Leah here from leah4sci.com and in this video we're going to look at practice problems for
Chirality and Stereochemistry. To solve these problems we're going to use the tutorials,
tricks and shortcuts that I teach in the chirality series which you can find on my website along
with the stereochemistry practice quiz leah4sci.com/chirality.
Let's start with some warm up problems where we want to draw the enantiomer of a given
chiral molecule. In this case we have a 4-carbon chain with a chlorine on carbon 2 coming forward.
That tells us the chiral carbon is carbon number 2. We have two easy ways to draw the
enantiomer and while it doesn't matter which one you use sometimes one will make more sense
than the other. We can either use the swap method so simply swap the chiral centers or
we can draw the mirror image by drawing a mirror plane next to the molecule and drawing
the other one in reverse. Let's start with the swap. To do that I copy the carbon skeleton
exactly as I see it and simply swap two of the groups on the chiral carbon. In this case
I'll be swapping chlorine which is coming forward with the invisible hydrogen which
is going back. I don't have to show the invisible hydrogen, I just have to recognize that that's
where I'm creating the swap and so when I draw up the enantiomer I put chlorine on dashes
to show that it's going into the page allowing us to understand that hydrogen even though
invisible is now coming forward out of the page.The other option is the mirror image
and this is where we do exactly what we see but right becomes left and left becomes right.
That means the carbon skeleton looks like this but chlorine is still coming up and out
of the page. This looks tricky, the instinct for many students is to draw the mirror image
but to put this in the back, to put it on dashes rather than on a wedge but you wanna
be very careful. Drawing the mirror image is like doing number 1, creating a swap but
we're not swapping this group, that still comes forward in the mirror, we're just swapping
whatever space in the mirror and whatever is away. So one other way to think about it
is that we really did option number 1 but what we swapped is the ethyl and the methyl
group so that the ethyl on the right plane on the page is now on the left in the plane
of the page and methyl which was on the left is now on the right. Let's prove that we did
the swap correctly by finding R and S for each stereocenter. When assigning priorities
we take the periodic table into consideration specifically these 10 atoms that i highly
highly recommend you memorize to help you with so many topics. On this molecule, we're
comparing chlorine to carbon, to hydrogen. Chlorine is going to be our highest priority,
hydrogen would be our lowest priority. And with carbon we'll look at what else is attached.
On the ethyl group we have a CH2 bound to a CH3. On the methyl group we have a carbon
bound to 3 hydrogen atoms. When we compare the carbon to the carbon and they cancel out
we move on the next highest priority which on the ethyl is another carbon and the methyl
is anyone of the hydrogen atoms. This makes the ethyl outranked the methyl or you can
simply remember the trick that a methyl group is always going to be your lowest priority
after a hydrogen atom. Chlorine is priority number 1, ethyl is number 2, methyl 3, and
hydrogen on the back is number 4. To cancel out number 4 trace a path from 1 to 2 to 3
giving us a starting starting molecule with the R configuration. Let's look at the mirror
image, chlorine 1, ethyl 2 , methyl 3. Hydrogen is number 4 but it's going into the back so
we don't really care we can just put dashes 4 to show that we know it's there, we're gonna
account for it by crossing it out and that's it. We don't actually have to show the hydrogen
and then trace the path from 1 to 2 to 3 giving us the S configuration. Let's do the same
thing for the molecule below, chlorine is still 1, ethyl 2, methyl 3. Hydrogen is now
a little bit of a problem because it's number 4 but it's not in the back. Same thing though,
cancel out number 4, trace the path from 1 to 2 to 3 but when you're arc goes clockwise
and number 4 is in the front we simply switch the direction to counter clockwise giving
us the S-configuration. If the starting molecule is R, the enantiomer should be S and we know
we did it correctly.
Same thing for the next problem, draw the enantiomer of this chiral molecule, go ahead
give it a try, let's see what you got. For a molecule like this with a ring, I find the
easiest thing to simply draw a mirror next to the molecule and draw the same thing in
reverse. So I draw everything I see but whatever is facing the mirror has to be facing the
mirror from the other side and that means bromine up and towards the right is still
going to be up but now it's towards the left because it's still facing the mirror. OH down
but towards the right is now going to be OH down towards the left once again facing the
mirror. For more practice try to assign R and S for the starting molecule and the enantiomer
to prove you did it correctly. Once again we want to draw the enantiomer this time looking
at a fischer projection. If you haven't studied fischers ignore this and jump to the next
problem, otherwise take a look and think about what you wanna do.
