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Swap Method for finding R and S Configurations Chirality Vid 4
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Leia here from leafersai.com
and in this video we'll look at how the
swap method will help you find RNS for
tricky molecules when priority group
number four is not in the back or facing forward.
forward.
You can find this entire video series
along with my practice quiz and cheat
sheet by visiting my website leafersai.com/kyality.
In the past videos, I showed you how to
rank priority groups and I also showed
you how to find the arnest configuration
when priority group number four was in
the back so that we cross it out trace
from 1 to 2 to three and in this case we
have S. Then I showed you how to find
RNS if priority group number four is
facing forward.
In this case you approach it the same
way but then you reverse it so that S
becomes R and R becomes S. since you're
looking at it backwards. But now what do
you do if priority group number four is
not facing towards the back? So you can
trace one to two to three. It's not
facing forward so that you can flip it.
How do you approach it? Now
your professors may teach you to redraw
the molecule which works except that it
wastes time and causes a lot of doubt on
your exams because you're not always
sure how to flip the molecule and if you
did it right. On exams, you can't afford
that kind of doubt. Some professors will
even allow you to use a model kit, which
works except that you have to waste time
building a kit, figuring out which
substituent is what, and that is also
going to waste time and cause doubt. You
can't afford that either. So, what you
can do on exams and what is guaranteed
to save you 5, 10, or even 15 minutes
and avoid any confusion is the swap
method. The swap method relies on your
understanding of the most basic
principles of kirality and identifying ananti
ananti
and recognizing that a pair of ananti
are the same exact molecule except that
they are the reverse the mirror image of
each other meaning that their groups are
oriented in the opposite direction.
Here we have two chyro molecules. I put
one on a gray center and one on a black
center so we can tell them apart. But
notice that they have four unique
substituents each.
And since they're perfectly superimposable,
superimposable,
they're the same exact thing, right?
They have the same kerality, same
substituents, same everything.
What we're going to do here is use the
gray one as a reference. And then we'll
take this molecule and grab any two
random substituents. I don't care which
ones they are because it's about the
fact that we're swapping them rather
than what we're swapping. We just did
one swap. Now, let's see what happened.
If we try to superimpose them,
it doesn't work. The white and red are
perfectly superimposable, but the green
and the blue are opposites. In fact, if
I put them side by side and attempt to
see a mirror between them, look at that.
They're perfect and anti of each other.
So, we took a molecule that was exactly
the same, swapped two random
substituents, and we got the anantimer.
Since we're using the gray one as a
reference, let's put it down. And now
we'll swap any two other random
substituents. It doesn't matter what we
swap second in reference to what we
swapped first. Now, let's see if we can
still have that mirror between the two.
We can't. The blue and white are
mirrored. The green and red are not.
Now, what happens if I try to
superimpose them? Look at that. They're
perfectly superimposable.
We started with a molecule, did one swap
to get the anantimer, but when we did a
second swap, we're back to the same
exact configuration.
Two swaps brought us back to the initial
kirality. Let's put our reference down
and we'll randomly do a third swap. As I
do this, I want you to predict what
we're going to see. One swap gave us the
anantimer. Two gave us the same thing.
Three swaps. Well, can we superimpose them?
them?
Looks like we can't. Red and green are
lined up. Blue and white are not. But if
we try to put a mirror between them,
they're perfect mirror images of each other.
other.
So, three swaps gave us an anantimer.
What do you think will happen if I do a
fourth swap?
Again, I'm choosing substituents at
random because it's about the fact that
we swap rather than what we swap. Four
swaps. Let's try to do that mirror
again. And it doesn't work. The blue and
green are mirrored. The white and red
are not. So, let's try to superimpose
them. And look at that. Voice swaps
exactly the same kirality. They are
perfectly superimposable.
So, let's quickly review what we saw.
When we had one swap, we got an
anantimer. When we did two swaps, we got
the same exact configuration. If we did
three swaps, we got the anantimer.
