Transcript for:
Stereocenter Configurations

inter chapter C is now going to address how to identify the configurations about a stereocenter and we already note that the configurations are R and s but now we're going to learn the tricks and the skill sets on how to actually assign those configurations and then we are going to learn about how to assign Z and E configurations to a double bond so the IU pack rules for assigning configurations that has some three simple steps that we need to take in order to do this and in order to there's a name right here that's not the button so the rules to figure out the configuration is however you pronounce that okay that's the special name you don't need to remember that name okay but you do need to understand these three rules or steps to assigning the R and s these are the basic steps so there's going to be additional steps as we progress through the lecture but these are the three basics to get us started so what it is telling us to do is if we have a molecule with a stereocenter so one that looks like this let's see a carbon with a chlorine with the nitrogen okay so if we have a stereocenter like this and we know it's a stereocenter because there's four different groups step one says that we need to prioritize the four groups about that stereocenter in a priority of one through four four being the lowest and one being the luck the number one priority who are the top priority and the way you prioritize these four groups the chlorine nitrogen oxygen and hydrogen is looking at the periodic table and looking at the atomic mass so if we look at chlorine we see we have an atomic mass of thirty five point four five and if we look at nitrogen its 14 if we look at oxygen its 16 and hydrogen is one so you just rank the groups based off if they're their weight now I'm not looking at the weight of the Oh H or the nh-2 I'm just looking at the weight of the atom that is directly attached to the stereocenter so the chlorine would get Priority One let's see what's bigger between oxygen and nitrogen oxygen is larger so that gets priority two nitrogen gets prior to three and then hydrogen gets the lowest priority with a four mmm now this is step two is critical for your success in being able to identify the configurations is that the lowest priority right here the lowest priority may not be a hydrogen it may be a different group the lowest priority group has to be pointing away from you which simply is saying it needs to be the - lowest priority groups need to point away from us in the - and then you take your pencil or pen and trace through numbers 1 2 & 3 you don't worry about number 4 so if I trace one let's make sure this let's clear that up if we go from 1 to 2 to 3 what direction is that arrow going in we see that it is going counterclockwise and when it goes through the numbers 1 2 & 3 in the counterclockwise direction that is going to be a s configuration now let's draw let's do another problem here what if we have this okay remember the steps that says prioritize things 1 2 3 4 step two we make sure the lowest priority is a - which it is and then we just trace through one two and three one two three we can see that that is going in the clockwise direction so that makes this stereocenter a R so those are the simple or not simple the basic steps for assigning R and s now you can see where the mistakes are going to can possibly come in so this is an ARFF no doubt it's an r but what if you made a mistake and assigned nitrogenous priority to an oxygen as priority three you would be like okay well one two three counterclockwise so that's an S and that's a hundred percent wrong and you won't get any credit for it because obviously I know you guys have the skill of tracing an arrow through one two three that's easy right that's it and it's easy to know that and to memorize that counterclockwise is s that's not hard okay but what gets tricky is the prioritizing that has to be done correctly guys so that one makes it are the other thing that gets students all the time it is let's say the hydrogen or the lowest priority is not in is not a - let's change it up to this what if we had this so it go through steps one which simply states we need to prioritize things so that's one two three and four okay and then just then if you forget and skip step two you'll go right to step three and go like this and say hey that's going in the clockwise direction so that's an R but you skip step two you have to get the lowest priority group point it away from you okay so let's go back to the slide before we go further into that so you can see what's going on so if we have yeah okay let's get to a site that we want to talk about let's see here okay let's talk about this molecule here so step one we have to prioritize all the groups and we see that we have prioritized things correctly bromine is the largest chlorines the next the carbon is third and hydrogen's the fourth step to states that we need to get the lowest priority group into the back or the - so you can do that by rotating about the bond so let's say here let's draw this molecule so what do we have at Carbon let's do a different color so we have carbon bromine methyl group hydrogen and then our chlorine so if this technique works for you this is great what you're going to do is grab the bromine with your hand and you're just going to spin the whole molecule by grabbing onto the bromine so look at the bromine as like a handle or a knob you're grabbing it and you're spinning the whole molecule and you're going to spin it in such a way that you're going to get the hydrogen to become a - so you rotate it and you get the hydrogen as the - but then the tricky part if you can't see it you need to make a model where is the chlorine going to be and where is the methyl group going to be if you spin it to get the hydrogen into a - you're going out the chlorine in the plane and you're going out the methyl group coming out at you okay and now you so that's like step so step one was to prioritize everything step two make sure the hydrogen's a - or the lowest priority - - so it this was