With Fischer Projections it's just as easy to draw a mirror or to do a swap. But remember
if you're doing a swap we have two chiral centers so we have to make sure to swap both
giving us the enantiomer rather than a diastereomer. Let's start with a mirror image, just draw
a mirror down center, copy out the skeleton adding all the achiral substituents then we'll
add the chiral substituents such that anything facing the mirror is still facing the mirror
in the opposite direction giving me 2 H groups on the left and 2 hydrogens on the right.
If you choose to do this with the swap method instead, you can swap any substituents, most
students like to swap what's red in left and wait a minute, doesn't I give you the same
exact thing? What do you think? Ready to go to more advanced problems? First give this
video a thumbs up and if you haven't already subscribe to my channel and then let's see
if we can assign R and S to a few more chiral centers. For this molecule, first we want
to identify the chiral centers. How many do you see here? Pay very very close attention
because if you said two, the answer is just one. Look at it, this appears to be a chiral
center and indeed it is! In order for a carbon to be chiral it has to have 4 unique substituents.
This carbon right here has a methyl group has an NH2 and an isopropyl group. That's
just 3 but don't forget we have an invisible carbon atom coming out of the page giving
us substituent number 4. This carbon is pretending to be chiral and I'm including this here because
professors love to put this on exams and students always fall for it but you're not going to.
Even though the methyl group is on a wedge it's pretending to be 3-dimensional, it's
pretending to be chiral, it's not. What do we have in this carbon? Let's first fill in
the invisible hydrogen going into the page towards the back and then counter substituents.
We have a hydrogen into the page, that's 1. This entire group, not even trying to name
it just recognize it's complicated but it's unique, that's number 2. And then we have
a methyl and we have a methyl. Just because they're drawn differently doesn't make them
different. Having 2 CH3 groups on the same carbon makes this an achiral carbon and therefore
gives us just one chiral centers on the molecule. Now that we have our chiral center, let's
find its designation. What do we have? we're comparing Nitrogen to Hydrogen, to 2 carbon
atoms. A quick peek at the periodic table tells us Nitrogen is highest priority, hydrogen
you don't even need the table just recognize it's lowest priority giving us a number 4
for hydrogen, a number 1 for nitrogen. And then isopropyl versus methyl, methyl is CH,
isopropyl is C bound to 2 Cs making it higher priority. Isopropyl is 2, methyl is 3, carbon
number 4 is not going towards the back so we're gonna reverse what we get, cross out
number 4, trace the path from 1 to 2 to 3 which appears to go clockwise but reverse
it to go counterclockwise because hydrogen coming forward requires us to reverse what
we found and that gives us the S-configuration.
Let's try another. In this case we have a ring but that doesn't make it more difficult.
A lot of students instantly freeze up when there's a ring but if you treat it the same
way as a linear molecule it'll be much easier. First thing we do is identify the chiral centers
and in this case we have 2. How do I do there are two? Let's not forget our invisible hydrogen
atoms on the upper right we have a hydrogen coming forward out of the page and on the
bottom of the ring we have it going down and into the page. Let's peek at the periodic
table starting with the lower chiral center. We're comparing sulfur to hydrogen to a carbon
that's a CH2 on the ring and carbon that's a CH2 on the ring. That doesn't help us. This
is where you remember the trick, if you can trace a boring path on the chain or around
the ring, the first one to hit the substituent is higher priority. We have a shorter path
in the counterclockwise direction making that the higher priority and this one which is
a much longer path to the next substituent. What is that tell us? Sulfur is always gonna
be higher than carbon, hydrogen is always number 4, the shorter path to this substituent
will be priority number 2, the longer path to this substituent will be priority number
3. Number 4 is conveniently in the back so cross it out, trace the path from 1 to 2 to
3 which gives us the S-configuration.
Let's do the same thing for the upper chiral center but here it get s a little tricky.
We automatically know that hydrogen is number 4 but the remainder of the groups are comparing
carbon. Carbon on the ring, carbon on the ring. This is where we look at the next atom
and the next atom. The substituent is a tert-butyl, it's a carbon bound to 3 other carbons making
it the highest priority. Even Though sulfur as a substituent is higher in priority than
anything on the tert-butyl, the fact is carbon hits the 3 methyl groups faster than this
carbon hits the sulfur because this carbon hits a carbon before it hits sulfur. However,
comparing the 2 directions around the ring, it's a shorter path the sulfur making this
number 2 and a longer path to the sulfur making this number 3 and now we're good to go. Since
number 4 is in the front, we're going to reverse our configuration so first cross out number
4, trace a path from 1 to 2 to 3, reverse the path because number 4 is in the front
and that gives us the S-configuration.
What do you think? Do you feel warmed up for stereochemistry? ready to tackle even more
practice? If so, make sure you visit my website for the chirality and stereochemistry tutorial
series, practice quiz and cheat sheet. Go to leah4sci.com/chirality.