4.4 Drawing Structural Isomers | Organic Chemistry

draw constitutional

isomers

which will be the topic of this lesson in my

organic

chemistry

playlist. Now we spend the entire first half of this chapter naming alkanes and almost the second half

drawing

different confirmations of alkanes and we'll start here by

drawing

constitutional

isomers

. but we will move on to Newman projections and then to different confirmations for cycloalkanes, including the most famous cyclohexane chair conformation. Now, if this is your first time coming with me, welcome to Chad's Prep. My name is Chad and the goal of the channel is simply to do science. Both understandable and even enjoyable now, this is my new

organic

chemistry

playlist.

I will be posting these lessons weekly throughout the 2020-21 school year, so if you don't want to miss any, subscribe to the channel by clicking the bell notifications. I will be notified every time I post a new video, so before we start drawing a little bit on constitutional isomers, we need to talk about hydrocarbons, we often talk about what is known as saturated versus unsaturated hydrocarbon and this applies to saturated fats versus unsaturated fats and all that really means saturated means you have as many hydrogens as possible so we know that carbon likes to form typically four bonds in most stable structures and if you fill every available bond position with a hydrogen, this is what happens with a hydrocarbon. then you'll get a certain formula associated with that and that formula looks like this one here where you take the number of carbons and if you double that number and add two, you'll get the saturated number, so in a little bit But let's take a look at a formula which has seven carbons and the most hydrogens you could have with seven carbons present, so in this case it is double that number fourteen, add an additional 2 to get to 16. and then the formula would be c7h16 and seeing this formula would be like oh yeah the amount of hydrogens is twice the amount of carbons plus 2.

That's a saturated hydrocarbon and we'll see where this comes from so if we actually take our level the structure of carbon e7 and turn it into a species straight chain. Here we will see that all the internal carbons to reach four bonds have room for two more hydrogens each, but the n2 carbons here have room not only for two hydrogens but also for one additional hydrogen each, since they were missing an extra bond there, so the same thing at the other end here and that's where the extra two come from, so each carbon in the chain has two hydrogens, but the ones at the end have three each and that's where the plus two comes from, so this would be a saturated formula, now where you get an unsaturated formula, the most common place is if you stick, you take and replace one of those single bonds with a double bond when you put a pi bond in a structure.

You'll notice that these two carbons are currently violating the octet rule. This is not a real structure, so each of those two carbons has five bonds, not four, so we have to remove a couple of hydrogens to incorporate that double bond into the structure and now we would find out by putting a double bond in there , a pi bond instead of having twice as many carbons plus two, it's just twice as many, so now, oh, I forgot to add that one in there, so in this case with seven carbons instead of having 16 hydrogens now we're just counting them , you only have 14 hydrogens in place, so every two hydrogens you are missing in that saturated formula will be one degree of unsaturation and the easiest way to do this is pi Bond notice: if I put a triple bond in there, I would have to kill two more hydrogens, so for each pi bond you will be missing a pair of hydrogens.

Now the other thing you could do is join the ends into a ring and so when you join a ring you would also have to kill a couple of hydrons in that sense so that these two carbons can bond together and that's your other form of unsaturation, we call them degrees of unsaturation, so it's a pi bond or a ring and for every pi bond and every ring that you have in your structure you will be missing two hydrogens from your saturated formula now someone has come out instead of look 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, then subtract the number of h, subtract the number of halogens, add the number of nitrogens and then divide by 2, which will basically give you how many elements of unsaturation or how many degrees of unsaturation you have, so let's say that in this one here we have 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 have 14 hydrogens, no halogens, no nitrogens, and then we divide by 2. So, in this case, 14 plus 2 is 16 minus 14 is 2, everything divided by 2 would be 1 and this formula of c7h14 would have a degree. of unsaturation which would be a pi bond or a ring so this is a quick way to get that number of degrees of unsaturation that's what that hydrogen deficiency index is cool now this is useful when you're asked to draw isomers constitutional because If you are given a formula like this here, we will draw the constitutional icebergs of c7h16 in a minute.

