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Ch.22 - Organic Chemistry

Chapter 22, Problem 41b

Name each alkane.

b.

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Welcome back every once in another video name this L Kane. We want to recall that the first step in naming L Keynes in systematic naming would be identifying the longest continuous carbon chain, which we call a parent chain. So what we're going to do is just identify our parent and we can do that by counting the number of carbon atoms in the longest continuous chain. So we have 12345, six and seven. Now we have to recall that a seven numbered L cane is called heptane. So that's our parent. Our next step is to identify the substitutions. What we notice is that we have two subscriptions bonded to carbon number two and carbon number six because this is where we have branching. Right now. Our subscriptions are CH three. We know that CH four is methane and its corresponding substituents. Ch three is called methyl, right? We're essentially adding a new prefix. We can also see that the two methyl substituents are bonded to carbons two and six. Before we move on, we simply want to check if we can minimize those locums. What if we actually enumerate our parent from? Right? To left. We get 123456 and seven. We still have our metal substituents bonded to carbons two and six. Meaning it doesn't matter if we enumerate the parent from left to right or right to left. Now, we have all of the information, we need to name the L cane. Let's recall that we begin by identifying the locus of the subscriptions. We're going to essentially go with two comma six in an ascending order. We're going to use a prefix dye because we have two equivalent methyl substituents. And now we're just naming the substituent which is methyl. So two comma six dimethyl means that we have two metal groups bonded to carbons two and six. And now we're going to finish our name with the parent, which is hap and that's our final answer. Thank you for watching.