The R and S Naming System

According to IUPAC protocol, each molecule must have a unique, unambiguous name – even stereoisomers.

Concept: Why stereoisomers need their own naming system.  

2m
Video Transcript

So now that we understand what a chiral center and we know how to recognize them on molecules, it's important that we know how to name them. So if you remember back to when I first taught you guys about just simple alkane nomenclature, we talked about the IUPAC naming system. And what I told you was that according to IUPAC protocol, every single molecule in the universe needs to get its own unique name. That's kind of the point of IUPAC that now we have a systematic way to name every single molecule.
Well, now that we've learned about stereoisomers meaning molecules that possess chiral centers, that means that those stereoisomers, since they have different shapes, they're going to need to be included in this system somehow because they have a distinctive difference.
Well, it turns out that all we have to do is we're going to use the same IUPAC names from before, but we're going to add just one extra step to account for those chiral centers and that extra step is called the Cahn-Ingold-Prelog nomenclature. I know that name blows. What we're going to be doing is we're not going to be using that name so much. That's kind of the technical name that's in your book. I like to just call it the R and S naming system. You'll understand why it's called R and S in a second, but for right now just kind of take my word for it that you don't need to say Cahn-Ingold-Prelog in class, you can just say R and S and everyone will know what you're talking about.
In order to learn it's one extra step, it's actually kind of a five-step process. What we'll be doing is I'll be teaching you one step at a time with a molecule so that you can apply it and see exactly how to use that rule. So let's go ahead and get started with our first step

Rules for the R and S Naming System

Step 1: Assign priorities to the four atoms on the chiral center according to their atomic mass on the periodic table. 

Concept: R and S Naming- Step 1  

4m
Video Transcript

All right, so our first step starts off pretty easy. What it states is that you're going to have to assign priorities to the four atoms on a chiral center based on their atomic mass in the periodic table. So let's break that down a little bit more.
First of all, notice that I underlined four atoms. Why specifically is it the number four? Because remember the definition of a chiral center was any atom that has four different groups. So what I'm basically just saying here is that you find those four different groups and you give them priorities. How? Based on their mass in the periodic table.
Let's just go ahead and look at this molecule to kind of get started with this rule. First of all, were you guys able to find an atom here that would count as a chiral center? Because that's always going to be the first step, locate your chiral center.
So do you guys see one? Maybe an atom that's four different – yeah, you found it. It's this one right here. Remember I can use a star to represent a chiral center. Awesome job. Notice that that carbon has four completely different atoms coming off of it. It has oxygen, nitrogen, hydrogen. What's that stick? Obviously, that would be a carbon. Perfect. So you've got four different atoms. This is definitely chiral center.
So we've got the first step down. The second step is that we need to assign priorities based on the periodic table. This might be the point of the video that you want to pause and grab a periodic table and rush back over here. But we could also use to remember our big seven. Remember that the big seven is just a method to remember some of the most important atoms on the periodic table.
Remember that the way you would do this is we just draw literally a big seven and then you'd split the big seven into seven boxes. And then we would just fill these out in order of the atoms on the upper right-hand side of the periodic table.
Remember that we always start off with carbon and then we go to nitrogen, oxygen, fluorine. Now we're in the halogens and we have to go down so that would be fluorine, chlorine, bromine, iodine. That's our big seven. Obviously, this isn't comprehensive, but it's a great reference point even for all the other atoms on the periodic table.
So now I'm just going to trust you guys that either you're looking at the big seven that I just drew or you have a periodic table. Let's go ahead and assign priorities. So my first question to you is, which one would get the highest priority out of these four different atoms, which one is the highest according to atomic mass?
It has to be the oxygen, right? Because we said that we have four different atoms: oxygen, nitrogen, hydrogen and carbon. So this is going to be my number one priority because it has the highest weight on the periodic table. I don't have any of the heavier atoms.
So now if that's number one, which one is number two? Number two has to be nitrogen because it's a little bit lighter so that's going to get my two. Which one's number three? It has to be carbon. So carbon's my number three. And just so you guys know if you ever see a hydrogen, that's always going to be your four because nothing is lighter than hydrogen, right? So you can always just draw four there immediately if you just want to get good at it.
So that was the first step guys. Just look at your periodic table. If you need to, draw your big seven and go ahead and compare priorities. Not so bad, right? Let's move on to the next step. 

