Understanding Stereochemistry and Configurations

In this video, we're going to talk about stereochemistry. We're going to talk about how to assign R-S configuration to chiral centers, including Fischer projections, and a lot of other examples as well. So let's start with this one. Assign a R or S configuration to each chiral center shown below. So first, what is a chiral center? A chiral center, or a chiral carbon, is a carbon atom that has four different groups attached to it. In this case... These are the four different groups. Now in order to assign that chiral center a RS configuration, we need to rank those four groups using the con ingall pre-log process. Group number one is going to have the highest priority. And the way we determine priority is based on atomic number. So looking at these four atoms, bromine, chlorine, carbon, and hydrogen, which of these four atoms have the highest atomic number based on the periodic table? So if you're using the periodic table, you'll find that bromine has the highest atomic number. So this is going to be group number one. It has the highest priority. Chlorine has the second highest atomic number relative to bromine. So that's going to be group number two. Carbon has an atomic number of 12. Hydrogen has the lowest atomic number of all of the elements in the periodic table. It has an atomic number of one. So it's always going to be group number four. Now, before we can count it, we need to make sure group number 4 is in the back, or it's in the hatch wedge, which it is in this case. So we can count it from 1 to 2 to 3. So notice that we're rotating in the counterclockwise direction. Whenever that happens, you need to assign an S configuration if group number 4 is in the back. So that's the configuration for this particular carocenter. Now let's focus on the chiral center on the right. And if you want to pause the video to try that example, feel free to do so. This bromine atom, we know it's going to be number 1. Chlorine is going to be 2. The methyl group will be group number 3, and H is 4. So counting it from 1 to 2 to 3, this time we are rotating in the clockwise direction. So this is going to have the R configuration. Now looking at these two molecules, what is the relationship between them? Would you describe them as constitutional isomers, enantiomers, diastereomers, mesocompounds? How would you describe these two compounds? Now notice that these two molecules are mirror images of each other. Therefore, they are known as enantiomers. Enantiomers are a special type of stereoisomer. A stereoisomer, like any other isomer, have the same chemical formula, they're connected the same way, but notice that the way the atoms are arranged in space, it's arranged in space differently. This carbon and that carbon, they're connected to the same four groups. But the way those four groups are arranged in space, they're not the same. Here, chlorine is on the left side. Here, it's on the right side. So they are arranged in space differently, which makes it a certain type of stereoisomer. Diesteromers, enantiomers, cis-trans geometric isomers, those are all stereoisomers. But enantiomers, they're mirror images of each other, and that's how you can distinguish them from diastereomers. Another feature of enantiomers is that they have the opposite configuration at the chiral center. Here, the configuration, we determine it to be S. Here, it's R. Now, let's work on some more examples for the sake of practice. So go ahead and assign the R and S configuration to each chiral center that I'm going to draw. So let's put an ethyl group here. Let's add a hydrogen, a bromine atom, and let's put a hydroxyl group. So go ahead and determine the configuration at this carbon. And so while you do that, I'm going to write up another one. So let's start with the first one. The first thing we need to do is determine which group has the highest priority. So we know it's going to be bromine. Next, if we compare oxygen to carbon and hydrogen, we know oxygen is going to have a higher atomic number. number for oxygen is 8, for carbon is 6. So this is going to be number 2, 3, and then hydrogen is always 4. Now hydrogen is in the back, which is good. So we're going to just go from 1 to 2 to 3. And so for this one, we... we have the R configuration. We're rotating clockwise, or in the direction of a clock. Now for the next one, hydrogen's gonna be group number four. Here we have oxygen, carbon, carbon. Oxygen wins, so that's gonna be number one. Next, notice that these two carbons, they have the same atomic number. So if we move on to the next group, here we have carbon versus hydrogen, carbon wins. So the ethyl group has a higher priority than the methyl group. So if we count it from 1 to 2 to 3, notice that we're rotating in a clockwise direction. But group number 4, it's not in the back. It's in the front. It's not on the dash. This time it's on the solid wedge. So