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Optical activity calculations

How to calculate specific rotation and % enantiomeric excess.

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  • piceratops tree style avatar for user John John
    How do we know that the 86% figure calculated refers to natural cholesterol and not its enantiomer? I guess what I'm asking is why do we assume cholesterol is in the higher quantity among the two entantiomers?
    (15 votes)
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  • blobby green style avatar for user Siew Woon Tan
    is %ee always be the natutal compound?
    (6 votes)
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    • piceratops tree style avatar for user Tombentom
      Think of this:
      We have : x the dominant Enan, Y the other enan
      Excess = Optical PURITY
      100 = x + y (1)
      Excess = x - y (2)
      <-----------------------------> (1) - (2): 100 - Excess = 2y => y = (100 - excess) / 2
      Hope this help!
      (8 votes)
  • orange juice squid orange style avatar for user LaFontaine
    Please correct me if I'm wrong , at do we take 14% as the value of racemic mixture because we consider that some amount of the enantiomer of cholesterol would have formed a racemic mixture with a conc. of natural cholesterol equal to its own concentration and this part would have had zero optical activity?
    thanx
    (4 votes)
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    • spunky sam blue style avatar for user Ernest Zinck
      That's correct. You have 86 % natural cholesterol and 14 % racemic mixture.
      The racemic mixture is 7 % (+)-cholesterol and 7 % (-)-cholesterol. It will have zero optically activity, so all of the observed rotation comes from the 86 % natural cholesterol.
      The natural cholesterol then accounts for 86 % + 7 % = 93 % of the mixture.
      (7 votes)
  • aqualine tree style avatar for user meenakshig097
    Jay says that %ee = [observed[a]] / [[a] of pure enantiomer] * 100.
    So my question is, what is the difference between the numerator & denominator? Jay said that the specific rotation was a constant in the last video...so how could there be "observed specific rotation" and "specific rotation of a pure enantiomer?" And between those two and "normal" observed rotation? (not specific).
    (5 votes)
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    • piceratops ultimate style avatar for user Darmon
      The specific rotation of a particular enantiomer is a constant. Here, "observed specific rotation" is referring to the specific rotation induced by a mixture of two enantiomers, and it depends on the percentage composition of each enantiomer in the mixture. Nonspecific observed rotation is the rotation measured by the polarimeter; the conditions of the experiment (sample concentration, temperature, light wavelength, and container length) are then factored in to yield the specific rotation of the sample. :)
      (3 votes)
  • leaf green style avatar for user Onwudiwe Natachi
    At , why is it said that the remaining 14% MUST be a racemic mixture? Why isn't the remaining 14% the enantiomer?
    (2 votes)
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    • piceratops seed style avatar for user RogerP
      He's calculated that the enantiomeric excess is 86% - this is the percentage that is responsible for giving the observed specific rotation of -27. It therefore follows that the remaining 14% is not responsible for rotating plane polarised light. Therefore, this 14% must be the racemic mixture of the two enantiomers.
      (6 votes)
  • male robot hal style avatar for user Yash Lal
    Why is the D line of sodium used? Why that wavelength specifically?
    (4 votes)
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  • leaf green style avatar for user ZT Guo
    Why would we convert the length of the polarimeter tube 10.0cm into 1dm? How is this consistent with the rest of the units?
    (2 votes)
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    • spunky sam blue style avatar for user Ernest Zinck
      The units certainly aren't consistent with SI. The units of measurement are:
      α = °
      c = g/mL
      l = dm
      [α] = α/(cl), so the units are °·mL·g⁻¹dm⁻¹, but this is usually shortened to °, with the rest of the unit being understood.
      So, you must express the length of the tube in decimetres if you want to get the same values as in the literature.
      (4 votes)
  • leaf green style avatar for user Soumil Sahu
    will using SI units yield the same result?
    (3 votes)
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  • aqualine ultimate style avatar for user jae choi
    I still don't get why that leftover 14% from the last problem is racemic mixture. There was literally nothing that indicates that whatever is left from 14% is racemic mixture
    (3 votes)
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  • leaf green style avatar for user cbmaughan1
    At , why is it that optical purity is equal to the percentage of one enantiomer minus the percentage of the other enantiomer? Where does the basis for that equation come from?
    (1 vote)
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    • piceratops seed style avatar for user RogerP
      It's logical. Take the last of his three examples where you have a solution containing 75% of one enantiomer and 25% of the other enantiomer. Let's call these enantiomers A and B and imagine there are 75 molecules of A and 25 molecules of B in the solution.

      The optical rotation caused by the 25 molecules of B will be cancelled out by the opposite optical rotation of 25 molecules of A - in other words, the 25 molecules of B and 25 of the 75 molecules of A form a racemic mixture.

