Enantiomeric Excess - Online Tutor, Practice Problems & Exam Prep

Enantiomeric excess is crucial for understanding optical activity. Specific rotation indicates the degree of rotation caused by pure enantiomers. A racemic mixture (50% S and 50% R) results in zero optical activity due to cancellation. Enantiomeric excess is calculated by subtracting the lower percentage of enantiomer from the higher. The observed rotation can be determined using the equation: $\alpha =\mathrm{[\alpha ]}\times \mathrm{ee}$, where [α] is specific rotation and ee is enantiomeric excess, representing the optically active portion.

How to calculate enantiomeric excess.

So now what I want to do is I want to talk about enantiomeric excess, which is actually one of the most important concepts of optical activity. Okay. So remember that I told you guys the specific rotation is the rotation that 100% enantiomers would produce. So if I had a 100 percent enantiomer, let's say my specific rotation is 10 degrees, then I would get 10 degrees. Well, it turns out that if I have the opposite enantiomer, meaning that this would be, let's say, the R. But let's say that I have the S enantiomer over here, then what would I get? I would just get the opposite rotation. Okay? So then that means that instead of getting a positive ten rotation, maybe I get a negative ten, something like that. Basically, it would be the same absolute value, just a different sign. So that's the first thing that you guys need to know.

The second thing is that we can mix enantiomers. It doesn't always have to be a 100% pure solution. And in fact, for most of these questions, it never will be. Your professor is going to mix it up and give you guys some kind of mixture and you're going to have to figure out what the optical activity is at the end of the mixture. Okay. So a perfect one-to-one ratio of enantiomers, meaning that I have 50% S and 50% R is called racemic. That's very important. That's such an important word for organic chemistry. You can never forget that. Okay? A non one-to-one ratio, that it's not pure but it's also not 50/50 is called scelemic. This one is not used quite as frequently, but it's still something that you should be aware of.

So now what I want to do is I want to take our polarimeter tubes, our sample tubes and I want to do some experiments. We're just going to make up some numbers and we're just going to see what happens. Alright. So for this first one, as you can see, I have my S enantiomer and it has a specific rotation, that means the alpha in brackets, of 20. Okay? And I want to know if I took a 100%, if I had a 100% of S in this tube, what would be the observed rotation? The answer is that the observed rotation would just be the same thing. It would be positive 20. I'm going to explain why in a second. Okay. Then the R enantiomer, let's say that my S enantiomer is positive 20, but let's say that I have a 100% of my R enantiomer in here. So now I just mixed it up, then what would that be? What that would give me is negative 20 because it has an opposite configuration, so my observed would also be opposite.

Now what I want to do is let's say that I had in this one, I had 50% of S and then I had 50% of R. What would be my observed rotation in that case? Well, just intuitively, the way you can think of it is well, I have 50% rotating it 20 degrees to the right. I have 50% rotating it 20 degrees to the left. What would be the observed rotation? The observed rotation or just the alpha symbol, not the alpha with the brackets, would be 0. Okay? Because they would cancel each other out and this is what we would call racemic. Okay. So a racemic concentration is always going to have an optical activity of 0 because it's going to perfectly cancel out. Does that make sense so far? Cool.

So now what I want to do is I want to talk about something called the enantiomeric excess. And what the enantiomeric excess says is it's just basically you take your highest percentage enantiomer and subtract it from your lowest percentage enantiomer. You only have 2 enantiomers, so you would take your highest minus your lowest and whatever you have at the end of that, that's going to be your enantiomeric excess and that's the amount that is actually optically active. So if we ever want to calculate observed rotation, it's actually really easy. All we do is say observed rotation or alpha equals the specific rotation, meaning the amount that that molecule would produce at 100%, times the enantiomeric excess, which is actually the optically active part because it's the part that isn't cancelled out by anything else.

Calculating Enantiomeric Excess and Observed Rotation

So my equation says that my observed rotation equals my specific rotation times my enantiomeric excess (EE). Okay? Well, in this, what was my specific rotation equal to? It was equal to positive 20. Okay? So now I just have to figure out what EE was equal to. EE is equal to my highest enantiomer minus my lowest enantiomer. In this case, it was 50 minus 50, which meant that I had an enantiomeric excess of 0. That means that my EE equaled 0, so it was 20 times 0, which means that even though I have this awesome specific rotation of 20, all of the enantiomers are getting cancelled out, so there's no overall rotation at the end. So that is why my observed rotation equals 0.

