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Ch.22 - Organic Chemistry

Chapter 22, Problem 38

Write structural formulas for any 6 of the 18 structural isomers of octane.

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Welcome back, everyone. Draw four of the eight constitutional isomers of hete. The first step in this problem is to understand that the heptane corresponds to an L cane with seven carbon atoms. So we are going to draw heptane, it has 123456 and seven carbon atoms. What we want to get in this problem is an isomer of heptane, right. And specifically, we want to get an isomer of heptane that is a constitutional isomer, meaning we want to have the same molecular formula, different linkage of our lal subscriptions. So what we can easily do is just shorten our chain. Instead of having a seven member chain within the longest continuous carbon chain, we can just have six and we can bond the remaining carbon to carbon number two, to introduce branching, right. This is one of the ways to get a constitutional isomer. So we have the first constitutional isomer because it still has the same molecular formula but different linkage. Now the longest continuous chain only has six carbon atoms while there's a methyl substi bonded to carbon number two. And that's our first isomer. So let's label it as number one and we can just continue and draw three more. What we can do now is shorten our chain even more. What if we just take a total of five carbon atoms in the longest continuous carbon chain? In that case, we want to bond the remaining carbon to any of the two carbon atoms within the middle of the chain. So we can just go with the terminal end of the chain. And this is how we get a five membered carbon chain with two methyl substituents at carbons two and four. And that would be our second iser. Now, we can choose another chain. What if we specifically choose a four member chain? So we are shortening our chain once again. Now we have a four member chain and we want to bond the remaining we carbon atoms two carbons two and we, so we can essentially bond two of them to carbon number two and the remaining one to carbon number three, this gives us our third constitutional isomer. In addition to that, we can get our fourth isomer by choosing a, let's say five member chain once again. So if we take our previously drawn structure with a five number chain, we can simply bond the two metal substituents to the same carbon number two instead of two different carbons. And this gives us an additional fourth constitutional isomer of HEP. So we have four of them and that's sufficient for this problem. Thank you for watching.