You may also want to know if I have to incorporate pi bonds or rings in my structures and in this case. with 16 being twice 7 plus 2, you would see that no, there are no degrees of unsaturation, it is a saturated hydrocarbon, so none of my structures will get a pi bond or a ring, on the other hand, if I had you draw all the constitutional isomers of c7h14 now all of a sudden every structure you would have to draw would have a degree of unsaturation, every structure would have a pi bond and you would have lots of different places you could put it or you would have to have one ring or another, and that's what you can get from this formula now most of the time in this section again, this is a whole chapter on alkanes which in all likelihood will ask you to draw the constitutional isomers of a saturated hydrocarbon, but with the odd chance that you are not, that's why we covered this, but I just want to make sure that we're verifying that we have saturated hydrocarbons before we start trying to extract them. structures, so now let's start drawing these constitutional isomers for c7h16 right now.

If you have the study guide on the border, they are already there, but it would probably be good practice if you made an effort to write them on your own and rather Instead of writing them in any kind of link line structure, I think it's more easy if we represent them in a more Lewis structure. Now there's one thing to keep in mind here, so my question to you and obviously I'll give you the answer in a The second is what is the bond angle between these two bonds here around this carbon and again, imagine that this carbon will end up with a couple more bonds as well, but my question for you is if you look at the two adjacent ones. bonds, what is that bond angle right there, but also if you look at the two opposite bonds coming out of that carbon, what is that bond angle?

So if you said 90 and 180, be very careful, this carbon is sp3 hybridized and all its bond angles are 109.5. regardless of how mr lewis makes it look now mr lewis makes everything look like 90 or 180, but regardless of how you look at it, this angle here is 109.5 and this angle is also 109.5 and the reason why I mention that's for a lot of the students will say yeah, chad, we're going to draw here some carbon structures that you know and come up with this one right here and we're going to fill in the hydrogens later, but if you get the carbon backbone and , based on the knowledge, that this is saturated, we know that we don't need any double bonds, so it will be much easier to get out than if you try to do this with a bond line structure from the beginning, but in this case there are many students will come and go for a second here, so one, two, three, four, five, six and seven, and they will do something like this and they will say this is different, Chad, I say no, it's the same as us.

I only did it because my question would be what is the angle between these two bonds. Well, it's 109.5. What is 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 complete all of them. the hydrogens will be exactly the same as this structure with all its hydrogens filled as well, so I just want to make sure that we realize that I can know zigzag and stuff, but as long as my longest continuous chain is There are seven carbons here in A linear chain, regardless of how it zigs and zags, is the same molecule, so we're going to take a systematic approach here to draw these constitutional isomers and what you want to do is not the only systematic approach.

Just the one I recommend is to start with the longest straight chain and then slowly but surely shorten it, so in this case it will be seven carbons on a straight chain, not that bad and eventually we would go back and complete the whole thing. the hydrogens, so if you're asked to do this in an exam, obviously start here, draw all the carbons and then just save space because you'll go back and draw all the hydrogens later and if you were to convert that into a bond line structure , you just go one, two, three, four, five, six, seven, when you're done, great, now to get the other constitutional ice first, what we want to do now is go back and say, well, let's shorten that longer chain of seven carbons, which is the entire chain here is reduced to six, so we would go one, two, three, four, five, six and then we ask ourselves a question: "Okay, where do I put that last carbon?", because now that I have six carbons left, I have one.

There's carbon left and the key is that you can't attach it to either end because if I attach it to either end, whether I put it here or here, it's going to lead me to a situation where I have something like that again. I have my longest chain that will be seven carbons long continuously, just like we did here and it won't be any different than when we draw, so when we start shortening the chain we'll make sure that we can't add. any of the carbons we have to fill them out to the ends or it will again be a longer chain of carbons than just the six that we have here, but in this case I have all these intermediate carbons and I can add that extra. carbon in any of them now, so let's say I decided to put it here, well, one thing you have to keep in mind is that at this point it's no different than putting it here because this angle is 109.5, that angle will be 199.5, is the same.

Either way, you also have to realize that it's no different than adding it here or here too, whether it's the second carbon from the left where I put that extra carbon, the seventh carbon or the second from the right where I put that. The seventh carbon is the same either way, just the whole molecule rotated around 180. But it's the same thing, so it makes no difference whether I put it here or here, that seventh carbon that will be one of my structures now moves down. the chain then is one, two, three, four, five, six, so I've considered adding it to one of these carbons, but then you could add it to one of these as well and if it's the third from the left or the third from the right .

