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2) (15 pts) Provide the IUPAC or common name, as indicated, for the following compounds_IUPACOHOHIUPACIUPACIUPACIUPAC C=NIUPACCommonOHHO...

Question

2) (15 pts) Provide the IUPAC or common name, as indicated, for the following compounds_IUPACOHOHIUPACIUPACIUPACIUPAC C=NIUPACCommonOHHO

2) (15 pts) Provide the IUPAC or common name, as indicated, for the following compounds_ IUPAC OH OH IUPAC IUPAC IUPAC IUPAC C=N IUPAC Common OH HO



Answers

Give the IUPAC or common name for each compound.

This is the answer to Chapter 22. Problem number four Fromthe Smith Organic chemistry. Textbook on this problem is just asking us to name six molecules that we've been given with either eye you pack or common names. And so Ah, these air all carbon Neil derivatives on there are some different rules and naming conventions depending on which derivative of car box l gassid we're talking about. And so is a nasal chloride. So the name is going to end in chloride Andi, we're going Teo name it by its longest chain, which have numbered here with the carbon Neil Carbon as carbon one on DH, then the longest chain that you can make from that is for so butin oil chloride and, uh so at the two carbon, there's also an ethyl group branching off. So when you put that all together, we call it to Ethel Butin oil chloride b is an Esther, and it's pretty straightforward. So we're going to name it. The bass name is Ben's away. Um and then before benzoate, we will put thie alco group That's on the oxygen of the Esther. So in this case, it's methyl group, so be is Mme. Ethel Benz away for sea we're talking about and a mite um and so the name is going to end in a mine. Uh, that's how immature named we will also will get What is Ah, the chain that's attached to the A mine. So starting at the carbon, you'll carbon We have 123 So it's going to be a pro pen a mod or prepared amid, um And then we also need to name the two substitutes on Are the nitrogen in the am I And so we use the designation and to mean that it's a substitute on nights chin. So we have a nephew group and a method group on this nitrogen. Andi will listen alphabetically, Ethel, before Mme Ethel. So this molecule becomes an Ethel and metal propane and mine for D. Um, we have a faur me derivative. So, uh, is similar to an Esther. But there's nothing no other carbons attached to the carbon Neil Carbon. Um And so whereas an Esther, we would name it. 08 Based on whatever was attached to the cardinal carbon off for this. When there's nothing attached, it's for me. And then again, we just start with the alcohol substitute went on, uh, the oxygen off the Esther. And so this becomes Ethel for me, for E we haven't anhydride, which remember, is a condensation of two carbon fuels essentially on DH. So well, just named them with awake for the carbons attached to either carbon Neil and again we'll list them in alphabetical order. So we have Ah, benzene on the one side and a propane group on the other on dso This becomes Ben's OIC, propane, OIG and hydride. Lastly for F, we have a night trial, and so the name is going to end in night trial. We're also going to base the name on the longest chain. Here it is a 123456 carbon chain as number I'm so it will be a heck sane night trial. And then again, we need to just account for the other branch. So it's coming from carbon three. It's an ethic group, and so this molecule becomes three. Ethel Vaccine night trial on that is the answer to Chapter 22. Problem

For auction E the RU The molecule is made up off and Toxie group attached with beauty in Jane the ups The name is one talk See beauty provided the common name is beauty in tight in tow for option B In this molecule on it, Toxie Group is attached to Propane Jing The R u B SC name is one it Toxie Propane The common name for this molecule is entire profile Ito auction See over here Um a toxic group is attached to propane chain Therefore are you B SC name is want mid Toxie dropping And the ups he name is one are you B sc name is metal profile Ito

In this problem from the chapter Organic Chemistry, we will discuss you pc names of organic compounds. Some of the structures of organic compounds are given here. We have to provide their I. U. P. C. Names. So for writing I PC name first, we will look for the longest chain of carbon atom and then we will provide numbering to the carbon atom in such a way that substituted carbon gets the lowest number. And then we will provide the names to the compound. So let's start with the first compound which is shown here here. If we see in this compound we can see that there are five carbon atoms in the street chain. And if we provide them bring to the cabinet um we can start from the right side. We are carbon having some constituent, that's the lowest number. So with the chain of five carbon atoms, carbon number two and carbon number three are substituted by metal groups each. So we can write the name of this compound as to three, die middle 19. Moving on to the next compound, we can have this structure where we have Six carbon atoms in the parent train with substitution at carbon three. When we start numbering from the right hand side, Carbon three is substituted by an entire group. So we can write the name of this compound as three tile eggs in. Then let's work on the third structure where we have this and if we see the number of carbon atoms in the parent train, there are seven carbon atoms with the carbon, the center substituted by isopropyl group. So we can start numbering from either side home providing numbering from either side. The carbon having a supervised group. That is a substituted carbon Gets the same number, that is # four. So we're going to now write the name of this compound has four. I saw profile have 15. Then we have this last structure. We are if we see There are eight carbon atoms in the parent chain, where we can provide numbering from either side. The carbon having substituted the same number. So we can start it from the right hand side. We are carbon three and carbon 6th having metal substitute. Get their priority. So we can write the name of this compound now as three six diamond tile octane. So these are the OPC names written for the compounds given in the problem.

