What's the relationship here? Are they the same? Enantiomers? Quickly finding R and S,
we see the alcohol is R and S or opposites, Enantiomers? The bromine is R and R, same
thing and what about this molecule? the blue chlorines are R and S, Enantiomers? The purple
chlorines are S and R, also Enantiomers? Actually, No, No, and DEFINITELY No. In previous videos
at leah4sci.com/chirality, linked below, we looked at how to find chiral centers and designate
R and S no matter how it's presented. In this leah4sci video, we'll look at what happens
when the molecule has more than one chiral center. If a molecule has more than one chiral
center, we use 2 to the N to find the maximum number of potential stereoisomers where in
this molecule, OCH3 and Chlorine both sit on a chiral carbon and if N is equal to the
number of chiral centers, we have a maximum of four potential stereoisomers. Let's find
them all using a simple copy and flip method, I start with the skeletons and then I copy
the substituent and flip the chirality. On the right, the starting molecule has two wedges
so we put both substituents on dashes. For this next one I'll keep the first substituent
the same and flip the other and then find its mirror by flipping both wedge becomes
dash and dash becomes wedge. Now that we have the structures, let's find the relationship
between them. The standard method of finding R and S for every single chiral center absolutely
works but takes SO MUCH TIME. Instead, we'll solve for one set and then find the rest using
the swap method that I teach in the videos linked below. If one swap gives me the enantiomer,
having OCH3 on the wedge for R, means OCH3 on dashes for S, and wedge of course is still
R. Same thing for chlorine, having chlorine on a wedge gives me R which means chlorines
on dashes gives me S. Not confident with the trick? Pause the video, Prove the Rs and Ss,
as for the relationships, if R to S is a mirror image or Enantiomer when the entire molecule
is mirrored, the molecules are Enantiomers to each other. What if it's only partially
mirrored where I have R R on one chiral center and R S on the other? In this case the molecule
is of 2 MINDS or Di-stereominds, giving us a Diastereomer. Even without the R and S designation
the quickest way to identify a diastereomer is to see that one substituent on the molecule
is exactly the same and the other is opposite. This gives me another set of diastereomers
and let’s not forget the corners, we have one that is different and one that is the
same for another set of diastereomers and the same thing here. This became so messy
so let's summarize. If you have two chiral centers for a maximum of four potential stereoisomers,
every molecule will tend to have one Enantiomer and two Diastereomers. Pause this video if
you need a minute for it to sink in and let's continue. Now let’s try it with 2,3-Dibromobutane
but before we dig in, make sure to give this video a thumbs up, subscribe to my channel
and hit that bell icon so you don't miss out on any new videos and let’s count. I see
two chiral centers, does that mean I have four stereoisomers and will each molecule
have one Enantiomer and two Diastereomers? We start with four skeletons and draw one
chiral molecule. I'm choosing to put both bromines forward and then we'll copy and flip
both bromines on dashes, keep one and flip the other and of course its Enantiomer. Next
we could find R and S for each one but what if we don't? What if instead we just look
at it the way we did before. These two molecules are mirrors at every chiral center, making
them Enantiomers, these two appear to be mirrors as well. Are they Enantiomers? Actually No,
these two are not Enantiomers. It's the exact same molecule. The green bromine is R, the
purple is also R, that makes their Enantiomers S and S and on the bottom we have the green
R, green S, purple S, purple R. Once again, if we have R and S then S and R, how are they
not Enantiomers? The difference between this and our last example is that both chiral centers
have the same exact substituents. We have bromine, a methyl group and don't forget that
invisible hydrogen. Here I have a molecule with two chiral centers, you'll notice that
they each have the same four unique substituents. White, red, green and the other side white,
red, green and the other side. If I rotate it, notice that internal plane of symmetry
where this half is reflected in this half right down the middle, it's exactly the same.
If I take this molecule and build its mirror image, notice the plane of symmetry going
down the middle. These are not Enantiomers because look at what happens when I rotate
it, they are perfectly lined up. They are the same exact thing. Let's draw it out with
this molecule and we won't look at the invisible hydrogen. We'll do a 180 degree rotation around
the central bond between the two chiral centers. We'll tilt the results slightly and redraw
the molecule as follows, the left half doesn't change with bromine coming out of the page,
this is where I had a rotation and that gives me for the right side of the molecule where
I have the purple bromine. A methyl group that is down in the plane of the page because
180 degrees from up in the plane of the page gives me down in the plane of the page. Bromine
is down and into the page, rotated 180 we get up and out of the page. Down becomes up,
dash becomes wedge and look at that, I have an internal plane of symmetry in a molecule.
