Optical Activity: Specific vs. Observed Rotation

One of the special features of chiral molecules is that they are able to rotate plane-polarized light. Unfortunately, this means that now professors have an excuse to ask you math problems. Let’s see how this works. 

The Concept of Optical Activity

Concept: Specific rotation vs. observed rotation.  

6m
Video Transcript

Now we're going to talk about one of the only topics in organic chemistry that's going to require you to use a calculator. I know that's disappointing because at the beginning of the semester I told you guys there's barely any numbers in this course, but this is one exception. There is this one topic and it's called optical activity.
Optical activity is a special feature of chiral molecules. What it basically means is that chiral molecules, when light is passed through them, they're able to rotate plane-polarized light. The machine that measures this is called – let me just write this down here – a polarimeter. What I'm going to do here is I'm going to show you how light travels through a polarimeter and show you guys where the numbers come into play.
First of all, we have some light bulb. That light bulb is a source of light. That is the ugliest filament ever. I'm just going to erase that. This light bulb is a source of multi-directional light. What do I mean by that? What I mean is that it doesn't just shoot light in one direction, it's obviously, scattering light throughout a space.
Then we have a polarizer. A polarizer is just a type of lens. Just like your polarized glasses how it filters light and makes sure that it's only going in one direction. Basically, the light's going to pass through the polarizer and it's going to turn into what's called plane-polarized light. It's only going on one plane.
Then it passes through the actual functional part of the polarimeter and that is this tube right here. This tube is going to carry a chiral concentration. It's going to carry a chiral mixture. The interesting thing is that chiral mixture is going to be able to – one of the cool things about chiral molecules is that as light passes through them, it rotates the light so that it basically changes it's angle after it has passed the chiral molecule. This is just something that a scientist discovered a long time ago and it's still used today.
Now what's going to effect, what kind of equations come into play to determine what the angle is of the rotation? What's going to effect it is a few things.
First of all, every molecule has what's called a specific rotation. The specific rotation is just a random number that has to do with the amount of rotation that you would get if you had 100% of that enantiomer or 100% of that molecule present, what's the maximum rotation that you could get.
Just so you know specific rotation is truly a random number. It doesn't have to do with the chirality necessarily. It doesn't have to do with the size of the molecule. Nothing. There's no way to predict it. You will always be given the specific rotation or you'll be given the other variables to solve for the specific rotation, but you're not supposed to know it. That's all I'm saying. Just so you know, the specific rotation could be a positive rotation or a negative rotation. We're going to talk about that in a second as well.
Then the next thing is the concentration of my reagent. The more my specific rotation and the higher the concentration in the tube, obviously, the more it's going to turn. The last thing is the length of the tube. That just makes sense. The longer the tube is, the more time that light has to rotate as it's passing through.
All these things are going to come together to equal my observed rotation. The observed rotation is just going to be the product of these three things combined. It's going to be the specific rotation times the concentration times the length of the tube. Does that make sense? That's going to effect what I observe at the end. If I make my tube twice as long, I'm going to get twice the amount of rotation. Cool? Awesome.
It turns out that sometimes we're not always going to solve for observed rotation. A lot of times we're going to be solving for specific rotation, so instead, in the problem they're going to give us the observed, the concentration and the length and then we're going to have to solve for a specific in which case we would just flip the formula, use a little bit of algebra and it looks like that. That basically says that your specific rotation is your observed rotation over concentration times length. Easy stuff. It's just a little bit of solving – we're just basically taking on a variable.
Now let's talk about the actual rotations. A clockwise rotation is called dextrorotatory. It's symbolized using a positive symbol. That's what I was talking about how you can have a positive rotation or a negative. A counterclockwise rotation is known as level levorotatory and that has a negative symbol. These are just words that were given to these rotations a long time ago. Just remember that if you see dextrorotatory that's positive, levorotatory, that's negative.
These positive and negative names have – what's this blank say? Nothing to do with the chirality of the molecule. Nada. Zilch. What that means is that a lot of people get confused and they think that positive means that you have an R. They think that positive means you have an R chiral center or a negative means you have an S chiral center because they see the clockwise and they think that it's the same thing. They're completely different. A clockwise rotation of light has nothing to do with a clockwise chiral center. The clockwise thing just has to do with how we name the chiral center. It doesn't actually have to do with what the chiral center looks like.
What I'm trying to say is that let's say an R enantiomer for some molecules that could be a positive, for other molecules that could be a negative. The only thing that it tells you is that it is chiral. But it doesn't tell you what type of chirality you have. Does that make sense? Just have to really emphasize that point. 