Calculating Enantiomeric Excess and Observed Rotation

Hey, everyone. So here it says, calculate the EE, which is our enantiomeric excess, and observe rotation for the following chiral mixtures where the S enantiomer has a specific rotation of plus 20 degrees. Alright, let's take a look at these two images.

In the first image, we're dealing with a racemic mixture because it has an equal amount of R enantiomer versus S enantiomer. So here, they would be 0 for our enantiomeric excess. Remember, the formula for enantiomeric excess is we subtract the larger percentage enantiomer by the lower percentage enantiomer. Since they're both 50%, they cancel out to 0. And remember, with the racemic mixtures, since the R and S enantiomers are completely neutralizing each other, there is no observed rotation. So, it'd be 0.

Over here though, we're no longer dealing with a racemic mixture because the R and S enantiomers are not equal in terms of percentages. So, in this case, let's calculate what our enantiomeric excess will be. Again, our enantiomeric excess is the higher percentage enantiomer; S here is 70%, minus the lower percentage enantiomer, which is R at 30%. So here, our enantiomeric excess would be 40%.

Next, they need us to find the observed rotation. So remember, your observed rotation is just α = σ ⋅ △, where △ is the enantiomeric excess in its decimal form. To find the decimal form, you just divide your percentage by 100. So that would be times 0.4. Here we have our specific rotation which is 20. Positive 20. So now you multiply the positive 20 degrees times the 0.4. So your observed rotation here should be 8 degrees.

Now, does this answer make sense? It does make sense because remember, when we're dealing with pure enantiomer, its full rotation is 20 degrees. We're no longer dealing with a pure enantiomer; we have a mixture. 30% of it is R. And if we want to think about it like this, out of the 70%, we had 30% originally, that was S. Over here we said that these percentages are equal, so they completely cancel one another out. The same thing happens here too. 30% of this R, which is all of it, cancels out with 30% of this S. So they're completely taken out. And what we have left here is this 40%. That is the enantiomeric excess. So, this 40% that's remaining, that's the observed rotation we're seeing. We're seeing it of that remaining 40%. So that's why we wouldn't expect a full positive 20 degree; we'd expect something much lower than that. We expect a positive 8 degree. So again, we're observing this portion, the portion that hasn't been canceled out by this 30% of R. Right? So, keep that in mind when dealing with the enantiomeric excesses. If you're dealing with a racemic mixture, there is no enantiomeric excess, and the observed rotation is 0. Once the percentages of the R and S enantiomers are not equal, there's going to be some type of observed rotation. And here, to determine that, it's important to be able to calculate your enantiomeric excess beforehand.

When 0.200 g of lactose is dissolved in 10.0 ml of water and placed in a sample cell 10.0 cm in length, the observed rotation is +2°. Calculate the specific rotation of lactose.