because it's on the solid wedge or it's in the front, coming out of the paper, we need to rotate it. So just switch the configuration and it's equivalent to putting it back to the to putting group number four in the back. So it was R but now it's S. So that's the configuration for this chiral center. So anytime H is in the front, simply count it from one two to three and then reverse your answer. Number two, name the following compounds using the RS system for stereoisomers. So first, let's identify the chiral center. So let's focus on this carbon. That carbon has a bromine, a hydrogen, a methyl group, and an ethyl group. So it has four different groups, which makes it a chiral center. Now let's rank those groups. Bromine has the highest atomic number, that's going to be number 1. We know ethyl has a higher priority than methyl. So this will be 2, that's going to be 3. H is going to be 4. So going from 1 to 2 to 3, we can see that it's going in the R direction. And when you count from 1 to 2 to 3, ignore 4. Now as you count it... We're going to focus on 4. 4 is in the front, it's not in the back, which means we need to reverse it. So it's going counterclockwise, which means the configuration is S. So let's put the S configuration here. Now, in order to name it, we need to count the carbons in the parent chain. So we're going to count it in this direction. So we have a 4-carbon parent chain. We're going to be dealing with butane. So first we need to write the configuration of the chiral center, which is S. We're going to close that in parentheses. And then it's going to be, we have a bromine on carbon 2, so 2-bromo. And for 4-carbon alkane, that's going to be butane. So it's S, 2-bromo, butane. That's how we can name this particular stereoisomer. Now let's move on to the next one. So here we have two chiral centers, and you could check it. Here we have one group, two, three. Relative to this carbon, this is the fourth group. If we focus on the second chiral center, this is one group, two, the ethyl is three, and that's the fourth group. So let's focus on the first chiral center, this one. So we're comparing a chlorine atom, hydrogen, carbon, carbon. So if we look at the first atom that's connected to that caryl center, we see that chlorine has the highest atomic number. We don't look at bromine because it's not the first atom attached to this caryl center. So we look at it from a one-to-one basis. So this chlorine has a higher atomic number than that carbon. So this is going to be group number one. Now, for group number two, we know H is number four, so we got to compare these two. Here we have a carbon, and then a carbon, and then after that... Carbon is a hydrogen. Here, there's a bromine. This is going to win. So this entire group is group number three, which makes the methyl group... Actually, take that back. This is two. This is number three. This has a higher priority than the methyl group. So now we can count it from 1 to 2 to 3, and so that's going to be the R configuration. So let's put R here, and now let's focus on the next chiral center. So bromine, hydrogen, carbon, carbon. Bromine is going to have the highest atomic number, so that's 1. H will be 4. So now let's compare these two. So those two carbon atoms are the same. Next we have a chlorine versus a carbon. The chlorine is going to win. So this whole group will be group number 2. And the ethyl group is going to be number 3. So going from 1 to 2 to 3, we're going in the R direction, but H is in the front. We're going to have to reverse it, making it S. So this is S. Now that we've assigned the configuration to each chiral center, we can name this theory isomer. So we need to number it from left to right. So the substituent will be on 2, 3, as opposed to 3, 4. Now... On carbon 2, we have the R configuration, so this is going to be 2R. And on carbon 3, we have the S configuration, so this is going to be 3S. Now we need a comma between them. And then we need to put the bromo and the chloro group in alphabetical order. So B comes before C. This is going to be 3 bromo. It's on carbon 3. Chlorine's on carbon 2, so we're going to have 2 chloro. And for a 5-carbon alkane, that's going to be pentane. So this is 2R3S3-bromo-2-chloropentane. So that's how we can name this particular stereoisomer. Now here's a question for you. How many stereoisomers are possible if you have one chiral center? And how many are possible if you have two chiral centers? So if you have one chiral center, you can have the S isomer or you can have the R isomer. So there's two possibilities. The equation that will give you the number of stereoisomers is 2 to the n, where n is the number of chiral centers. So in this case, n is 1. 2 to the first power is 2. So for one chiral center, there are two possible stereoisomers. Now what if we have two chiral centers? How many stereoisomers are possible? Well, it's going to be n is 2, so it's going to be 2 squared. 