Optical Activity - Specific Rotation & Enantiomeric Excess - Stereochemistry Youtube - By The Organic Chemistry Tutor
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 . |
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.
OVERVIEW:
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