Optical activity calculations

- [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.