Optical Activity - Specific Rotation & Enantiomeric Excess - Stereochemistry Youtube - Free Educational videos for Students in K-12

Optical Activity - Specific Rotation & Enantiomeric Excess - Stereochemistry Youtube - By The Organic Chemistry Tutor

Transcript


00:00	in this video , we're going to talk about optical
00:02	activity . We're going to discuss whether molecule is optically
00:07	active or optically inactive . We're going to talk about
00:11	specific rotation and how to get it from observe rotation
00:16	. We're going to discuss terms such as optical purity
00:19	and financial market access . So let's begin our discussion
00:23	with financial matters . And the tremors have a Cairo
00:28	carbon . A Cairo carbon Is basically a carbon with
00:35	four different groups attached to it . The first group
00:40	that I drew was a method group . The second
00:43	group was hydrogen , the third was the carbonic acid
00:47	function group and the last was an alcohol . To
00:52	draw the indian summer of this molecule , we simply
00:55	need to draw it's a mirror image . So we're
01:01	going to draw its reflection across that vertical dash line
01:08	here . The O . H . Group is pointing
01:10	towards the left . Now it's going to be pointing
01:13	towards the right Now let's assign RNS configuration to each
01:23	carl center to do that . Using the con angle
01:27	prolonged process , we need to rank the priorities based
01:31	on atomic number , oxygen has the highest atomic number
01:35	. Next is carbon . Here we have two carbon
01:39	atoms . But this is an oxygen versus the hydrogen
01:42	. So the auction will win over the hydrogen .
01:45	This will be group number two , three and H
01:48	is always four . So counting from 1 to 2
01:51	to 3 . This is going to be the S
01:54	. I . Summer because we're rotating it in a
01:56	counterclockwise direction . Now , for the molecule on the
02:05	right , This is going to be group number one
02:08	number two three ages four . So if we counted
02:12	from 1 to 2 to three , ignoring four ,
02:14	we see that we're rotating it in the clockwise direction
02:17	. Given us the r is summer . So the
02:25	molecule on the left is called S . Lactic acid
02:33	. Where is the molecule ? Right . This is
02:35	called are lactic acid . These two instruments which are
02:44	mirror images of each other . They have different optical
02:51	properties , but they have the same physical properties .
02:55	So naturalists have same physical properties like boiling point melting
02:59	point density , solid ability . It's the same for
03:02	these two I summers . However , their optical properties
03:06	are different . The way they rotate plane polarized light
03:09	is not the same . It turns out that the
03:12	S . I . Summer rotates light towards the right
03:18	and the artist Summer rotates plane polarized light towards the
03:23	left . Now when it rotates light towards the right
03:28	, there is a positive sign that is associated with
03:31	that . When it rotates light towards the left ,
03:34	there's a negative sign associated with that . So this
03:38	is known as the D . I . Summer .
03:40	The D . Stands for dextre row to Torrey .
03:48	The prefix dextrose tells us that it's a rotating light
03:52	towards the right . The negative sign is associated with
03:58	the AL is summer , which is known as level
04:03	regulatory . Hopefully I said that right but it rotates
04:07	plane polarized light towards the left . Now the S
04:11	. I . Summer is not always associated with the
04:14	december . The S . Can have , it can
04:18	rotate like towards the right or I could do it
04:21	towards the left , so S . Is not always
04:23	associated with the positive dextre revelatory molecule . So sometimes
04:31	the system could be positive of the times it could
04:33	be negative depend on the molecule dealing with same thing
04:36	for art . The ri summer could be positive or
04:40	can rotate light towards the left . So you can't
04:45	really determine if it's positive or negative . Just by
04:48	looking at the configuration of the caramel center . So
04:52	just because you have the size and what doesn't mean
04:54	you have the the december , the allies make .
04:58	They're not associated the S and R Tells you the
05:02	configuration of the carl center , but the plus or
05:05	minus tells you how that particular I simmer rotates plane
05:08	polarized light . So S . Doesn't always rotate plane
05:12	polarized light to the right . For some molecules were
05:16	rotated towards the left , which it varies from molecule