2 squared is 2 times 2, which is 4. So we have four possibilities. The first possibility is if both chiral centers are R. The second one is if one is R, I mean if the first one is R and the second one is S, or if the first one is S, the second one is R, or if they're both S. So those are the four possibilities, or the four stereoisomers that are possible, if you have two chiral centers. So the number of stereoisomers is going to be 2 to the n chiral centers. Number three, the chemical structure of cholesterol is shown below. How many stereoisomers are possible for this molecule? So feel free to take a minute and try this problem. So this right here is a chiral center. We have an OH group. There is an invisible hydrogen, but now it's visible. And then this side is different than that side. This side is closer to the double bond. This other side has a methyl group. So, these two sides are different, which makes this a chiral center. Now, this carbon is also a chiral center. It has four different groups. Here we have a methyl group. Here is a secondary carbon, a tertiary carbon, and this tertiary carbon. This tertiary carbon is close to the double bond. This one is not, so they're different. Which makes this a chiral center? Here is another chiral center. This is one group, this is another, this one's another, and there is a hydrogen attached to it. And we could put that hydrogen here. This carbon here is not chiral. Whenever you see a carbon with two bonds and that's it, it's a CH2 carbon, which means it doesn't have four different groups because these are the same. So that is not a chiral carbon. So it's best to focus on the tertiary carbons and the quaternary carbons. This is a tertiary carbon. It's attached to three other carbons, and there's a hydrogen attached to it. So that is another chiral center. Here's another tertiary carbon. It has three other carbons, all of which are different, and there's a hydrogen here, too. Now, this carbon here is quaternary. It's attached to four different carbon atoms. So that's another chiral carbon. This tertiary carbon is chiral, too. There's a hydrogen here. And this carbon is tertiary, and it's chiral. This is group 1, this is another group, this is another group, and then the hydrogen is the fourth group. So let's put that hydrogen on the hatch wedge. Now this carbon here is tertiary, but it's not chiral, because it doesn't have four different groups. These two methyl groups are the same. Thus, cholesterol has a total of 1, 2, 3, 4, 5, 6, 7, 8 chiral centers. So because cholesterol has 8 chiral centers, it's going to have... 2 to the 8 stereoisomers. 2 to the 8 is 2 to the 4 times 2 to the 4. 4 plus 4 is 8. If you multiply 2 four times, 2 times 2 is 4, times 2 is 8, times 2 is 16, you get 16 times 16, which is 256. So this is the number of stereoisomers that are possible for this particular molecule. Number four, determine the absolute configuration of each chiral center and name each Fischer projection shown below. When dealing with Fischer projections, it's easy to determine the chiral center. So we have two chiral centers, which means that we have four possible stereoisomers for this particular structure. Now before we determine the absolute configuration of each chiral center, let's talk about Fischer projections. So let's say if we have this particular Fischer projection. Let's make this an ethyl group and let's say this is hydrogen, this is OH. What we need to realize is that the groups on the horizontal part of the Fischer projection, they're in the front. So you can redraw the structure like this if you want. They're coming out of the page, and so they're on the solid wedge. These two groups, they're on the dash or the hatch wedge. So they're going into the page. you can view them as being in the back. So make sure you understand that. Now hydrogen is typically on the side which means that as group number four you're going to have to reverse it when determining the configuration so keep that in mind. Anytime group number four is in the front you need to reverse it. So let's begin by determining the configuration of this Carol Center. The hydroxyl group is going to be group number one. Oxygen beats these two carbon atoms. Now, comparing those two carbon atoms, they're the same, but here we have a bromine, here is a hydrogen, so this group has a higher priority than the methyl group. So this will be group number two, the methyl will be three, and then hydrogen is always four, if it's there. So going from one to two to three, we're going in a clockwise direction, so that's R. But H, remember H, when it's in a horizontal part, it's like being on a solid wedge. H is in the front, so we got to reverse it. So we're going to get S. So we have the S configuration. for that chiral center. Now for the next one, so we're going to focus on this chiral center. Bromine is going to be group number one. It has the highest priority. Next