5

Bond: recuce Ientl7iCcs nil 0ijn Ineesmem Pertiono? nave ir foljairadrinHOnontcomoule[ vAtlaEtiI=LneeElifdedenatomDaretaConiuaCh-branea intera AntFolt= KartaneuerJu...

Question

Bond: recuce Ientl7iCcs nil 0ijn Ineesmem Pertiono? nave ir foljairadrinHOnontcomoule[ vAtlaEtiI=LneeElifdedenatomDaretaConiuaCh-branea intera AntFolt= KartaneuerJuUcDt Lrtne t camman 757 U [ne retuitt[~0 Funo Alarcurhnie ul tlc #tore tund_Tueer ranq" tinduehlanced tundnoetnan DrzMnennncemALsSneentunbci,Cnmntt .Do_Hiliant Vallationah mnid [eacaLeeh IccuinatAuaaonul retuttu Che siect Iurd fas Iowsl carn Jmi o Tgn CEaleit [Jatn unn'elun enftaadetWane ttudsnMretWrad [email protected] 3 D'conpue

bond: recuce Ientl7i Ccs nil 0ijn Ineesmem Pertiono? nave ir foljairadrin HOnont comoule[ v Atla EtiI= Lnee Elifde denatom Dareta Coniua Ch-branea intera Ant Folt= Kart aneuerJu UcDt Lrtn e t camman 757 U [ne retuitt [~0 Funo Alarcu rhnie ul tlc #tore tund_ Tueer ranq" tin duehlanced tund noetnan Drz Mnennncem ALs Sneentunbci, Cnmntt . Do_Hiliant Vallation ah mnid [eaca Leeh Iccuinat Auaaon ul retuttu Che siect Iurd fas Iowsl carn Jmi o Tgn CEaleit [ Jatn unn 'elun enft aadet Wane ttudsn Mret Wrad a [email protected] Pra 3 D 'conpue Kattad $



Answers

The corrccr ordcr of bond cncrgics for the C-H bond is (a) $\left(\mathrm{CII}_{3}\right)_{3} \mathrm{C}-11<\left(\mathrm{ClI}_{3}\right)_{2} \mathrm{CI}-11<11_{5} \mathrm{C}-11<$ $11_{3} \mathrm{C} . \mathrm{Cll}_{2}-11$ (b) $11_{3} \mathrm{C}-11<\left(\mathrm{Cll}_{3}\right)_{3} \mathrm{C}-\mathrm{H}<\mathrm{H}_{3} \mathrm{C} . \mathrm{Cl} \mathrm{I}_{2}-11<$ $\left(\mathrm{Cl} \mathrm{I}_{3}\right)_{2} \mathrm{Cl} \mathrm{I}-\mathrm{H}$ (c) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{H}<\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{H}<\mathrm{H}_{5} \mathrm{C} . \mathrm{CH}_{2}-\mathrm{H}$ $<\mathrm{H}_{4} \mathrm{C}-\mathrm{H}$ (d) $\mathrm{H}_{3} \mathrm{C}-\mathrm{H}<\mathrm{H}_{3} \mathrm{C} . \mathrm{CH}_{2}-\mathrm{H}<\left(\mathrm{CH}_{y}\right)_{2} \mathrm{CH}-\mathrm{H}<$ $\left(\mathrm{CH}_{3}\right)_{5} \mathrm{C}-\mathrm{H}$

The bond order is calculated by the following formula. Oh, number of bonding orbiters minus number of anti bonding orbiters. These are the number of bonding orbiters, bonding orbital's and number of anti bonding orbiters and the bonding orbiters. So according to this formula, we can see that it can assume that is the bond or formula can be assumed the value as positive integral or fractional fractional values and zero values also zero values. Okay, for example, let us take an example of die atomic molecule of helium. So here the molecular orbital electronic congregation for the that technology will will be oneness two sigma minus two sigma and the wounding one is two, then sigma us too Stigma and the Wounding two West. So here when I calculate one order than it is given by half the number of bonding orbiters, the number of bonding electrons. Okay, so these are the number of wanting orbital electrons. Right? We are counting what number of bonding and anti bonding orbital electrons. So here the number of bonding orbital electrons are two and two. So these are four. Similarly in the anti bonding orbital electrons, these are 4, 2 and two. So this will be zero. So here you can see that the bond order is zero. And this is the reason that helium, the atomic molecules do not exist because bond order is zero and such a molecule will be highly unstable. But yes, as you can see that the bond order can be zero. So the assertion given here is correct. Now the reason which is given to us is higher the bond order. higher the bond order. The bond order shorter. The borderland shorter the bond land. So yes, of course this reason statement is also got it shorter. The bond length. This is because higher the warned order more will be more will be full of electrons full of electrons more will be full of electrons more will be the overlapping of orbital's of the bond in world. So that's why I start, the shorter will be the born land shorter will be the born land shorter will be more like now since you're born length is short. So that this makes the born length, the born very strong this mix the bond very strong. Such a strong bond cannot be broken easily and requires very high energy to be broken requires very high energy hi energy to be broken. Hi energy to be broken. And this is the reason why the bond energy will be very high, that's why therefore higher will be the born energy higher will be doc bond energy for breaking down bone bond energies. What bond energy is the energy needed to break the bonds? So since the cool off with electrons is more efficient, that's why the bond line is shorter. The bond length is shorter. That's why the bond is very, very strong since the bond is very, very strong. So the high energy is required to bring the bomb. So the reason statement is also correct. So here, both the statements, both the statements for exertion and reason are correct as a Russian and reason are correct. Listen are correct. But the explanation, the reason given to us is not the correct explanation for the given assertion, but the But doc reason is not the correct explanation, correct explanation for the assertion. Explanation for dark as ocean. So the correct option will be what the correct option will be. Option me. Hence option B is correct. Hands of shin. We is panic.