05:19	to molecule . So I want to make sure you
05:21	get this because this is a typical test question .
05:25	So remember S . Can be positive or it could
05:30	be negative dependent on the molecule . Sometimes it will
05:33	rotate light towards the right and other times the S
05:36	. I . Smell will rotate light towards the left
05:38	. Same thing for art . For some are molecules
05:41	it will rotate light towards the right and for others
05:44	for other our eyes immerse it can rotate light toward
05:47	the left . So our doesn't always mean that it's
05:50	positive , nor doesn't always mean that it's negative .
05:53	It just it varies . They're not direct , there's
05:55	no direct association between our and plus or minus .
06:01	Now let's see if we have a light bulb ,
06:04	a light bulb through the process of incandescents emits light
06:10	in all directions . So we're gonna have light along
06:15	the vertical axis , light along the horizontal axis and
06:19	just going in all directions . This is normal light
06:23	. Now , if we pass through normal light ,
06:26	if we take normally and pass it through , let's
06:27	say like a polarizer , the polarizer is going to
06:31	act as a filter . Most of the light is
06:35	going to absorb , but it's going to allow a
06:39	certain type of light that is in the direction of
06:43	the polarizer to pass through . So in this case
06:47	only the vertical portion of light will pass through .
06:51	So this is known as plane polarized light . It's
06:54	been polarized by the polarizer . Now , what we're
06:58	gonna do is we're going to pass this through a
07:00	sample tube . Now , let's say this sample tube
07:12	contains an A Cairo molecule and kyra molecule does not
07:19	rotate plane polarized light , sorry . So as light
07:23	passes through the tube , the molecules in this sample
07:27	tube will have no effect on the plane polarized light
07:33	. So the light will emerge the sample tube without
07:37	any rotation . So there won't be no rotating this
07:42	plane polarized light . If you have a car molecules
07:44	in the sample tube , now , let's see what
07:48	happens if we put a Cairo molecule and a sample
07:54	to . So we're going to follow the same process
07:57	. We're going to take light traveling in all directions
08:01	. We're going to pass it through a polarizer .
08:07	We're going to obtain plane polarized light . We're going
08:10	to filter out all of the unwanted light traveling in
08:14	different directions , But the light that is in line
08:17	with the polarizer , that's going to come through now
08:20	, just like before we're going to pass it through
08:22	a sample tube this time inside the sample to we're
08:29	going to have Cairo molecules as opposed to a car
08:32	molecules . So if we have a car molecule ,
08:36	let's say like uh one form of the unanswerable ,
08:39	let's say the S . And asthma . This is
08:43	going to rotate the plane polarized light so as a
08:46	light passes through , it will begin to change direction
08:54	. And so when it emerged from the sample tube
08:57	, it's not going to be at the same angle
08:59	in which it was . Notice that a rotated bye
09:04	Some degree may be rotated 15° or 20°, , but
09:08	there was some rotation here . So we need to
09:11	know is that Cairo molecules , they are optically active
09:19	. They show optical activity . They can rotate plane
09:22	polarized light . A chara molecules are optically inactive .
09:30	They do not rotate plane polarized light . So how
09:36	can we determine if a molecule is optically active or
09:39	optically inactive ? Let's work on some examples . So
09:44	I'm going to draw a few molecules and I want
09:45	you to determine whether it's optically active or optically inactive
09:54	. Feel free to pause the video as you work
09:55	on these examples and I'm going to draw one more
10:32	molecule , so this one's gonna be trans instead of
10:40	cysts . So which of these molecules are optically active
10:49	and which ones are optically inactive . In order to
10:51	determine the answer , we just got to determine which
10:54	one is chiron , which one is a Cairo .
10:56	Let's start with this one . If you have a
10:58	molecule with no carl centers that is with No carbon
11:02	atoms with that don't have four groups . That molecule
11:06	is going to be a Cairo every carbon atom ,
11:10	there's no carbon atom in hexane that has four different
11:13	groups . So there are no carl centers , which
11:15	means the entire molecule as a car . And if