we're comparing the methyl group with this group. So we have carbon to carbon and then hydrogen to oxygen. Oxygen wins. So this entire group is going to be number 2. Method was 3. H is 4. So going from 1 to 2 to 3, it looks like we're going in the S direction. But if we reverse it, it's going to be R. So we have the R configuration here. So now that we've assigned the absolute configuration to each cryocenter, we can go ahead and name this particular molecule. So the hydroxyl group is going to have priority over the bromine group, so we're going to count it in a way that we give the hydroxyl group the lower number. We want to give it a 2 instead of a 3, so we're going to count it in this direction. So on carbon 2, we have the S configuration. So this is going to be 2S. And on carbon 3, we have the R configuration, so 3R. And then... We have a bromine on carbon 3, so this is going to be 3-bromo, and the alcohol is going to be part of the parent name. And it's on carbon 2, so it's going to be 2-butanol, because we have a 4-carbon chain. So 2s3r3-bromo-2-butanol. That's how we can name this particular fissure projection. Now let's try another example. Go ahead and name this fissure projection and determine the configuration at each call center just like we did before. So let's start with this one. Bromine is going to be group number one. This group with the chlorine atom, we know that's going to be two. Methyl is three, H is four. So going from one to two to three, ignoring four, this is going in the counterclockwise direction, so that's S. But H is in the front, so we need to reverse it. So we're going to get R for the first one. Now for the second chiral center. Chlorine is going to be group number one. This entire group is number two. If we compare carbon to carbon, it's a tie. And then carbon to bromine, bromine wins. So this is number two. Ethyl is number three. H is four. So going from one to two to three, skipping four, it appears to be R. But H is in the front, so we reverse it and we get the S configuration. So now that we have the configuration at both chiral centers, we can name it. So we don't want to count in this direction, because this would be 3 and that would be 4. Rather, we want to count in this direction. So this would be 2 and this would be 3. So we have a 5-carbon chain, so we're dealing with pentane. But first, let's focus on the configuration. So the configuration is R at carbon 2, so we're going to have 2R. And then it's S at carbon 3, so 3S. And then we have a Br on carbon 2. And we need to put it in alphabetical order. B comes before C. So this is going to be 2 bromo. And then we have a Cl on carbon 3. So 3 chloro. And then... the five carbon chain that's going to be pentane so that's the nomenclature or the IUPAC nomenclature for this particular fissure projection it's 2r3s 2-bromo-3-chloropentane now let's work on a more challenging problem so let's say this Carol Center has an ethyl group a methyl group in the front and let's put a chlorine group in the back and the hydrogen go ahead and assign the configuration to the chiral carbon is it RS now this problem is different because Hydrogen, group number four, is neither in the front nor is it in the back. So how do you assign the configuration in this situation? Well, let's begin. We know that chlorine has the highest atomic number, so that's going to be group number one. Ethyl has more priority than methyl, so ethyl is two, methyl is three, H is four. In a situation like this, there's a technique that you could use to assign the configuration. Whatever group is in the back, put it in a circle. So I'm going to put one in the circle. And you could put a small subscript b to indicate that whatever in the circle is in the back. Sometimes it could be in the front. But for now, whatever's in the circle in this problem is in the back. Now, the other numbers, 2, 4, and 3, you want to arrange them in the form of a triangle. Now, if you notice, 2 is at the top, so let's put that at the top. 4 is at the bottom left with respect to 2, so let's put that here. 3 is on the bottom right. Now what we're going to do is we're going to rotate this molecule in such a way that number 4 will be at the top where 2 is. So we're going to rotate it 120 degrees clockwise. So 1 is still in a circle. 4 has replaced 2. 2 is now where 3 was. And 3 is where 4 was. Now the next thing we're going to do is we're going to flip the molecule. As we flip it, 2 and 3 will move to the top. Group number 4 is going to expel group number 1. So now 4 is in the circle, 1 is beneath it, and then 3 and 2 are at the top. Now 4, group number 4 is in the back, because what's in the circle represents what's in the back. And so we can count it from 1 to 2 to 3, and this gives us the S isomer. So the configuration at this Carroll Center is S. So that's the technique that you could use whenever group number 4 is not in the front or in the back.

