Entgegen-Zusammen naming scheme for alkenes examples
Entgegen-Zusammen Naming Scheme for Alkenes Examples. Created by Sal Khan.
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- Is Bromine considered with a higher priority then the ethil group (5:03) becouse the total atomic mass of the group is smaller then the Bromine or just becouse of the Carbon = 6?(12 votes)
- When determining priority, the total molecular weight of a functional group has nothing to do with it. The only thing to compare is the singular atom of the highest mass in each group. e.g. if you were comparing a fluoro- group [~19g/mol] and an isopropyl- group [~43g/mol, CH(CH2)2], the fluoro- group still has a higher priority. It helps me to think in relatable metaphor, so I think of it like this: If each one of the functional groups were a food, E/Z and cis/trans indicate which food is tastier, whereas the molar mass is just the caloric content.(14 votes)
- What if there are two double bonds? How do you use the E-Z naming system in this case?(5 votes)
- Each double bond is assessed separately to see if it is E or Z, then this information is put at the beginning of the name. For example if you have a molecule that has a Z double bond on the 3rd carbon and an E double bond on the 6th, you would name the molecule as (3Z,6E)-restofmoleculename.(16 votes)
- How can you label the compound at3:20a Z isomer when the functional groups on either side of the compound are different (propyl and ethyl)? i still don't understand i thought they have to be the same functional group to be an isomer? (e.g. methyl and methyl etc.)(5 votes)
- This is a good question. We should review two things to understand what's going on here:
1. Isomers are two different molecules that have the same molecular formula. This video is dedicated to teaching the difference between E and Z isomers (similar to cis and trans isomers, only more general). The molecule drawn at3:20is the Z isomer of 4-methylhept-3-ene—the corresponding E isomer is a completely separate molecule and is not drawn. This is why the propyl and ethyl functional groups don't have to match. Your suspicion was good, though—if Sal were trying to draw the E isomer to the molecule at3:20, he'd definitely have to match the functional groups.
2. To actually determine that it was Z, Sal used the ICP convention (see Stereochemistry > Chirality and the R,S system > Cahn-Ingold-Prelog System for Naming Enantiomers) to determine the priority of the two functional groups on either carbon. Since the highest priority functional groups were one the same side of the double bond, it happened to be Z. A mnemonic you may find helpful for Z isomers is "On Z same side," spoken in a German accent. Obviously, if it's not Z, it's E.
Check out the video again and see if it makes sense.(8 votes)
- Why are some methyl groups named H3C whilst others are named as CH3 - aren't both of these the same thing?(4 votes)
- Good observation! They are exactly the same thing! This is more for technical consistency than anything. Usually, when H3C is used, it is because it is drawn on the left side of the molecule, the reason for this, is so that you can clearly see that the bond comes from/goes to the CARBON atom, not a hydrogen. It is still perfectly correct to use the CH3 every time, but it can sometimes get confusing when you have a complex molecule and you need to know what bonds are where.
for example, dimethyl ether can be written as such:
the two methyl groups are the same, but since they are written in this way, you can clearly see that the bonds are between the C and the O atom. Hope this helped, feel free to ask for any other clarifications!(7 votes)
- Hi, is it possible to have an E,Z configuration with cyclic molecules. As a specific example could you use 1-methyl, 4-isopropyl cyclopentene? Thanks!(3 votes)
- It is not. Try drawing it: the bonds would be exceptionally strained and unstable if they went into a "trans" configuration.(4 votes)
- What would the naming be for the last molecule if there were 2 bromines attached to the left hand side carbon in the last molecule ("3:52")? Would the E-Z naming apply at all in this case?(3 votes)
- If there's two then you won't have to convey which bromine is where using the E-Z convention because you have no choice but to draw both in the correct position.(3 votes)
- can alkenes with a chiral sp3 hybridised carbon show optical isomerism?(3 votes)
- Yes, although any carbon that is part of the double bond will not be be sp3 hybridized and therefore is not chiral. The chiral sp3 hybridized carbon will have to be found on some other part of the molecule.
Another possibility is to have a polyene with two adjacent double bonds. This is known as an allene and forms a stereogenic center.
Check out wikipedia for more details http://en.wikipedia.org/wiki/Allene(2 votes)
- Hi, so if you had an alkene with two substituents that were only different enantiomerically i.e. one was an R group and the other an S group - how would you prioritise them?(2 votes)
- My professor seems to not like the double bond number being within the alkene name. Would it be acceptable to call the last molecule in the video (E)-3-bromo-3-octene?(2 votes)
- I'm a bit confused:don't you need to indicate the position of the ethyl and propyl group as well in the name? i.e. (Z)-3-ethyl-4-propyl-4-methylhept-3-ene? or am i wrong?
- Yes, you have to indicate the position of all the groups in the name.
But your compound (Z)-3-ethyl-4-propyl-4-methylhept-3-ene doesn't exist, because you can't have two substituents on C-4 of hept-3-ene.
You can have 3-ethyl-4-propylhept-3-ene or 3-ethyl-4-methylhept-3-ene, but neither of them is E or Z.
You could, however, have (Z)-4-ethyl-3-methylhept-3-ene.(2 votes)
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.