drawing constitutional isomers that'll be the topic of this lesson in my organic chemistry playlist now we spent the entire first half of this chapter naming alkanes and almost in the latter half drawing different confirmations of alkanes and we'll start here with drawing constitutional isomers but we'll move on to newman projections and then different confirmations for cycloalkanes including the most famous the cyclohexane chair conformation now if this your first time with me welcome to chad's prep my name is chad and the goal of the channel is simply to make science both understandable as well as even enjoyable now this is my brand new organic chemistry playlist i'll be releasing these lessons weekly throughout the 2020-21 school year so if you don't want to miss one subscribe to the channel click the bell notifications you'll be notified every time i post a new video all right so before we start actually drawing constitutional isomers a little bit we got to talk about with hydrocarbons we often talk about what's referred to as a saturated hydrocarbon versus an unsaturated and this applies to like saturated fats versus unsaturated fats and all that really means saturated means you have the greatest number of hydrogens possible so we know that carbon likes to make typically four bonds in most stable structures and if you fill every available bonding position with a hydrogen so on a hydrocarbon then you're going to get a certain formula associated with that and that formula looks such something like this here where you take the number of carbons and if you double that number and add two that'll get you the saturated number so in a little bit we're gonna be taking a look at a formula that has got seven carbons so and the greatest number of hydrogens you could ever have with seven carbons present so in this case is double that number fourteen add an extra 2 to get to 16. and so the formula would be c7h16 and seeing this formula be like oh yeah the number of hydrogens is double the number of carbons plus 2. that's a saturated hydrocarbon and we'll see where this kind of comes from here so if we actually take our level e7 carbon structure and make it kind of in a straight chain here we'll see that all the internal carbons to get to four bonds have room for two more hydrogens each but the n2 carbons here they have room not just for two hydrogens but for one extra each since they were missing a an extra bond there so same thing on the other end here and that's where the extra two comes from so every carbon in the chain has two hydrogens but the ones on the end have three each and that's where that plus two comes from and so this would be a saturated formula now where you get an unsaturated formula from the most common place is if you stick and take and replace one of those single bonds with a double bond when you put a pi bond in a structure you'll note that these two carbons right now are violating the octet rule this is not a real structure so those two carbons each have five bonds now not four so we have to kill a couple of hydrogens in order to incorporate that double bond in the structure and so now we'd find out by putting a double bond in there a pi bond instead of having double as many you know carbons plus two it's just double as many so now oh i forgot to add that one there so in this case with seven carbons instead of having 16 hydrogens now we're only counting them up you only got 14 hydrogens instead and so every two hydrogens you're missing from that saturated formula is going to be a degree of unsaturation and the easiest way to as a pi bond notice if i put a triple bond in there i'd have to kill two more hydrogens so for every pi bond you're going to be missing a couple of hydrogens now the other thing you could do is you could join the ends together in a ring and so when you join a ring i'd have to kill a couple of hydrons in that sense as well to make make it so these two carbons could attach to each other and so that's your other form of unsaturation we call them degrees of unsaturation so it's either a pi bond or it's a ring and for every pi bond and every ring you have in your structure you're going to be missing two hydrogens from your saturated formula now somebody's come out instead of looking at that way with what's called the hydrogen deficiency index and so in this case if you take the number of carbons and double it plus two so and then subtract off the number of h's subtract off the number of halogens add the number of nitrogens and then divide by 2 what that will give you essentially is how many elements of unsaturation or how many degrees of unsaturation you have so like let's say in this one right here we've got the seven carbons one two three four five six seven so two times seven plus two minus the number of hydrogens well in this case we've got 14 hydrogens no halogens no nitrogens and then divide by 2. and so in this case 14 plus 2 is 16 minus 14 is 2 all divided by 2 would be 1 and this formula of c7h14 would have one degree of unsaturation that would be either one pi bond or one ring so this is one quick way to get you that number of degrees of unsaturation that's what that hydrogen deficiency index is cool now this is useful when you're being asked to draw constitutional isomers because if you're given a formula like this one here we're going to draw the constitutional icebergs of c7h16 in a minute which you might want to know as well do i have to incorporate any pi bonds or rings in my structures and in this case with 16 being double 7 plus 2 i would see nope there are no degrees of unsaturation it is a saturated hydrocarbon and so none of my structures are going to get a pi bond or a ring now on the other hand if i was having you draw all the constitutional isomers of c7h14 now the sudden every single structure you'd have to draw would have one degree of unsaturation every structure