Leah here from leah4sci.com and in this video we’re going to look at E and Z Isomers for
alkenes. In the last video, we looked at cis and trans isomers, if you haven't seen that
yet, click the link below or visit leah4sci.com/cistrans
And were left off with a question of what happens when you're trying to find cis or
trans for a pi bond where the substituents are set up in a way where they could be cis
to each other or they could be trans to each other. And when that's the case you need a
higher level of ranking and this is where the E and Z configurations come in. You typically
use E and Z when you have more than one carbon coming out of the same sp2 carbon atom. For
example, here we have an sp2 carbon with both a methyl and an ethyl so we can't rank based
on just one. We also use it when we have none carbon substituents like halogens, alcohols
or larger priority groups. For example if you have to name a molecule that look like
this we would have to find a way to imply that the chlorine is on the same side as the
methyl where as the methyl on the left is opposite to methyl on the right. The first
thing we have to do is rank the substituents coming off of that carbon atom, using the
Cahn - Ingold - Prelog system. This is something I teach in detail in the stereochemistry series
which you can find the linked below or visit leah4sci.com/ranking
Now we skimp on this, it's very important. As a quick review, we are ranking based on
atomic number, now you'll typically have a periodic table on your quiz or exam but this
is where those ten atoms that I have you memorized in the beginning of orgo, this is another
place where they really come in handy so if you haven't memorize that yet, do so now.
We have hydrogen, then carbon, nitrogen, oxygen fluorine. Below that we have phosphorus, sulfur,
chlorine, and then bromine and iodine. These ten atoms are the most common that you're
going to see especially in a topic like and E and Z. As a reminder, Iodine is your highest
priority, then bromine, then chlorine, sulfur, and phosphorus, then fluorine, oxygen, nitrogen,
carbon, and hydrogen is always your lowest. If there is a deuterium thrown in, remember
that it outranks hydrogen because it's just a heavy isotope of hydrogen. If you're comfortable
with ranking lets move one. Let's try the chlorine example and if you haven't memorize
the atoms yet we'll keep it on the side for you. Here's the method that I find to be simplest
and least overwhelming, first, identify and highlight your pi bond, make it very obvious
which two carbons you're looking at and then cover half of the molecule. This is so important
because when you have a lot of substituents, it's very easy to get overwhelmed and it's
very easy to forget what you're looking at but if you cover it by putting your hand or
a piece of paper or something over half of the molecule, you're only focusing on one
sp2, one pi bond carbon at a time and keeping it simple. Next, you want to identify the
atoms that you're ranking specifically, in this case we have a chlorine versus a carbon
and that methyl group, and we see that chlorine outranks carbon that makes chlorine priority
number one and carbon priority number two. Finally, draw an arrow perpendicular to the
pi bond in the direction of your higher priority, in this case my pi bond goes in this direction
so I will draw an arrow facing up towards the chlorine. We have one side, now we cover
it so we're not distracted and do the same thing on the other side. Here's my sp2 carbon
but I only have one substituent, what am I ranking? Don't forget that in line structure,
hydrogen atoms are invisible but they're still there. We have a hydrogen atom on this carbon
of the pi bond and that means we're ultimately comparing carbon and hydrogen. Carbon will
outrank hydrogen, making the methyl group priority number one, and the hydrogen priority
number two. Since the higher priority is facing up from the pi bond, we'll draw an arrow perpendicular
to the pi bond facing priority number one and then uncover both sides. Having the arrows
allows you to avoid the groups, it allows you to avoid being overwhelmed and all you
have to do is identify, are your arrows facing in the same direction as we have here, or
do you have one arrow facing up and one down in opposite directions. Now be careful, don't
call them cis trans, this is not cis trans, we're simply looking to see if the two higher
priority groups are facing in the same or opposite direction.
