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41) Draw all pssibe isoress f8r each mdlecular frmula aCzHqa (2 isovoc) 6)C2H40 ( 3 isomes) CzHan (4 isonen) c)...

Question

41) Draw all pssibe isoress f8r each mdlecular frmula aCzHqa (2 isovoc) 6)C2H40 ( 3 isomes) CzHan (4 isonen) c)

41) Draw all pssibe isoress f8r each mdlecular frmula aCzHqa (2 isovoc) 6)C2H40 ( 3 isomes) CzHan (4 isonen) c)



Answers

Draw structures for the following: (a) 2 -Methylheptane (b) 4 -Ethyl- 2,2 -dimethylhexane (c) 4 -Ethyl-3,4-dimethyloctane (d) 2,4,4 -Trimethylheptane (e) 3,3 -Diethyl- 2,5 -dimethylnonane (f) 4 -Isopropyl-3-methylheptane

So in this question, we're asked to draw some psycho al canes so cyclo in its prefix name means a cyclic compound and an out. Cain, of course, is a, um, carbon molecule with all single bonds connecting all of our carbon, um, bath zone. So starting with our first, um, structure, we have cyclo butane. So Butte refers Thio four carbons. So if we draw out our skeletal model like this, remember, each vertex of this model represents one carbon, So 1234 and then each line represents a bond. Now, when we draw these, um, skeletal models like this, it's assumed that each carbon is going to have a complete octet if drawn like this, and it has no other side chains or our groups. And since, um, we have a cyclic compounds like this each each of these carbons is going to have to hydrogen attached to it to complete. There are Tet moving on to our next one. We have psycho obtain helped refers to, uh, seven carbons in our parent change. So let's go ahead and draw that out to have one, 23 45 six, and then seven right here. This is kind of hard to tell the way I've drawn it, where the vert disease are. But, oh, count out all the carbons for you. Label them all here, as we can see There, seven of them here moving on to our next, um, compound. We have metal cyclo painting. So, uh, pent means our parent chain is going to have five different carbons, and we have a attachment group A metal. And since there's no, um, number indicator, um, labeling where it is on our parent chain, it's assumed that this method was going to be on the first carbon. So let's go ahead and draw out our cycle plantain, a five carbon chain, and then on our first carbon, we're going to have a metal group. So we're just gonna draw that out here and our method group All well, this in green. So this is one carbon right here. So this is a cyclo plantain. Now, for our next one, we have 14 dimethyl cycle. Heck, sane. So our parent chain has the prefix hex, which means six carbons, and we also have to attachment groups represented here by 14 dime Ethel. Now, this time we have number indicators of which, um, carbons are, um, metal groups are attached to We have a method group at one. And the Method group at four. So let's go ahead and draw this out. So we have our six carbon chain right here. Let me label are carbons. The one to 3456 And we're going to have to metal groups attached one at one and one of four. And this labeling here that these are both one carbon each. Now for our last one. We have one Ethel, three profile cycle octane walked right here. Our parenting has eight carbons, and we also have two different attachment groups who have, um, Ethel Group at one. And a proper group at three. Ethel means we have two carbons, and then a pro pole means this is three carbons. Yeah, so let's go ahead and draw that out. So we have one, 23456 seven and eight. Go ahead and label are carbons one through eight. And now on our first carbon, we have an ethyl group. So again, that's two carbon. So 12 and then on our third carbon, which is right here we have a profile group, which is three. So one, 23 and I'll go ahead and label the number of carbons on each of those. So we know. So our 1st 1st carbon on our parent chain has Unethical Group, which is two carbons and then our third carbon chain or third Carbon on. Our parent chain has three carbons. One, 23 and that sums up this problem.