would either have one pi bond and you'd have lots of different places you could put it or you'd have to have a ring one or the other so and that's kind of what you can get out of this formula now most of the time in this section again this is a whole chapter on alkanes you're going to pretty much be in all likelihood asked to draw the constitutional isomers of a saturated hydrocarbon but on the odd chance uh that you're not that's why we kind of covered this but i just want to make sure we're verifying that we've got saturated hydrocarbons before we start trying to draw these structures all right so now we're going to start drawing these constitutional isomers for c7h16 proper now if you got the study guide in frontier they're already there but probably good practice if you did make an effort to write these on your own and rather than writing these in any kind of bond line structure i find that it's easiest if we kind of represent them in more of a lewis structure now one thing to note here so my question for you and i'm gonna give you the answer obviously in a second is what is the bond angle between these two bonds right here around this carbon and again just picture that this carbon is going to end up with a couple more bonds on it as well but my question for you is if you look at the two adjacent bonds what's that bond angle right there but also if you look at the two opposite bonds coming off that carbon what's that bond angle so and if you said 90 and 180 so be super careful this carbon is sp3 hybridized and all of its bond angles are 109.5 regardless of how mr lewis makes it look now mr lewis makes everything look either 90 or 180 but regardless of how you look at it this angle right here is 109.5 and this angle is 109.5 as well and the reason i bring that up is a lot of students will say yep chad we're going to draw some you know carbon backbones here and they come up with this one right here and we're just going to fill in the hydrogens later so but if you get the carbon backbone and based on knowing that this is saturated we know we don't need any double bonds so it'll be a lot easier to come out than if you try to do this with like a bond line structure from the start anyways so but in this case a lot of students will come and go for a second one here so one two three four five six and seven and they'll do something like this and they're like so this is different chad i'm like nope that's the same thing as the one we just did because my question would be what's the angle between these two bonds well it's 109.5 what's the angle between these two bonds well it's 109.5 no matter how we draw it like this i still have seven carbons in a straight chain and if i fill in all the hydrogens it'll be exactly the same thing as this structure with all of its hydrogens filled in as well so just want to make sure we realize that i can you know zig and zag and stuff but as long as my longest continuous chain kind of is seven carbons here in a straight chain regardless of how i zig and zag it's the same molecule all right so we're going to take a systematic approach here to drawing these constitutional isomers and what you want to do it's not the only systematic approach it's just the one i recommend is you start with your longest straight chain and then slowly but surely shorten it up so in this case this is gonna be seven carbons in a straight chain not too bad and eventually we would just go back in and fill in all the hydrogens so if you're asked to do this on an exam obviously start here draw on all the carbons and then just save room because you're gonna go back in and draw in all the hydrogens later and if you had to turn that into a bond line structure you just go one two three four five six seven when you're done cool now to get the other constitutional ice first what we want to do now is go back and say well let's shorten that longest chain from seven carbons which is the entire chain here down to six and so we'd go one two three four five six and then we ask ourselves a question say okay where do i put that last carbon because now that i'm down to six carbons i've got one carbon left and the key is you can't attach him to either end because if i attach him to either end whether i put him here here or here it's going to lead me to a situation where i've got like this again where i've got my longest chain is going to be seven carbons long continuously just like we did up here and it'll be no different than when we draw so when we start shortening the chain up we're going to make sure we can't add any of the carbons we've got to fill in to the ends or it will go back to being a longer carbon chain than just the six we got here but in this case i've got all these middle carbons and i can add that extra carbon onto any one of them now so let's say i decided to put it here well one thing you should realize is that's no different at this point than to putting it here because this angle is 109.5 that angle is going to be 199.5 it's the same thing either way you should also realize that it's no different than adding it either here or here as well so whether it's the second carbon in from the left where i put that extra carbon that seventh carbon or the second one in from the right where i put that seventh carbon it's the same thing either way just the whole molecule rotated around 180. but it's the same thing so there's no difference there so whether i put it here or here that seventh carbon that's going to be one of my structures now moving right on down the chain then so one two three four five six and so i've considered adding it to one of these carbons but then i could add it to one of these as well and whether it's the third from the left or third from the right that'll be the same thing and so as long as i put that carbon in any one of these four positions that'll be another constitutional isomer again we would