Assign a R or S configuration to each chiral center shown below. So first, what is a chiral center? A chiral center, or a chiral carbon, is a carbon atom that has four different groups attached to it.

In this case... These are the four different groups. Now in order to assign that chiral center a RS configuration, we need to rank those four groups using the con ingall pre-log process.

Group number one is going to have the highest priority. And the way we determine priority is based on atomic number. So looking at these four atoms, bromine, chlorine, carbon, and hydrogen, which of these four atoms have the highest atomic number based on the periodic table? So if you're using the periodic table, you'll find that bromine has the highest atomic number. So this is going to be group number one.

It has the highest priority. Chlorine has the second highest atomic number relative to bromine. So that's going to be group number two.

Carbon has an atomic number of 12. Hydrogen has the lowest atomic number of all of the elements in the periodic table. It has an atomic number of one. So it's always going to be group number four.

Now, before we can count it, we need to make sure group number 4 is in the back, or it's in the hatch wedge, which it is in this case. So we can count it from 1 to 2 to 3. So notice that we're rotating in the counterclockwise direction. Whenever that happens, you need to assign an S configuration if group number 4 is in the back. So that's the configuration for this particular carocenter. Now let's focus on the chiral center on the right.

And if you want to pause the video to try that example, feel free to do so. This bromine atom, we know it's going to be number 1. Chlorine is going to be 2. The methyl group will be group number 3, and H is 4. So counting it from 1 to 2 to 3, this time we are rotating in the clockwise direction. So this is going to have the R configuration. Now looking at these two molecules, what is the relationship between them?

Would you describe them as constitutional isomers, enantiomers, diastereomers, mesocompounds? How would you describe these two compounds? Now notice that these two molecules are mirror images of each other. Therefore, they are known as enantiomers. Enantiomers are a special type of stereoisomer.

A stereoisomer, like any other isomer, have the same chemical formula, they're connected the same way, but notice that the way the atoms are arranged in space, it's arranged in space differently. This carbon and that carbon, they're connected to the same four groups. But the way those four groups are arranged in space, they're not the same. Here, chlorine is on the left side.

Here, it's on the right side. So they are arranged in space differently, which makes it a certain type of stereoisomer. Diesteromers, enantiomers, cis-trans geometric isomers, those are all stereoisomers.

But enantiomers, they're mirror images of each other, and that's how you can distinguish them from diastereomers. Another feature of enantiomers is that they have the opposite configuration at the chiral center. Here, the configuration, we determine it to be S. Here, it's R. Now, let's work on some more examples for the sake of practice.

So go ahead and assign the R and S configuration to each chiral center that I'm going to draw. So let's put an ethyl group here. Let's add a hydrogen, a bromine atom, and let's put a hydroxyl group. So go ahead and determine the configuration at this carbon. And so while you do that, I'm going to write up another one.

So let's start with the first one. The first thing we need to do is determine which group has the highest priority. So we know it's going to be bromine.

Next, if we compare oxygen to carbon and hydrogen, we know oxygen is going to have a higher atomic number. number for oxygen is 8, for carbon is 6. So this is going to be number 2, 3, and then hydrogen is always 4. Now hydrogen is in the back, which is good. So we're going to just go from 1 to 2 to 3. And so for this one, we...

we have the R configuration. We're rotating clockwise, or in the direction of a clock. Now for the next one, hydrogen's gonna be group number four. Here we have oxygen, carbon, carbon.

Oxygen wins, so that's gonna be number one. Next, notice that these two carbons, they have the same atomic number. So if we move on to the next group, here we have carbon versus hydrogen, carbon wins.

So the ethyl group has a higher priority than the methyl group. So if we count it from 1 to 2 to 3, notice that we're rotating in a clockwise direction. But group number 4, it's not in the back.

It's in the front. It's not on the dash. This time it's on the solid wedge.

So because it's on the solid wedge or it's in the front, coming out of the paper, we need to rotate it. So just switch the configuration and it's equivalent to putting it back to the to putting group number four in the back. So it was R but now it's S. So that's the configuration for this chiral center.

So anytime H is in the front, simply count it from one two to three and then reverse your answer. Number two, name the following compounds using the RS system for stereoisomers. So first, let's identify the chiral center.

