Finding Constitutional Isomers and How to Draw Them | Organic Chemistry

Finding Constitutional Isomers and How to Draw Them | Organic Chemistry

November 12, 2019 14 By Bernardo Ryan


Constitutional isomers have the same chemical formula but different structures. This type of isomer differs in its bonding. For example, this methyl group is on carbon 3, while this methyl group is on carbon 4. So a difference in bonding just refers to a substituent being bonded to a completely different carbon. If we wanted to draw the constitutional isomer for C4H10 we would start with the typical skeletal formula then start moving carbon chains around and make sure they are still the same chemical formula. So let’s check if this has 4 carbons and 10 hydrogens. Each point is a carbon, and we will place all the hydrogens in this structure. Remember carbon can make 4 bonds so this carbon has 1 bond here and 3 hydrogens same goes for this top carbon and the carbon to the right the center carbon has 3 bonds so it needs a hydrogen here. We have 1,2,3,4, carbons and 1,2,3,4,5,6,7,8,9,10 hydrogens. If we were to move this methyl group here, it wouldn’t have been a constitutional isomer because it is the same structure as the first one just flipped around. So we only have 2 constitutional isomers. Sometimes we will given a chemical formula that can have constitutional isomers that are either an alkane, an alkene or a cyclic ring. These two formulas will help you determine which type of structure you will draw. An alkane has this formula In the previous example we had C4H10 where n is equal to 4. We’ll plug n into the formula of 2n+2, so multiply 2 times 4 to get 8 plus 2 is 10 so this is how we knew it was an alkane meaning everything had a single bond. For an alkene or cyclic ring, this is our formula Let’s say we had C3H6 Where n is equal to 3 and we will plug this into the formula of 2n to get 2 times 3 which is 6 so since it fits with the formula the constitutional isomers can include either an alkene or a cyclic ring. Let’s do two common test questions. The first one asks us to find all the constitutional isomers for C6H14 . So start with identifying if this is an alkane or an alkene or cyclic ring. We’ll use the formula for an alkane first where in this case n is 6. Plug 6 into 2n+2, 2 times 6 is 12, plus 2 is 14 so yes this is an alkane meaning everything will have a single bond. Now start with drawing the simplest structure, in our case it’s the typical skeletal structure. Next start moving a carbon chain to a different location and double check that we still have the same amount of carbons and hydrogens as the given chemical formula. So I’ll place all of the hydrogens where they need to be. And we have 3, 6, 7, 8,9,10,11,12,13,14 hydrogens and 1,2,3,4,5,6 carbons. Now let’s move this carbon chain here, and double check our hydrogens, so 3,6,9,10,11,12,13,14, and 6 carbons. We can keep going so we’ll move this carbon chain here, recount and yes we have all 14 hydrogens and 6 carbons, and next, we’ll move this carbon chain here. And we again have 14 hydrogens and 6 carbons. If we were to move this back it would revert to one of the isomers we already found so that’s how we know to stop. And we found all 5 constitutional isomers. The second question asks us to find all the constitutional isomers for C4H8 So we’ll check and see if this is an alkane or an alkene or cyclic ring. I’ll start with the alkane formula. Where n is 4 and if we were to plug in 4 for n, we have 2 times 4 is 8 plus 2 is 10 so no it’s not an alkane since we have 8 hydrogens and not 10. I’ll try the alkene/cyclic ring formula Where n is 4 and we’ll plug in 4 for n so 2 times 4 is 8 so yes this is an alkene and cyclic ring. Since we know it is an alkene, that is the first structure we will draw. For the second isomer, we can move this double bond over. If we move it again to the end carbon it is actually the same structure as the first original structure we just drew. But we can instead move this bond down. We now have a cis and trans isomer, where trans refers to the substituents being on opposite sides of the double bond and cis refers to the substituents being on the same side of the double bond. Since there are no other ways to draw this alkene we can now see if we can make this into a cyclic ring. This would be our cyclic ring, it works because there are 4 carbons and 8 hydrogens. There are two more possible isomers, we can rearrange this to form a cyclic ring with a methyl group on top and once again we would have 4 carbons and 8 hydrogens. We could also put this back into its alkene form and have the substituents going this way. This last isomer is commonly missed by most students so don’t miss this one. And these are all 6 possible constitutional isomers. If you would like additional help like online tutoring or homework help or other resources all of the links are in the description box. And other helpful videos are right over here. And remember stay determined you can do this!