just then go back and fill in all the hydrogens and stuff cool but that's it as far as six carbons i've exhausted all the places i could put that extra carbon and still have my longest chain b6 carbons long now if i needed to go back and you know draw the bond line structures so one two three four five six so there's six carbons long and the second one from the end gets another carbon or this one here and now it's the third one in from either end that gets a carbon and whether i put it here or here would have been the same thing either way so if at the end you've got to translate these back into bond line structures like for based on the instructions on your exam or something like that great but i would highly recommend setting it up like lewis structures and just set up the carbon backbones because these will be easy enough to draw after the fact all right so we did seven carbons in a straight chain then we made our longest chain six carbons and added that extra carbon now we'll drop this down to five carbons as well all right so now we make our longest chain one two three four five now we've got two carbons remaining that we have to add onto this structure and again i can't add it onto the ends or end up with something longer than a five carbon chain we've already considered all the possibilities of something longer than a five carbon chain continuously so i've really got to put it on the middle carbons here i've got these six positions to work with to put those remaining two carbons now i've got a choice i can do two methyl groups two separate methyl groups in two different of these six locations that's an option or i could do a single ethyl group now we've got to be careful though because you can't just put that ethyl group anywhere and the truth is i can't put the ethyl group either here or here at all because if i do like let's just say we made it an ethyl group two carbon chain of two carbon branch now my longest continuous chain would not be these five carbons it would be these one two three four five six carbons and this would actually be the same thing as this guy right here so drawn a little differently but it's exactly the same thing so we couldn't just put an ethyl group on either again we can't put anything on the ends but we couldn't put a two carbon chain one carbon in as well or we get a longer continuous carbon chain than the one we're intending to get so but i could put that ethyl group on this guy right here and that would indeed be something new this could still be my longest chain or this you know or this so but my longest carbon chain is still five carbons but that's the only place whether i put it down below or up above same diff but that's the only place i can put that two carbon chain so now i've got to draw again five carbons and get once again these are my six positions where i have an option to put something other than a hydrogen and i could do two methyl groups as well now one way to do this would be to put both methyl groups attached to the same carbon so whether i put it both on this carbon or both on this carbon here that would be exactly the same thing so but i could also put them both on the middle carbon and that would be a unique structure as well all right so however we could consider also putting them on two different carbons let's make ourselves just a little bit of room here all right so once again i've got these six positions to work with so and with two methyl groups on two different carbons well i could put one here and one here so and again it doesn't really matter if i put one up here or down here and then up here down here whether i put that there or there it's all the same thing so as long as there's one each on these two carbons that's one way to do this now we've got one more way to pull this off as well so instead of doing these two carbons i could either choose these two or these two but essentially they'd just be the same thing flipped 180. and again whether i put them both on top or both on bottom or one on top and one on bottom it's all the same thing cool and this would be the last kind of carbon backbone i could set up and once again once we get the carbon backbones we can just go back in and fill in all the hydrogens and if a lewis structure is acceptable according to your exam instructions great but if you got to go back and translate these into bond line structures that's easy enough to do once you've got your carbon skeleton as well all right so now we've gone from seven carbon straight chain down to two possibilities we had with a six carbon uh longest chain and now we've got five different possibilities when we've got a five carbon longest chain so now let's shorten this down to four carbons so we're starting to run out of room on my little board here so but we're gonna start running into some major problems here so because with four carbons here i've got three carbons remaining and again you can't put anything on the ends or you end up with something longer than a four carbon chain so you only got these four positions in which to put things and i've still got three carbons remaining now again with three carbons remaining i could just use one propyl group one three carbon branch but that's not going to work because if i just put a three carbon branch here so all of a sudden this becomes my longest chain and that's one of the possibilities we already considered so if i want to put a three carbon branch on something and not have it be something i've already covered it's got to be at least three carbons in from the end and that's not possible here so and if i want to put even an ethyl group on something it would have to be on a carbon that's at least two carbons from the end and that's not possible here because these each are only one carbon from an end so the only thing we can put on these in any of these four positions are a single carbon methyl groups