And when we did four swaps, we got the
same exact thing. If I were to continue
this pattern and let's say have 27 swaps
once again I would expect to have the
anantimer. If I had enough time and
swaps once again I would get the same
exact thing. And that's because as you
saw it didn't matter which groups I
took. It only mattered the number of
times that I broke and swapped bonds. So
if I do any odd number of swaps, I will
get the anantimer of the molecule I
started with. And if I do any even
number of swaps, I'm going to get the
same exact configuration of the chyro
molecule that I started with. Odd gives
me the anantimer and even gives me the
same exact thing. Let's prove this with
a very simple example. Remember that we
rank according to atomic number. If you
haven't memorized the atoms that I have
on the cheat sheet, I'll let you follow
along real quick. Over here, we have
HCNF, then phosphorus, sulfur, chlorine,
bromine, and iodine, where iodine is the
highest and hydrogen is lowest. We're
ranking a chlorine, bromine, hydrogen,
and carbon. Chlorine, bromine, hydrogen,
carbon. Roman is the highest priority
followed by chlorine followed by carbon
and hydrogen is the lowest priority.
Step one, prioritize. Step two, ensure
number four is in the back. We are good
to go. Step three, cross it out. Step
four, trace an arc from 1 to 2 to three.
In this case, we get S. What I have here
is the same exact molecule. But now,
what I want to do is take any two random
substituents and swap them. We just want
one swap from the original. And we want
to see if that comes out to be R or S.
So I will randomly swap the methyl group
with a bromine. We'll put a CH3 down
here and a Br up here. Keep track of the
swaps by writing swaps and just one. One
is odd. We expect it to be the
anantimer. Let's prove it. Bromine 1,
chlorine 2, methyl 3, hydrogen 4. Cancel
out number four. Trace the path from 1
to 2 to 3. And lo and behold, we have R.
We did one swap and we got R where our
starting molecule was S. Let's do this
again. First, we'll number we have
bromine 1, chlorine 2, methyl 3,
hydrogen 4. We'll keep track of our
swaps on the side. And let's just do two
swaps. And we want to prove that an even
number of swaps gives me the same exact
kirality. It doesn't matter what I
switch. For the first swap, I'll go
ahead and switch bromine and chlorine so
that I have bromine number one on the
right, chlorine number two on the left.
That's one swap. For the second swap,
I'll swap methyl and bromine so that I
have CH3 number three coming out of the
page. And I have bromine number one up
in the plane of the page. Two swaps.
That's an even number. That means I
should have the same exact kerality as
this molecule. I expect it to be S.
Number four is in the back. We cross it
out. Trace the path from 1 to 2 to
three. And look at that. We have S. The
molecule looks completely different from
what I started. But because I did two
swaps, I retained the chyality and once
again I have S. But this was a simple
molecule. Group number four was a
hydrogen, hydrogen was in the back. How
does this work in real life when you
have a molecule where group number four
is not in the back and not in the front?
So you can reverse it. Let's go back to
our initial example and see how to solve
that. I'm going to draw it twice because
I want to show you a trick that once you
get it makes it even simpler. We've
already prioritized the groups. So, what
we want to do now is keep track of the swaps.
swaps.
Anytime you're doing swaps, you want to
follow these rules. Swap number one is
to put group number four in the back
because the reason for setting it up the
way we do is having number four out of
the way and ranking one to two to three.
If number four is not in the front or
the back, the first swap, the entire
purpose of swapping is getting that in
the back so that we can prioritize
everything else around it. But doing
that one swap is an odd number, it gives
us an anantimer. And if you're anything
like me where you're absent- minded and
you'll do step one, get so excited,
forget about step two, you're going to
lose points. So step two or swap number
two is to add a second swap. any other
swap other than number four cuz that has
to stay in the back and you do that
second swap to get a total even number
so that your kirality stays the same. So
let's start by putting number four in
the back. We'll swap the hydrogen with a
group that happens to be in the back. So
we have hydrogen on dashes and O up in
the plane. That's my first swap. Keep
track of it. For the second swap, I need
to undo the anantimer. I need a total
even number of swaps and I can do
anything at random. I can swap the Cl
and the CH3. I can swap the Cl and the
O. It doesn't matter. You don't believe
me? Try it both ways and check both
ways. You'll see it's the same. In this
case, we have a CH3 moved left and a
chlorine that moved right. But we just
lost all the numbers. So, what we have
to do is put the numbers back based on
the priority that you see. Chlorine is
still number one. O is still number two.