our original problem so step two would tell us hey we need to spin the molecule and orient it in which to get that lowest priority into the - and then we do what we are good at tracing the numbers one through three one two three we are going in the clockwise direction so that would make this both of those ours but not everyone is good at taking molecules and spinning them and rotating them in three-dimensional space so there's some tricks for those people who have a hard time seeing things in their head okay here's the first trick that you can do the so when the first step is to prioritize things which the slide is done for us very nicely step two says hey the lowest priority isn't a - but I don't like rotating molecules because I get confused so what you can do is just leave it alone don't do anything and proceed to step 3 and you do you trace it like you normally would and you can see going from priority one two and three we get the s we're going counterclockwise which is an S but then you just simply take the opposite of what you just calculated so we know that this is a R that has an R configuration yes the way I'm tracing it through the priorities one through three that's going counterclockwise and that is an S that is true but because we're violating step two we just take the opposite of what we just calculated so that has a our guy do you see how when I go here that's the exact same molecule the molecule on the slide is this molecule here I just rotated it exact same off - and it has an R so by golly this one has to be R right because they're the same molecule but this is just a second another way to find the configuration just leave it alone but here's the cautionary tale this trick only works only only only works if the hydrogen or the lowest priority is a wedge okay you can only do this opposite trick when the lowest priority is a wedge if the hydrogen or the lowest priority is in the plane okay if the lowest priority right here we can rank them 1 2 3 if it's the lowest priority is in this plane and you know it's in the plane by looking at this bond it's not a wedge or a dash if it's in the plane that trick the opposite trick that I just showed you does not work you have to take this molecule that I just drawn and you have to rotate it ok you have to have to have to or if there's another trick so let's say we have let's look at this molecule here so I'm going to rotate it let's get that hydrogen into the plane ok so we need to make sure we're doing this right so that means the chlorine would now become the wedge and the methyl group is on the dash okay so remember or just notice that this molecule sir in blue right here this guy right here is the exact same off to over here so we have to get the same configuration it has to be an R but what if this was what was given to you on the exam if that was given to you on the exam you had no idea about this guy to my left so what you have to do now step one says we prioritize one two three four step two we clearly see that the lowest priority is not a dash so that's a problem we can't do the opposite trick because the lowest priority is not a wedge so what can you do you can do a switcheroo trick okay what you can do is you identify this molecule as the original and then you redraw the molecule but you do a switch and what you're going to switch is you're going to take the lowest priority and switch it with the dash group so we can see now the hydrogen is there and the methyl group is here and the chlorine just didn't move it just stayed now this molecule that I just drew is not the same thing as the original and the reason is because I took those two bonds broke them off and switch them when you break two bonds and switch them you're going to change the stereo you're going to change the configuration so this one right here is our trick molecule this is to help us okay so now with the trick molecule go through steps 1 through 3 step 1 prioritize step 2 is the lowest priority group a - yes it is now we just do our go through the network go 1 through 2 1 2 3 and what direction are we going - are we going in counterclockwise and that makes us have an S configuration so the trick molecule has an S configuration but we are interested in what is the configuration of the original molecule and the way you figure out the original is you just take the opposite configuration of the trick molecule so that makes this guy an R ok so you see how this molecule circled in red as in our configuration and it's exactly the same configuration is this one and it has to be right because molecule a is the same thing as molecule B they're the same molecule they're identical ok we're just looking at the same molecule in different orientations this one we have the methyl or the hydrogen as the - and here it's in the plane so this is the third way of figuring out configurations is the switcheroo trick but this can get confusing sometimes that students will look at the trick molecule and do it correctly and they're full of confidence and they put the S down as the answer they fail to realize that this trick was is use the trick molecule is used so you can find the opposite of it for the original so watch yourself if you choose to do that trick now there's there's three ways of doing this some people like to just stick with the first way which is rotate the molecule some people like to mmm if the lowest priorities is the wedge they like to do the opposite trick and or you can do the switcheroo trick okay you find what works best for you now when you name a molecule and you know that you have a stereocenter you need to identify in the name if it's an R or an S so if we see this molecule here we see that there's one stereo Center and we know that the stereo Center is a R and so you just stick the are at the beginning of the name so r1 bromo one chloro ethane really really straightforward now what if you have a molecule with two stereo centers how do you name that okay so I'm gonna skip ahead here so we have this molecule shown and we see that there's two stereo centers okay so this is how you would name it with