and i've just got to choose three out of the four and so in this case with three out of these four getting chosen to have a carbon uh there's two and whether i put one here or here it's just the same thing flipped over so in filling three out of those four positions it turns out no matter how you do it it actually all ends up being equivalent there's only one ultimate way to pull this off and so there you've got your one four carbon longest chain as well so here we had a seven carbon chain two versions where the six carbon chain was longest five versions where we had a five carbon chain is the longest continuous chain and then finally one option where a four carbon chain was the longest but these would now represent all of the constitutional isomers for c7 h16 okay so the last one was all on your study guide so but this one is totally on you i want i want to make sure you have to draw this one out instead of just being like yep i put all these uh chat already put them on the board but they're already on the handout so life is good but in this case you're going to draw these out so and i put an auction in there just in case it's not necessarily something you will come across here but many classes would be on the hook for something like this now it turns out that auction is not part of your calculation for determining degrees of unsaturation or the hydrogen deficiency index we can just kind of ignore it pretend it's not there and for four carbons the greatest number of hydrogens you could get well double four is eight plus two is ten and we have a saturated molecule again and again what that means is no pi bonds and no rings in our structure whatsoever now one thing you should keep in mind for oxygen is oxygen tends to make two bonds being two electrons shy of a filled octet and so with oxygen making two bonds that means it also could go in the middle of the carbon chain and something we might have to consider for a possibility so but we'll start in drawing constitutional iceman for this with the longest carbon chain we could have a straight four carbon chain so one two three four and now we've got to figure out where to put the oxygen how many different places there are there for putting the oxygen well in this case i could put the oxygen on either end like so or i could put on either of the middle carbons one carbon in from either end and that would be a unique location and then from here we get you know go back in and fill in all the h's and just keep in mind that oxygen needing two bonds would also need one more h on this structure as well cool but as long as we don't put any pi bonds rings in there we'd end up with room for exactly ten hydrogens being again that this is a saturated molecule cool now for a four carbon straight chain those are my options for where i could place the oxygen as long as it's not incorporated into the middle of the chain now let's shorten this down to three carbons so and with three carbons my fourth carbon is going to have to go as a branch on that middle carbon whether it be up or down is totally doesn't matter and the question is now where can i attach that oxygen well on a structure like this whether i attach the oxygen to this carbon this carbon or this carbon they're all equivalent just rotate this thing around a little bit and so i'm just going to make one of the structures with the oxygen here but the other place the oxygen could have gone was that carbon which is definitely different than these three and so that's another option and we'll find out here we can't actually shorten the chain down any shorter we went from four carbons down to three carbons but you can't really go down to two carbons because then both are the ends and you can't add anything to the ends or you get a longer chain however we can consider possibilities where the oxygen is in the middle of the chain so every single one of these is going to have an o h you'll add an h to the oxygen every case these are all alcohols but if you put the oxygen in the middle of the chain and bond it to two carbons we'll have ethers instead and so in this case we could do this one of a couple of different ways we could have three carbons on one side of the oxygen and one carbon on the other side we could have two carbons on one side of the oxygen and then two carbons on the other side and notice if i made it then one on this side and three and this would be no different than what we did up here and then you've got to consider well can either of these carbon chains have any kind of branched equivalent as well so well with two and two there's no way for that to be any different than what it is and for a one carbon brain you know chain here there's nothing you can do with that as well but for three carbons you can bond that through one of the end carbons but you could also potentially bond the oxygen through the middle carbon as well and so what you could also then do is something like that as another possible structure as well and so in this case with putting the oxygen in the middle of the chain we've now created three new possibilities as well and again we just had to look at putting in the middle of a chain breaking up and saying how many carbons can go on both sides and we can either do it three and one or two and two and then with three and one there were two different ways of pulling this off and so in this case we found out for c4h10o there are indeed seven different constitutional isomers four that were alcohols three that were ethers and again from here you just fill in the hydrogens or draw the corresponding bond line structures now if you found this lesson helpful consider giving me a like and a share pretty much the best things you can do to support the channel and if you're looking for practice problems or if you're looking for the study guide that goes with these lessons check out my premium course on chadsprep.com