5

Rank the carbon-mzizl bond the following orgunonietallic nilgems orer ol aecre sing polant stening #ith the mast pola;,CH CH_Mebr CH CH,Li (CH CH WCuLi Yhch oltr. f...

Question

Rank the carbon-mzizl bond the following orgunonietallic nilgems orer ol aecre sing polant stening #ith the mast pola;,CH CH_Mebr CH CH,Li (CH CH WCuLi Yhch oltr. follow ing siatemenls 1but orginonselillic rexpentsJnlc?Orgnnomel_Ilic reagents conlzin cabon Mlun honiel [ mche Tc more polar tha carbn-nxtal bonc, ule Extr nenting the orancmculic IcICCM Omaranciallic rcagents react as bxses Jd nuclecphiles Organorelallic rcagents jr sbrong acids thar rexdily donnta polon 1hict.uheusi : miar Or2nIC F

Rank the carbon-mzizl bond the following orgunonietallic nilgems orer ol aecre sing polant stening #ith the mast pola;, CH CH_Mebr CH CH,Li (CH CH WCuLi Yhch oltr. follow ing siatemenls 1but orginonselillic rexpents Jnlc? Orgnnomel_Ilic reagents conlzin cabon Mlun honiel [ mche Tc more polar tha carbn-nxtal bonc, ule Extr nenting the orancmculic IcICCM Omaranciallic rcagents react as bxses Jd nuclecphiles Organorelallic rcagents jr sbrong acids thar rexdily donnta polon 1hict. uheusi : miar Or2nIC Froxdluct of bz lollowing Tejctica? Chlou rRp ^LB) I_C) IlI DM Whxl E Ulic IliIor 0xg1Iltc" product ol Ibc (ollowing IcKLoU? JCH | CuLi ALLW " C)[ D' ^



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) $\mathrm{H}_{3} \mathrm{C}-\mathrm{H}<\mathrm{H}_{3} \mathrm{C} . \mathrm{CH}_{2}-\mathrm{H}<\left(\mathrm{CH}_{y}\right)_{2} \mathrm{CH}-\mathrm{H}<$ $\left(\mathrm{CH}_{3}\right)_{5} \mathrm{C}-\mathrm{H}$

This is the answer to Chapter 22. Problem number 42 Fromthe Smith Organic Chemistry textbook. And in this problem, we're asked to rank to Siri's of compounds from least reactive to most reactive for a nuclear Phillips substitution reactions. And so remember, as I've drawn here on the right in green for this, uh, tem ple e for car box, silly gas and derivatives, the weaker of a base that the Z Group is, the more reactive the molecule was going to pay on. So you can look at each of these molecules, look at their Z groups on DH, then rank them Ah, from strongest base to weakest base. And that will be ranking them from least reactive to most reactive. And so, for a we have ah de pro nated ammonia. We have n each to as the secret, so that's a super strong base. So that's going to be the least reactive. So next we have, uh, pro pock side so warm to three carbons on oxygen de pro donated. So it's ah, it's now cock side face. And so that's that's a moderately strong base. It's not going to be a strong is end each too. But it's a moderately strong base on then. Lastly, we have chlorine on DSO. Corinne is the congregate base of hydrochloric acid. And so chlorine is going to be a very weak base because, remember, the congregate base of a strong acid is a weak base on DH. So we've ranked thes from strongest base to weakest base in terms of the Z group. And so that's actually also ranking them from least reactive, most reactive. So then we can do the same thing for B S O. Again. Now we have this preposterous I this Alcock side on that za strong base and so that's ah, that's going to be our first are least reactive. So then we have the sand hydride. Another way to think of this. After after reading this chapter, you should know that anhydride is going to be more reactive than Nestor and so you could write them that way as well. Or you can look at this and say, this is the you know, one of the sides of this anhydride is the cognac, it base of pro bono Gassid on DH. So since it's the consequent base of a fairly strong acid. It's going to be Ah, moderately week based on DH, so it should be more reactive than this Alcock side, which would be a moderately strong base in terms again of of when it's Z Group here on then. Lastly, we we have this try Floro acetic anhydride and again, you might just look at this and know that this is going to be super reactive or you can look at this and say Okay, well, half of this anhydride would be the consequent base of trifle, or OSI, to gas, which is a ridiculously strong acid. And so it's conflict base is super weak and remember again, the weaker the base, the more reactive molecule is when we're talking about the Z Group off this template card oxalic acid derivative. And so, since this is such a weak base because it's the con ticket base of such a strong acid, it's going to be a very reactive molecules. And so, again, that's how I would go about ranking these on. That is thie answer to Chapter 22. Problem number 42

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.