the skill set that you already have up to this point but now with that there's two stereo centers we need to designate if there are or s in the name so what we do is we assign what's our and what's s ok so we will figure out how to assign R and s in a minute I'm just focusing my attention on the name how to name it okay so this stereocenter right here at carbon one is a R and the stereocenter at carbon four is an S ok so just bear with me here so there's two ways you can name this molecule you can see going back there's the original name so how can you incorporate the R and s into this name well there's two ways the top one is where you put the OP the configuration next to the substituents so you can see that the one bromo one chloro is referring to these groups right here which are attached to carbon one so you just stick the are right next to those substituents and you can see the fluoro and the Nitro group are right here and that is on carbon 4 and that has a s configuration so it could be written that way I prefer the bottom way that way I just keep the name the same as here so you just keep the name the same and then to the left you just specify what carbon is the stereocenter at well carbon one has a stereocenter and at carbon one the configuration is our carbon four is a stereocenter and its configuration is an S and you put those into parenthesis I like the Bottomley better that's just a personal preference if you did this one right here as an answer choice you'd get a hundred percent of the points okay so it doesn't matter to me you pick what works best you in when you name molecules that now we I've been showing you how to find or designate the configuration of the stereo centers when there's only one but organic chemistry is more complex than that and you're going to have multiple stereo centers in a Meyer in a molecule so the trick is you just look at each stereo Center one at a time treat the entire molecule as if there's only one stereo Center okay and you just follow what you've learned so step one so let's look at the stereo Center at carbon one and figure out what it is what is the first step prioritize so we can see we're looking at this carbon what are the four groups that are attached to directly attached to it well we have a hydrogen a chlorine a bromine and then this carbon right here okay now we prioritize the groups bromine would be one chlorine two carbon would be three in hydrogen four and in this particular example lucky for us the lowest priority step two says the loss priority needs to be a - and it is so lucky for us step three just trace through the priorities one two three we see we're going in a clockwise direction gives us an R and then you do the exact same process for the second stereo Center follow that all the same steps and we can see that the lowest priority is once again a - so we've lucked out it's pretty nice okay so you just treat each stereo Center as if the it's the only one okay and you just follow the same steps that we've we've been doing now in this example we find the stereocenter where is the stereocenter it's right here that's our stereocenter because we have 1 group 2 group 3 group 4 group there's 4 different groups so that's our stereocenter so we've identified that that's great now we go to step 1 which says to prioritize we see we have a chlorine that's really large and we see a hydrogen and that's the smallest so we assign priorities 1 & 4 but looking here we have a carbon and a carbon so there's a tie which one takes priority over the other they're both carbons so we have 2 we have another trick to help us break ties there's going to be a tiebreaker and here's the rules it's just easier and better I feel just to show you the how to do it okay so here's the molecule again so we've already assigned the highest and the lowest priority so what you do is you can see that this carbon well right here's a carbon and right there's a carbon it's the same priority so what you have to do is you have to expand you have to then look and see what is directly attached to this carbon ok and we can see those atoms in red it's the hydrogen's the two hydrogen's and this carbon now we don't count this cut the carbon that's attached to the stereo the stereo Center okay because all work all we care about is what is directly attached to this carbon except what's before it so at this carbon we see that we have a carbon hydrogen hydrogen and you write those down ok so you can write them down let's just say we have a hydrogen carbon hydrogen let's say I wrote it in that order okay that is a mistake once you find all the atoms that are directly attached to this carbon here which are shown in red you then take that list that you made and you prioritize them in by size and so you can see how we actually wrote it like that CH H okay so after you've identified the atoms that are attached prioritize them by order of size right and we do the exact same thing for this carbon right here what's directly attached to it we see three hydrogen's and so that's an H HH h okay then once you have that data you are looking for the first point of difference among that set so we have this set and that set and we just go from left to right I see a carbon and a hydrogen there's my first point of difference at that point of difference which one has the largest atom a hydrogen or a carbon clearly carbon is larger than the hydrogen so that breaks the tie and this carbon right here because that breaks the tie it's going to get the second priority and that's how you break tiebreakers you see what's attached to it sometimes you don't break the tie on the first try right so if we have a molecule that looks like this and here's our stereocenter right there we can clearly see we have the chlorine a hydrogen carbon and a carbon they are these carbon groups clearly are different groups because look at it there's this thing and that group they're completely different so we know it's a stereocenter because we have four different groups but when you start assigning the configuration we see that it's a carbon and a carbon so we have to break