      This leaves 50 molecules of A whose optical rotation is not cancelled by B molecules. These 50 are reason for the observed optical activity. Hence, the optical purity is defined as 50% which, you will notice, is also the enantiomeric excess (= 75 minus 25).
      (3 votes)

Video transcript

- [Narrator] Let's do some calculations using optical activity. So for our first problem, let's say we have .300 grams of natural cholesterol. So here's the dot structure for natural cholesterol, it's an optically active compound, and we dissolve our cholesterol in 15.0 milliliters of chloroform. And we put that solution in a 10.0 centimeter polarimeter tube, the observed rotation at 20 degrees C, using the D line of sodium, it turns out to be negative .630 degrees. And our goal is to calculate the specific rotation of cholesterol. We saw how to do this in the last video. The specific rotation is equal to the observed rotation, divided by the concentration times the path length. So let's plug in some numbers, here. The specific rotation is equal to the observed rotation, which is negative .630 degrees, so we put that in. Negative .630 degrees. We divide by the concentration, which is in grams per mL. So that's .300 grams, divided by 15.0 mLs. So .300 grams divided by 15.0 mLs. We multiply that by the path length, and the path length needs to be in decimeters. So we have a 10.0 centimeter tube, 10.0 centimeters is 1 decimeter, so that makes our math easy, here. So this would be 1.00 decimeter. All right, let's do the math. So let's get out the calculator, and let's solve for the specific rotation. That would be negative .630 divided by, we have .300 divided by 15.0. And then we multiply that by one. I don't really need to do that, but I'll go ahead and do it anyway. So that's multiplied by 1.00, here. And we get negative 31.5. So that is our specific rotation. So let's write that down, here. So we have our specific rotation at 20 degrees C, so we put a 20 here, using the D line of sodium, so we put a D here, and this is equal to negative 31.5. Now, sometimes you see this with a degrees sign, so sometimes you'll see it written like that, but I'm going to take that out, because normally, we don't have any units for our specific rotation. So it just depends on what book you're looking in. For our next problem, problem two, let's talk about percent enantiomeric excess, or optical purity. This is where you take the percentage of one enantiomer, and from that you subtract the percentage of the other enantiomer. So for part A, let's calculate the percent enantiomeric excess for a solution that contains a single enantiomer. So if we have only one enantiomer, this is like the first problem that we did, with natural cholesterol. That means you have 100% of this enantiomer, and obviously 0% of the other one. So the percent enantiomeric excess would just be 100 minus zero, or 100%. So we have 100% optical purity, so this is an optically pure solution. For part B, let's do this for a solution that contains equal amounts of both enantiomers. So when that happens, it's called a racemic mixture. So if we have equal amounts of both, that must mean we have 50% of one enantiomer, and 50% of the other. So the percent enantiomeric excess would be equal to 50 minus 50, which of course is equal to zero. So this has an optical purity of 0%, and a racemic mixture is not optically active. You get a net rotation of zero if you have equal amounts of both enantiomers. For part C, we have a solution that contains 75% of one enantiomer, and 25% of the other. So the percent enantiomeric excess is equal to, this would be 75% minus 25%, which of course is equal to 50%. So we have 50% excess of this enantiomer, and we have a 50% optically pure solution. For our last problem, we have a mixture of natural cholesterol and its enantiomer. And our mixture has a specific rotation of negative 27. Our goal is to calculate the percent enantiomeric excess of this mixture, and we can do that using this equation up here. So the percentage enantiomeric excess is equal to the observed specific rotation, divided by the specific rotation of the pure enantiomer. And to get a percentage, we multiply it by 100. So the percent enantiomeric excess is equal to the observed specific rotation, which is negative 27, so we write that in here. So negative 27. We divide that by the specific rotation of the pure enantiomer. And for natural cholesterol, we saw what the specific rotation of the pure enantiomer was in the first problem. We got negative 31.5. So I'll write in here, negative 31.5. And we multiply it by 100. So that gives us our percent enantiomeric excess. So let's get out the calculator, here. We don't need to worry about negative signs, so we can just take 27 and divide that by 31.5, and multiply it by 100, and we get 85.7. And let's round that to 86%. So our percent enantiomeric excess is 86%. So we're done with our calculation, here. Our next question is, what percentage of the mixture is natural cholesterol? Well, 86%, this was our enantiomeric excess. So if we think about this as being 86% of natural cholesterol, so let me write this down, here. 86% of natural cholesterol. And the remaining 14% must be a racemic mixture. So if the remaining 14% is a racemic mixture, that means half of it is natural cholesterol, and half of it is the enantiomer. So that means that 7% is our natural cholesterol, and 7% is the enantiomer. So seven plus seven is, of course, equal to 14. So what's the total percentage of natural cholesterol in our mixture? That would be 86 plus seven, which of course is 93%. So that's our answer. So 93% of our mixture is natural cholesterol. This can get a little bit confusing sometimes, so you can check this answer to make sure it's correct. You know that the total of natural cholesterol and its enantiomer should be 100%, so if natural cholesterol is 93%, and its enantiomer is 7%, obviously 93% plus 7% is 100%. Also, we know from the previous problem that the percentage enantiomeric excess is equal to the percent of one enantiomer minus the percent of the other enantiomer. So we can say that the percent enantiomeric excess is equal to 93% minus 7%. And 93 minus 7 is 86%, which is what we got in our calculation down here. So that's just a nice little check to make sure you did the problem correctly.