Based on what we learned in the
last video, let's see if we can name these two
molecules here. Let's start with
this blue one. So the first thing we notice,
it is an alkene. It has a double bond
right there. So we can, right from the get
go, write that it's going to end with an ene. And then we need to figure
out what the longest carbon chain is. And it looks like it's
this one here. One, two, three, four, five,
six, seven carbons. So it's going to be hept as
our prefix on the alkene. Let me write it out a little
bit further to the left. And then we have to figure out
where the double bond is. And we always want to start
numbering our carbons from the direction that bumps into the
double bond first. And so the double bond is closer
to this end of the chain than that end. So we're going to start
numbering here. So one, two, three, four,
five, six, seven. And so the double bond starts
on the three carbon. So it is hept-3-ene. And we also have a
functional group. We have a methyl group sitting
on the number four carbon. So it is 4-methylhept-3-ene,
which is correct. But we still haven't considered
what's going on, on either side of the
double bond. Whether the higher priority
functional group is closer to each other on both sides of
the double bond or whether they're farther away
from each other. And to do that we first have
to identify the higher priority group on each side
of the double bond. So on the right side of the
double bond-- let me circle it in magenta on each side-- so
on the right side of the double bond, we only
have one functional group, right over here. You could view this
as an ethyl group. That's pretty obvious. There is no other functional
group bonded to the carbon. There's only a hydrogen
over here. Now on the left side it's a
little bit less obvious. We have two functional groups
from the point of view of this carbon. You have the methyl that we
already pointed out, and you also have these three carbons
right over here. You could view that
as a propyl group. And what we need to do to
identify the highest priority group is to use the
Cahn-Ingold-Prelog namings or priority scheme that we learned
several videos ago. And there you literally go from
this carbon, you look at what it's bonded to, and you
compare the atomic numbers. But in both cases it's
a carbon to a carbon. This is a carbon to a carbon. So their atomic numbers
are the same. So then you go one bond further
away and you see which one is bonded to a higher
atomic number atom. This carbon bonds to a
carbon, which is a higher atomic number. This carbon only bonds
to three hydrogens. This one does two hydrogens
and one carbon. Because it's bonded to another
carbon, it takes priority. This propyl group is a higher
priority functional group. So now when we're trying to
decide whether it is entgegen or zusammen, we look
at these 2 groups. And we see that they are sitting
on the same side of the double bond. They are both above
the carbons. They are closer to each other. So this molecule is zusammen. Which on some levels, you can
think of as the same thing as cis, but cis and trans stops
applying when you start having more than two functional
groups. In this case, we have three. So we would call this
Z-4-methylhept-3-ene. And that's because the higher
priority functional groups are on the same side of
the double bond. Now let's do this
one over here. And someone pointed out,
rightly, that I had misspelled zusammen in the last video. It's actually spelled
like this, zusammen. I had spelled it with
two s's and one m. I guess you can forgive me. I don't speak German. But anyway, I thought I
would point that out. Now let's try to label this
thing right over here. So the first thing, this once
again is an alkene. Let's identify the longest
carbon chain here. So it looks like one, two,
three, four, five, six, seven, eight carbons. Double bonds are closer
to the left hand side. So we will start
numbering here. One, two, three, four, five,
six, seven, eight. So just the main chain is oct--
let me make sure I have some space here--
it is oct-3-ene. And then we have, well we have
one functional group sitting off of the main chain. We have this bromine sitting
right over there on the third carbon. So we would call this
3-bromooct-3-ene. And now we have to figure out
is it entgegen or zusammen. So if we look on the carbon on
the right hand side, it's pretty obvious that this is
the only functional group. We just have a hydrogen there. So let me circle it
in the magenta. And then on the left hand side
we have two functional groups. We have this [UNINTELLIGIBLE] bromo or we have a bromine
sitting right there. And then we have this
propyl group. And because this propyl group
is bigger, you might be tempted to say it takes
higher priority. But remember, in the
Cahn-Ingold-Prelog system, you give higher priority to
the atom that has a higher atomic number. Bromine has an atomic
number of 35. Carbon has an atomic
number of only 6. So Bromine is actually
higher priority. So this is the higher
priority functional group right over here. So now for deciding whether it's
entgegen or zusammen, we see that our higher priority
groups are apart. They're on opposite sides
of the double bond. This one is on top. This one is below. We are apart. So this is entgegen. Or we would write this is
E-3-bromooct-3-ene. And E is for-- just as a bit
of a refresher-- it's for entgegen, a word that I enjoy
saying, entgegen. There you go.