Give the IUPAC name for each of the following: (11.3)

Question

c.

Accepted Answer

Hi, everybody. Let's take a look at our next problem. Provide the Iupac name of the molecule below. And we have a condensed structural formula here, it shows six carbons with dashes in between them representing six carbons in a line. The third carbon along has a methyl group attached to it and an ethyl group attached to it. The fourth carbon along has a methyl group and the fifth carbon along has an Appy group. Then our four answer choices are a three ethyl, 234 tri methyl heptane B 35, diethyl, 34, dimethyl hexane C 24, diethyl, 34 dimethyl hexane or D three ethyl, 345 tri methyl hept. So our first job of naming a compound is find the longest carbon chain. Now, this can be a little trickier than it sounds because of course, sometimes the structure is presented to you in a way that one area looks like the main chain. But if you instead follow along a branch, you'll find a longer chain going off in that direction. So when we look at this molecule, I'll use a blue highlighter to identify and count carbons. So first, I'll go along this main horizontal chain. So highlighting each one in turn, I have 123456 carbons in that chain. However, note that on carbon five, there's an ethyl group of two carbon group. So if I erase that sixth highlighting and instead at carbon five, go down the ethyl chain that would be 56, seven carbons. So by going down the chain, I end up with a longer overall carbon chain that if I continue along what looks like the main chain, the way this structural uh the structure is presented to me. So that's definitely something to keep an eye out. So if you have long chains, double check that or long branches, double check that going down the branch doesn't give you a longer main chain there. So we have here our main chain which has seven carbons. So our base name here will have the prefix hep or seven. We have no double bonds, triple bonds, any funny functional groups. So it's a HEP D. Now note when I look at our answer, choices, that choice B has Hexane as does choice C. So both of those named it based on the main horizontal line of carbons here and not the actual longest chain. So choice B and Choice C will be incorrect based on labeling the longest chain as being six carbons. But choice A and Choice D have the proper heping. Now that I have my main chain, I need to start looking at substituents. So I have a chain of carbons that can be numbered from either end. So I need to look at the numbering. So I'll say number chain to give the lowest possible substituent number. So if I go from left to right, I'll do that with little blue numbers. I go along counting carbon one, carbon two, carbon three and then I come to two branches. I've got a methyl group and an ethyl group on carbon three. So in the blue numbering, all right, blue, my first substituent is at carbon three. Now a number in red starting from the other side of the chain. So on that, what looked like an ethyl branch, but I've now made my main chain carbon one, carbon two, carbon three and on carbon three, I have a methyl group that was originally the last carbon in my horizontal chain there. So my red numbering system also has my first substituent on carbon three. So how do I choose which nut break system to use? Given that I don't have a symmetrical molecule. So it will matter how I name this. Well, in my blue system, I have two substituents on carbon three while in my red numbering system, I only have one. So using the blue system will give me the lowest overall substitute number since I'll have two substituents at that lowest C three. So I'm going to cross out red C three and erase those numbers, so as not to get confused. So I'll add more blue numbers to help me number my substituents. So I'll add 456 and seven to help me keep those straight. I've numbered my carbon chain. I'm ready to number and name my substituents. So I have, I look overall at my molecule, I have one ethyl group and 123 methyl groups. So my ethyl group is on carbon three. So I will be naming that as three hyphen ethyl. So that will describe one group, I have three meals and they're on carbons, 34 and five. So that will be three comma four, comma five hyphen. So note, we use commas in between numbers hyphens when we're separating numbers and names, I have three metals. So I'll give it the prefix. Try. So that will be try methyl. Well, how do I decide how to put them in order when putting them in front of my name? Well, I use alphabetical order so I'll write ABC order, not including the prefix, all right, not prefix. So I don't count that try in there when I'm putting them in alphabetical order. Well, Ethel will come before methyl. So my name will be three ethyl coming first. So both A and D have that and then 345 tri methyl heptane. And that's what I have in choice D. That will be my answer. When I look at choice A, I have 234 tri meyl heptane. So my methyl groups are given the wrong numbers for where they're located. So that's why choice A is incorrect. So once again, the IOP name of the molecule given is choice D three ethel 345 triethyl heptane. See you in the next video.

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Give the IUPAC name for each of the following: (11.3)

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Struggling with GOB Chemistry?

Give the IUPAC name for each of the following: (11.3)c. <IMAGE>

Watch next

Naming Alkanes with Substituents practice set

Give the IUPAC name for each of the following: (11.3)

c. <IMAGE>