the tie okay so let's look at this so we look at this atom here and see what's directly attached to it CH H we do the exact same thing for this one and we see we have a CH H so when you analyze this you see a carbon and a carbon same hydrogen and a hydrogen hydrogen and hydrogen there's no tie but there is no tiebreaker it's still a tie so we have to take another step so at the point if we get a tie once again you just go to the next atom so now we move to let's see here it's not working so here we have to tie because we can see these red atoms and these red atoms are exactly the same so now we go one step further away or one atom away from the stereo Center so now we are focusing on this carbon and we're focusing on this carbon and we are asking what's directly attached to that ignoring everything before it so you can see that we see we have two hydrogen's there and a carbon there and we prioritize it right there Carbon H H and then we see here on this carbon right here we have a chlorine and two hydrogen's and we list them in priority and then we just go through the salmon analysis as we did on the left side over here carbon carbon hydrogen hydrogen hydrogen hydrogen tie carbon fluorine oh those are different which one's larger the chlorine so that means this carbon right here is going to take priority number two and now they are ranked or prioritized properly and then we just go through and do our arrows so if that's one that's two that's three and four we see that we go one two three so we are going clockwise so that would make that a our and that's what we that that's what this slide is showing us here okay so this is the the flow chart of how we did it okay this is our original we then look at the next carbons we or we look at the carbons and what's attached to it we can it tie so we have to move to the next carbons and then we see if we can break the tie and if you can't you just if you can't break the tie on the third try you just keep going out further and further and further until you find a difference and if you don't find the difference then it's actually not a stereocenter the groups are exactly the same and so you made a mistake okay so what I mean by that is let's say we have a molecule looks like this let's just now this is not a stereocenter because we have two groups that are exactly the same but let's just say you made the mistake and thought it was a stereocenter you would go to this carbon and and that and that's a carbon so how would you break the tie and prioritising well you would see that this carbon right here is attached to a carbon and a hydrogen and a hydrogen the same thing with this guy you have a carbon hydrogen hydrogen so there's a tie so then your instinct tells you well you got to go to the next carbon and that carbon is attached to a hydrogen hydrogen hydrogen and that also has the same thing so you have a tie and now you're at the end of your rope there's no more carbons to jump to and now you see that it's a tie all the way through and that's just a that's evidence for you to say Oh oops I made a mistake in that this is not a stereocenter because the two groups right here are this same okay now let's look at this molecule here we have a stereocenter right there well how do we prioritize this because look at the groups they're clearly different we have a methyl we have an alkene and we have this propyl group and a hydrogen four different looking groups so we know it's a stereocenter but how do we prioritize it because we have a carbon carbon carbon so the only thing that we know at this point in the game is that the hydrogen is the lowest priority we know that for a fact but the three carbons we need to figure out what what to do now when you have alkenes in the molecule there's a trick that we use to help determine the priorities and that's shown in the these panels here so if you have something like this okay an alkene so this right here represents our situation right here to see how this bond right here is attached to the stereo Center that's what this bond is showing that this bond right here is a test of the stereo Center okay and what we do is we fictionally draw another molecule which we take this carbon right there and put it there and then this carbon we take it and put it there so another way of saying it is that if I'd loved numbered that carbon a and that carbon B we are taking carbon a and sticking it there and this carbon right here would be carbon P do you see how we're just doing a switch so we get rid of the double bond in this structure and we take those carbons and we turn them into single bonds that so if you have in this example right here mm-hmm when prioritising you get rid of the double bond mm and you take this atom right here and and you switch it you see how we took the nitrogen and stuck it on that carbon and then stuck a carbon on that nitrogen now we're doing this just so we can prioritize okay we're not actually doing this in the lab we're not changing the molecule we're just doing this as a trick okay so this would be treated as this guy and so you can go through the all these examples I will mention this one right here when there's a triple bond you then add nitrogen twice and then you would add carbon to nitrogen twice shown there okay so now you take this molecule right here and redraw it using these rules that are I've shown you above so let's take our molecule this is right here we have a stereocenter and now we replace that duck those double bonds with what we've shown and since there's two carbons in between we just add carbons and that's what we see two carbons okay now we start to prioritize just like we normally have here's our stereocenter so let's look at this carbon right here what's directly attached to it what's directly attached to this carbon that I've looked pointing to well we see that we have these two hydrogen's and this carbon and that's what we see here okay now we just do that for every single carbon what about this guy well that's only attached to three hydrogen's okay and then this guy we see that it is attached to this red carbon a hydrogen