Seven alkynes have the formula C₆H₁₀. Draw them using line str...

Question

Seven alkynes have the formula C₆H₁₀. Draw them using line structures.

Accepted Answer

Welcome back, everyone. Here's our next problem. It says there are a total of three alkins with the chemical formula C five H eight give their structures. So we're talking about having a triple bond in our molecule. So let's start out, we have five carbons. So we'll start with looking at a molecule that has five carbons all in a row. And I'm writing out the letter C just to think through where this triple bond is going to be, I'll eventually draw a line structure for each of them. So we have five carbons here. The first thing to note is that when they're all in one chain, this is going to be a symmetrical molecule, it will have this line of symmetry through it when we have no particular branches. And therefore anything we do on one side will have sort of a identical molecule created by doing the same thing on the other side. So that rules out certain options in terms of how many structures we have, we are given a little tip that there are only three possible, all kinds. So when we get to three, we can stop. So our first possibility would be with a triple bond between carbons, one and two. Again, our plane of symmetry shows us that if we put that bond between four and five, we'd have the same molecule. So we don't have to worry about that one. Another possibility with five carmens in a row. So we'll write that uh I'm going to do that bond highlighted in blue as one possibility. Another possibility of placing this triple bond would be between the carbons two and three. So sort of internal to my molecule. And so that would be another possibility. And again, that will be identical to actually, I'm gonna highlight that in a different color. I'll highlight that in red to show there are two different possibilities. I wouldn't obviously have two triple bonds next to each other. And that obviously will be the same molecules if I put the bond between carbons three and four. So that gives us two possible arrangements. And that exhausts the possibility of our five carbon chain. How about if we had a four carbon chain? So 1234 carbons, well, we could put a branch on here, obviously not on either end or we just make a five carbon chain. So we could put 1/5 carbon up on carbon two again, with a four carbon chain. Anything we do on carbon two will be the same as carbon three due to the symmetry there. Well, where could his triple bond be on this molecule? Well, this carbon too, in this case, cannot have no triple bond, cannot have any triple bond because it already has three bonds, it can't have a triple bond, that would mean adding two more. So there can be no triple bond on our methyl group nor on carbons, one or three. So there's only one possible location for triple bond on this molecule. And that's that terminal carbon on one end. So that's yet another possibility, I'll highlight that in green. So there we go, we have three different structures. So two possibilities for a five carbon chain and one possibility for a four carbon chain, obviously a three carbon chain wouldn't be possible one because we already know there's only three. But if you think about a three carbon chains jot this real quick, any branch we would put on here could only be in the center. And of course, that that central carbon already has three bonds. No triple bond is possible. In that case, you might say, well, what about a triple bond up at, if we put an ethyl group in the middle of that three carbon chain, could we put a triple bond at the end? Well, then we've just made our mo we've just made a four carbon chain and our molecule will be the same as the one we've already drawn. So cross that one out. So we have our three possibilities. Let's just draw our structures for them. So on our first one, we have our five carbon chain with the blue highlighting our triple bond at the end. Now, which one key we have to remember the geometry of triple bonds is linear. So we're going to draw this long line to represent three carbons. So the triple bond in place between carbons, one and two and then we'll have carbon three in a straight line. I'm just gonna put sort of a little dot There to remind me and then I'll race a little tricky to know how long to make that line. So redraw that. So we have carbons 12 and three. Now we can start our zigzag pattern but always remember that that um CC triple bond has a linear geometry. So we've got carbons 123, we just need carbon four and five. So here we go BC five H eight. It's got a triple bond at the end between carbons one and two. Now we draw our other five carbon chain where our triple bond is in between carbons two and three. So we'll draw an even longer line because we've got this sort of medial triple bond. So we have all in a row, carbons 123 and four and we just stick off carbon five there. These are kind of funny looking structures. They really look odd, but they help us remember that that triple bond is a linear geometry. So we've got our two structures with five carbons in a row. Now, we just need to do our third possible structure with a four carbon chain and a branch. So I'm going to sort of reverse how I drew it again to have our triple bond in that straight line. So we have our straight line, our triple bond between carbons, one and two. So 123 is the end. There's going to race a little bit, make it a little shorter. We got carbons 12 and 3123. And then we need a little branch there, essentially two methyl groups. So just double check that we have 1234 carbons in a, in a our longest chain and a methyl on carbon three. So here we go are three formulas. We have 25 carbon chains with one with a triple bond between one and 21 with a triple bond between two and three and then 14 carbon chain with a metal branch on carbon three and the triple bond between carbons one and two. See you in the next video.

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Seven alkynes have the formula C₆H₁₀. Draw them using line structures.

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