So let's focus on this carbon. That carbon has a bromine, a hydrogen, a methyl group, and an ethyl group. So it has four different groups, which makes it a chiral center.

Now let's rank those groups. Bromine has the highest atomic number, that's going to be number 1. We know ethyl has a higher priority than methyl. So this will be 2, that's going to be 3. H is going to be 4. So going from 1 to 2 to 3, we can see that it's going in the R direction. And when you count from 1 to 2 to 3, ignore 4. Now as you count it...

We're going to focus on 4. 4 is in the front, it's not in the back, which means we need to reverse it. So it's going counterclockwise, which means the configuration is S. So let's put the S configuration here.

Now, in order to name it, we need to count the carbons in the parent chain. So we're going to count it in this direction. So we have a 4-carbon parent chain. We're going to be dealing with butane.

So first we need to write the configuration of the chiral center, which is S. We're going to close that in parentheses. And then it's going to be, we have a bromine on carbon 2, so 2-bromo. And for 4-carbon alkane, that's going to be butane. So it's S, 2-bromo, butane.

That's how we can name this particular stereoisomer. Now let's move on to the next one. So here we have two chiral centers, and you could check it. Here we have one group, two, three.

Relative to this carbon, this is the fourth group. If we focus on the second chiral center, this is one group, two, the ethyl is three, and that's the fourth group. So let's focus on the first chiral center, this one.

So we're comparing a chlorine atom, hydrogen, carbon, carbon. So if we look at the first atom that's connected to that caryl center, we see that chlorine has the highest atomic number. We don't look at bromine because it's not the first atom attached to this caryl center. So we look at it from a one-to-one basis.

So this chlorine has a higher atomic number than that carbon. So this is going to be group number one. Now, for group number two, we know H is number four, so we got to compare these two.

Here we have a carbon, and then a carbon, and then after that... Carbon is a hydrogen. Here, there's a bromine.

This is going to win. So this entire group is group number three, which makes the methyl group... Actually, take that back.

This is two. This is number three. This has a higher priority than the methyl group. So now we can count it from 1 to 2 to 3, and so that's going to be the R configuration.

So let's put R here, and now let's focus on the next chiral center. So bromine, hydrogen, carbon, carbon. Bromine is going to have the highest atomic number, so that's 1. H will be 4. So now let's compare these two.

So those two carbon atoms are the same. Next we have a chlorine versus a carbon. The chlorine is going to win.

So this whole group will be group number 2. And the ethyl group is going to be number 3. So going from 1 to 2 to 3, we're going in the R direction, but H is in the front. We're going to have to reverse it, making it S. So this is S.

Now that we've assigned the configuration to each chiral center, we can name this theory isomer. So we need to number it from left to right. So the substituent will be on 2, 3, as opposed to 3, 4. Now... On carbon 2, we have the R configuration, so this is going to be 2R.

And on carbon 3, we have the S configuration, so this is going to be 3S. Now we need a comma between them. And then we need to put the bromo and the chloro group in alphabetical order.

So B comes before C. This is going to be 3 bromo. It's on carbon 3. Chlorine's on carbon 2, so we're going to have 2 chloro.

And for a 5-carbon alkane, that's going to be pentane. So this is 2R3S3-bromo-2-chloropentane. So that's how we can name this particular stereoisomer.

Now here's a question for you. How many stereoisomers are possible if you have one chiral center? And how many are possible if you have two chiral centers? So if you have one chiral center, you can have the S isomer or you can have the R isomer. So there's two possibilities.

The equation that will give you the number of stereoisomers is 2 to the n, where n is the number of chiral centers. So in this case, n is 1. 2 to the first power is 2. So for one chiral center, there are two possible stereoisomers. Now what if we have two chiral centers?

How many stereoisomers are possible? Well, it's going to be n is 2, so it's going to be 2 squared. 2 squared is 2 times 2, which is 4. So we have four possibilities. The first possibility is if both chiral centers are R. The second one is if one is R, I mean if the first one is R and the second one is S, or if the first one is S, the second one is R, or if they're both S.

