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Provide an IUPAC name (ignore cis-trans stereoisomerism) of each ofthe following molecules draw chair conformation, flip the conformation except for (d)) and determ...

Question

Provide an IUPAC name (ignore cis-trans stereoisomerism) of each ofthe following molecules draw chair conformation, flip the conformation except for (d)) and determne which contormation more stable

Provide an IUPAC name (ignore cis-trans stereoisomerism) of each ofthe following molecules draw chair conformation, flip the conformation except for (d)) and determne which contormation more stable



Answers

Name the following alkenes, and tell which compound in each pair is more stable:

This problem is from organic chemistry and it's based on the U. P. C. Names of the organic compounds here we are given this compound for which we have to ride at UBC name. Also we have to draw a constitutional consumer of this compound and that are you pc name of that compound. So first of all we will expand the structure to see how different items are arranged. So we can expand it like CH three C. H. Having metal group CH two CH two ch three. I'm providing numbering to the carbon in terms of the streets and we can write numbering from the left side. We are in carbon second is substituted when a metal group. So we can name this compound as two metal bending. Then we will draw a constitutional is more of this compound. Constitutional. I summers are dry summers which have the same molecular formula but different order of atoms in the bond sold the molecular formula that's given components C. Six H. 14. Now we will draw another compound with the same molecular formula, which will be the constitutional assuming of the given compound. We can make the compound as soon here CH three, see having to metal responded to it, then see it too, CH three. And this compound, we can see four carbon atoms in the chain when carbon number two substituted by two metal groups. So we can write the name of this compound as to two bio metal built in.

So Hey, we're just looking. Are ice Amazon? We're gonna be drawing Iceman's for C three h 80 So, firstly, we could have our primary alcohol. We could also have a secondary alcohol for this structure where we still have the exact same molecular formula present on now to look at our second example, we're during an elder hide on a key tone with the molecular formula C for age 80 So our carbon chain is full carbons long. That's a bute tiles on for our key tone. We're adding a carbon I'll to any carbon that isn't the end carbon, how that one drawing and all the hide we are adding a carbon I'll to a terminal carbon.

All right, So this question asks us to draw. Part of the question asks us to draw all of the structural I summers possible for the C three h eight. Oh, formula so administered out by my first compound is drawing my three card in skeleton, and I'm gonna go ahead and put one oxygen right there. Now, by doing this, we can see that will have on h there in h here. Two legends coming off that middle carbon and then three coming off that end. When we count up all those hard pigeons, we only have seven. So we know that that last time that he needs to go right here on this oxygen with this being the case, we know that we can't make a compound that is an alga hide or key tone because that double bonded oxygen on one of the carbons would take away one of the hydrogen. So we only have seven instead of eight. So all of our compounds are going to be alcohol's. You know, when we go ahead and name this molecule, we see that it's a the re carbon compound. So you know that it's propane and we see that this O age group right here is on carbon one. So this would just be groping all and it is an alcohol great melody. Want to our second compound again? I'm gonna go ahead and draw my three carbon skeleton, and it's sort of putting the h on one of these either. And carbons. I'm gonna go ahead and put my O H group in the middle. Now we can just go ahead and add in. Hydrogen is making sure that we have eight. And again this molecule Haslett. Ohh. Here. So it is in alcohol and it's again a three carbon cornpone. But that only true was on that second carbon. So this would be named too Probe in all when it again is an alcohol? No, For our third compounds, we can't add an O h group. So any of the other carbons about repeating one of the molecules you've already drawn But we can, however, at in in oxygen in the middle of this molecule. By doing this, we can add our head regions and see that there are indeed eight of them. No, this molecule falls under the either category because it has this oxygen and are coming off his oxygen. We have one functional there appear and one functional group here. Now, when we go ahead and name those functional groups, we can see that on this side right here is just a metal group and on the other side, but left side of my oxygen We have an effort. So when we go ahead and name, this molecule would just be a soul math. All be sir, you Can we put the I feel first because he is comes alphabetically before the letter m So those you have three compounds you can draw of your C three h h o molecular formula My moving on a part Be of this problem. It asks us to draw both an alto hiding acute tone with the molecular formula C for H a o and the name. Hm. Now we know that in Aldo Hide partial group looks like the sea double wanted Oh, no one hydrogen and then the rest of the molecule. And we know that we were naming and Aldo hide it has a Suffolk's of A and A So I'm gonna go ahead and start by first drawing my Aldo hide functional group that uses up one of our carbon. So I'm just gonna go ahead and add in the other three. Well, you see that We already have the oxygen in there, so I'm just gonna go ahead and at hydrogen is to all of the other carbons making sure that each carbon has four molecular bombs and we can see that this molecule does, in fact, have eight hydrogen in one oxygen and four carbons. So now when we go in a minute, we see that it's a four carbon compound means it isn't butane molecule. And you see, that has an all the hard functional group right here. So has the Suffolk's a n A L. So when you go ahead and name this, it's just butin all and again, we don't have to put a number in front of it. Beautiful, like one, because we know that when it's just a word like this, usual that that functional group falls on the first carbon of the molecule. So now we can go ahead and draw our key tone. Now we know a key turn functional group looks something like this. It just has a C double barnet. Oh, in the middle of a molecule with two different function orders on either side of the carbon it's attached to You know, that a key tone has the suffix a n o e and no. So when we go ahead under our key tone, I'm gonna start off by drawing or key tone functional, See, Double 10 and all they have to do is add three more carbons And it doesn't not matter which side those apartments go on. It just can't be the key to have. You can't be one of the end carpet that has to be in the middle. So now I'm gonna go ahead and draw in all of my other hydrogen molecules, Adams. And we see that we also have a hydrogen is here. Now we go ahead and name this key tone. We see that the key tone functional group falls right here on this card, which is carbon too. So this would just be named too, you know? Well, we could if we choose you, we can juice, huh? Not include that too. And it would just be view to known. But we can do this with those four carbon molecule because We know that the key tone functional group has to fall somewhere in the middle of the molecule. And because there are only four carbons, that functional group can either fall on this carbon right here or this carbon right here in either way. Whichever carbon you put it on, it would be to you to know. So both of these answers are