Methyl is number three. Hydrogen is
number four. We have a total of two
swaps. That's an even number. That means
the kirality is the same as what we
started. But now hydrogen number four is
in the back. We cross it out. Trace the
path from 1 to 2 to 3. It goes towards
the left counterclockwise. And we know
that our molecule is S. We didn't build
a model kit. We didn't try to rotate it
in our brains or on paper. We simply
went by the anantimer principle or the
number of swaps and we got S. But if you
look at this, it starts to get messy,
especially when you're dealing with more
complex molecules. So an even simpler
way to do this is to pay attention to
what happened when we moved hydrogen to
this position. We then had to prioritize
hydrogen and we got a number four in
that spot. The purpose of swapping was
to move the numbers so that we can
prioritize numbers. So at the end of the
day, does it matter if I move the atom
or just the number that came with the
atom? When your molecules are bigger, it
gets so much easier to just focus on the
numbers and nothing else. I want to show
you how this works on the molecule on
the right. Once again, I'll keep track
of my swaps because number four is up
and that doesn't help me. For the first
swap, I need to put number four in the
back. So, all I'm swapping are the
numbers, not the atoms. I'm not saying
that O is priority number four. I'm not
saying that hydrogen is priority number
two. I am simply swapping the numbers,
keeping in mind that I would have
swapped the atoms with it, but I'm
saving time. That was swap number one.
That's odd. It's an anantimer. We can't
stop here. We have to swap again. The
second swap we had done was for the two
lower groups. So, here I'll swap just
the number one with the number three.
Placing the number one to the right, the
number three on the left. Keep track two
swaps. That's an even number. Should
have the same exact kirality. And
looking at just the numbers, cross at
number four. Trace the path from 1 to 2
to three. And just like in the previous
version, we have S without having taken
the time to swap all the atoms
themselves. And a question I often get
is, okay, so do I have to swap the first
two and the next two? No. Why did I swap
number four and number two first? Number
four had to go to the back. Number two
happened to be in the back. That is why
I had to do that as my first swap. The
second swap, why did I choose these two?
Cuz I felt like it. I could have chosen
any other groups to swap as long as once
four is in the back. I don't touch it.
Because the principle is the same. Just
like you saw with the model kit. It
didn't matter which substituents I
grabbed, which color I grabbed. As long
as I swapped any two an odd number of
times, I got the anantimer. Any
substituents for an even number of swaps
gave me the same exact thing.
And this also works for molecules that
are written in line structure, chair
confirmation, cycllohexane projection,
fisher projections, anything. The
principles apply everywhere. So, for
example, if I give you a molecule that
looks like this, where I have a mythoxy
coming forward and an O going to the
back, the first thing I want to do is
rank my substituents. If you're not
comfortable ranking when you have the
same atom on both sides, but then they
differ, make sure you see the next
video, which shows you how to rank
longer chains. For this molecule,
mythoxy is number one, hydroxy is number
two, and that's because mythoxy is OC.
Hydroxy is O. C outranks H. Then on the
top I have an isopropyl which is going
to rank number three and the ethyl ranks
number four. That's because both carbons
are outranked by both oxygens. But the
carbon with two carbons attached. The
isopropyl outranks carbon with one
carbon attached, the ethel. And once
again we have number four in the plane,
it's now forward or back. The few
seconds it takes you to write out the
word swaps A will avoid confusion and B
is still so much faster than whatever
method your professor has been teaching
you. For the first swap, I need to put
number four in the back. So, I will
simply swap the number. I'm not even
starting to redraw this chain. It's
going to get so messy. All I swap is the
number. I put number four on the dashes,
putting it back, and number two where I
grab the number four. So, that's one
swap. For the second swap, I can swap
whatever as long as I don't touch number
four. So, I'm going to choose to swap
one and two. One goes on top, two goes
on bottom. I have a total even number of
swaps. That means the kirality is the
same thing as what we started with, and
we're good to go. Cancel out number
four. Trace the path from 1 to 2 to 3,
and our molecule is S. If you need help
ranking molecules that are deeper with
multiple atoms on the substituent, be
sure to see the next video where we look
at how to compare substituents that have
the same starting atom, but then
something changes down the line. If you
feel confident with this information, be
sure to try my stereochemistry practice
quiz, which features medium to tricky
questions to make sure you know all the
tips and tricks to look out for.
Consider yourself warned, it is not an
easy quiz. You can find it along with
the entire series and my stereochemistry
cheat sheet by visiting my website leafersai.com/kyality.
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