5

Resonance15,42 Draw resonance structures for each radical:...

Question

Resonance15,42 Draw resonance structures for each radical:

Resonance 15,42 Draw resonance structures for each radical:



Answers

Draw resonance structures for each radical.

This is the answer to Chapter 16. Problem number five from this Smith organic chemistry textbook. Ah, and this problem asks us to draw additional residents structures for each of the four ions that were given in the problem. I'm so in order to do that, we're going to move electrons on DSO for a We can draw two more residents structures. So the 1st 1 the electron movement ah, would look like this. Um, so this loan pair can add in here, uh, these electrons can go to this carbon. And so our first resident structure is going to this. So a single bond here now, a double bond here now. And, um, our loan pair and negative charge are now located on this carbon. Ah, and so we can also go the other way so these electrons can add back in here. Um, these electrons can go here, and these electrons can go to this carbon. And so the result of that electron movement is going to be a resident structure where we have a double bond here, a single bond here. He double bond here, and a single bond here with the loan pair and the negative charge on this terminal carbon. Okay, s so then for B, we can do the exact same thing so our loan pair can add in here. These electrons can go to the oxygen. Um, and so that is gonna give us our only resident structure here. Uh, where, uh, the oxygen has a single bond to the carbon on. There is a carbon carbon double bond here now. Ah, and this would be considered. Ah, more stable structure. Um, we do always want to put extra electrons and charges on Electra, native Adams and oxygen is more lecture negative than carbonates. Okay, s So that's b. So then for C, we can have one of the lone pairs on this chlorine ad in here. And the result of that is going to be the single resident structure that we can draw here. And so when we do this, um, are chlorine is now doubly bound to the carbon, Uh, and on Lee has two lone pairs on, and so it's gonna have the positive charge on it. Um, but again, it's it's always better to put positive charges on Electra. Negative. Adam's eso that that looks fine. Us then for d we're gonna be able to draw two residents structures here. Um, and they're both gonna be just just movement of electrons, as always. And so the first thing we can do is a move these electrons over that double bond. And so that's gonna give us our first structure, which is gonna look like this. Okay. And so are positive charges now on this carbon. Um, And then we can also, uh, move these electrons to here. And so that's gonna give us a resident structure. Uh, where are positive? Charge is on the ring, and we have double bonds here and here. Ah, and so this would, um, also be a good residents structure, because you can see, we now have three double bonds that are conjugated in this system s So that's good. Um, okay. And so that's the answer to this problem again. The way to draw our resident structures is just to move electrons. Nothing else about the molecule changes on sometimes when, as a result of the electrons, moving charges will move as well. But really, the way to to visualize this is just movement of electrons. And that's the answer to Chapter 16. Problem number five

This is the answer to Chapter 11. Problem number nine, fromthe Smith Organic chemistry textbook. And this problem asks us to draw an additional resident structure for each cat eye on. Um And so one thing to notice here is that each of these cat ions um, each of these carbo cat ions is bound to a new electro negative Adam with a loan pair or more than one lone pair to spare. I mean, so all that we have to do is add uh, the electrons from one of these lone pairs into the bond between the carbon and the hetero Adam. Ah, and that will seat this positive charge on the Electra native Hetero Adam. Um, yeah. And so that's Ah, that's the resident structure. So for a that looks like this. So our chlorine eyes now down to two lone pairs, and it has a positive charge on it. So for be, it's gonna be the oxygen. So lone pair from the oxygen add in there to give us this structure. And now the positive charge is seated on the oxygen. The oxygen's down to one lone pair. Ah, and then similarly, for see, the nitrogen is lone pair can add in to the carbon nitrogen bond, which will take us to you. Here and now are the hoops. Can't forget the hydrogen. Never forget your hydrogen is on your nitrogen. Sze Um it is a pet peeve of every chemistry teacher and every chemistry grad student greater exam greater that I've ever met, including myself. Don't Don't forget, your hydrogen is on nitrogen sze. Um And so the positive charge is now seated on the nitrogen and the nitrogen, uh, no longer has a loan pair. Um and so these are all good resident structures. Um, it's always best to try to seat your charges on electro negative, Adams. And so chlorine, oxygen and nitrogen are all much more electro negative and carbon. So these air good residents structures. Um, And again, the key to doing this is to add in the lone pair of these adjacent Electra native Adams. Um, And that shifts the positive charge to the hetero Adam. And that's the answer to Chapter 11. Problem number nine

