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.