Okay. So as to draw the chair confirmation of the following compounds and identify Carol centers and label them R R s. So the first compound is this? Yeah. Okay. We have a turf battle group. We're carbon bound to three methods here. 123 and then a carbon four. We have Hey, Mitchell. Yeah. Yes. So they draw a chair. Confirmation, use draw. Two lines go down and then up, then same here, too. So I'll put Axle in red. It's up. Down, up, down, up, down That equatorial and green. This is six. So we want to put the turbulent group in the equatorial position, and I'll explain why in a second? Yeah, so that's down. The methods should be down as well, so it'll be axial. So it doesn't matter whether it's axial or equatorial as long as there both pointing down or up. If they're assists. And if the trans they should be opposite, then the other chair confirmation, we can put it in the actual position. Yes, Mhm. So this is going to cause if we put the tribunal group of the axle position, it's going to interact with the hydrogen. Is that our three carbons away. So this one counting with red, we have one, 23 and then counting. With green, we have one, 23 So, yeah, this carbon, too. The hydrogen are going to interact with the whatever carbon group is in the actual position. And this is known as 13 dioxido stream. So it occurs when a carbon group is in the actual position and the hydrogen is on a carbon that's three away will interact with it to the second. So therefore, this is going to be one on the left is more stable. We also have to identify the carol centers, too. So, um, over here we have a very beautiful hydrogen, and these two sides are actually the same. So we don't have Carl center in this molecule, and it be the same with this, sir, is it? These two sides are the same. The second one is this compound to broom in here. Mhm. Okay, Yeah. Okay. So we can put Mm hmm. So it's number at first. This is one. This is two. This is three and four. If we did it the opposite way and read to 12345 maybe too many numbers. So let's put the broom in in the actual position. So it's up. Which means carbon four should be up as well. It'll be equatorial because there, sis. Mhm. And carbon two is down. Because there and it's anti from carbon one and four to carbon two is down an axle. And this is more stable. Yeah, I'll explain why in a second? Mhm. Yeah, it's Yeah. Mhm. Okay, so we can let's put the broom in down, which means this one has to be up. Mhm. So this is down because they're cysts. So you notice that these two Burmese are very close to each other. They interact, and this is going to cause strain. So therefore, this is less stable. It's due to the interaction. Okay, so let's look at the Carroll centers too, So we get rid of these, Okay? So all right. So looking at this one in red, we have a brimming. We also have a hydrogen are here. This side is that this side is different from this side. So this berman is one. The hydrogen is four, and now we're looking at ch br verse ch two ch two C H p R first ch two c h to the ch br was gonna win the Burmese a higher atomic number than the second hydrogen. So the starts to this is three. So going from one or two and then 2 to 3 is going to be our So you get there. It's pretty red. So this is our than this one again. The Bruins won. Hydrogen is four. So we've siege to siege too. And then it's a siege, too. This star is going to win. So if you look over here, we have ch two. And the second carbon has a a booming attached to it. But this side has We've siege too. And then a siege, too. Then the Berman wins against the second hydrogen. Yes. So the sides to over here this is three. So from 2 to 3 or from 321 And that one or two eventually skipping four. We have s they Finally, this this carbon and blue. This is one. The side over here is to weave a ch br instead of a siege too. This is too. The hydrogen is for and besides three so 1 to 2 to three or from 321 skipping four. It should be s but the hydrogen is on the on the wedge me redraw that it's on the wedge. So it's going to be instead of, um, yes, it's gonna be our So if we Okay, um it's everything. Yes, we are R. R. S. You want me to put the the Bethel here too?


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