Step 2: When there is a tie between atomic weights, compare the next set of adjacent atoms (playoffs!)

Concept: R and S Naming- Step 2  

6m
Video Transcript

So step two deals with a really common issue that we face with R and S naming, which is that sometimes you're going to be comparing your atoms and you have a tie because two of them have identical atomic weights. Well, what do you do in that situation? Well, if you do have a tie between your atomic weights what we're then going to do is we're going to compare the next set of atoms that they're both attached to and basically have a playoff system.
In the playoff system, we look at the next immediate atoms that they're both attached to and then we compare those atomic weights and we say which of these is bigger, which of these is heavier and that's going to be the winner.
So let's go ahead and apply this to an example. First of all, I really hope that you guys can identify the chiral center here. Yeah, so you should be getting a little better at this. That's it. And now we have to figure out first of all, are there any issues with step one in terms of are we able to name these priorities easily. Let's kind of just do the first step which is just figuring out what all the atoms are.
Notice that right away I have an issue because two of the atoms that are attached to my chiral center are the same. I have an oxygen and an oxygen. Now if I compare both of those on the periodic table, I'm obviously going to get equal atomic masses so that means I'm definitely going to have to use a playoff there no matter what. Let's continue. Maybe the other atoms, we can figure out what order they're in.
Well, this is a carbon because that's a carbon group, an R group. And Et – look that up on the periodic table, you're not going to find it because that stands for ethyl. That really just stands for CH2CH3. I've got another two-way tie guys. I've got another two-way tie between this carbon and this carbon, so I'm definitely having some issues with rule number one.
Now there is one thing that we can do even before this playoff system. What I can safely say is that my oxygens are going to get positioned one and two and my carbons are going to get positions three and four, right? Because oxygens are bigger than carbon. That's just common sense. But, I don't know which order they're going to be in. In order to figure out the order, I have to use playoffs.
So let's do the oxygen playoff first. This is for spots one and two. This will be my red oxygen. This will be my blue oxygen. So what is my oxygen attached to, the red one? It's attached immediately to an H. What is the blue oxygen attached to? That's immediately attached to Me. What does Me stand for? That's a methyl group, so a methyl group is CH3. So it's attached to a C.
So in the playoff system, I look at what's in the bracket. It's almost like March Madness or something, but it's just a lot more boring and you can't win money from it, unfortunately. At least I haven't developed that way to make money yet. But which one is going to win inside of that bracket? It's going to be the carbon because carbon is heavier than hydrogen, right? So you've got a clear winner. This is my number one and this is like number two. Make sense so far? Both my oxygens beat my carbons, but the methyl beats the H.
Let's move on to the carbons. I'm going to do the same thing. I'm going to do a color system. This will continue to be my blue carbon. I'll make this one green. So what I want to do is that notice that for oxygen, oxygen is only attached to one more thing, but carbon always has three bonds. So I'm going to draw all three.
So for my blue one, what are the three additional bonds, besides the chiral center because you never go backwards, you only go forwards. What are the three atoms that that carbon is attached to? Well, it's attached to one carbon on one side. Let's just say that that is the left side. It's attached to another carbon on the right side. And then it's also attached to an H that wasn't drawn. So I'm going to draw that. I'm going to put that here. That's what's in my bracket there. Those are the three additional atoms. That's actually a carbon.
Now let's look at green. Green, once again, I'm not looking at the chiral center I'm just going to move forward. Green is attached to two H's because it's CH2 and then CH3, it's also attached to this carbon here because obviously it's a chain, so it needs to be attached. What we tend to do for brackets is we put the biggest atoms first. So instead of writing this as H, H, C, I would actually write this as C, H, H. And that just makes it easier to compare the brackets and see who wins.
So now I know that I had a tie between the carbons. If I move them to the bracket, is there a winner? Well, if you look at the first atom, they're still a tie. They're both attached to carbon. But if you look at the second set, this is where your winner is determined. It came down to the wire. It was a close game, but I'm sorry green, you lost.
So that means that my blue is going to be number three and my green is going to be the biggest loser at number four. And see how we systematically did that with the playoff system. That's all there is to it. We'll practice it more. Don't worry. But those are the basics. 