11:18	the molecules in Cairo , then this molecule is said
11:22	to be optically inactive , it will not rotate plane
11:25	polarized light . Now when you have a molecule that
11:30	have one carl center , like this one , that
11:33	carbon has four different groups , that molecule is going
11:36	to be Cairo overall . So if it's Cairo that
11:41	means that it's going to be optically active , this
11:45	will rotate plane polarized light . Now when you have
11:51	to carol centers like this one , you need to
11:56	look at symmetry . If there is a plane of
12:00	symmetry , what you have is a missile compound .
12:03	And so it's going to be a Cairo , which
12:06	means it's optically inactive . Here's another example here we
12:15	have a molecule with two carol centers and there's a
12:19	plane of symmetry . So this molecule is going to
12:21	be a Cairo and it shows no optical activity .
12:32	Now for this one here , we have to Cairo
12:34	centers , but we don't have a plane of symmetry
12:38	. The O . H . Is on the front
12:40	here , this one is on the back that's going
12:42	into the page . So this is a Cairo molecule
12:46	. And so this molecule will show optical activity .
12:50	It will be optically active . Now for the last
12:56	one , this is not a carl center . We
12:59	have a hydroxy group , there's a hydrogen , but
13:02	notice that the left side is the same as the
13:05	right side . So there are no carl centers for
13:07	that molecule . Therefore this is an A carbon molecule
13:11	, which means it's optically inactive . It will not
13:14	rotate plane polarized light . Now , let's talk about
13:17	some formulas that you need to be aware of and
13:22	the specific rotation is equal to the observe rotation represented
13:29	by the simple alpha divided by the path life times
13:34	the concentration so alpha without the brackets is the observed
13:40	rotation . And it's measured , it's an angle measured
13:44	in degrees . The same is true for the specific
13:49	rotation , which is in the bracket . See is
13:54	the concentration of the sample or the solution which is
13:57	typically ingram's from a leader . Sometimes it might be
14:01	represented in literature as g per 100 ml . Oh
14:08	, is the left of the sample tube ? So
14:11	earlier we drew the sample tube like this . So
14:15	L . Is a pathway for the life of the
14:17	sample tube . That's how far light has to travel
14:21	through the Cairo or the Cairo molecules . Now the
14:26	length of the sample tube is measured in deaths in
14:29	meters . So you want to make sure your unit
14:32	is in Destin meters . Now one decimate er Is
14:36	equal to 10 cm and one cm is 10 millimeters
14:43	. So those are some conversion factors that you're gonna
14:46	need to know when dealing with problems . They ask
14:49	you to calculate the specific rotation . Now the subscript
14:55	of the specific rotation is the wave left and the
14:58	wave life lambda is in units of nanometers . The
15:04	superscript is going to be the temperature in Celsius .
15:09	Now sometimes you may see the specific rotation with a
15:14	subscript of D . When you see that , what
15:18	that means is that this is the sodium d .
15:20	line with a wavelength of 589 nm . And this
15:25	particular specific rotation is calculated at a temperature of 25°C.
15:31	. So that's the formula that you need . If
15:33	you wish to calculate the specific rotation of a pure
15:36	ice , more or certain an instrument . Now the
15:41	next thing that we need to talk about is something
15:43	known as and natural Merrick access . So this occurs
15:51	when you have both . I smith . The city
15:54	are in the system and solution but they are unequal
15:58	one exists and the concentration that's more than the other
16:04	. A natural Merrick excess can be calculated by taking
16:09	the observed rotation of the solution more of the sample
16:13	and dividing it by the specific rotation of one of
16:18	the pure and natural gas . And then you would
16:23	multiply this by 100% . So if you have the
16:26	observe rotation of the solution and the specific rotation of
16:29	one of the enhancements , then you can calculate the
16:32	non traumatic access with those values . Another way in
16:36	which you can calculate the natural excess is if you
16:39	know the relative percentages of the R . N .
16:42	E . S . Items . For instance , Let's
16:45	say if a certain sample is 80 are And 20
16:49	s . The next American access is the difference between
16:54	the two . It's the absolute value of ar minus
