Entgegen-Zusammen naming scheme for alkenes examples

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.