Cahn-Ingold-Prelog System for Naming Enantiomers Cahn-Ingold-Prelog System for Naming Enantiomers
Cahn-Ingold-Prelog System for Naming Enantiomers
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- Right now, based on what we know so far, if we wanted to
- name this molecule, we would say, well, what's the longest
- carbon chain here?
- Well, we have a two-carbon chain, and there's all single
- bonds, so we're dealing with an ethane.
- Actually, I'll write it all at once.
- And then we have on the one carbon, we can call this the
- one carbon, and call this the two carbon, we have a bromine
- and a fluorine.
- So we could call this 1-bromo, and we're putting the bromo
- instead of the fluoro because B comes before F
- 1-bromo-1-fluoro, and then we're dealing with an ethane.
- We have a two-carbon chain, all single bonds,
- That's the name of that molecule there, just a review
- of some of the earlier organic nomenclature
- videos we had done.
- Now, we know immediately, based on the last few videos,
- that this is also a chiral carbon, and if we were to take
- its mirror image, we would get another enantiomer of this
- same molecule, or that they are enantiomers of each other.
- So what is the mirror image of this
- 1-bromo-1-fluoroethane look like?
- Well, you'd have the carbon right here.
- I want to get all the colors right.
- You would still have the bromine up above.
- You would have this methyl group that's attached to the
- carbon now pointing in the left direction, CH3.
- The fluorine would now still be behind the carbon, and now
- the hydrogen would still pop out of the page, but it would
- now pop out and to the right.
- That is the hydrogen.
- Now, based on our naming so far, we would name this
- 1-bromo-1-fluoroethane, and we would also name this
- 1-bromo-1-fluoroethane, but these are fundamentally two
- different molecules.
- Even though they have the same molecules in them; they have
- the same molecular formula; they have the same
- constitution in that this carbon is connected to a
- hydrogen, a fluorine, and a bromine; this carbon is
- connected to the same things; this carbon is connected to a
- carbon, three hydrogens; so is this one; these are
- These are stereoisomers, and they're mirror images of each
- other, so they're enantiomers.
- And actually, they will, one, polarize light differently,
- and they actually can often have very different chemical
- properties, either in a chemical or biological system.
- So it seems not good that we have the same
- names for both of these.
- So what we're going to focus on in this video is how do you
- differentiate between the two?
- So how do we differentiate between the two?
- So the naming system we're going to use right here is
- called the Cahn-Ingold-Prelog system, but it's a different
- Cahn, it's not me.
- It's C-A-H-N instead of K-H-A-N.
- Cahn-Ingold-Prelog system, and it's a way of differentiating
- between this enantiomer, which right now we would call
- 1-bromo-1-fluoroethane, and this enantiomer,
- It's a pretty straightforward thing.
- Really, the hardest part is to just visualize rotating the
- molecules in the right way and figuring out in which
- direction it's kind of-- whether it's kind of a
- left-handed or right-handed molecule.
- We're going to take it step by step.
- So the first thing you do in the Cahn-Ingold-Prelog system
- is to, one, identify your chiral molecule.
- Here, it's pretty obvious.
- It's this carbon right here.
- We'll just focus on this left one, the one we started with
- first. It's bonded to three different groups.
- And then what you want to do is you want to rank the groups
- by atomic number.
- So if you go up here, out of bromine, hydrogen, fluorine,
- and a carbon, this is what is bonded directly to this
- carbon, which has the highest atomic number?
- Bromine is over here-- let me do this in a darker color.
- We have bromine at 35, we have fluorine at 9, we have carbon
- at 6, and then we have hydrogen at 1.
- So of all these, bromine is the largest. We'll just call
- this number one.
- Then after that, we have fluorine.
- That is the number two.
- Number three is the carbon.
- And then hydrogen is the smallest, so
- that is number four.
- So now that we've numbered them, the next step is to
- orient this molecule so that the smallest atomic number
- group is sitting into the page.
- It's sitting behind the molecule.
- Right now, this hydrogen is the smallest of all of them.
- Bromine's the largest, hydrogen is the smallest, so
- we want to orient it behind the molecule.
- The way it's drawn right now, it's oriented in
- front of the molecule.
- So to orient it behind the molecule, and this really is
- the hardest part is just to visualize it properly.