If you identify that your two arrows are facing in the same direction, this would be the z
configuration, and the way I remember that is by thinking that the two higher priority
groups are located on "ze zame zide"(the same side) , all spelled with the Z, "ze zame zide"
tells me that it's the Z configuration. If the two high priority groups are located opposite
to each other, you get the E configuration and the way I remember it is that these two
groups are located "Eeposite" (opposite) from each other, opposite spelled with an E to
remind me they are on "Eeposite" sides of each other. Let's go back to our example,
in this case, we know that the chlorine and methyl, both higher priority are facing "ze
zame zide", that makes this the Z configuration, to name this molecule, we start with the basic
naming rules where we see a total of four carbons in the parent chain, for a first name
of "but", and the pi bond sitting in the middle of the molecule. I will number this molecule
from the left so that chlorine and the pi bond are both on number two rather than three,
remember we want the total lowest set numbers which gives me a last name of "2-ene" and
then a chlorine substituent on carbon two, for "2-chloro", and lastly, we want to include
the E Z configuration. In this case, we have Z. To put the name together, we start with
the Z at the start of the name, and put it into parenthesis followed by the substituent
which is a chlorine on carbon two, we drag the two in front of the "but", giving me "2-butene".
The reason we put the Z in parenthesis is to make sure that we don't confuse the letter
Z with the number two, look at how I've drawn them, they kind of look similar but having
the Z in parenthesis reminds us "no, this is not a designation of where something shows
up, this is Z, telling me about the stereochemistry or the geometry about that pi bond." For our
final name of "(Z) - 2 - chloro - 2 - butene". Let's go back and name our first example,
we have a total of six bonds, giving me a first name of hex, we have an option to number
it from the right or left, if I numbered it from the left, I hit the pi bond to three
and the methyl group at four. If I number from the right, I hit the pi bond and the
methyl group at three for a total lower set of numbers so we're gonna choose the red numbering,
with a pi bond starting at carbon 3, we have "3-ene", a methyl group at carbon 3 gives
me "3-methyl", but now we have to specify what's going on with the pi bond. We don't
use cis trans because we have a methyl and an ethyl, instead we'll use the E Z system.
We'll start with the group on the left, identifying the sp2 carbon and recognizing because the
substituent is invisible, there is a hydrogen atom. If you don't have them memorized yet,
use this chart as a guide we're comparing carbon to hydrogen, carbon outranks hydrogen,
making the ethyl group priority number one, and the hydrogen priority number two, or simply
the higher priority in the up direction. Now cover the left side and look at the sp2 carbon
on the right, coming off of this carbon, we have a methyl and an ethyl, methyl is a ch3
that's carbon, ethyl is a ch2 ch3 that's also carbon, this is where it really helps to understand
priority ranking. As a reminder we're dealing with methyl, which is ch3 and ethyl which
is ch2 ch3. We look at the very first atom which is carbon, since it's the same, we rule
it out and go to the next highest priority. The methyl group has three hydrogens, so any
one hydrogen is the next high priority. The ethyl group has two hydrogens and one carbon
but carbon is the highest priority, so now we compare carbon to hydrogen, carbon wins,
making the ethyl group priority group number one and the methyl priority number two. We'll
draw an arrow in the direction of number one, and then uncover the molecule. Looking at
the two arrows, they are away from each other or "Eeposite" from each other, telling me
that this molecule is in the E configuration. Let's put the name together starting with
E in parenthesis, followed by the substituent "3-methyl". Pull the 3 in front of hex, giving
me "3-hexene". For our final name of "(E) - 3 methyl - 3 - hexene". Your professor may
choose to leave the three where it is, in that case the name would be "(E) - 3 methyl
hex - 3 - ene", both are correct. What about a molecule that looks like this? We have a
pi bond with two unique substituents, if we look at the sp2 carbon on the left, chlorine
outranks ethyl, making the chlorine higher priority, but on the right we don’t appear
to have anything. Meaning, we have two hidden hydrogen atoms, but once again like we saw
on the last video, hydrogen is the same as hydrogen. If you have two equal priority groups,
one doesn't outrank the other and that means there is no stereoisomerism here, because
there is no way to differentiate the top or the bottom, so this is not going to be cis
trans and it's also not going to be E or Z because of that hydrogen. The trick to remember?
Terminal alkenes without anything fancy on the end are not cis trans and not E Z.
For even more examples on cis trans and E Z, click the link below or visit my website,
leah4sci.com/cistrans
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