So the question reads like this. When 0.2 grams of lactose is dissolved in 10 milliliters of water and placed in a sample cell of 10 centimeters in length, the observed rotation is positive 2. Calculate the specific rotation. So in this case, what I would do is I would use the type of formula that puts the specific rotation on one side and gives me all of the other variables on the other. So all I'm going to have to do here is plug in variables. Now the only thing that is kind of tricky here is that the units for my concentration and my length are not the units that are given to me in the problem. In fact, in my problem, I'm not even told the concentration. I'm going to have to figure out the concentration based on what I know about concentration, how it has to do with a certain amount of mass and a certain amount of volume. Okay? So we're going to go ahead and figure out concentration first and pay attention to units the whole time. So the α, would just be positive 2 because this is my observed rotation. That one was easy. What's concentration equal to? Well, the units we're going to use are grams per milliliter and thankfully I've actually been given grams and milliliters, so this actually turned out to be pretty easy. I'm just going to put them over each other. 0.210 and now when I go ahead and I simplify this because I don't want to have a 10 at the bottom, what I'm going to wind up getting is that I have 0.02 grams per milliliter. That's my concentration. Okay? So I'm going to put that right here. 0.02 grams per milliliter. So that one is also pretty easy. Now let's go to this last one. Sample cell is 10 centimeters in length. Is length measured in centimeters? Not for this question. For polarimeters, we measure in decimeters. Okay? So we have to think, okay, we're going to have to use a conversion. So I would say I have 10 centimeters and I want to multiply it by something that has centimeters at the bottom and decimeters at the top. Right? Because I want the centimeters to cancel out. Alright? So I just have to figure out what's the relationship here and it turns out that you have 10 centimeters in every 1 decimeter. Alright. So I go ahead and do the math and wow, this just cancels out so beautifully. What I wind up getting is 1 decimeter. Okay. So I'm just going to put that down here and now we're ready to solve. Okay. So I'm just going to put these variables over. 20.02×1 And what is that equal to? So at this point, pull out your calculator and we're going to solve this together. So it could just be your iPhone if you want because this is a really easy problem. 20.02 and that's going to give us a specific rotation equal to positive 100. Okay? So the answer here is positive 100. Alright? So I hope you guys got that. Hope that makes sense. This is a type of question that I have seen asked before, so you should just be able to convert really simple units. It's usually not that big of a deal. Alright, so let's go ahead and move on to the next question.

Calculate the observed rotation of a chiral mixture that contains 65% (S)-stereoisomer where the [α] of pure (S)-stereoisomer = -118.

Alright. So for this one, we just have to look at the variables that we've been given and see how we can solve for the one that we need. So here it says calculate the observed rotation of a chiral mixture that contains 65% of the S stereo isomer, where the rotation of the pure S stereo isomer equals -118. Okay? So what are we given here? Well, what we're solving for, what we're trying to solve for is α. Okay. That's our variable. And what we've been given is that the specific rotation is equal to -118. We haven't been given enantiomeric excess, length of the tube, or concentration, so I know I'm not going to use that equation. The closest equation that I can think of that relates specific rotation and observed rotation is this equation right here, where it talks about how the specific rotation times the enantiomeric excess is going to be the observed rotation. In this case, instead of using the equation that has to do with cell length and concentration, I'm not given any of that, so that would just be a dead end. I should use this formula instead. So now I'm thinking, do I have an enantiomeric excess? Actually, no I don't, but I have been given one of the percentages of an enantiomer. So I could use this equation first and calculate what the enantiomeric excess is and then I would just plug that into my equation. So all I need here is to figure out what the enantiomeric excess is. Let's use the first equation. It says that the percent of the higher enantiomer minus the percent of the lower enantiomer is the enantiomeric excess. So first of all, I have 65% of one enantiomer. That means that I must have 35% of the other enantiomer, because 100 makes a whole. So what that's going to do is that's going to give me 35. So what that means is that I have 65% S and that must mean that I have 35% R. Does that make sense? Because I need them to equal 100 at the end. So now all I have to do is subtract these to get the enantiomeric excess. So 65 minus 35 equals 30. And this is going to be 30% of S. That's actually going to be important later. We have an enantiomeric excess of 30% S. And now all I have to do is plug that into my equation. My equation says that my observed rotation is equal to α = −118 0.3 . And what that's going to give me after I calculate it, so go ahead and type this in into your calculator, you get -118 times 0.3 and what I'm getting is that the rotation is going to be -35.4. Now I was saying earlier, why does it matter which enantiomer I had the excess of? Notice that I had excess S here and now I said that my observed rotation is going to be -35.4. That's absolutely correct. But what if I had excess R instead? What if instead of having 30% of S, I would have had 30% of R? Well then what that means is that I would have flipped the sign. It still would have been the same exact value. It still would have been 35.4, but instead, it would be positive. Why is that? Well, because think about it. Remember that by definition, opposite enantiomers have opposite rotations. So what that means is that if my S enantiomer has a specific rotation of -118, that means that my R isomer has to have a specific rotation of +118. So whichever one the enantiomeric excess is of, if I have excess S, then it's going to be a negative answer. If I have excess R, then it's going to be a positive answer. The number doesn't change, but the sign does. In this case, we got lucky because it happened to be S, my excess, and S was the one that I actually had the value for. So I didn't have to worry about changing the sign. But if it would have been R, then I would have changed the sign. Does that make sense? Cool. So let's go ahead and move on to the next.