- Remember, this fluorine is behind; this is right in the
- plane of the paper; this is popping out of the paper.
- We would want to rotate.
- You could imagine we'd be rotating the molecule in this
- direction so that-- let me redraw it.
- We have the carbon here.
- And now since we've rotated it like this, we've rotated it
- roughly 1/3 around the circle, so it's about 120 degrees.
- Now, this hydrogen is where the fluorine was.
- So that's where the hydrogen is.
- The fluorine is now where this methyl group is.
- These dotted lines show that we're behind now.
- This shows that we're in the plane.
- And the methyl group is now where the hydrogen is.
- It's now popping out of the page.
- It's going to the left and out.
- So this methyl group is now popping out of the page, out
- and to the left.
- That's where our methyl group is.
- So all we've done is we've just rotated this around about
- 120 degrees.
- We've just gotten this to go behind, and that's kind of the
- first step after we've identified the chiral carbon
- and ranked them by atomic number.
- Of course.
- The bromine is still going to be on top.
- Now, once you put the smallest atomic number molecule in the
- back, then you want to look at the rankings
- of one through three.
- And we have four molecules here.
- We look at the largest, which is bromine, number one.
- Then number two is fluorine, number two, and then number
- three is this methyl group.
- That's the carbon that's bonded to this carbon, so it's
- number three right there.
- And in the Cahn-Ingold-Prelog system, we literally just
- think about what would it take to go from number one to
- number two to number three?
- And in this case, we would go in this direction.
- To go from number one to number two to number three, we
- would go in the clockwise direction.
- We're just ignoring the hydrogen right now.
- That's just sitting behind it.
- That was the first step, to orient it so it's sitting in
- the back, the smallest molecule.
- And then the three largest ones, you just say what
- direction do we have to go to go from number one to number
- two to number three?
- In this case, we have to go clockwise.
- And if we go clockwise now, then we call this a
- right-handed molecule, or we use the Latin word for right,
- which is rectus.
- And so we would call this molecule right here not just
- 1-bromo-1-fluoroethane, this is R, R for rectus.
- Or you could even think right, although we'll see left is
- used as S, which is sinister, so the Latin is really where
- the R comes from.
- But this is (R)-1-bromo-1-fluoroethane
- That's this one right here.
- So you might guess, well, this must be the opposite, this
- must be the counterclockwise version.
- We can do it really fast.
- So same idea.
- We know the largest one.
- Bromine is number one.
- That's the largest in terms of atomic number.
- Fluorine is number two.
- Carbon is number three.
- Hydrogen is number four.
- What we want to do is put hydrogen in the back, so what
- we're going to have to do is rotate it to the back to where
- fluorine is right now.
- So if we had to redraw this molecule right here, you'd
- have your carbon still.
- You still have your bromine sitting on top.
- But we're going to put the hydrogen now to the back, so
- the hydrogen is now where the fluorine used to be.
- The hydrogen's there.
- This methyl group, this carbon with the three hydrogens, is
- going to be rotated to where the hydrogen used to be.
- It's now going to pop out of the page, because we're
- rotating it in that direction, so this is our methyl group
- right there.
- And then this fluorine is going to be moved where the
- methyl group was, so this fluorine will go right here.
- And now, using the Cahn-Ingold-Prelog system,
- this is our number one, this is our number two, just by
- atomic number, this is number three.
- You go from number one through number two to number three.
- You go in this direction.
- You're going counterclockwise.
- Or we are going to the left, or we use the Latin word for
- it, which is sinister.
- And the word sinister comes from the Latin word for left,
- so I guess right is good, and people thought either
- left-handed people were bad, or if you're not going to the
- right, it's bad.
- I don't know why sinister took on its sinister meaning now in
- common language.
- But it's now the sinister version of the molecule.
- So we would call this version, this enantiomer of
- 1-bromo-1-fluoroethane, we would call this S, S for
- sinister, or for left, or for counterclockwise:
- So now we can differentiate the names.
- We know that these are two different configurations.
- And that's what the S and the R tell us, that if you have to
- go from this to this, you would literally have to detach
- and reattach different groups.
- You'd actually have to break bonds.
- You actually have to swap two of these groups in some way in
- order to get from this enantiomer to this enantiomer.
- They're different configurations, really
- fundamentally different molecules, stereoisomers,
- enantiomers, however you want to--
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