So we're going to be drawing out our bond nine formulas for these, um, names for molecules. So starting with a So we see that we have octane so talked, meaning there's gonna be It's carbons. Een means that there is going to be a double bond. It's all right, d B. And it's going to be cysts. So we want to keep in mind that when we are looking at the double bond, um, our double bond, it's substitue INTs are going to be on the same side and we can see, um, that our double on will be on the third carbon. So let's do that now. So let's start. We're gonna have one to three. So here is gonna be our third carbon. And if we know it's going to be cysts, we want to make sure drawing are substitue INTs on the same side. The 34 five 67 h Let's just double check that 12345678 or rights. And then again, we want to put our double bonds on third carbon 123 You put a double bond right there, and that is going to be our um, molecule. So let's move on to be so be, we are going to have trans to heck scene. Um, so in this case, we have hex. So we're gonna have six carbons again we have seen. Which means we're going to have that double bond. And the information we're given about this double bond is that the double bond will be trans. Um, so we don't have to worry about it being on the same side. We can destroy our normal zigzag, and then it'll be on the second carbon. So let's just draw out our six carbons. 123 456 Let's double check. 123456 And then on the second carbons, the 12 we're going to cuts our double bond. So when? Six. Okay, um, let's move on to see so again, we look at the end of the know the name to check what we're going to be dealing with. So we have pence as our prefix, So we're going to have five carbons again you haven't seen, so we know we're gonna have a double bonds. Um, and then we have closest to that. We have our double bond is going to be on the second. Carbon were also given a couple of constituents subsidence. Um, so we have two methyl groups, you know, die means to, and they're going to be on, um, carbons two and four. So let's first just draw out our parent chain here, so we're gonna have pent in. So that's one, 234 five sets are five carbons and again or double bonds gonna be in the second carbon. So, um, let's just pick, um, aside to assign carbon one, I'll pick this one. So 12 three, or sighs. Well, what are double von here? And then again, we're going to put our, um substitutes. So our dimethyl, which is gonna be on carbon too, and then carbon four. So, um, that is the molecule for C. Um, Now we can move on to D. So same thing is, before we're going to be looking at the end of the name first. So we see here we have Butte to mean. That means we're going to have four carbons with a double bond on the second carbon. And we also told that the double bonus trans So let's draw our parent chain right now. That is going to be 12 three for, um Then let's pick a side it to assign carbon number one. So I'm gonna put this is carbon number 12 three and four. Um, and then we can add our double bond on carbon number two. Ah, And as you could see, the substitue INTs of the double bonds are across from each other. So that looks good. That means they're trends. And now we can add our chlorine and chlorine is going to be on carbon number one can see here one claro Um, so we will put that over here on carbon number one. Really? It wouldn't matter which side you put it on, since it would give the same name either way. But since I picked this carbon is carbon number one all, just leave it there. So that looks good. We can now move on, Teoh the next molecules. That's gonna be molecule e so pent again. We've seen this before. It's gonna be five Corbyn's. Then again mean so again, we're gonna have that double bond that's gonna be on carbon number one. That's all the information that we were given about our parent chain, so let's just draw it out for now. So we have 12 345 Um so let's take a carbon to assign is carbon number one. So let's say I pick this one. So 12 three or five. So, um, we can now place are double bonds on carbon number one and then we can place our other substitue INTs. So we're told that we have die. Brahmos. That's to Brahmos. And they're living located on carbons four and five. That's gonna be right here. Brahmos, bro mean right here. And that is the molecule for East. So moving on, let's take a look at F. So f is a lot