To determine which is the greatest bowler character. We need to first determine the election negativity between the atoms. So for Korean oxygen, oxygen has an electro negativity value of 3.44 Carve our guys and always a foreign. But carbon carbon has an Electra negativity value of 2.55 So 3.44 minus 2.55 is your A 0.90 point 89 for a silicon oxygen. Again, oxygen has an Electra negativity value a 3.44 minus the lichen, which is three. Car 1.90. So 3.44 minus 1.90 is 1.54 for Germaine IAM and oxygen oxygen has a value of 3.44 remaining has a value of 2.1 So 3.44 minus 2.1 That's gonna be a value of 1.43 For carbon and chlorine. Lauren has an Electra naked city value 3.16. Carbon has a two point by five. So 3.16 minus 2.55 is 0.61 And then for carbon and bro mean roaming has a value of 2.96 and carbon has a value of 2.55 2.96 Minus 2.55 is a value of zero point for one. So for Koller co, Vaillant bonding is gonna have to have an Electra negativity difference. What? Between 0.4 and one find seven. So we're going to go from the least value, so the smallest value to the largest value. So in this case, the smallest value will be E, which is our chlorine ambro mean, Next is D, which is Are you gonna keep saying, Gloria are carbon and bromide and bro mean? And then it's d with the SEC, the next, um, greatest Electra negativity difference, which is carbon and chloride. And then it will be a which is carbon and oxygen that no BC with Germaine iam and oxygen and at last be silicon and oxygen will have the greatest Electra negativity difference. So it's gonna be the one with the most fuller co Baylin character

Here we have toe under the mounds from leads to greater polar crater. So here, e party will be it first mean, then party, then a then see and then be okay.