An optically pure (R)-stereoisomer of a molecule has a specific rotation of – 20°. What specific rotation would be observed for a mixture of the (R) and (S) stereoisomer where there is an enantiomeric excess equal to (S) 60%.

Alright, so this question was actually even easier than the last one because if you'll notice, what we're given is the specific rotation and we're actually given the enantiomeric excess. We don't have to calculate it. Okay? All I need to figure out is the observed rotation. Okay? So this is actually just like I'm just going to plug in numbers. That's it. So remember that my equation was α, basically observed, equals specific rotation times enantiomeric excess. So what I'm going to do here is I'm going to say that it's going to be what's my excess? It's 60%, so that's 0.6. Okay. And what's my specific rotation? Well, it says here that my specific rotation is negative 20. Okay. You guys agree with that? But now check it out. My specific rotation is negative 20, but that's of the R stereoisomer. Okay? So if I have 100 percent R, then I would get negative 20. Now let's look at my enantiomeric excess. What type of enantiomeric excess do I have? Is it S or R? It turns out that it’s actually S. Okay? So is that specific rotation going to be fully accurate for this? No, because that would work. That would be my specific rotation if it had an excess of R. But since I have an excess of S, my specific rotation is actually the opposite. So my specific rotation for the S stereoisomer should actually be equal to positive 20. Does that make sense? Because the S was supposed to be different from the R. So which of these two values? Should I put positive 20 in my equation or negative 20 in my equation? And the answer is I should put positive 20 because I have an excess of S, so I should use the S specific rotation, not the R. So I'm going to put here instead of negative 20, I'm going to put positive 20. So now this just became 20 times 0.6. Okay? Equals α. Let's put this into our calculators. So we've got 20 times 0.6 and that is equal to positive 12. Okay? So positive 12 here was the right answer because I have an excess of the S stereoisomer, which is the positive one since the R is the negative one. Alright? So that was tricky. I know a lot of you guys thought it might have been negative 12, but that's if it would have said R here. But since it says S, it has to be the opposite. Cool? Alright, so hopefully that made sense to you guys. Let's go ahead and wrap this one up.

Here’s what students ask on this topic:

What is enantiomeric excess and how is it calculated?

Enantiomeric excess (ee) is a measure of the purity of a mixture of enantiomers in terms of its optical activity. It is calculated by taking the percentage of the major enantiomer and subtracting the percentage of the minor enantiomer. The formula is:

$\mathrm{ee}=\%\mathrm{major}-\%\mathrm{minor}$

This value represents the proportion of the mixture that contributes to optical activity, as the rest cancels out.

How does a racemic mixture affect optical activity?

A racemic mixture contains equal amounts of both enantiomers (50% S and 50% R). Because each enantiomer rotates plane-polarized light in opposite directions by the same magnitude, their effects cancel each other out. As a result, the optical activity of a racemic mixture is zero.

What is the relationship between specific rotation and observed rotation?

Specific rotation ([α]) is the rotation caused by a pure enantiomer at 100% concentration. Observed rotation (α) is the actual rotation measured for a mixture and is calculated using the formula:

$\alpha =\left[\alpha \right]\times \mathrm{ee}$

Here, [α] is the specific rotation, and ee is the enantiomeric excess. This equation accounts for the proportion of the mixture that is optically active.

What is the significance of a racemic mixture in organic chemistry?

A racemic mixture is significant in organic chemistry because it represents a 1:1 ratio of enantiomers, resulting in no net optical activity. This concept is crucial for understanding stereochemistry and the behavior of chiral molecules in reactions. Recognizing racemic mixtures helps in determining the purity and optical properties of compounds.

How do you determine the observed rotation of a mixture of enantiomers?

To determine the observed rotation (α) of a mixture of enantiomers, use the formula:

$\alpha =\left[\alpha \right]\times \mathrm{ee}$

Here, [α] is the specific rotation of the pure enantiomer, and ee is the enantiomeric excess. The enantiomeric excess is calculated by subtracting the percentage of the minor enantiomer from the major enantiomer. This formula gives the actual optical rotation observed for the mixture.