longer than the other molecules we're looking at. But let's break it down. So let's first look for our parent chain of the very end of our name. So we have cyclo pence and between now is going to look like a Pentagon. So here's their cycle up insane. Um, now, let's just assign what we want to be. Carbon one to three, four and five. So looking at our name. So we have one, two and one hell, that's a very long names. Let's break that down. It's gonna be on carbon one. So it's gonna be on talk here. We have pro pope, so we know that there is going to be three carbons on this Malkiel group. We know it's a local group. You can just see the Y l at the end that indicates to now kill group eso It's going to be just sees in ages. Um, so let's just draw in those the re carbons for now, said one to three. So we have that and were also given that it's two n so to e n. That means we're going to have a double bond on these second carbon. So if this is carbon number one, which is the carbon that is attached to the parent chain, we know that that's carbon number one, cause there's a one here in front of the Y el. I'll even write that in So criminal one to three. If we know that, then we know that carbon number two will be. This carbon right here in our double bond is going to be right there, So that is what this first substitution looks like. Now let's move on to the next substitue in. I'm just going to go, um, continuing to the left. We have to metal. So put a metal group here. And then finally you have our, um our first substitue int one borough, mo. And I'm going to put that right here, which is gonna be carbon number one. So that is going to be the molecule for F. So let's move on to G. So Fergie again, we're going to look at the end of our molecule zero parent chains of cyclo pence een So we know that Pantene is going to be a Pentagon. It doesn't give us a number for where the double bond is going to be. Um, so in this case, since the methyl groups of substitue INTs or carbons great and four, let's just assume that our double bond is going to be on carbon one. And now that we have that, we can add in our substitue INTs again, uh, on carbon 74 we're gonna have Dime Ethel. So carbons three and four will have these methyl groups hanging off of them. Um, so that looks good. We can now move on to H. So again four h we want to look at the end to see our parent, Jane. So we have cyclo pension again. So we're gonna have our cycle a plantain. And now we went to a sign. Um, actually, since we only have one subsidiary and we don't need to assign the numbers for the carbons Ah, that's my bed. But if we actually take a look at the beginning of the name here we have ah, vinyl. Um, and remember that vinyl means that it will be in the position that is attached to the double bonds. Um, between carbons. So, um, we would draw that onto our molecule like this says you can see our, um our substitue int here. The double bond between these carbons is attached to our parent chain, Um, in a position that is where it's directly attached to the carbon participating in the double bonds. Friends. That's the vanilla positions. That's why it's called. Um, that's why it's called Rydell. So we can now look at I ah, this time we have cyclo hexane scene, so we know that vaccine is going to look like a hexagon again similar to G. We aren't told where the double bond is going to be because we have e any, but we can assume it's going to be between carbons one and two. So 12 three, four, five, six So we'll put it right here. So I used the wrong color. There we go. So now we can add in our sub situ INTs and it's die Claro. So to Claros on carbons one and two so we can write them in right now. So we'll have our seal right here and her seal right here. And that is going to be the molecule for I. So let's finish up with molecule J. So molecule J we look at the end of the molecules were pent mean, and we know that the double bond is going to be on the second carbon and it's going to be Trans, so he could destroy our zigzags. So 123 45 Now let's just pick a Corbyn to assign carbon one. So I'll say this is one, 23 for five on that. We can put in our double bonds on carbon, too, for here. Um says you can see we have our double bond and the substitutions are across from each other. So it's Trans. That's good. Um, no. We can add our other subsequent store parent chain. So we had die Claro again, which is to Kouros Corbyn's one and force. Let's add those. And now we'll have a quarry in carbon one and then we'll have a chlorine on carbon four right here. So that is the last molecule that we had to draw in. I'll just scroll back to the top in case he wanted to look at our answers again. So again, here is the answer for molecule number one or a sorry, um, molecule B mark. You'll see molecule day molecule e mark your f molecule G mark your age molecule I and lastly molecule J and that is all.