Okay. This problem is asking us to draw the structures given these compound names. Okay, so let's zoom into the 1st 1 For the 1st 1 we have to to dimethyl four isopropyl octane. So whenever I'm dealing with these types of problems in which I'm given the name and to figure out the structure, what I'm gonna do is I'm gonna start off in the back, okay? This is just personal preference. I mean, obviously adding what can do whatever they want, But I personally prefer toe start from the back and then work my way to the left just because we always know that in the very end of my molecule name, we're gonna have my longest carbon chain. And I would prefer to start with that longest carbon chain and then work my way from there. Okay, So starting live octane. I know that oct correspondence to eight carbons and then ain't corresponds to the Al que nature or single bonds. So what I'm gonna do first is draw out my eight carbon chain two ch three ch two ch two, and then I have five more. So while I'm doing this, I need to ensure that each Carbon has a full ahead by having four bones. Okay? And I think I have two more. And then ch three. Okay, lets just ensure that that's eight. Carmen, it's 12345678 Perfect. Okay, now we can move on to the left side. So again, I'm working this way. Okay, so octane, that's done. OK, now I still propose before isopropyl so therefore isopropyl corresponds to at carbon number four. So 1234 and then obviously would have 5678 on that way. So carbon number four I haven't isopropyl group, so every time we see ailed so right here l just means that is a substitute went off of my carbon chain, so they'll isopropyl isopropyl is a substantial went off of my carbon chain. Get in that position for I just draw that out. So what I'm gonna do, I always need to make sure that each carbon has a full attempt and four bonds, So I need to have hopes too big of an eraser. Okay, so ch two I'm racing one of those Hodgins and making it a ch. And then I'm going to connect my isopropyl group to that carbon. So isopropyl that is a C H two a ch three and a ch three. So, propofol pope means three carbons, and then the cell means that we have a branch point. That is my isopropyl group on carbon number four. Okay, now we can move on over here. We have to to dimethyl. So the two to die metal correspond to the second carbon. Okay. At carbon number two, we have two methyl groups, and then yes, we do have to include two twice. Okay? It's just to depict that we have to metal groups, and they're both on carbon number two. Okay, so what I'm gonna do is right now I have this carbon, this one we're here that has a full of that because it has four bonds, but I need to attach those two muscle groups onto it. If I attached to metal groups, then I would have an exceeded octet on the carpet. So I'm going to erase both hydrogen so that I can, um, make two muscle groups come off of that carbon. Okay, so I'm just gonna make my carbon a little bigger so I could make this a little bit neater. Erased my to and then connect a ch three and another ch three ch three. That is just a metal substitute. Okay, so I'm gonna clean this up now. 134 And then this should be the final answer in which we have to to die metal for isopropyl octane. Okay, next up, moving on to three dimethyl Hexi. So again, I want to start from the very back and then work my way to the left. So starting from the back, I have Hexen. Hex corresponds to six carbons and ain't corresponds to the Al que nature or single Bonds only. Okay, so I'm gonna have see you three ch two ch two and then I have two more carbons like this. Okay? So, as of right now, that is just Hexen. Great. There are no substitutes. That is just heckling. But now I have to add my subsistence. And that is all according to my name of my compound. So I have to three dimethyl. So the two or three that corresponds to to the second and third carbon of this chain, so obviously have 456 and then at position two in position three. I have a methyl group on each of those, so I'm going to erase a hydrogen on both them. So now I just have those bonds. And obviously, between each of these, there's a bond, Okay? And then I'm gonna have a method group coming off of this carbon and a method group coming off with this carbon. Okay, so that is my 23 dimethyl hexi. Okay, next up, we have 44 die fo decking. Okay, So, decking, that's something that we don't obvious don't always come across. But deck just corresponds to 10 and in the main again corresponds to the al que nature so it might take a while. But we have 10 carbons, so right as fast as I can. But obviously, we're always making sure that we have the correct number of carbons and the correct number of bonds. Each one must have four. Okay, I think I need maybe one more Lester count. Let's go. 123456789 10. Perfect. Okay, so at four. Position for this one right here. I have to ethel groups, and that's depicted by 44 die F o Okay. So that means that I have to put two ethyl groups. The ethyl corresponds to a carbon chain of two carbons. So I'm going to erase those hydrogen. So, like, I can come date that bond. So in fact, I'll make this carbon a little bit bigger. And then I'll put my two effort groups. So ch two after two exceeded three. And then on the bottom, I have the same exact thing. See you, too. And then just to clean up a little bit, I have my series three. Okay, so that should be 44 die fo Decade decade is 10 carbons and then the 44 dot Ethel That corresponds to two sets of ethyl groups attached to carbon number four. Okay, moving on. Here we have 245 triumphal for one metal. Ethel. Captain, get that's a lot to handle. But let's see if we can attack this the same as always. Start from the back. Obtain Stepped corresponds to 7 a.m. Corresponds to the al que nature. So I need to have only single bonds and then I need to make sure that I have seven carbons total, at least in my parents game, so I have six so far, and I need one more. Okay, so each of these currently has a full looked at. Each of them has four bonds. This is currently he obtained with no substitutes. Okay, so what I'm gonna do now is work my way backwards. Okay. So obvious. Um, if we remember to algebra pretty algebra, we remember Penn does soap. Endo's deals with parentheses first, right. The P penned us parentheses and exponents, etcetera. But when we're dealing with aiming, we don't necessarily follow that rule and said we look to what has ever left of my pregnancies. So this is a case where we