Step 3: Double bonds count twice. Triple bonds count three times

Concept: R and S Naming- Step 3  

4m
Video Transcript

Step three is pretty straight forward, it just says if you're using the playoff system and you run into a double bond or a triple bond, the double bond is going to count twice and a triple bond is going to count three times. So I think the easiest way to relate to this is just to do an example.
So our chiral center is where? Right here. And let's kind of figure out those four different atoms that we've been doing so far. So we have an H, so obviously, we can already tell what priority that's going to be, right? Out of the four numbers, which one is it going to be? Four, automatically. Good job.
But these are other three, let's figure out, I've got carbon. I've got carbon. Oh man, I have a three-way tie guys. So remember that if we have a tie where the atoms don't have different atomic weights, we have to use the playoff system. But now on top of that, I have a slightly more complicated situation because some of these carbons have double bonds on them. So let's figure out how to do the bracket playoff system like before.
So let's do I'd say the easier one first which is the blue carbon. So the blue carbon, I would say what are the three atoms that are attached to it? Well, I definitely have two H's coming off of it. And I have another carbon. So I'm going to put in order of atomic mass carbon, H, H. So that wasn't so bad. We're used to doing that.
Now let's look at the red one. The red carbon is a little bit tricky because the red carbon, notice it only has one H coming off of it and it has a carbon over here. But notice that it's double bonded to that carbon. So basically we're going to count that as two separate bonds. Why? Because we need to account for all three bonds that the carbon makes. So, to me, a double bond to carbon is just as good to two bonds to carbon, so I'm going to put carbon-carbon. So both of those count as the double bond and hydrogen.
And then finally I've got green, which is even a little bit crazier because notice that green is actually attached to two carbons, but one of them is single bonded and one of them is double bonded. So you guessed it, you're just going to add that up together. The green is attached to all carbons, C, C, C. Two from the double bond, one from the single bond. Does that make sense? And notice that I'm always moving forwards, I'm never moving backwards towards the chiral center.
So now that we've applied that double bond rule, of course, this would also apply with triple bonds, I just don't have them drawn. What are our priorities going to be? Well, green is going to be number one. Red would be number two. Blue would be number three. I'm just going to write these here. One, two, three and then obviously your hydrogen is number four. 

Example: Determine priorities for the following chiral center.

2m
Video Transcript

So this example should have been really easy for you guys. First of all, the chiral center was right in the middle. We did have four different groups because even though we have two carbons, meaning that we're going to have to use a playoff system, they're not exactly the same. One is a methyl group and one is an ethyl group, so this does count as a chiral center.
Because we have a tie, we should use the playoff system for those, but right away I know that my number four priority is going to be my H. I also know that my number one priority is going to be the O. So really I just need to do the playoff system for priority two, priority three.
Now some of you guys might already be able to do this in your head, but I just want to be real careful since we just learned this to do it really clearly. I've got a blue C. I've got a red C. The blue C is attached to 3 H's because it's a methyl group. The red C is attached to one carbon and two H's and I'll just draw those H's out so you can see.
So which one won? The red carbon won. This is my two and this is my three. So we've got our priorities. Let's move on to the next rule. 

