent that I have. So that's 50% of S, which is my enantiomeric excess. Is that cool?

Awesome, guys. So we're going to do one more. Same exact concept, just different numbers. So go ahead and try to do it and try to get the right answer.

The [α] of pure S-epinephrine is +50°. Calculate the ee of a solution with an observed value of -40°. Calculate percent of each enantiomer. Then sketch the approximate mixture in our sample polarimeter tube.

Alright guys, so for this one, let's just use the same exact equations. So my enantiomeric excess is equal to negative 40 over 50. K? When I divide that out, what I'm going to get is negative 0.8, or if I were to convert that into percentages, what that would equal is negative 80%. K. Now what the heck does negative mean? Okay. It's as easy as this. If you get a negative enantiomeric excess, that means that you have an excess of the other enantiomer, not the one that's stated for your specific rotation. Okay? So what that means is that since it's negative, I have 80% excess of the other enantiomer because that's what the negative sign means. Negative means excess of other enantiomer. K. If it was positive, then it would be an excess of the same enantiomer, which is the answer that we got in the last question. Okay? So now all I have to do is use my equation to figure out which one is which, or what my numbers are. Some of you guys might be able to do this in your head. E = 100 - 80 2 + 80 Okay. So what that's going to give us is 20 divided by 2, which is 10 plus 80, that equals 90%. And which one are we going to have 90% of? Well, if we have 90% of 'r', that means you have 10% of the 's'. So the 90% must be of the 'r' because of the fact that it's an excess of 'r', and then 10% must come from the 's'. Okay? Now one thing that I forgot to do and that I want you guys to do for both of these is to sketch what that would look like in the tube. Okay? So for this first one, what would that sketch look like? Well, what I would say is that I have let's use red for 'r' and blue for 's'. So I would have 25%, That's basically right up to around here. 25% of my 'r'. And then I would get 75% of my 's'. Okay? So this would be my 25. This would be my 75. Now the reason that this works out to have an enantiomeric excess of one half is because if you guys remember, this 25 from the 'r' would cancel out with 25 of the 's', making all of this part right here racemic. Okay? So since 50% of it is racemic, only 50% is available as my enantiomeric excess to actually rotate it. And that's why my observed goes down from 25 to 25 from 50 because the only part that's really doing any rotating is the part that's in the enantiomeric excess which is 50%. Okay? Now let's look at this next one. This next one I'm going to do the same thing. 'Red' will be 'r' and 'blue' will be 's'. So in this case, how much of the 'r' do I have? I have tons. I have all the way up to 90%. Okay? So all of this is 'r'. And then how much of 's' do I have? Only 10%. So that's right here. So it's like a little sliver. Okay? So how much of this is the enantiomeric excess? Well, not all 90 because remember the 's', even though I have a small amount of it, it's going to cancel out with a small amount of the 'r'. Okay? So what that means is that altogether, 20% of this is going to be racemic or not optically active, racemic. And then that means that my enantiomeric excess is going to be 80%. Okay? And also, just so you guys know, you were given these data points but you should have also written that you have positive 50 here and that now my observed is negative 40. And the reason my observed is negative 40 is because I have a 80% enantiomeric excess of the other enantiomer, so it's going to rotate it almost fully in the other direction. Does that make sense? So guys, I hope that by visualizing it with these tubes, it's helped to take something that's really abstract. A lot of students have a really hard time understanding how this works. Maybe you're still struggling. That's okay. You can ask me questions. Okay. But I'm hoping that by looking at these tubes, at least it's giving you an idea of what this looks like and how these things tend to cancel out or if they don't cancel out. Okay? So anyway, that's it for that question. Let me know if you have any questions.

Here’s what students ask on this topic:

What is enantiomeric excess (ee) and how is it calculated?

Enantiomeric excess (ee) is a measure of the purity of a chiral compound in terms of its enantiomers. It indicates the excess of one enantiomer over the other in a mixture. The formula to calculate ee is:

$ee=\frac{\mathrm{observed}}{\mathrm{specific}}$

where $\mathrm{observed}$ is the observed rotation and $\mathrm{specific}$ is the specific rotation of the pure enantiomer.

How do you determine the percentage of each enantiomer in a mixture?

To determine the percentage of each enantiomer in a mixture, you first calculate the enantiomeric excess (ee) using the formula:

$ee=\frac{\mathrm{observed}}{\mathrm{specific}}$

Then, use the following equations:

$\mathrm{Higher}=\frac{100+ee}{2}$

$\mathrm{Lower}=100-\mathrm{Higher}$

where $\mathrm{Higher}$ and $\mathrm{Lower}$ are the percentages of the higher and lower enantiomers, respectively.

What is the relationship between observed rotation and specific rotation in calculating enantiomeric excess?

The relationship between observed rotation and specific rotation is crucial for calculating enantiomeric excess (ee). The formula is:

$ee=\frac{\mathrm{observed}}{\mathrm{specific}}$

Here, $\mathrm{observed}$ is the rotation measured for the mixture, and $\mathrm{specific}$ is the rotation for the pure enantiomer. This equation helps determine the purity of the mixture in terms of its enantiomers.

How do you sketch the approximate mixture of enantiomers in a polarimeter tube?

To sketch the approximate mixture of enantiomers in a polarimeter tube, first calculate the percentages of each enantiomer using the enantiomeric excess (ee) and the relevant equations. Once you have the percentages, represent the mixture visually by dividing the tube into sections proportional to these percentages. For example, if the mixture is 70% of the higher enantiomer and 30% of the lower enantiomer, draw a tube with 70% of its length labeled as the higher enantiomer and 30% as the lower enantiomer. This visual representation helps understand the composition of the mixture.

What are the three key equations used in calculations with enantiomeric percentages?

The three key equations used in calculations with enantiomeric percentages are:

1. $ee=\frac{\mathrm{observed}}{\mathrm{specific}}$

2. $\mathrm{Higher}+\mathrm{Lower}=100$

3. $\mathrm{Higher}=\frac{100+ee}{2}$

These equations help determine the enantiomeric excess and the percentages of each enantiomer in a mixture.