and that carbon so two carbons and hydrogen and now you just compare these three and see which one has them most which one's the heaviest really so if you compare these two together you can see that there's a point of difference at the first carbon so this one takes priority this one takes priority and then if you compare these two you can see same same up there's a point of difference right there so this carbon right here takes priority one priority one priority two and priority three and now you can go so when we're going through our steps steps one is to identify or prioritize the four groups when you have double bonds and triple bonds it takes a little bit more work okay and when we do that what is what are we going to get well the answer is shown at the bottom of the screen but I like to just I like to put my numbers here 3 2 & 4 and then I trace it 1 2 3 so we are going in the counterclockwise direction which means that is a s now we have to be also have to be able to describe the configuration of the stereo centers in Fischer projections ok so when we look at the Fischer projection right here we need to identify the stereo centers and there they are there's two stereo centers now looking at the top stereo center we see that the lowest priority group which is this one right there is a - because that's what Newman projections tell us right because if we look at a generic Fischer projection here okay let's go X Y a and B what does that tell us it's actually telling us it's looking like this right so you can see that the lowest priority here is going to be a - and that's what we want so it's oriented just right and so all we have to do now is prioritize the groups around that stereo Center as shown there on the slide we go boom so that's going to be a our configuration because we're going clockwise but now looking at this stereocenter at the bottom we see that the lowest priority is now a wedge so that's a problem but you cannot rotate and flip Fischer projections you have to just leave them alone but because the lowest priority here is a wedge we can do the opposite trick so we just price we prioritize we do our or arrow here and we see that we're going in the clockwise direction so that should be our but we take the opposite because the hydrogen is a wedge and it gives us a s configuration and so you can see that this is a very simple Fischer projection when you start looking at sugars and other molecules that are larger it's there's just going to be more to it but just because it's large doesn't mean it's impossible or it needs to be daunting you just take one stereo Center at a time so if you have four or five or six stereo centers just relax do it one at a time so I talked about this previously in which if you didn't if this wasn't here and I asked you what is the relationship between these two molecules and you didn't have this information here okay you didn't have the names so all you had was the molecules and I asked you what is the relationship between these well the first way that I taught you is well rotate the molecules orient them to see if this molecule B is a mirror reflection of a and if it's a mirror reflection and it's super imposable then they're identical if they're a mirror image and they're not superimposable than they are in antemer if they if B is not a mirror image and not superimposable than their diastereomers okay that requires a lot of mental manipulation of these molecules okay and for some people that is rather difficult so here is a another way that you could do it you can name them name both of them okay so now we have our names and then you identify in the stereo centers and write down what the configurations are okay so this one would be the are s s and this would be the s r r now you can see that on net molecule a or smell like you'll be that this the state the configurations are completely opposite and when the configurations are all completely opposite then you have a man tumors but when you look and you see some are opposite like these right here and then some stay the same then you know you have a diastereomer so if you're really good at assigning configurations to the stereo centers and are comparable at manipulating molecules mentally in three-dimensional space then this might be a good technique for you to use to identify enantiomers diastereomers and whatnot so if we go back to this molecule what if okay now I'm making a modification here what if this molecule didn't look like this and it was twisted and distorted and just was crazy-looking and I asked you what's the relationship between a and B what you could do even if this molecule is distorted and all crazy-like okay well let me show you what I mean so let me go to a program here that I have and let me double check okay so let's draw that molecule how many carbons are in there 1 2 3 4 5 6 so I got one too many carbons ok now we have all dashes okay - - and the - we'll add the chlorines ok so we have that molecule here I'll put it there and that's copy and paste this put it there but now let's replace those Alta wedges so we already know the answer right what's the relationship there in an tumors okay so if we call this one on the Left molecule a the same molecule on the right molecule B we see a and B are enantiomers of one another but what if I did something like this took B and did this clearly you can see that they're identical right because they're super imposable but what if I took this molecule on the right and I changed the way it looks what if I did that so all I'm doing is I'm just orienting it in different three-dimensional space here okay let's I want to try to get it to look a little bit more crazy and it's just not doing it for me okay hmm but if I make it look crazy-looking how do I know both of these are identical well you would go and you would see that if you can follow my cursor here let's does this have a draw function I don't think this maybe it does okay it doesn't have a really good draw function here but what you can find is that this Sara sign right here is an are s s and this is an RSS if you have the same