  • Clockwise rotation = dextrorotary (d) or (+)
  • Counterclockwise rotation = levororatory (l) or (-)

These random names/signs have nothing to do with the chirality of a molecule!

Enantiomeric Excess

Concept: How to calculate enantiomeric excess.  

5m
Video Transcript

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.
Remember that I told you guys the specific rotation is the rotation that 100% enantiomers would produce. If I had a 100% enantiomer, let's say my specific rotation is 10 degrees, then I would get 10 degrees. It turns out that if I had the opposite enantiomer, meaning that this would be the R, but let's say I have the S enantiomer over here, then what would I get? I would just get the opposite rotation. Then that means that instead of getting a positive 10 rotation, maybe I'd get a negative 10, something like that. Basically, I'd do the same absolute value, just a different sign. That's the first thing that you guys need to know.
Second thing is that we can mix enantiomers. It doesn't always have to be a 100% pure solution. 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.
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. A non-one-to-one ratio, meaning that it's not pure but it's also not 50/50 is called scalemic. This one is not used quite as frequently, but it's still something that you should be aware of.
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.
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. I want to know if I took 100%, if I had 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.
Then, the R enantiomer – let's say that my S enantiomer is positive 20, but let's say that I have 100% of my R enantiomer in here. 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? Just intuitively the way you can think of it is I have 50% rotating at 20 degrees to the right, I have 50% rotating at 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 zero because they would cancel each other out. This is what we would call racemic. A racemic concentration is always going to have an optical activity of zero because it's going to perfectly cancel out. Does that make sense so far? Cool.
Now what I want to do is I want to talk about something called the enantiomeric excess. 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 two enantiomers. 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. That's the amount that's actually optically active.
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 molecule would theoretically produce at 100% times the enantiomeric excess, which is actually the optically active part because it's the part that isn't canceled out by anything else. 

Some more facts about optical activity:

  • Specific rotation [α] is the rotation that 100% pure enantiomers produce.
  • Opposite enantiomer = opposite rotation.
  • Racemic: A perfect 1:1 ratio of enantiomers
  • Scalemic: A non-1:1 ratio of enantiomers

Example: Calculate the ee and observed rotation for the following chiral mixture where S-enantiomer has [α] = +20.

1m

Example: Calculate the ee and observed rotation for the following chiral mixture where S-enantiomer has [α] = +20.

3m
Video Transcript

All right, I'm going to go ahead and take myself out of this video so that you guys can see the full image. First of all, we had to figure out the enantiomeric excess. The way we figure that out is by saying, 70 minus 30% equals what percentage? 40% ee. That means that my 33 is 40%.
So what's my equation? My equation says that my observed rotation should be equal to my specific rotation which is still positive 20, times my ee, which is now going to be 40%. Now we don't really want to work with percentages if we're multiplying, so we would turn this into 0.4, which is just I'm turning my percentage into a decimal, which is very common.
Now I would just multiply these together. What I'm' going to get is that 20 times 0.4 is going to equal positive 8. My enantiomeric excess here should have been 40% and my observed rotation, instead of being positive 20 which is what it would be if it was 100, it's going to be positive 8.
How does that intuitively make sense? Actually, it makes a lot of sense. If you think about it, remember what I said about racemic, racemic isn't optically active. If I were to look at my tube as broken up into parts, what I would see is that the 70% is actually broken up into two parts. There's 30% here S that is perfectly canceled out by 30% R. What that means is that this entire area right here is racemic.
The racemic portion adds up to – I'm just going to put here racemic. The racemic portion adds up to what? It adds up to 60% because I have 30 and 30. That means that the only part of this that's optically active – let's put that back. That means that the only part of this that's optically active is the extra remaining S or the enantiomeric excess and that part equals 40%. That means that even though my S would love to rotate this positive 20, there's only 40% of it available that can actually rotate it because the other 30% is being canceled out by 30% of R.
What that means is that only 40% is functional and that's why it's only 40% of the total rotation that's happening. Out of the total rotation that I could have, only 40% of that is actually taking place, so it's positive 8.

We can also use these equations to calculate specific and observed rotation when other variables are given to us. Just plug the numbers in! Let's try a few. 

Problem: 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.

5m

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

5m

Problem: 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%.

3m

Solving for Enantiomeric Percentages

Sometimes professors will ask us to solve for exact the percentage of each enantiomer in solution. For that we’ll need some new equations. 