Problems with eight. We're still on our part, naming. So let's start with port, eh? As usual, let's go and find the longest chain we could find. And thanks for this, for this month's pretty obvious, all right. And here we have a double bun, right? And the rule is, whenever we have a double bond or a triple bond, we count from the side. We're we're closer. Ooh, that's fun. So what I mean by that is this double parked right here looks like it's gold or to the right side. And I mean, you don't really have to estimate you could, um, check, but I think it's pretty clear to see that there's only two carbons on the right side of the bond, and there's three on the left, so it's closer to the right, so we count flirting on the right side. What's going count? We've got one, two, three or five on our main main chain again. Let me give a space for Part eight. We got five, meaning we have a pen teen, my so pent for five and 19 because we have, um, a double bind in that double bond starts on carbon, too. right, Sergeant. Carbon tube. So this, this would actually be to painting. Okay. Just kind of round out of space there, but Okay, so we have to bend next, is rechecked our bench point. And we see here on carbon four that we have a metal group. I'm branching out, so we have a four metal foreman. Hey, so now that we have all the information we need, we could combine all of it and come up with our engine. So we have only one, um, one group we need to take care of. We got four method for metal. Two Fantine still Trib. Your answer for part eight. Go to part B. No, Part B. All right. So, again, as usual, let's find the, um, longest chain in this. Looks like our longest Jane right here. Right. Corbin's up here compared to just one. If you go, um, along a straight line and again account from where our double bind is closer, and for this one, it's pretty obvious that it's on the right side. In fact, it's out the end. The right side here. Okay, so we're counting from the right side here. 12 three or five Hey, So, uh, just like part A, We got Fantine. But this time, our double bond starts on carbon one. Right? So we have one Fantine bent because there's five carbons and pantin e n e. Because we have double bun. So next as we take care of our branch points and for this one, we actually only have one. We have a metal group connected carbon three. So we have I mean, it's a different color pen here. Um, we have three methods, three metal. So I think that's all the information we need. And we could proceed to combining all these information. So we have three muscle. The name our, um, branching group first. So three metal, one Fantine. And that's her. Answer or part be for Chino. We have a benzene ring and connected to our benzene ring or two chlorine. Adam. Hey, So got account. One on the carbon wear. Our marshal a group is or our Adam is. So this would be one and we gotta count two on the closest I'm branching group as well. So this has to be to hey is if we count to hear than this Korean beef carbon sick and we don't want that. So this Korean would be on carbon too. Okay, so we have die Koro. Right? So we have one too. One two. Since that's where we have our Corinne Adams and gotten die Claro for having to Korean Adam's, right, Like quarrel Benzie. Alternatively, the way that their positions since they are each other, we could also name that for so forthem. Shall we? Right? Oh, for Ortho. Oh di Toro, then. Hey, so there's two possible I APAC names for court. See? All right, we've been on to Port Party, as usual. Let's find the longest chain. And I think for this one that looks like it doesn't matter, right? So even if you get this metal or this metal, it's going to be the same length since it's just one method for both choices. All right, so let's just take the straight Jane, I think it's simpler and the count and see where we hit our double bond first. Okay, so if you count from the left, you got one too. 34 And if we come from right, we have one two me or Okay, So either way, like, either way, you know, it's kind of getting messy here, but either way, our double bind is gonna hit carbon to first. Hey, but if we count from the right and follow the green numbering once we hit carbon to not only do we hit our double bond, we also hit this branch point. You're good. So kind of on a shooting. Two birds in one stone right there. So we are gonna come from the right on this one, Scott, from the right. I got one. Who? Three four. And since we have or that's a beaut mean because of the double bind. And it starts when carbon too. Right? Double bond starts income number on carbon number two and gives us too beauty. Okay, I didn't write that well, but get what I mean, that's true beauty. And we have this metal group sticking out from carbon to as well. So we have a two metal. So combining all those information, we have two Bethel two. Too much too beauty. Right? And that would be your answer for party. All right. Finally, we are on part E here and again. Find the longest Jane, and it looks like it's pretty obvious whole thing here is our longest. Jane. So again, we are accounting from where our triple bond is closes. Okay, so this applies to both double and triple bond. So it's closer to the left Overcounting from the left. First, I want to Hey, 45 maybe write that of it in eight. So we get a and our triple bond starts on carbon three. So we got three. What? Right that I got it for eight. And why n e? Because of the triple bond. So we got three about a three off time, Okay? And last thing that we need to do here is to account for our method groups thinking out from carbon seven. Okay, so you got a seven metal. Hey. All right, so time to combine all the information that we have. We got seven metal three. Oh, time. Okay, so that's your answer for party. And finally, we're done here. We have names, all the molecules. We have four problem. 58


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