might work the left to right. But I like to see before I know that that four corresponds to carbon number four. So 1234 We're dealing with this carbon right there. So that four is where I'm gonna have to deal with my parentheses. So as far as the princes go, and in dealing with these types of problems, I look at whatever whatever number is to the left of my prince sees. I know that that is the position in which my parentheses will be contained. Okay, so at position before a carbon number four, I have my one methyl ethyl. That might be a little confusing because we don't usually see two substitutions within the same set without being separated by dashes. But what that means is I have my effort. So again, this is working backwards. So I have Ethel, I'm gonna write down effort. That's just to carbon chain. So see, it's three. And then we deal with the meth apart so that the metal part, the one the one metal is corresponding to the ethyl substitute, so that might not make sense immediately. But what it means is that here is my ethel subsequent when I destroyed here and then the one metal is corresponding to that Ethel, to pretend that this Ethel's my parent chain. The one method is just, um, a substitute on the ethyl chain. So if this is one and this is to Ethel subsequent than at carbon number one, I have my methyl group something to write down ch three. Okay, so that's that. And now let's just clean up a little bit. Okay? Now moving on backwards again. Okay, So I already covered covered happening have covered four methyl ethyl for one methyl ethyl. And I need to deal with this. 245 tri metal. Okay, So what does this mean? This means that at carbon number two So this wonder here, this carbon at carbon number. Oh, by the way, in the previous one, the 41 methyl ethyl I need to make sure that I have a full attention this carbon. So I need to erase one of those Hodgins, That is now just a bond. Okay, now I have 245 tri metal. So I have 245 That corresponds to cover number two. That corresponds to number four. And that corresponds to number five, which is right here. Okay. And what that means is, each of those has a method group attached to it. So I'm gonna erase one of those Hydrogen is on the card number two and then put a ch three ch three metal. Okay. And then at carbon number four. So this one right here, I'm gonna race one those hydrogen, so I can accommodate an extra set of bonds. When I write down another ch three. That is my metal group and then Carbon number five right there. I'm going to write one of those hundreds to accommodate a new bond, and that is a methyl group. So C h three. Okay, so just cleaning up here. I have my 245 tri methyl for one methyl ethyl helped ain't and that is it right there. Sorry if it's a little messy, Um, this, too should not be there, but there we go. Okay. Next up, we have 25 dimethyl four to metal propofol octane, another mouthful. But the same principle applies to the previous one. Opting we have eight carbons. Seems three ch two ch two, etcetera. Okay, so I'm doing this. Let's try to consider what is going to be attached to my for two methyl propose. What will that be? OK, so siege to and last one. Okay, so now we can consider this for two metal propofol on my octane chain. Okay, So what I would do first is recognized that this four corresponds to the fourth carbon in my chain. So 1234 And then, of course, I have 5678 Okay, so at carbon number four, I have What is in parentheses, and again, I'm going to work backwards. So first up, I'll deal with my Pro Bowl. So propofol is what is immediately attached to my carbon number four. So I'm going to erase that to to accommodate a new bond. And we're just gonna draw that bond, actually. And right here, when draw my purple. So that corresponds to three carbons. Okay. And then at carbon number two. So now I'm dealing with this two methyl group right now, I have one too. Three. This is my proposal at carbon number two of that purple group. I have a method group, so I'm going to erase this hydrogen to accommodate a new bond. What, to raise that to Andhra out my ch three. Okay, so that is not to four to metal. Purple attachment. Okay, Next up, let's deal with the 25 dimethyl so to five dimethyl. That means that carbon over two and a carbon number five. We have a metal group on each of those. Okay, So what's zoom in? Erased that hydrogen to accommodate a new bond, draw bond, and then we have my method groups. So there's one and then at carbon number five. So what I just did was Carver number two. Now I'm doing dealing with carbon number five case. The here is carbon number five. I need to accommodate a new bond. So that means I'm going to erase what I previously had, which were ch two. And then we're going to accommodate that new bond. One drop below a C. H three ch three is just a metal group. Okay, so now to clean up there, we have my 25 dimethyl four to metal propose octave. Okay. And last but not least, we have my 411 dime. Ethel. Ethel Octane. Okay, octane. Same thing is always eight carbons. Okay. And then the main course wants to the lk nature, which is all just single ones. We'll get more into al canes, which are double bonds. But for now, we're just dealing with these types of problems. Get how many's that for one more? Okay, next up, I still this 411 and di methyl ethyl. So first up, I ignored my parentheses and just focus on what's immediately to the left of it. Which is Carver number four. So here's 12345678 at carbon number four. I have whatever is in parentheses, so let's just accommodate that right now. But it reeks in one of those Hodgins making a bond there. Okay, So what I have right now is my 11 die metal SL. So I'm gonna work backwards, starting with my ethel. Have two carbons like this. Oops. Not three ch three. OK, so that's one set. And then next up, I have my 11 dime a thoughtful. So if I were toe label, this ethnic group have carbon whenever one and carbon number two. Okay, so I'm gonna do this. I have to write out my di methyl my dime. Ethel, is that karma number one? And, uh, that Ethel group I have in my finger. But I have two of them, so I'm gonna write down See you three and see if three. But that means is I have to accommodate that those bonds. So I have to erase It was hydrogen. Okay. And I'll just extend this one to make room. Okay. And then that should be the final answer. So that is 411 done. Methyl ethyl octane. Okay. And then let's just clean up. There we go. And I was about to show off. It