It's going to be the same thing, and whenever you put that carbon in any of these four positions, it's going to be another constitutional isomer again, so we would go back and fill in all the hydrogens and stuff, but that's it as far as we can. six carbons. I've exhausted all the places I could put that extra carbon and I still have my longest chain of b6 carbons now if I were to go back and you know, draw the bond line structures, so one, two, three, four, five, six, so there are six carbons. long and the second from the end receives another carbon or this one here and now it is the third from either end that receives a carbon and if I put it here or here it would have been the same either way, so if at the end you have to go back to translate them into bond line structures like according to your exam instructions or something cool, but I highly recommend setting it up as Lewis structures and just setting up the main carbon columns because they'll be pretty easy.

To draw after the fact, okay, so we made seven carbons in a straight chain, then we made our longer chain with six carbons and added that extra carbon. Now we'll reduce this to five carbons as well. Alright, now we make our chain longer. one two three four five now we have two carbons left that we have to add to this structure and again I can't add them to the ends or end up with something longer than a five carbon chain, we have already considered all the possibilities of something longer than one five carbon chain continuously, so I really have to put it in the middle carbons.

Here I have these six positions to work with to put the remaining two carbons. Now I have a choice I can do. two methyl groups, two methyl groups separated in two different of these six locations,that's an option or I could do just one ethyl group, but now we have to be careful because you can't put that ethyl group anywhere and the truth is, I can't put the ethyl group here or here at all because if I like , let's say we convert it to an ethyl group, a two-carbon chain with a two-carbon branch. Now my longest continuous chain wouldn't be these five carbons, it would be these. two three four five six carbons and this would actually be the same as this guy here, drawn a little bit differently but it's exactly the same So we can't just put an ethyl group on either one, we can't put anything on the ends, but We can't put a two carbon chain with one carbon too or we get a longer continuous carbon chain than I intend to do, but I could put that ethyl group on this guy here and that would be something new.

This might still be my longest chain or this, you know, or this, but my longest carbon chain is still five carbons, but that's it. the only place whether I put it below or above the same difference, but that's the only place where I can put that two carbon chain, so now I have to draw again five carbons and get once again these are my six positions where I have an Option to put something other than hydrogen and I could also make two methyl groups. Now one way to do it would be to put both methyl groups attached to the same carbon, so if I put them both on this carbon or both on this carbon here.

That would be exactly the same, but you could also put them both on the middle carbon and that would be a single structure, so however, we could also consider putting them on two different carbons, let's do a little bit of There's room here, so once plus I have these six positions to work with, so with two methyl groups on two different carbons, well, I could put one here and one here, so over and over again, it doesn't really matter if I put one on top. here or down here and then up here down here whether you put it there or there, it's the same thing as long as there's one on each of these two carbons, that's one way to do this, now we have one more way to take this out. also, so instead of doing these two carbons, I could choose these two or these two, but essentially they would be the same thing flipped 180. and again, if I put them both on the top or both on the bottom or one on the top and one at the bottom is the same thing great and this would be the last type of carbon backbone that you could set up and again once we have the main carbon backbones we can go back and fill in all the hydrogens and if there is an acceptable lewis structure according to your exam instructions, great, but if you have to go back and translate them into bond line structures, that's easy enough to do once you have your carbon skeleton too, okay, now we have gone from the straight chain of seven carbons downwards. to two possibilities that we had with a longer six carbon chain and now we have five different possibilities when we have a longer five carbon chain, so now let's shorten this to four carbons so we start running out of space. on my little board here, but we're going to start having some major problems here, because with four carbons here I have three carbons left and again you can't put anything on the ends or you'll end up with something longer than a four carbon chain, so you only have these four positions to put things in and I still have three carbons left now again with three carbons left, I could use a propyl group, a three carbon branch, but that's not going to work because if I just put a branch of three carbons here so suddenly this becomes my longest chain and that's one of the possibilities that we've already considered, so if I want to put a three carbon branch on something and not make it something that I've already covered I have to There should be at least three carbons from the end and that is not possible here, so if I want to put even one ethyl group on something, it would have to be on a carbon that is at least two carbons from the end and that is not possible. here because each of these has only one carbon on one end, so the only thing we can put in any of these four positions are single carbon methyl groups and I only have to choose three of the four, and in this case with three of these four chosen to have a carbon, there are two and if I put one here or here it is the same flipped, so when filling three of those four positions it turns out that no matter how you do it, in reality everything ends up being equivalent, there is only a definitive way to achieve this, so there you also have the longest four-carbon chain, so here we had a seven-carbon chain, two versions where the six-carbon chain was the longest, five versions where had a five carbon chain which is the longest continuous chain and finally an option where a four carbon chain was the longest but now they would represent all the constitutional isomers for c7 h16 okay so the last one was in your guide to study, but this one is totally on you I want to make sure you have to draw this one instead of just saying yeah, I put all these uh chat, I already put them on the board but they're already in the booklet, so life is good, but In this case, you're going to take them out and I put an auction there in case it's not necessarily something you come across here, but a lot of classes would be committed to something like this.