Example: Determine priorities for the following chiral center.

3m

Step 4: If the last priority group is in the back, then trace a path from highest to lowest priority

- Clockwise = R, Counterclockwise = S                            

- Always Ignore group 4

Concept: R and S Naming- Step 4  

4m
Video Transcript

So finally, in step four, we're going to learn how to name these guys. So step four starts off by saying that if your last priority group is in the back – I'm not even going to read the rest yet. Let's just define that first part. The last priority group is always going to be which group? Group number four. So let's actually just redefine that as four because that's the easiest way to say it.
So if four is in the back – now what does the back mean? In 3D structures that always means a dash. So this is actually starting off by saying that if 4 is on the dash, then we're going to trace a path from the highest priority to the lowest priority.
Now what that literally means is that we're going to draw arrows from my number one priority to my number two and from my number two priority to my number three. But remember – so that's from highest to lowest. Now remember, there's four groups. What happens to number four? Do I also draw an arrow to that one? No, because you're always going to ignore group number four because it's on the dash. Since it's in the back of the molecule, I don't look at it. I only look at groups one, two and three.
So now I trace that path and if that path happens to look like a clockwise path, meaning does it look in the direction that a clock would move in. For this, you're going to need to know what an analog clock looks like. I know that for maybe some of you younger folks maybe you're not used to seeing those too often but we're just going to have to stretch ourselves a little bit. Clockwise rotation is going to get an R letter. And a counterclockwise rotation is going to get an S letter, hence the name R and S.
Let's just use – bring that down, that example that we did before. Those priorities are going to be the exact same priorities and let's see if we can assign an R or an S to this chiral center. Those priorities should be correct. This is my chiral center.
So now the first thing I have to ask myself is my number four on the dash? Perfect. It is. So that means I can use this rule. So this next rule says I draw arrows from one to two, from two to three, and then finally back from three to one. What about number four? I ignore it because it's in the back. I don't want to use number four.
So is this a clockwise rotation or a counterclockwise rotation? This is clockwise folks. And since it's clockwise, this is going to be considered an R isomer. So we haven't learned how to incorporate this into the full name yet, but this is considered an R. Awesome, right? Awesome. So let's go ahead and move on to step number five. 

Step 5: If the last priority group is NOT in the back, swap that group with the group that is on the dash.

- Trace path as always, but this time swap the sign since you swapped groups.

  • R becomes S
  • S becomes R

Concept: R and S Naming- Step 5  

4m
Video Transcript

Step number four is really easy to use, but it has a major limitation, which is that it only works if you're number four is on the dash. And guess what? Your professor is often going to give you molecules where your number four is not on the dash. It might be on the front or it might be on the side. Why would he do such a thing? Obviously to make your life harder.
In that case we're going to have to move down to step five, which step five is kind of like your contingency plan. If your number four, last priority group, is not in the back, which, I'm sorry to tell you, is most of the time. Then we're going to have to make an adjustment because it turns out that the method we use for step four is actually the easiest way to figure this out, but my number four needs to be on the dash in order to use it.
So what we're going to do is we're going to cheat a little bit. And we're going to swap number four with whatever is on the dash already. Meaning that whatever number is on the dash, let's say it's two or three, we're going to redraw those numbers and swap them so that we can pretend like number four is on the dash even though it really isn't on the dash. We're just going to pretend.
Now there's no free lunch. We're cheating here. I just told you this is like a game of make-believe. We're going to pretend like the four's on the back even though it isn't. In order to make up for that, we're going to have to swap whatever sign we get to make up for the fact that we swapped the groups. This is really easy it just means that if you trace your path and it looks like an R, you're actually going to give it an S. If it looks like an S, you're actually going to give it an R.
So let's go ahead and do an example where you bring down the priorities from before and see how it works out. Remember that this was number one, number two, number three and number four.
So is my number four on the dash? No, it's not. It's actually on the front, so what are going to do? Well, in order to use the system from before I have to swap out. The rule says the number four has to swap with whatever's on the dash. What's on the dash? One.
Now I want to warn you guys against other methods of doing this. Some online tutorials like YouTube, whatever, or even your professor will sometimes tell you to redraw the molecule with this pretend switch. I think that's really unnecessary and that's really confusing. It's a lot easier just to swap out the numbers.
So I'm going to scratch out four. We know that that's moving to the back. We're going to scratch out one. We know that's swapping. And I'm going to redraw just the numbers. Now that my four's in the back, I can ignore it and I can redraw my path. My path goes one to two, two to three, three to one. Remember that I ignore four.
What does that look like to you? It looks like an R. Is this actually going to be an R? No, because I had to swap that group in the beginning, so I'm actually going to consider this to be an S isomer. And that's the final answer.
So this really isn't that hard. You just have to remember to swap at the end to make up for the fact that you swapped at the beginning. Great. Let's move on to an example. 