So those are the four possibilities, or the four stereoisomers that are possible, if you have two chiral centers. So the number of stereoisomers is going to be 2 to the n chiral centers. Number three, the chemical structure of cholesterol is shown below.

How many stereoisomers are possible for this molecule? So feel free to take a minute and try this problem. So this right here is a chiral center.

We have an OH group. There is an invisible hydrogen, but now it's visible. And then this side is different than that side. This side is closer to the double bond. This other side has a methyl group.

So, these two sides are different, which makes this a chiral center. Now, this carbon is also a chiral center. It has four different groups.

Here we have a methyl group. Here is a secondary carbon, a tertiary carbon, and this tertiary carbon. This tertiary carbon is close to the double bond.

This one is not, so they're different. Which makes this a chiral center? Here is another chiral center. This is one group, this is another, this one's another, and there is a hydrogen attached to it. And we could put that hydrogen here.

This carbon here is not chiral. Whenever you see a carbon with two bonds and that's it, it's a CH2 carbon, which means it doesn't have four different groups because these are the same. So that is not a chiral carbon. So it's best to focus on the tertiary carbons and the quaternary carbons.

This is a tertiary carbon. It's attached to three other carbons, and there's a hydrogen attached to it. So that is another chiral center.

Here's another tertiary carbon. It has three other carbons, all of which are different, and there's a hydrogen here, too. Now, this carbon here is quaternary.

It's attached to four different carbon atoms. So that's another chiral carbon. This tertiary carbon is chiral, too. There's a hydrogen here.

And this carbon is tertiary, and it's chiral. This is group 1, this is another group, this is another group, and then the hydrogen is the fourth group. So let's put that hydrogen on the hatch wedge. Now this carbon here is tertiary, but it's not chiral, because it doesn't have four different groups. These two methyl groups are the same.

Thus, cholesterol has a total of 1, 2, 3, 4, 5, 6, 7, 8 chiral centers. So because cholesterol has 8 chiral centers, it's going to have... 2 to the 8 stereoisomers.

2 to the 8 is 2 to the 4 times 2 to the 4. 4 plus 4 is 8. If you multiply 2 four times, 2 times 2 is 4, times 2 is 8, times 2 is 16, you get 16 times 16, which is 256. So this is the number of stereoisomers that are possible for this particular molecule. Number four, determine the absolute configuration of each chiral center and name each Fischer projection shown below. When dealing with Fischer projections, it's easy to determine the chiral center. So we have two chiral centers, which means that we have four possible stereoisomers for this particular structure. Now before we determine the absolute configuration of each chiral center, let's talk about Fischer projections.

So let's say if we have this particular Fischer projection. Let's make this an ethyl group and let's say this is hydrogen, this is OH. What we need to realize is that the groups on the horizontal part of the Fischer projection, they're in the front.

So you can redraw the structure like this if you want. They're coming out of the page, and so they're on the solid wedge. These two groups, they're on the dash or the hatch wedge.

So they're going into the page. you can view them as being in the back. So make sure you understand that. Now hydrogen is typically on the side which means that as group number four you're going to have to reverse it when determining the configuration so keep that in mind.

Anytime group number four is in the front you need to reverse it. So let's begin by determining the configuration of this Carol Center. The hydroxyl group is going to be group number one.

Oxygen beats these two carbon atoms. Now, comparing those two carbon atoms, they're the same, but here we have a bromine, here is a hydrogen, so this group has a higher priority than the methyl group. So this will be group number two, the methyl will be three, and then hydrogen is always four, if it's there. So going from one to two to three, we're going in a clockwise direction, so that's R. But H, remember H, when it's in a horizontal part, it's like being on a solid wedge.

H is in the front, so we got to reverse it. So we're going to get S. So we have the S configuration. for that chiral center.

Now for the next one, so we're going to focus on this chiral center. Bromine is going to be group number one. It has the highest priority.

Next we're comparing the methyl group with this group. So we have carbon to carbon and then hydrogen to oxygen. Oxygen wins. So this entire group is going to be number 2. Method was 3. H is 4. So going from 1 to 2 to 3, it looks like we're going in the S direction. But if we reverse it, it's going to be R.