I have an Enantiomer within the molecule itself where one half reflects on the other. This
is called a Meso Compound and while we do have two chiral carbons the molecule is optically
inactive because the symmetry cancels each other out. If I do the same thing for this
molecule where I keep the left the same and for the right, methyl goes down in the plane
of the page. Bromine wedge down is now dash up and once again, internal symmetry. If we
bring them side by side they appear to be mirror images of each other but if I flip
this molecule as I did with the model kit, they are perfectly superimposable on each
other and that makes them the same exact thing. Going back to our original example, no, these
are not Enantiomers, it's exactly the same thing and if they're exactly the same, we
have a total of three rather than four stereoisomers. So what about this equation? Two to the N
equals four max but the meso compound costs us one. How do you recognize meso compounds?
One, the chiral molecule has the same exact substituents on each chiral center and two,
they have the opposite chirality. One is R the other is S or one is S and the other is
R. But this is the long way, the faster way, which isn't always possible, is to look for
that internal mirror, take these chair conformations for example the molecule has two OH groups
but we can't necessarily see an internal plane of symmetry. What I can’t tell is that the
two molecules are mirror images of each other and so they are potential Enantiomers. When
in doubt, flatten it out. If you didn't immediately recognize that the OH is going up means that
they're internally symmetrical, just flatten it out, we'll make these carbons one and two
and put both OHs on a wedge since they're coming up and out of the chair and look at
that, internal plane of symmetry meso compounds. Or you can go back to option two and find
R and S on your chair conformation or on your flattened structure. R and S with identical
substituents once again confirms that this is a meso compound. For even more on chirality
including videos, practice quiz, and cheat sheet, visit leah4sci.com/chirality, linked
below, again that's leah4sci.com/chirality .
it notice that internal plane of symmetry where this half is reflected in this half
right down the middle it's exactly the same if I take this molecule and build its Mir
image notice the plane of symmetry going down the middle these are not an antimers because
look at what happens when I rotate it they are perfectly lined up they are the same exact
thing let's draw it out with this molecule and we won't look at the invisible hydrogen
we'll do a 180° rotation around the central bond between the two chyro centers we'll tilt
the result slightly and redraw the molecule as follows the left half doesn't change with
bromine coming out of the page this is where I had a rotation and that gives me for the
right side of the molecule where I have the purple bromine a methyl group that is down
in the plane of the page because 180° from up in the plane of the page gives me down
in the plane of the page bromine is down and into the page rotated at 180 we get up and
out of the page down becomes up down becomes wedge and look at that I have an internal
plane of symmetry in the molecule I have an enan within the molecule itself where one
half reflects on the other this is called a miso compound and while we do have two chyal
carbons the molecule is optically inactive because the Symmetry cancels each other out
if I do the same thing for this molecule where I keep the left the same and for the right
methyl goes down in the plane of the page bromine wedge down is now dash up and once
again internal symmetry if we bring them side by side they appear to be mirror images of
each other but if I flip this molecule as I did with a model kit they are perfectly
superimposable on each other and that makes them the same exact thing going back to our
original example no these are not a an humors it's exactly the same thing and if they're
exactly the same we have a total of three rather than four stereo isomers so what about
this equation 2 the N equals 4 Max but the misoc compound costs us one how do you recognize
miso compound one the chyro molecule has the same exact substituents on each chyal Center
and two they have the opposite chirality one is R the other is s or one is s and the other
is R but this is the long way the faster way which isn't always possible is to look for
that internal mirror take these chair confirmations for example the molecule has two o groups
but we can't necessarily see an internal plane of symmetry what I can tell is that the two
molecules are mirror images of each other and so they are potential ananth when in doubt
flatten it out if you didn't immediately recognize that the O is going up means that they're
internally symmetrical just flatten it out we'll make these C 1 and two and put both
O's on a wedge since they're coming up and out of the chair and look at that internal
plane of symmetry miso compound or you can go back to option two and find RNs on your
chair confirmation or on your flaton structure RNs with identical substituents once again
confirms that this is a miso compound for even more on chirality including videos practice
quiz and cheat sheet visit ly.com chirality link below again that's layer for.com kyal