Problem: Provide the full name for the following molecule, taking stereochemistry into account.

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Problem: Provide the full name for the following molecule, taking stereochemistry into account.

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Problem: Provide the full name for the following molecule, taking stereochemistry into account.

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The R and S Naming System Additional Practice Problems

(2S,3R)-2-Bromo-3-methylpentane is

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Shown below are chemical structures of an anti-cholesterol medication Crestor (left) and a narcotic analgesic hydrocodone, also known as Vicodin (right). Label the stereocenters indicated by arrows with appropriate (R) or (S) designators. Note that the question asks for configurations of only 3 out of 4 stereocenters in Vicodin.

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Draw the structure for (R,E)-5-bromo-6-methyloct-4-en-7-yn-4-ol

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Determine the R/S configuration of the following alkane. If there is no chiral center put an X. (Hint: Think about the definition of a chiral center)

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Provide a structural formula for (S)-3,6-dimethyl-3,4-octadiene. Note that any correct form of a structural formula is acceptable. Be sure to identify stereoisomers properly.

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Provide a name for the following pure compound. Be sure to name stereoisomers properly, designating configurations when necessary.

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Draw the structure for the following IUPAC name.

(2R,3S)-2-chloro-3-methylheptane

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Write an acceptable IUPAC name for the following two molecules. Where appropriate, use E and Z or R and S.

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Examine the following structures. For each molecule with a chiral center, assign the stereochemistry then write "R" or "S" as appropriate in the box provided below each structure. For all molecules that have no chiral centers, leave the box blank.

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The 3 stereocenters in penicillin V, a broad spectrum antibiotic, are labeled “a,” “b,” and “c” in the structure shown below. In each part of this question, the four bonds to the stereocenter are reproduced in the same orientation as they are written in this structure. Assign priorites to each of the four groups attached to the stereocenter (1 = highest priority; 4 = lowest priority), and write the number associated with each group in the appropriate circle. Then, determine whether the absolute stereochemical configuration is (R) or (S).

θ

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Give an IUPAC substitutive name for each of the following molecules. Use E/Z or R/S as needed to properly indicate stereochemistry

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Applying the correct rules for IUPAC nomenclature, select the correct name for the following compound. Make sure to include correct stereochemistry when appropriate.

A) (R,Z)-3-chloropent-4-en-1-yne

B) (R,Z)-3-chlorohex-2-en-5-yne

C) (S,Z)-3-chlorohex-2-en-5-yne

D) (S,Z)-3-chlorohex-4-en-1-yne

E) (R,Z)-3-chlorohex-4-en-1-yne

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Applying the correct rules for IUPAC nomenclature, select the correct name for the following compound. Make sure to include correct stereochemistry when appropriate.

A) (R,Z)-6-fluorohex-4-en-3-ol

B) (S,Z)- 6-fluorohex-4-en-3-ol

C) (R,Z)- 7-fluorohept-5-en-3-ol

D) (S,Z)- 7-fluorohept-5-en-3-ol

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Name the following compounds

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Assign (R) or (S) configuration to the chiral center in each molecule.