We're going to be drawing the bond line formulas of all of these molecules, so let's just jump right into it, starting with a So with a. We have a parent chain of plantain, which is five Karpin's, so that's going to be one, too. 345 I'm going to label the carbons as carbons, one to three for and five based on what you label that you can add the substitue INTs. In this case, both the situations are the same, and they're going to be on Carbons one and four and its book Going to be Koreans. So that's the final molecule for part A. So let's move on to Part B. In this case, our parent chain is going to be this beautiful group, even though it's not at the end of the name. Um, there isn't really a parent tune given at the end of the names, so this isn't an eye you pack name, but, um, the beautiful group is the only group that is going to have carbons in it. That's how we know it's the parent chain. So a beautiful group is going to have four carbons, so let's just draw that out first, so that's gonna be carbon one to three and four. So the sec insect beetle means that whatever is being attach, whether it's on the secular group for the second little group is the attached to something else. The attachment is going to be on the secondary carbon. In this case, carbons two and three are both secondary carbons. Um, and technically, you can put it on either one. Um, but if you whichever one you put it on will become carbon to based on Are you packing? Uh, so I'm just going to put it on carbon to that's where I serve. Sichuan is gonna go, and that's going to be bromide. So moving on to part C, our parent chain is going to be obtained, which is seven carbon. So 12 three, 4567 And then I'm going to name for number them. So 12 three for five, six, 17 And then we see here that on carbon number four is where substitution is going to be. And we have an isil poeple group which looks like this. So isopropyl proposal means there are three carbons and then I saw ah the attachments right in the middle. Um, so we could move on to party. So party, our parent chain is going to be plantain. So pen tain has five carbons, so that's going to be 12345 I'm going to number them 12 three for five on carbon number two. We have to of situ INTs. They're both going to be medical groups. So a catch stomach so and then on carbon three, we're going to have one methyl group, so I couldn't draw that right there. Moving on to Part E per e. We have a parent chain of Heche saying that's going to be six carbons. So 12 3456 number them again. 12 34 five and six. And then we can start adding our substitue int. So I'm just going to go in order of looks right. What the name gives us. So on our carbon number three, we're gonna have an ethnic group that's going to be to Corbyn's and then on carbon number two, we're going to have a method group, which is one carbon for a part F. Our parent chain isn't a ring. It's cycle plantain, which looks like a Pentagon, and I'm just gonna pick one carbon to net number as carbon one and then I'll go clockwise. So here are my five carbons. And then I'm told that I have both of my substitute wince on carbon number one and that these subsitute mints are Corinne's. So we'll have to Koreans on carbon number one. So moving on to part G hurt G. I have another ring for my parent chain. In this case, I have psych glow propane. So propane means we have three carbons. This is a little like a triangle, and I'll pick two carbons to be carbons 12 and degree. Um, as you can see here are substitutes are going to be Sisto one another, meaning they're going to be on the same side of the ring. So I want to give them the same type of line they're going to be on carbons 12 and they're going to be methyl groups, so I want them to be on the same side. In this case, I could just draw them with a bold wedge is my first monthly group. Then here's my second medical group again has to be on the same side. It's also gonna have a bold wedge moving on. We have a very similar looking molecule for part age. So again we have the cyclo propane. Um, so we're going to have the same triangle, and I'm going to use the same number, is there? You can see the comparison. So 123 in this case are substitue. INTs are trans to one another. That means they're going to be on opposite sides of the ring. So I have to make sure that I give them different types of lines. Um, it doesn't matter which one you give which type of line as long as they're different. So in this case, I'll just make carbon one this absurd chewing, which is metal, make it bold and carbon to I'll make it dashed. And there you go. Moving on to part I or a parent chain is gonna be Penton All so Pence and all has pent in it. It is going to be our five carbons. That's 12345 So there are five carbons and now we can utter substitue INTs. So on carbon four, we're going to have a methyl group and then on carbon to we're going to have the O H group. The reason that