So we have the R configuration here. So now that we've assigned the absolute configuration to each cryocenter, we can go ahead and name this particular molecule. So the hydroxyl group is going to have priority over the bromine group, so we're going to count it in a way that we give the hydroxyl group the lower number. We want to give it a 2 instead of a 3, so we're going to count it in this direction.

So on carbon 2, we have the S configuration. So this is going to be 2S. And on carbon 3, we have the R configuration, so 3R. And then... We have a bromine on carbon 3, so this is going to be 3-bromo, and the alcohol is going to be part of the parent name.

And it's on carbon 2, so it's going to be 2-butanol, because we have a 4-carbon chain. So 2s3r3-bromo-2-butanol. That's how we can name this particular fissure projection.

Now let's try another example. Go ahead and name this fissure projection and determine the configuration at each call center just like we did before. So let's start with this one.

Bromine is going to be group number one. This group with the chlorine atom, we know that's going to be two. Methyl is three, H is four.

So going from one to two to three, ignoring four, this is going in the counterclockwise direction, so that's S. But H is in the front, so we need to reverse it. So we're going to get R for the first one.

Now for the second chiral center. Chlorine is going to be group number one. This entire group is number two.

If we compare carbon to carbon, it's a tie. And then carbon to bromine, bromine wins. So this is number two. Ethyl is number three.

H is four. So going from one to two to three, skipping four, it appears to be R. But H is in the front, so we reverse it and we get the S configuration. So now that we have the configuration at both chiral centers, we can name it.

So we don't want to count in this direction, because this would be 3 and that would be 4. Rather, we want to count in this direction. So this would be 2 and this would be 3. So we have a 5-carbon chain, so we're dealing with pentane. But first, let's focus on the configuration. So the configuration is R at carbon 2, so we're going to have 2R. And then it's S at carbon 3, so 3S.

And then we have a Br on carbon 2. And we need to put it in alphabetical order. B comes before C. So this is going to be 2 bromo. And then we have a Cl on carbon 3. So 3 chloro.

And then... the five carbon chain that's going to be pentane so that's the nomenclature or the IUPAC nomenclature for this particular fissure projection it's 2r3s 2-bromo-3-chloropentane now let's work on a more challenging problem so let's say this Carol Center has an ethyl group a methyl group in the front and let's put a chlorine group in the back and the hydrogen go ahead and assign the configuration to the chiral carbon is it RS now this problem is different because Hydrogen, group number four, is neither in the front nor is it in the back. So how do you assign the configuration in this situation?

Well, let's begin. We know that chlorine has the highest atomic number, so that's going to be group number one. Ethyl has more priority than methyl, so ethyl is two, methyl is three, H is four. In a situation like this, there's a technique that you could use to assign the configuration. Whatever group is in the back, put it in a circle.

So I'm going to put one in the circle. And you could put a small subscript b to indicate that whatever in the circle is in the back. Sometimes it could be in the front.

But for now, whatever's in the circle in this problem is in the back. Now, the other numbers, 2, 4, and 3, you want to arrange them in the form of a triangle. Now, if you notice, 2 is at the top, so let's put that at the top. 4 is at the bottom left with respect to 2, so let's put that here.

3 is on the bottom right. Now what we're going to do is we're going to rotate this molecule in such a way that number 4 will be at the top where 2 is. So we're going to rotate it 120 degrees clockwise. So 1 is still in a circle. 4 has replaced 2. 2 is now where 3 was.

And 3 is where 4 was. Now the next thing we're going to do is we're going to flip the molecule. As we flip it, 2 and 3 will move to the top.

Group number 4 is going to expel group number 1. So now 4 is in the circle, 1 is beneath it, and then 3 and 2 are at the top. Now 4, group number 4 is in the back, because what's in the circle represents what's in the back. And so we can count it from 1 to 2 to 3, and this gives us the S isomer.

So the configuration at this Carroll Center is S. So that's the technique that you could use whenever group number 4 is not in the front or in the back.

Transcript for:Understanding Stereochemistry and Configurations

Transcript for:
Understanding Stereochemistry and Configurations