Concept: How to solve for the percentage of each enantiomer.   

4m
Video Transcript

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. It has to do with enantiomeric percentages.
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 what are the percentages of each enantiomer. I know that sounds like you should be able to figure it out, and I'm sure a lot of you guys could, but it can get surprisingly tricky, so I want to go ahead and go over this now.
It turns out that there's 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. That's the same equation, I just rearranged it so that we're solving for e, for enantiomeric excess now. That makes sense. The observed over the specific is your enantiomeric excess.
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 to 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 enantiomers. The way that it works is that if you're given the enantiomeric excess, so let's say that my enantiomeric excess is 50% and you're asked to figure out how much each enantiomer represents in that, then you would use this equation. What this equation says is that you take your percent highest enantiomer, and that would be 100 minus the ee over 2 plus the ee, 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.
Let's go ahead and get started with this question. I'm going to go ahead and read it for you guys, give you guys some hints and then I'm going to have you guys solve it on your own.
It says that the specific rotation of pure S-epinephrine is positive 50. So I've been given the specific rotation that would be right here. I'm going to say it's positive 50. Calculate the ee, the enantiomeric excess, of the solution of the following observed value. The following observed value is going to be this guy right here. That means that I'm going to put here for my observed, what I'm actually getting out of it, even though I have a 50% theoretical, I'm only getting a 25%. That 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 was 50 and 50, then I would want you to draw that. These numbers are obviously wrong. I just want you to draw something like that so that you guys can kind of understand why we're only getting half of the efficacy from this 50. Instead of rotating it 50, it's only 25, why is that?
Go ahead and try to just work with the equations above. See if you can get the right answer. Fill in those three blanks below, right here. Then I'll go ahead and solve it for you guys. 

Example: The [α] of pure S-epinephrine is +50°. Calculate the ee of a solution with an observed value of +25°. Calculate percent of each enantiomer. Then sketch the approximate mixture in our sample polarimeter tube.

4m

Problem: 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.

6m

Optical Activity: Specific vs. Observed Rotation Additional Practice Problems

Given that (S)-2-butanol has a specific rotation of +13.52 and (R)-2-butanol has a specific rotation of -13.52o, what is % composition of a mixture whose specific rotation was found to be +6.76o?

 

A) 75%(R) 25%(S)

B) 25%(R) 75%(S)

C) 50%(R) 50%(S)

D) 67%(R) 33%(S)

E) 33%(R) 67%(S)

Watch Solution

One of the following isn't true about a racemic mixture.

a) It is optically inactive

b) It has equal amounts of enantiomers

c) It doesn't rotate plane polarized light

d) Its ee is 50%

e) all are true

Watch Solution

The following statements describe some important properties of molecules. Choose from among the following three types of isomers and in the space provided write the letter of the one or more types of isomers for which the statemet is TRUE.

A. A single enantiomer          B. A meso compound          C. A racemic mixture

A solution of this will rotate the plane of plane polarized light.   _________________

A solution of this will NOT rotate the plane of plane polarized light.   _____________________

Might be produced when an alkene with no chiral centers reacts with a non-chiral reagent such as H-Br or Br2.  ___________________

The LEAST desirable option when developing a new drug  ___________________

Watch Solution

All S enantiomers rotate the plane of polarized light anti-clockwise ( True or False 

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When 0.300 g of glucose is dissolvd in 10.0 mL of water and placed in a sample cell 10.0 cm n length, the observed rotation is +3.1°C. Calculate the specific rotation of glucose.     

 

Watch Solution

Which of the following molecules is optically active?

Watch Solution

A forensic chemist (chemist who deals with crime-related investigations) analyzed a sample of a poisonous alkaloid (generic term for a naturally occurring nitrogen-containing molecule) and determined that its specific rotation [α] at 25 °C using sodium D light was + 12 °. What can the chemist properly conclude? 

a) there is one enantiomer in the sample

b) the enantiomer present has the R configuration

c) the sample is optically pure

d) the butler did it

e) none of the above

Watch Solution

Pure (S)-2-butanol has a specific rotation of +13.5 degrees. You have made and purified a sample that has a calculated specific rotation of -6.76 degrees. What can you conclude about this sample?

 

 

 

 

 

 

Watch Solution

Which of the following is true about any  R enantiomer?

 

(1) It has a (+) rotation in a polarimeter.

(2) It has a (-) rotation in a polarimeter.

(3) It is a mirror image of the S enantiomer. 

(4) When mixed with the S enantiomer, it forms a meso compound.

Watch Solution