I know it's no age group is because this ends in all, which means an alcohol. So we must have a no age group. And since this, too is closest to this, all that means that's telling us where the O a troop is gonna be just going to be on carbon number two. So you go and so we can move on to part J Hurt J. We have another ring. So here we have a hexagonal. So ignoring the O all the ol For now we have a cyclo hexane, which is a hexagon. And then I'm just going to pick. This is carbon one to three or five and six. Um, so we can see here are substitue INTs are going to be transferred to one another. So the first substitue in that I'm going to look at is going to be four ice, a beautiful that's going to be on carbon number four and an isil. Beautiful group Looks like this where we have one carbon and then we have and, um, a branch over here, so in total, you have 1234 groups, which is where the beetle comes from. And then it's I so because we have this kind of looks like isopropyl um group. So that's what it looks like. And so are other substitute A int is going to be on carbon number one. So since their trans, they have to be on opposite sides of the ring. So I'm just gonna make this substitute bold. It doesn't matter which ones. Bolden, which ones dashed as long as its opposite or other substitue in, is going to be the O H group, which we know is on the wall is from the all, and we know that this is going to be on carbon one. So which is going to be on carbon one because it doesn't specify if it's on another carbon. And if it doesn't specify, that means that you can assume it's probably going to be in carbon one. So here have my dashed line, and then I have my witch, So that looks good. We could move on to part K parte que we have, um, um part. Okay, we have ah, vaccine. This is going to six carbons. 123456 So, in this case, um, we're going to have to substitute mints. So the first substitution is going to be on carbon one. It's going to be a cyclo purple group, which is just a triangle. So it's gonna look like this, um and then we're going to have another one of these on carbon number four. So those are die cycle purples, and then it's going to be on the hexane parent chain. So it's good. Now we can move on to part l so part l. We have neo pencil alcohol. So neo pencil, it's not until you pack name, but it looks like this. So the central carbon has for Michael grips attached to it. And since it's an alcohol, we know it's going to need to have an O age group so you can put the, oh, a troop on any of the carbons that are not the central carbon because the central carbon has its octet already filled. So it's a public with this eso. Now we can move on to part m hurt family have by Saiko 2 to 2 octane. So when we have ah by cyclo, um, molecule. That means that we have essentially two rings connected to each other. And these numbers in the middle are going to tell us how money carbons number of Corbyn's um on each bridge. So in this case, we have two carbons on each bridge of this molecule. So let's say that my rings will connect here, right? So then I'm going to have two carbons coming off of this connecting area. So 12 it's going to go that way. Then I'm going to have another two carbons going upwards. And then finally, it will have another two carbons going this way. Then I'll just draw this dash so it looks like it's behind. So, as you can see, you can kind of see the two rings where they would be fused over here. Um, and then there are three bridges, each with two carbons. So we have our you to men, too. And then if you actually count all the carbons that are in this molecule 12345678 you would see that we have eight and that fits with their octane. That's perfect. So let's do the same thing for N so for n Let's first draw our connecting bridge for our rings. So then let's start with the three carbons Love. It may be the hardest one to draw. Um, so we'll have one carbon here to three. Loan connects right there. That's three carbons. And then we'll have one carbon on the other two bridges, so one forget to color code the three and then we'll have one on this last bridge. Over here it goes downwards. So again, as you could see, we have the three carbons, the one carbon and then the one carbon. And then if we count all the carbons, that's 1234567 and we have had obtained. So that works. Finally, we come to the last part of this problem, which is going to be, Oh, so oh, we have a parent chain of cyclo plantain. So it's going to be a Pentagon. As they said before, there's only one substitue int on our Pentagon, and that is cyclo pencil. We know it's a substitute because it ended that. Why else so it's incalculable constituent, So I mean this. There's just another cyclo plantain attached to this cyclo plantain. So that's two Pentagon's connected to each other, and that's how you get the last molecule. So that is how you can draw The structure is for all of these molecules.