16:57	S or s minus R . In this case it's
17:00	ar minus S because this is bigger So it's going
17:03	to be 80 minus 20% . So we get an
17:09	unnatural in excess of 60 in this example . So
17:16	that's how you can get it in the form of
17:17	a percentage . Now the natural american excess is equivalent
17:22	to the optical purity but in decimal form so the
17:26	optical purity It's gonna be 0.60 Which is equivalent to
17:32	60% . Now another formula that will help you to
17:36	calculate financial market access , particularly if you have a
17:40	sample where R and s . R . Listen ,
17:43	grams or moles as opposed to a percentage . You
17:47	could use this formula to calculate the benchmark excess ,
17:51	it's equal to the absolute value of ar minus s
17:54	or s minus art divided by the some of the
18:00	RDS I summer times 100% . So using these values
18:05	, Let's say if we have 16 g of art
18:08	four g of s . If we were to subtract
18:13	these two we will get 12 we won't get 60%
18:18	. Nevertheless 16 is 80 of the total of 20
18:24	. So as a percentage these are still the same
18:30	. But if we were to use this formula we
18:32	can still get 60% . So if we take 16
18:37	g of the ri summer , subtracted by four g
18:39	of the S . S . Summer And divided by
18:41	the total 16 plus four is 20 . We're gonna
18:47	get 16 -4 which is 12 , 12/20 Times 100%
18:54	. 12 , divided by 20 points , 60 times
18:57	100% . Which will give us 60% . So that's
19:00	another way in which we can calculate the national park
19:03	excess if we know the relative amounts of RNS and
19:07	units of let's say grams or moles . But if
19:10	we're given a percentage , we can simply just subtract
19:13	them directly . And that will give us DNA traumatic
19:16	excess as percentage as well . So now you know
19:21	how to calculate the specific rotation , the natural market
19:24	access and even the optical purity . So now let's
19:28	work on some practice problems . Number one .5 g
19:33	of a Cairo stereo . I summer was dissolved in
19:36	20 ml of solution . The observed rotation measured by
19:41	a polaroid meter Was found to be negative 3° at
19:45	25°C. . Using the wavelength of 589 nm . The
19:51	sample tube of the perimeter is 20 cm long .
19:55	What is the specific rotation of the stereo iceman ?
20:00	Well , let's begin by writing the former . So
20:03	the specific rotation , it's going to be equal to
20:06	the observer rotation divided by the path length times of
20:10	concentration . Let's write down what we know . We
20:14	know . The observe rotation is negative 3°. . Now
20:20	what about the path life or the length of the
20:23	sample tube ? The left of the sample tube is
20:27	20 cm . Now we need to convert that to
20:31	Destin meters . Remember one detonator is equal to 10
20:35	centimetres . So if we divide 20x10 we get to
20:41	so L is going to be too desi meters .
20:48	Now we are also given the mass and the volume
20:51	so we can calculate the concentration and units of grams
20:54	per milliliter . So if we take the mass of
20:57	the Cairo sterile ice marine grams , which is 0.5
21:00	g And divided by the volume of the solution ,
21:03	which is 20 . This is gonna be .5 divided
21:10	by 20 . Given us a concentration of point 0
21:15	to 5 grams familiar . So right now we have
21:20	everything that we need to calculate the specific rotation of
21:24	the stereo ice mark . So let's go ahead and
21:26	do that . So first let's replace the wave life
21:36	with 589 nanometers which corresponds to the wavelength of the
21:41	sodium D . Line . So we could just put
21:42	D . For that . The temperature Is 25°C. .
21:48	So that's gonna be the superscript . And then alpha
21:52	is negative 3° L . Is to desi meters .
21:58	And the concentration 0.0 to five g for me later
22:06	, two times .025 . That's going to be .05
22:11	, -3 , divided by .05 . That's negative 60
22:16	. So the specific rotation Of this stereo iceberg is
22:19	gonna be negative 60°. . And that's the answer for
22:22	this problem .

Summarizer

DESCRIPTION:

This organic chemistry video tutorial explains how to calculate the specific rotation of an enantiomer given the observed rotation, the pathlength, and the concentration of the solution. It discusses how to determine if a molecule will show optical activity based on its chirality. It also provides the formulas needed to calculate the optical purity and enantiomeric excess. Those formulas can be used to calculate the relative percentages of the R and S isomers.

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