configuration of all the stereocenters then you know you have the same molecule okay let's take a look at this molecule right here not that there's an alkene what is that a trans or assists we know that's a trans what about this alkene we see that as assists okay and we've gone over how to figure that out before but what about this alkene let's I want to go to my other writing program it's easier what about this one is that sis or trans well remember we have two then how do we do how do we figure out if this was sis or trans we drew a line through the double bond we see that there's a hydrogen and a hydrogen okay and then we circle the largest group on each half so that's the larger group and that's a larger group and then we ask ourselves what's the relationship between those two large groups if they're on opposite sides of that alkene then they're trans right so fleet attempt to do that for this guy we can see that this right here is the larger group and this is the larger group okay so those two groups are trans to one another makes sense right but if we do this analysis again what is the relationship between the methyl group on the left and the methyl group on the right the relationship between those two groups is cysts so how do you describe and discuss this molecule with with someone it is going to get very confusing are you talking about the bromine and the methyl group or trans or are you talking about the two methyl groups are they sis and so it's going to get very very confusing really really quickly so we have to use a different system to describe the configuration of a double bond and we use what's called a Z and E system okay and here's the rules on how to designate that if an alkene is e or Z well let's just show you okay so what you do is you look at a alkene okay and you split it in half again like like I've said okay and then on the left side you identified the highest priority based off the size and then you do the same thing on the right side what's the highest priority and if the highest priorities are on the same side of the alkene we call that a z configuration z standing for same side do you see the pun there same side same side I know it's stupid but it works and when you look at the figure on the right here you see the highest priorities are on opposite sides of the double bond so we call that an e configuration East adding for episode side so we got the same side and the episode sites and then really really silly but by golly it works and it has never left me for these 15 years of being in school okay so that's how you keep track of it so if you look at this example right so let's just go back to this molecule right here so what type of isomer is this is that a e or a Z well we do our process divide the molecule half identify the highest priority okay those are on the opposite side so that makes this a II if you wanted to look at the diastereomer of this molecule but the Z diastereomer what would it look like it would look like that because these these groups right here are on the same side no okay let's see here now on this slide but I want to show you is that when you name an alkene we see that the correct name but now you see how pent two een pent two een the same name but they're clearly diastereomers of one another there are different molecules they're going to have different physical properties but they have the same parent name so how do we distinguish between these two well we have to use the Z and the e and so you just stick the Z in parenthesis at the beginning of the name so we have the Z pentene and then the epen to eenz okay you just have to so we're just adding another layer of understanding to how to name molecules Z and e but what happens when you have multiple E or multiple double bonds how do you name molecules like that the isomer on the Left we see that there's two double bonds but this double bond right here at carbon 1 and 2 does not get a a or z assignment because it has the two hydrogen's on carbon 1 so there's no such thing as a II or Z for this guy but the alkene of carbon 3 and 4 does have a can be assigned an E and as Z and we see that it is going to be an e because when you take and do our trick draw up on there and find so we have a hydrogen there and a hydrogen there there's our methyl and then there's this piece and you can see that they are on opposite sides of one another and so we assign the e and in the name we just put an e at the very beginning because there's only one alkene in the name that can be your z configuration so this e right here does not apply to that guy we already know that because we see well if we just look at just the name we see that there's a dying so that means that there's two alkenes and this e clearly hat is be talking about the third alkene because the first one cannot be a a or a Z that's just implied but when you look at the molecule on the right you can see that this double bond can be a a or a Z and this double bond could be an e or a Z you can you have to figure out what it is and we see that at this this carbon this double bond right here is a e so we in the name at the beginning of the name we specify where the alkene is with the number it starts at carbon one and that alkene is AE and then the second alkene starts at carbon three and we see that that is a Z okay now why is the alkene at carbon three Izzie if you draw your plane there you can see the methyl group and this group this whole thing actually is on the same side okay so it takes practice looking at molecules when they're larger and just bringing it back to the baby steps or at the very simple simple way we started looking at things so this is the very simple the most basic way of looking at the e and the Z configurations and you just apply that to when two molecules that look a little bit nastier okay always go back to the basics okay when you're solving organic chemistry problems and that is the end of chapter or inter chapter C so if you have any questions as always please reach out I'm more than willing to work with you in groups or one-on-one I'm here for you I'm here to help