Okay, So for part aid, we have Prentice e ch 32 ch ch two ch three. Right? So this presidency's siege three and then the two outside in front tch means that there are two ch trees attached to this carbon. Right? So if we were to write something out like this So ch there's two. Ch three is attached to this carbon. Now we've we've done, he's already. Now we see that this CH three is connected to a CH two. Let me see a dad. Ch two is connected to a ch three, right? But they want the bond line formula. So all we gotta do is we don't draw any hydrogen or we don't and we don't draw any carbons. All we do is draw lines. And then if there are elements other than hydrogen and carbon is Demi strong men. So let's start with this carbon. So we have discovered right here, right? And it's connected to one ch 32 sch three. So once each three two ch three I remember we don't draw the carbons or the hydrogen is. We know that at the end of the line, there's always a carbon. If no element is specified. So right here, this carbon right here is this carbon right here. And we see that is attached to a C H two n a. C H. Three. So it's attached to a ch two and a ch three. Right? And that is the bond line formula. So if I were to draw the hydrogen in, so let me draw the hydrogen to see why this works. All right, so run he had, and we know it's a ch three, right? There's there's 300 is coming off in their ch two. There's a hydrogen here. It's not. Let me draw it for some reason. But there's a hydrogen out there, it's not. Let me draw it. And then there is a hydrogen here. Then there is 300 coming off here. And all of these in all of these all the carbons meaty octet role. So this is the answer for part A. There for part B, we have heredity ch 32 ch two in the no age. Okay, so so again, we have to to parentheses ch three. So this means that they are attached to this carbon and then any any Ah helmet that we right after a carbon that is not another carbon means that is attached to that carpet. All right, so we have carbon. It's attached to their just each three attached to it. And then there is another siege three attached to it, right, cause there's two here and remember, any element that is behind a carbon that is not a carbon means that it's attached to that carbon. So the hydrogen, they're so that means there's a hydrogen attached to it now a carbon that means that that carbon is attached to this carbon. Here. What I should say is that any element after a carbon means that is attached to the previous carbon pretty much okay, So Ah, you have a carbon there and then this carbon has to hydrogen behind it. So it's attached to one hydrogen and another hard you and then we have another carbon behind it. I mean, that is attached to this carbon. And then the 100 d h two is attached is behind this carbon. That means they are attached to discover in right. So I didn't have to. And then we see that there is an option here. So we write an option, and there is a hydrant behind it. So it's attached to this oxygen now they want the bond line formula. So we have. Ah! 1234 So we have four carpets and roll a strike. That first one too. 34 Right, Because at the end of every line, 12344 garments. All right, now we see that this carbon right here, the second carbon has a CH three coming off of Mr Literal DAP. And then we see that the last carbon has an oxygen coming off of it. Now, remember, we have the draw the Adam if it's, ah, not a carbon or hydrogen. But when the hydrogen is not attached to the carving, you have to draw the, uh, the hydrogen so you can dry like that. Or, uh, I would. I usually just writes hydrogen like this and said Okay, now for C, we have ch three to Corbyn Double Bon sees which ch two ch three. Now again, we have this carbon. And then there is the two metals inside the parentheses. So that means they are attached to that carbon. So we have this this carbon. All right, And then we have this metal. So ch three ch three. Now we see that there's double bond, so we have to put a double bond here. And then what? Piers acid A is a carbon. So right there. And we see this hydrogen appears behind this carbon. That means that that hydrogen is attached to this carbon. So hydrogen, and then we see the carbon immediately. So that is attached here. And then we have aged two, and that appears after this carbon. So it's attached to that carbon, and then we have ah, see, appear so that is that carbon ist actually previous carbon. And then we have three. Hydrogen is coming off Now. I saw the, uh, phone line structure. So we have 123455 carbons in a row for 12345 Now we see that the second carbon has a metal coming off. And then we see that one. Do you one to the second bond? Right? As a double bond. So this is bond one. This is bond too, right? So we put a double bond. Yeah, and that is the answer. You can also draw the double, but on the bottom, it doesn't really matter. Frd, We have ah ch three ch two ch two ch two and ch three. Well, this is just a chain