Now we will discuss the problem from structures of organic compounds where we are given some names of the compounds and we need to draw the structures of these compounds. So the first one is this. The name is E three Ben's I'll to five type LaRue for metal. Three exempt. This compound has six carbon atoms in the parent chain, with a double bond between carbon three and carbon four, so it can be shown like this. Now there is a chlorine atom at second in the fifth, carbon each at a seal at second Carbon atom and see that fifth carbon atom. And there is a mythical group at fourth Carbon Atom, which is shown as this. Now we have another group that is Benzel Group at the third carbon atom so it can be shown like this. This is Ben's I'll group in this. The higher priority items at the double fronted Cabinet terms are on the opposite side to each other, so it has e configuration where this is carbon number three and this is carbon number four. The second compound given to us is one retinal for nitro benzene. So in this compound there is a benzene ring at which, in the first position, a denial group is present that as this group is present and at the fourth position with respect to internal group, you know to that is nitro group is present. So this is the structure of second compound. The Taliban is trust 14 Brahma Final two metal cyclo hugs in. So this compound is shown like this. There is a cyclo hugs in ring containing four room final at the first cabinet. Um, let us draw it like this. This is drawn facing towards the viewer. That is for Ramo nine, then at second position. With respect to this group, there is a metal group. Since this is a trance compound trans configuration from metal group is pointing away from the viewer. This is a structure required for the given compound. So these all other structures for the given organic compounds

Hi, everyone. So first up, arrange serial number. Then you can send name and the last 20 sculpture structures. So serial number A for me type period in poor myth child, you're reading. So we know the structure people Allene. So this sort of structure of period then we'll give him, um brain one toe clean, then fall then this is fine. This is six. So I You know this foot carbon there will be a mental group ch tree. The next one is to Clara Payroll How I roll. So we're not in structure. Yeah, no news. So you can see this is 12 three, 45 the next. So see one wonderful die I drop really on four die No! Hey drove! We really that is she in h So they sort of unfold ahead of purity Their next de even Peter Bruno Peyron paid 12 he brought home by Iran. So data Brahma viral. The next one needs nicotine. Nicosia I see it. You did structure. There is the building ring is there along with the range group and pollution


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