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Give the IUPAC name, including stereochemistry of the compound below

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Give the IUPAC name, including stereochemistry of the compound below

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Name the following compound. 

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Draw the following compounds. 

a)  (S) – 2, 2 – dibromo – 1,4 – nonanediol

 

 

 

 

 

b)  (2R,3S)–3–bromo-2-butanol 

 

 

 

 

 

c)  (R)–1,2–dibromobutane 

 

 

 

 

 

d)  (S)–3–methyl-1-hexene 

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Determine the R and S configuration of the following compound.

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Determine the R and S configuration of the following compound. 

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Determine the R and S configuration of the following compound. 

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Determine the R and S configuration of the following compound. 

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Assign the Cahn-Ingold-Prelog priorities directly on each structure and determine the R/S configuration. 

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Draw the molecule using bondline structure.

(4S,5R)-4,5-dimethylhept-1-yne

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Name the compound given below.

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Meropenem is an injectable, broad spectrum antibiotic that is frequently used for the treatment of bacterial meningitis and pneumonia. The line-angle structure of Meropenem given below is drawn in a manner that shows the absolute stereochemistry clearly and unambiguously. This molecule contains several chiral centers, three of which are labeled 13, 14, and 15. The one alkene double bond present is labeled 19.

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Configurational isomers. Assign the follow chiral centers as R or S.  Show your logic by drawing the groups around a circle & assigning priority.

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Assign the follow chiral centers as R or S.  Show your logic by drawing the groups around a circle & assigning priority.

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Configurational isomers. Assign the follow chiral centers as R or S.  Show your logic by drawing the groups around a circle & assigning priority.

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Assign the follow chiral centers as R or S.  Show your logic by drawing the groups around a circle & assigning priority.

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Assign the following chiral center as R or S. Show your logic by drawing the groups around a circle & assigning priority.

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Draw the appropriate line angle drawing for the following IUPAC name.  

(2S,6S)-1-bromo-1,2,6-trimethylcyclohexane

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Which of the following is the acceptable structure for ( )-5-bromohept-2-yne?

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Give the proper IUPAC name that corresponds to the following structure.

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Which of the following is the correct structure for the compound ( R)-2-pentanol?

 

 

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Name the following compound correctly using accurate  R or S and E and Z nomenclature.

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Provide a structural formula for (R)-oct-7-en-3-ol. Be sure to designate each configuration properly for this single stereoisomer. You may give any acceptable type of structural formula.

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The other half of the 2015 Nobel Prize in Medicine  was won by Youyou Tu. Youyou Tu discovered Artemisinin, a drug that has significantly reduced the mortality rates for patients suffering from Malaria. 

a. List your priorities (1, 2, 3 etc.) in the boxes provided surrounding the chiral centers indicated with a black dot (·) and determine if each is either R or S.

 

 

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What is the absolute configuration of the chiral carbon shown below? 

The absolute configuration is ______________

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Draw the following molecule. (1S,3S,4S)- 4-fluoro-3-methylcyclohexan-1-ol

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Assign (R) or (S) configuration to the chiral center in each molecule.  

 

 

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For the following compound.

a) Identify absolute configurations of all chirality centers.

b) Draw its enantiomer

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Assign the priority to the four groups attached to the chiral center compounds A and B. Then assign R and S configuration to the chiral center of molecules A and B.

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Assign configuration for the following compound (Adrenaline) and draw its enantiomer.

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The Cahn-Ingold-Prelog stereochemical designations used for the following substance are:

a. (1S, 2R, 3R)

b. (1R, 2R, 3R)

c. (1R, 2S, 3R)

d. (1S, 2S, 3S)

e. (1R, 2S, 3S)

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Assign the correct IUPAC name to the following structure (include any stereochemistry)

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