of carbon. So we have 12345 So 12345 That's the answer to that part, right? Because we have 12345 carbons. That's it for we. We have ch three ch two c h o h c h two ch three. Okay, so we have ch three ch two c h. And then remember this, Oh, ages in a parentheses after Ah, a carbon. So this means that this O H. Is attached to this carbon. So let me just write the O. H here. And then we have another carbon so that carbon is attached here. And then we have another carbon. And that is right here. Now let's show the line. Phone lines. We have 123455 carbons in a row straw. So we have 12345 And now we see that the dirt carbon has an o. H coming off though a witch. And that is your answer now for part F, we have ch two double bond C and then ch two ch 32 So we initially have a CH two that is double bonded to a carbon, and these are in parentheses, So that means that they are attached to the carbon previously appearing before that. So we have a ch two ch three and we have a ch two ch three. Remember this to me is that there are two of them. All right, analysis of the bone line. So we have 1234 That is the longest change that we have. 1234 And we see that the despond right here is a double bond. So is put a double bond there. And if we come from the last, you have one to so the second carbon splits up. Right? So this is a second carbon right here. This splits up, so we have to add 12 more on and that is the answer now for part G. Give me a second for party. We have ch three c. Let me have a double bon. Oh ch two, ch two ch two and ch three. So seas three bonded to a carbon. And then they tell us that this oxygen is double bonds to that carbon. And we have another carbon appearing ch two, ch two and ch three on the shot, the bone structure. So we have 1234566 carbons in row. So 1234 by six carbons in a row. Now we see that one to the second Carbon from the left has a double bond. So one to this one has a double bond oxygen off of it. You can draw the oxygen on the bottom to it doesn't matter now for part HD last one, it's ch three ch si el Sisi eight ch two ch ch three and then two. Okay, so we have a CH three, then we have a C H. And then we have a C L appearing before. Uh, so we have the CEO appearing behind after a carbon. That means that that CEO is attached to the carbon before it. So it's attached to this carbon. So let's draw CEO down here. And now we have a carbon appearing right after, so we have a C h two. Then we have another ch. We have ch then parentheses. This means that print on apprentices means that they're attached to the carbon on the previous one. So we have a ch three ch three. Now let's draw the longest chain. First we have 123455 carpets in a well, one to be for five now, one to the second carbon from the left. So 12 has a chlorine off of it. So we try a chlorine. And now Ah, 123 44th Carbon has another metal coming off here. So 1234 So another metal coming off of it. I'm just going to draw a straight up, and that should be


Similar Solved Questions

5 answers
Point)Find the integralVzrs +1 dx using substitution The resulting integral Is &;^} [ vi8 3 / Vu du6 Vu duNone of thesef" svu du5 / Vu du
point) Find the integral Vzrs +1 dx using substitution The resulting integral Is &; ^} [ vi 8 3 / Vu du 6 Vu du None of these f" svu du 5 / Vu du...
5 answers
Ouiqieluaj sidluaje dnojb jowdnojn ainu3 MulouJerkultiwuan(uun)(roquunu)(uun)(jqunu)S 3S STOIB(uun)(jqunu)(uun)(jqumu)Uuanjins Jo 508u3_ suun Jq 138 01 Aq uunjins JO SWO)E , Apdnitt nof PInOM IB4Mtvopsonb 44401 D9pNa4 * nIeA Junuodwi %57998 0142Ju343p24 341 %sn1 Isoquolelou] sajdo] Molaou]
ouiqieluaj sidluaje dnojb jow dnojn ainu3 Mulou Jerkultiwuan (uun) (roquunu) (uun) (jqunu) S 3 S STOIB (uun) (jqunu) (uun) (jqumu) Uuanjins Jo 508u3_ suun Jq 138 01 Aq uunjins JO SWO)E , Apdnitt nof PInOM IB4M tvopsonb 44401 D9pNa4 * nIeA Junuodwi %57998 0142Ju343p24 341 %sn1 Isoquolelou] sajdo] Mol...
5 answers
Question [10 points]If T2_R2 is a transformation and the action of T is as given find formula for T '(V), where V is any vector in 2 2. 1-2y #]-C -31+7yTl]
Question [10 points] If T2_R2 is a transformation and the action of T is as given find formula for T '(V), where V is any vector in 2 2. 1-2y #]-C -31+7y Tl]...
5 answers
5 COUoUMGLS 2 1 0Hockey Istiastons) breaing strenath Composile Compos 1 7 8 388 " 3361 38 = 33 1 dimeten 8678 prapil 318513 ;0 [email protected] /& @ ?
5 COUoUMGLS 2 1 0 Hockey Istiastons) breaing strenath Composile Compos 1 7 8 388 " 3361 38 = 33 1 dimeten 8678 prapil 3185 1 3 ; 0 2 ldato 0.01 @ /& @ ?...
5 answers
Sigmund Freud's theory of the unconscious mind(A) was revolutionary because it was the first comprehensive explanation of human thought and behavior(B) resulted from discoveries about the human brain obtained by cadaver dissection.(C) is outdated and has no relevance for modern psychology.(D) focused entirely on human males' sex drive(E) depends on the idea that humans can remember events but not be consciously aware of the memory.
