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Course: Organic chemistry > Unit 4
Lesson 2: Enantiomers- Drawing enantiomers
- Cahn-Ingold-Prelog system for naming enantiomers
- R,S system
- R,S (Cahn-Ingold-Prelog) naming system example 2
- R,S system practice
- More R,S practice
- Fischer projection introduction
- Fischer projection practice
- Optical activity
- Optical activity calculations
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Optical activity calculations
How to calculate specific rotation and % enantiomeric excess.
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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)
At0:00, the problem tells you that the sign of the rotation from natural cholesterol is -.
The sign of the enantiomer must be +.
If a mixture of the two enantiomers is -. the natural cholesterol must be in excess.(31 votes)
is %ee always be the natutal compound?(6 votes)- 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)
Please correct me if I'm wrong , at6:51do 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)- 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)
- 6:22How do we know that the remaining 14% MUST be a racemic mixture?(5 votes)
Enantiomers are referred to as optically active because they interact with polarized light. One enantiomer will cause a rotation with a positive angle, which is known as dextrorotatory. The other will cause a rotation with a negative angle of the same magnitude, known as levorotatory. Since they rotate the polarized light with the same magnitude, if you put equal amounts together they cancel each other out and no rotation will be observed, a racemic mixture.
If it were enantiomerically pure cholesterol, we would expect it to interact with polarized light and produce a certain rotation of -31.5 at that temperature and wavelength of light. But with the enantiomerically impure sample we only observe a rotation of -27. This tell us two things, that the dextrorotatory enantiomer is present and that the levorotatory enantiomer is the dominant enantiomer. The enantiomeric excess tells us how much more of one enantiomer we have after equal amounts of the two enantiomers have canceled out each other's effect on the polarized light. Essentially we expect a larger rotation of the polarized light but we observe less than that because a minor amount of the enantiomers are cancelling each other out in a racemic mixture with is optically inactive. Hope that helps.(4 votes)
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)
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. :)(4 votes)
At6:22, why is it said that the remaining 14% MUST be a racemic mixture? Why isn't the remaining 14% the enantiomer?(2 votes)
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)
Are we just assuming in this case that the enantiomer in excess is nat. cholesterol? How do we know it's not 86% excess of the other enantiomer?(3 votes)- Enantiomers are either dextrorotatory or levorotatory, referring to how they rotate the polarized light and the sign of the angle. Dextrorotary refers to a clockwise rotation and a positive angle, while levorotary refers to a counterclockwise rotation and a negative angle. If we have a mixture, then we can know which one of the enantiomers is in excess based on the sign of the rotation, since the rotation is dominated by the one with a greater concentration.
For cholesterol, the natural enantiomer, known as nat-cholesterol, is levorotary with a negative angle. We know that from question #1 which has solely natural cholesterol and had a negative observed angle. The other enantiomer is known as ent-cholesterol and is dextrorotary and has a positive angle. Since the mixture produces a negative angle, nat-cholesterol is the excess enantiomer.
As a side note, technically there are 256 stereoisomers (2^(8)) because there are eight chiral carbons in cholesterol, but only nat-cholesterol and ent-cholesterol have an biological significance.
Hope that helps.(4 votes)
Why is the D line of sodium used? Why that wavelength specifically?(4 votes)- 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)
- 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)
- will using SI units yield the same result?(3 votes)
0:00Video 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.