Draw the condensed structural formula, or line-angle formula i...

Question

Draw the condensed structural formula, or line-angle formula if cyclic, for each of the following: (12.1)

c. 2-methyl-3-pentanol

Accepted Answer

Welcome back. Our next problem says, give the condensed or line angle structure of a cyclic compound of four ethyl to hex and all. And we have three condensed structures labeled a through C. We'll go through each one in detail as we compare it to what we come up with. And then choice D says none of the above since this is a cyclic compound. So first, when we look at a name and are generating a structure, we need to look at what is the base compound. And in this case, our base compound is Hexon all. So we're talking about Hex and all, we're talking about a six carbon chain because we're based on hexane with an oh substituent. We have an alcohol here. So we're going to need to draw a condensed structure. So choice D is incorrect. We do not have a cyclic compound. A cyclic compound would have a name like cyclohexane if it were a six carbon ring benzene, if or an aromatic ring, phenol anneline. So in this case, we just have a straight six carbon chain. So we're going to draw a condensed structure, not a line angle structure. So we'll draw six CS. I'm going to leave space in between each one to make room for the proper number of hydrogens. But then we draw bonds between each of them in a line since we have a chain here. So there's four carbons, 56. So we've got our six carbon chain. We know that we have this oh substituent. So we need to look, is there a number in front of our hex and all to let us know where the oh group is? There's no number, we assume it's carbon one, but we've got this number two in front of the name Hex and all. So our oh group must be on carbon two. So we'll just number this left to right. So carbon one is the first on the left. So carbon two, second one there add our oh group. I'll just finish numbering my chain up to number six and then actually were to change the numbers to red. We have our substituents in blue. That way we don't have any conflict, conflict about what or confusion about what is part of my structure versus what are my numbers? So changed my little numbers to red. Those are just the numbering system for the carbon helping me to locate things. So we have our base compound, we've put on the alcohol. That's our main functional group. Now, we just need to locate and properly label the substituents. So we've only got one substituent. That's an ethyl group. So that would be a two carbon branch and that's on carbon four. So from carbon four, we draw a line upward and write CH two C three, that's our ethel branch. And now we just need to fill in the correct number of hydrogens. So again, our following steps after our base compound were properly place the substituents and then last of all fill in the hydrogen atoms. This is a condensed, not expanded structure. So we just label the hydrogens in here in letter groups. So ch three ch two, we're not drawing out individual bonds to hydrogens. So carbon one is just a terminal carbon with no substituents. So we add three hydrogens to that. So we have ch three carbon two has a substituent and then two carbons in the chain. So it just gets one hydrogen, carbon three, no substituents, it's internal. So it gets two hydrogens. Carbon four has that ethyl group plus being in the middle of the chain. So just one hydrogen, carbon five, no substituents, internal two hydrogens and carbon six is terminal with no substituents. So three hydrogens. So we've got the structure, we need condensed structure. Let's look at our answer choices and see which matches. So in choice a do we have the correct number of carbons? Well, let's look at this chain. 123456. So we have six carbons is our oh, in the correct place, it is on carbon two and our ethyl group on carbon four. So right away with choice A, we have a matching structure to what we determined. So I'm going to circle A as my answer to be thorough. We'll just look and see what's wrong with our remaining B and C structures. So choice B, we look at our number of hydro carbons and we have 123456. So correct number of carbons move on to the oxy group. It is now it's drawn on the other side. So it's drawn on the second carbon from the right. So that is OK. It's still on carbon two, which is where it's supposed to be just numbered from the other side. So that would be OK. But then we need to look at the ayl group. So if our oxygen is on carbon two, so we've labeled 12, here's our ayl group and instead of being on carbon four, it's on carbon three. So the ethyl group is in the wrong place. So that's why choice B is incorrect. Then we look at choice C count our carbons 12345. So choice C only has five carbons in the main chain. It's missing that last methyl group because it has an O group right on that fifth carbon. So its main problem is that it has only five carbons. The oh group would be on the correct carbon if there were another carbon in that chain. But it's missing. And then uh we have two methyl groups rather than an ethyl group. So to c has several things wrong with it. Definitely not our answer. So once again, the proper condensed structure for four athol two Hexon all is choice A. We have C three C with an oh group, C two, C with AC two C three ethyl group, then CH two and then CH three. See you in the next video.

Draw the condensed structural formula, or line-angle formula if cyclic, for each of the following: (12.1)
c. 2-methyl-3-pentanol

Key Concepts

Condensed Structural Formula

Line-Angle Formula

IUPAC Nomenclature

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