Draw all the ketones you can with a chemical formula of C₈H₁₆O whose longest chain is eight carbons. Name each using both its IUPAC and common name.

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Hi, everyone. Welcome back. Here's our next problem. Which of the following structures represents a ketone with the longest chain of five carbon atoms and a chemical formula of C seven H 14 0. What is its Iupac name and common name? And were presented with three different skeletal structures labeled 12 and three. I'll talk about and describe each one when we go back and evaluate them. But meanwhile, let's look at our answer. Choices. We have four different answer choices. A structure one, Iupac name two methyl pentane, three own common name, ethyl isopropyl ketone. Choice B structure two, Iupac name 44 dimethyl pentane, two own common name methyl neal ketone. Choice C structure three, Iupac name three ethyl Pentel, common name ethyl pal ketone and choice D, none of the above. So really important to note are four different answer. Choices A B and C, each one has a different structure offered as a correct structure. So there's no case where the name of the compound is going to make a difference in our correct answer. You might think. Well, choice D is none of the above. But it's just answering the question. Which structure represents this. So you have either structure, one structure two, structure three or none of them. So this is why it's really good to look at the multiple choice options before you start to work through a problem, because you aren't going to need to figure out the name of the compounds to distinguish between the correct answers. So with that behind us, let's look at our structures. So structure one shows a carbon group in the center of this molecule. There's an isopropyl group on one side of that carbon and an ethyl group on the other. So our first condition says our structure has to represent a ketone. And as we recall, a ketone is ac o double bond where that carbon has two R groups attached to it. Al Kar Aal groups. So structure one fits this condition. We look at it co double bond that carbon has two different R groups. I've highlighted that co double bond in blue. Now, let's look at structure number two. Here we have the co double bond they'll highlight in blue. It has a methyl group on one side and it has a 123 carbon chain with two methyl groups on the second carbon attached to it. So that's two R groups. So this is a ketone as well. Now, let's look at structure number three, structure, number three has a co double bond, but it's on the end of the molecule, it has two hydrogens, they're not on the structure, it's a line structure. So they're not drawn explicitly, but this terminal carbon would have two hydrogens attached to it along with its co double bond. And then it has a 1234 carbon chain coming from it with an ethyl group attached to the second carbon away from the co double bond. So this is not a ketone, since it doesn't have two R groups, it only has one R group. This is an aldehyde, not a ketone. So structure three cannot be our structure. And therefore we eliminate choice C which has structure three as our structure. So we've eliminated structure three. Let's move on to our next condition. It says that the longest chain must have five carbon atoms. So I'm going to highlight that in red, we'll look for our longest chain. Let's look at structure one. Well, I think it's always easier to start with a non branched um our group if you have one. So I'll start with that ethyl group and trace the longest carbon chain possible. So I have one to come to my ketone on carbon three and then four and then five down on one of the isopropyl branches. So 12345 carbons in a chain with the one methyl group left over. So structure one fits that condition. Now, let's look at structure two again, I've got my branched four carbon chain on one side and my methyl group on the other. So I'll start with my methyl group. 12, there's my carbonel carbon 34 and then up onto one of the methyl branches makes five. So I have a five carbon chain with two methyl groups left over. So both one and two fit that condition. So I've got a ketone. I've got structures with a longest chain of five carbon atoms. Now let's move to the next condition. A chemical formula of C seven H 14 0. Well, I know each has one oxygen. How many carbons do we have? Structure? One? I have a five carbon chain which I've just counted and one methyl group left over. So I have only six carbons in structure. Number one, there's only one carbon outside of my five carbon chain. So structure one cannot be correct since it wouldn't have ac seven chemical formula. Let's look at structure number two, I have my five carbon chain. What's left over two methyl groups? Two more carbons giving me that C seven. So I just need to check the number of hydrogens. Well, I know that if I have just a plain old alkane, only a hydrocarbon, my molecular formula will be C subscript NH subscript N plus two as long as it's just an alkane. So no rings, no double bonds, which in this case, there are no rings and no double bonds in my structure. The only exception is I have that C double bond O that will take away two hydrogens because two of the bonding spots on the carbon are taken up by that double bond. And so we'd expect C 714 as indeed we have. So we've got our chemical formula for structure two matches. And since that's the only structure left that works, I know that choice B must be the one, it's not one or two. And then I see that it fills all the conditions so I can eliminate choice D none of the above. Since all the conditions are met by structure number two. So choice B is going to move my answer just to be thorough. We'll double check this name for my Iupac name. My main chain of course is Pentane. As we see, five carbons, the ketone takes top priority as a functional group. It's on carbon two because we want to give it the lowest possible number. So we designate the methyl group as number one. The carbon carbon is number two. So Pentane two own as we have. And then we have the rest of our five carbon chain with two methyl groups on the fourth carbon. So 44 dimethyl pentane 20. So our iupac name is correct. The common name is a little bit trickier because we have this group with five carbons, our 44 dimethyl um pentane there, that whole R group has five carbons total. And most of us are not as familiar with the common names for five carbon branches, we usually know our isopropyl, we know our different beetles, not the pal group. So that's a little tricky. But again, we don't really have to distinguish this because we just have the one correct structure. We can recall in general that the common names of ketones involve naming the R group, the second R group with a space between them and then the word ketone. So when we look at the common and given we have methyl, which we know is our one, our group. And then Neo Penal, well, we might not remember that neo penal is the correct name for this sort of branch structure, but we know we have five carbons in that branch. So it makes sense, that would be correct. And then ketone, but again, we don't really have to distinguish our correct answer from our incorrect answer by the name. So don't stress too much about that. So the structure that represents a ketone with the longest chain, five carbon atoms chemical formula of C seven H 14 0 with its names is Choice B structure two, with that Iupac name of 44 dimethyl pentane tuum and the common name of methyl neopet ketone. See you in the next video.

Draw all the ketones you can with a chemical formula of C₈H₁₆O whose longest chain is eight carbons. Name each using both its IUPAC and common name.

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Draw all the ketones you can with a chemical formula of C₈H₁₆O whose longest chain is eight carbons. Name each using both its IUPAC and common name.