Now it turns out that bidding is not part of your calculation for determining the degrees of unsaturation or the hydrogen deficiency index, we can just ignore it, pretend it's not there and for four carbons, the most hydrogens you could get is the double four, eight plus two, ten and we have a saturated molecule over and over again, what that means is that there are no pi bonds or rings in our structure. Now, one thing you have to keep in mind for oxygen is that oxygen tends to form two bonds, two electrons less than a full octet, so with oxygen it occurs. two bonds, that means it could also go in the middle of the carbon chain and it's something we might have to consider for a possibility, but we'll start drawing the constitutional iceman for this, with the longer carbon chain we might have a chain of four straight carbons. so one two three four and now we have to figure out where to put the oxygen.

How many different places are there to put oxygen? In this case, you could put the oxygen on either end like this or you could put it on either end. of the middle carbons, one carbon from each end and that would be a single location and then from here we know it goes back and completes all the h and just keep in mind that the oxygen that needs two bonds would also need one more h. in this structure is also cool, but as long as we don't put any pi bond rings in there, we would end up with room for exactly ten hydrogens, again being that this is a saturated molecule.

Great now for a linear four carbon chain, those are My options for where you could put the oxygen as long as it's not incorporated in the middle of the chain. Now let's shorten this to three carbons, and with three carbons, my fourth carbon will have to go as a branch on that middle carbon. whether it's up or down doesn't matter at all and the question now is where can I attach that oxygen well in a structure like this, if I attach the oxygen to this carbon, this carbon or this carbon, they're all equivalent, just turn this around.

This changes a little bit, so I'm just going to do one of the structures with the oxygen here, but the other place where the oxygen could have gone was the carbon, which is definitely different than these three, so that's another option and Find out here we can't actually shorten the chain any further. We went from four carbons to three carbons, but you can't really go down to two carbons because then they are both the extremes and you can't add anything to the extremes. or you get a longer chain, however we can consider possibilities where the oxygen is in the middle of the chain, so each of these will have an o h, you will add an h to the oxygen in each case, these are all alcohols, but if you put the oxygen in the middle of the chain and attach it to two carbons, we will have ethers instead, so in this case we could do this a couple of different ways where 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 do that then one on this side and three and this would be no different than what we did here and then you have We have to consider whether any of these carbon chains can have some kind of branched equivalent, so 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 too, but for three carbons you can bind it through one of the end carbons, but you could also bind the oxygen through the middle carbon, so what you could also do is something as that is also another possible structure and in this case by putting the oxygen in the middle of the chain we have now created three new possibilities as well and again we just had to consider putting the oxygen in the middle of a chain breaking and saying how many carbons can go on both sides and we can do three and one or two and two and then with three and one there were two different ways to achieve this and in this case we discovered that for c4h10o there are seven different constitutional isomers, four that were alcohols, three which were ethers and again from here just fill in the hydrogens or draw the corresponding bond line structures.

Now, if you found this lesson useful, consider giving me a like and sharing the best things you know. can do to support the channel and if you are looking for practice problems or if you are looking for the study guide that accompanies these lessons, check out my premium course at chadsprep.com