Similar Solved Questions

5 answers
705 6wj0 tvuJt s50vSC - Kowpn 69*M919+5" To v "4 % O1 7 0GDosb4 8( F0nyn hos bm Tdw 4lm 69 '11 6641 GAE 609 (bolFON)SW- (OS"H ~UoF(ON)eH + (H#OSSWFont bw Tw 9m-#os6v J0 Shqw(sdt) 32[durs 341 tOSBWJO JupJuOX Jupjuad 341 JE4M "OSBH Jo S9L- Jo UOQu diouud JupinsaJ *Z(EONJed Ssjor7 4Wm p318jJ} PUT Joiet U! POA[OSSIP 'OSIW Juuugruuo) JuMXl PloS ;-0*0S kvfrov ks 5 t 79h0*0 TJH WWh IH IQ1u sLbiq: Inuu =3 (€
705 6wj0 tvuJt s50v SC - Kowpn 69*M 919+5" To v "4 % O1 7 0G Dosb4 8( F0nyn hos bm Tdw 4lm 69 '11 6641 GAE 609 (bolFON)SW- (OS"H ~UoF(ON)eH + (H#OSSW Font bw Tw 9m- #os6v J0 Shqw (sdt) 32[durs 341 tOSBWJO JupJuOX Jupjuad 341 JE4M "OSBH Jo S9L- Jo UOQu diouud JupinsaJ *Z(EONJ...
5 answers
Place these functions in ascending order of asymptotic growth rate (slowest to fastest) Assume all logarithms are base unless otherwise stated,In each spot write the number of the function occupying that spot That is if function fs (n) occupies that spot vrite{i (s)Hu+Aln}3(3} Adn) Jae) 3sh kgl)fe'=) 1010SlowestSecondThirdFourthFtt
Place these functions in ascending order of asymptotic growth rate (slowest to fastest) Assume all logarithms are base unless otherwise stated, In each spot write the number of the function occupying that spot That is if function fs (n) occupies that spot vrite {i (s) Hu+ Aln} 3(3} Adn) Jae) 3sh kgl...
5 answers
Find the determinant of matrix D:abCD= 3d + a 3e+ b 3 f + c g ha b If IA-| d e f =3 g h 1
Find the determinant of matrix D: a b C D= 3d + a 3e+ b 3 f + c g h a b If IA-| d e f =3 g h 1...
5 answers
Question 105 ptsThe following intermediate tableau for dual maximum problem was obtained using the simplex method for optimizing minimum problem_RHS 3 3 1 18~10Find the pivot element(s) in the above tableau; and perform all the necessary pivot operations to obtain the-findltableau: Then; using the final tableau; answer the following questions:The minimum function value is:The value of 31 in the minimum problem isThe value of T2 in the minimum problem is:
Question 10 5 pts The following intermediate tableau for dual maximum problem was obtained using the simplex method for optimizing minimum problem_ RHS 3 3 1 1 8 ~10 Find the pivot element(s) in the above tableau; and perform all the necessary pivot operations to obtain the-findltableau: Then; using...
5 answers
Points HoltLinAlg2 6.4.011_Find the general solution for the system_ Y' 1 = Y1 + 16y2 y'2 = Y1 + Y2(Y1(t), Yz(t))~11 points HoltLinAlg2 6.4.017 .Find the general solution for the system_ y'1 7y1 + 3Y2 8Y3 Y' 2 -2Y1 2Y3 Y' 3 6Y1 3Y2 7y3(Y1(t), Yz(t) , Y3(t))
points HoltLinAlg2 6.4.011_ Find the general solution for the system_ Y' 1 = Y1 + 16y2 y'2 = Y1 + Y2 (Y1(t), Yz(t)) ~11 points HoltLinAlg2 6.4.017 . Find the general solution for the system_ y'1 7y1 + 3Y2 8Y3 Y' 2 -2Y1 2Y3 Y' 3 6Y1 3Y2 7y3 (Y1(t), Yz(t) , Y3(t))...
5 answers
Question 12Let y = arccos (3x) , where 0 < x < 4/3. Write sinl) as an expression in terms of x.V1 + 9x2b) 0 V9x? ~1OV1+ 9x1 _ 9x29x2None of these _
Question 12 Let y = arccos (3x) , where 0 < x < 4/3. Write sinl) as an expression in terms of x. V1 + 9x2 b) 0 V9x? ~1 OV1+ 9x 1 _ 9x2 9x2 None of these _...
5 answers
L! @1a "Lx"S (1l,'3'CLly] "1lte"3 6s216y3 r ly6) 216} 9035-5-4 2163 5Y0l Y'lg] 2,"1Ly} 7ay>8421Cy>091y}1' (4'HeBt&7054
l! @1a "Lx"S (1l,'3'CLly] "1lte"3 6s216y3 r ly6) 216} 9035-5-4 2163 5Y0l Y'lg] 2,"1Ly} 7ay> 8421Cy> 091y} 1' (4'He Bt& 7054...
5 answers
What is the Voltage across a 11.8 ,F capacitor with 108 UC of charge on each plate? AV 9.153 unitHow much energy is this capacitor storing? U 0.0004 unit
What is the Voltage across a 11.8 ,F capacitor with 108 UC of charge on each plate? AV 9.153 unit How much energy is this capacitor storing? U 0.0004 unit...
5 answers
Let €1 T2 T2k-1 T2k be 2k consecutive points on the real line. Describe the Riemann surface of T1) T2k) . Show that this surface is topologically a (a) sphere for k = 1, (b) torus for k = 2_ What it will be for & general k? (Any compact closed Riemann surface is topologically sphere with handles Thus torus is sphere with one handle: _
Let €1 T2 T2k-1 T2k be 2k consecutive points on the real line. Describe the Riemann surface of T1) T2k) . Show that this surface is topologically a (a) sphere for k = 1, (b) torus for k = 2_ What it will be for & general k? (Any compact closed Riemann surface is topologically sphere with h...
5 answers
1 have birds and fish circugtoreyy sentences_ and H both have explain how -finneci and H fish why these 1110 to describe the structural Write systems differences there are suited are 3 to the lifestyle of ahorunctina difference betwees structural an each. spJOM functional 3 points M Wsgsteous the
1 have birds and fish circugtoreyy sentences_ and H both have explain how -finneci and H fish why these 1110 to describe the structural Write systems differences there are suited are 3 to the lifestyle of ahorunctina difference betwees structural an each. spJOM functional 3 points M Wsgsteous the...
5 answers
Calculate the root mean square velocity and kinetic energy of CO, CO2, and SO3 at 298 K. Which gas has the greatest velocity? The greatest kinetic energy? The greatest effusion rate?
Calculate the root mean square velocity and kinetic energy of CO, CO2, and SO3 at 298 K. Which gas has the greatest velocity? The greatest kinetic energy? The greatest effusion rate?...
5 answers
How many moles of a gas are ina47.1 container at 311.6 Kand0.9660atmmoles
How many moles of a gas are ina47.1 container at 311.6 Kand0.9660atm moles...
5 answers
SHOW ALL WORK(6 points) Solve the initial-value problem24' 3u (1+0) cost,!(O) =4v() =0
SHOW ALL WORK (6 points) Solve the initial-value problem 24' 3u (1+0) cost, !(O) =4v() =0...

-- 0.024618--