Sigmund Freud's theory of the unconscious mind (A) was revolutionary because it was the first comprehensive explanation of human thought and behavior (B) resulted from discoveries about the human brain obtained by cadaver dissection. (C) is outdated and has no relevance for modern psychology. (...
5 answers
Find the values of $oldsymbol{b}$ such that the function has the given maximum or minimum value.$$f(x)=x^{2}+b x-25 ; ext { Minimum value: }-50$$
Find the values of $\boldsymbol{b}$ such that the function has the given maximum or minimum value. $$f(x)=x^{2}+b x-25 ; \text { Minimum value: }-50$$...
1 answers
Solve. $$2 x \leq 5-7 x<7+x$$
Solve. $$2 x \leq 5-7 x<7+x$$...
5 answers
Determine whether the sequence is increasing, decreasing, or not monotonic. Is the sequence bounded?$a_{n}= rac{1}{2 n+3}$
Determine whether the sequence is increasing, decreasing, or not monotonic. Is the sequence bounded? $a_{n}=\frac{1}{2 n+3}$...
5 answers
For r k ci + yj + 2k; evaluate a)V x (kXr)b)(F)
For r k ci + yj + 2k; evaluate a) V x (kXr) b) (F)...
5 answers
Giving a test to a group of students, the grades and gender aresummarized belowABCTotalMale1811736Female1691237Total34201973Let p represent the population proportion of all malestudents who received a grade of B on this test. Use a 99% confidence interval to estimate p to four decimalplaces. ______< p < ______
Giving a test to a group of students, the grades and gender are summarized below A B C Total Male 18 11 7 36 Female 16 9 12 37 Total 34 20 19 73 Let p represent the population proportion of all male students who received a grade of B on this test. Use a 99% confidence interval to estimate p to four...
2 answers
I) EvaluateK [dydc . Vr? +ii) Find the volume of the space region which is bounded above by below by the plane region R : y = 1l3, y=I,I <0, y < 0 (in the third quadrant) .V1-y' and
i) Evaluate K [ dydc . Vr? + ii) Find the volume of the space region which is bounded above by below by the plane region R : y = 1l3, y=I,I <0, y < 0 (in the third quadrant) . V1-y' and...
5 answers
Consider the following reactions: A =B,K,-7.35 A = C,K2-2.00 What is K for the reaction C = B?pull up for additional resources
Consider the following reactions: A =B,K,-7.35 A = C,K2-2.00 What is K for the reaction C = B? pull up for additional resources...
5 answers
The X-intercept of the line whose graph passes through (3, 5) and has slope of 5 is,Select one:b. 2d.4
The X-intercept of the line whose graph passes through (3, 5) and has slope of 5 is, Select one: b. 2 d.4...
5 answers
UnaOrL f(z) necomnrnchen mudctcnnin Aeuncdcilt Inearaly Inmqlelliricy matlincluding the endnoinb;""(r)Finduksinlcnl-Khichfunchonconcat down Wme [Intcrvals wbctc I(r) Is concave UP;FuetieyIntcrual wbcic /() ennnnaaenEnotemuFentpotuleiee
unaOrL f(z) necomnrnchen mudctcnnin Aeuncdcilt Inearaly Inmqlelliricy matlincluding the endnoinb; ""(r) Finduksinlcnl- Khich funchon concat down Wme [ Intcrvals wbctc I(r) Is concave UP; Fuetiey Intcrual wbcic /() ennnnaaen Enotemu Fentpotuleiee...
5 answers
Problem(10 points) Find the particular solution of the initial vale problem:CUS(1 +x) 1 3y = dx (1 +1)?2(0) = 3.
Problem (10 points) Find the particular solution of the initial vale problem: CUS (1 +x) 1 3y = dx (1 +1)? 2(0) = 3....
5 answers
Find the limit lim 10022 + l52 + 14 10022 + 9x + 12 I-00 Note- Give the exact answer but not the decimal approximation (for example, write 4/5 and not 0.8).
Find the limit lim 10022 + l52 + 14 10022 + 9x + 12 I-00 Note- Give the exact answer but not the decimal approximation (for example, write 4/5 and not 0.8)....

-- 0.018908--