Well, everyone, today we're doing Chapter 23 problem for in this for masses to draw additional resident structures for each of these animals. So the features start from and we close on a We see that we have on a carbonara here is negative church. And then we have these Esther groups. You're Ethel Esters on either side. So we know that residence is the movement of pilot tranche along collection. So south. Here's some lone pillar and electrons on this carbon nucleus. We know that this can move to form a new pie bond here. And when make a bomb, you need a break of monsters. Carbon Opie Bond. Kimberly can go up to the auction nucleus giving us of one resident structure, which is illustrated with double headed heroes, show that it is truly a resident structure. Now we have the negative charge now on the auction nucleus with an extra long parallel actions. You're a new double bond that formed Let me have this one resident structures. But on the other hand, because marking symmetric this an ounce and a and a carbon with the same probability Khun formidable bond on this side of molecules making a second resident structure that can be illustrated like this. So now for B. If you use that sing mentality in same logic for big, we can see that we have and I on a carbon here. And once again we have two sides in which there's adjacent carbon also weaken former resident structure on either side. But now, unlike the 1st 1 these two sides are asymmetric, so we would get to unique resident structure's. Unlike this, these are actually identical Gabi's flip them around. You get the same molecule. So if you do the same thing so one could be moving the an antic long parallel transport this bond to making your pies. One breaking this cardinal. Yeah, you get resin structure like this, but at the same time is nothing stopping it from going on the right side to form another resident structure like this. And now for the very last problem. He have a molecule that looks like this, and we have the We have our own pair carbon here on now. Once again, we have two sides in which there's a Jason Pie bond so we can have resin. Structure's going on either side. Well, sir from the left, just like the previous examples. But at the same time is nothing stopping this and I are going on the right side. We came on the make a Bonny break about this sign Our group will break and we'll form now You mean negatively charged anything. And on this night, vision has to appear electrons as a formal negative charge.

All right. So what? This problem? We're looking at the three different residents form of the Cyclone Texel diagonal radical, which I've shown here. So you can see that, Um, here's just the cycle extraditing me of the radical here. And we won't explore the residents structures of this compound when it comes to residents, structures of radicals we want to be looking for are radicals next, two carbons or other molecules that have pie bombs, pure riddles that air filled lone pairs, something like that. So we can move this electron density around, remembering again that residents structures don't change the Sigma framework off these molecules. So in this case, we're working with radicals, meaning we official carols and are radical is next to a pile on which we can break two. It's two separate electrons generating our first resident structure by moving the electrons in. That way, we're going to get is a molecule that looks like this are radical, has moved to this carbon here, and we have this new cycle actual dying anal radical again are radical, is next to ah pie bond which can break apart again, generating 1/3 resident structure of the cycle Texel diagonal radical, and our radical now ends up on the carbon at the bottom of the cycle. Actual ring is I've drawn here, and what we can see is that we've made one, 23 resident structures of the radical. If we wanted to move this radical any further, uh, we've actually kind of already returned to the starting material. If you took the starting material and just turned it, you'll notice the bonds are identically the same until we've completed all the residents structures of this molecule and that answers this problem.


Similar Solved Questions

5 answers
Point)Find r(t) and v(t) given acceleration a(t) = t4j,initial velocityv(0) = 31,and initial positionr(0) = ~2kv(t) = r(t) =Usage: To enter a vector; for example (€,Y, 2) , type "< x,Y,z >"
point) Find r(t) and v(t) given acceleration a(t) = t4j, initial velocity v(0) = 31, and initial position r(0) = ~2k v(t) = r(t) = Usage: To enter a vector; for example (€,Y, 2) , type "< x,Y,z >"...
5 answers
Suppose that E a diverges with cach Gnand let $r2 &Show thatan/S, diverges. [Hint: Use the Cauchy condition ] Show that 2 a,/s' converges. [Hint: On/s; ~/s; 1/* 1/sa for n 2 2and € (/5,-1 I/sa) converges:|
Suppose that E a diverges with cach Gn and let $r 2 & Show that an/S, diverges. [Hint: Use the Cauchy condition ] Show that 2 a,/s' converges. [Hint: On/s; ~/s; 1/* 1/sa for n 2 2and € (/5,-1 I/sa) converges:|...
5 answers
QUESTION 8Calculate the arc length ofthe indicated portion of the curve r(t)r(t) (2cos 3 '6t) j (2sin 3 6t) k; {nsts 12
QUESTION 8 Calculate the arc length ofthe indicated portion of the curve r(t) r(t) (2cos 3 '6t) j (2sin 3 6t) k; {nsts 12...
5 answers
Question 8What is the IUPAC name of the following compound?OHT T IiT ParagraphArial3 (12b02
Question 8 What is the IUPAC name of the following compound? OH T T IiT Paragraph Arial 3 (12b02...
5 answers
Apply Newton's Method approxlmate the X-value(s) of the indicated point(s) of Intersection the two graphs: Continue the iteratlons until two successive approximations differ by less than 0.001 _ [Hint: Let h(x) 7(*) 9(x) ]g(x) cos(-) 0.89055 (smaller value) 89079 (larger value)
Apply Newton's Method approxlmate the X-value(s) of the indicated point(s) of Intersection the two graphs: Continue the iteratlons until two successive approximations differ by less than 0.001 _ [Hint: Let h(x) 7(*) 9(x) ] g(x) cos(-) 0.89055 (smaller value) 89079 (larger value)...
5 answers
INTERACTIE EXAMPLE IJIt _ knnrnirt oreac tin OnnnnnCommcnts EXITIOSCiIQuantum Mechanical AtomThe orbital quantum numb-r for thc ciectronhydrogcn atomWhat the smallest possibl- valuc for tha tota cncrgy of this clcctron?Emin ~0 3778 03778 No (-03778)Enter Help
INTERACTIE EXAMPLE IJIt _ knnrnirt oreac tin Onnnnn Commcnts EXIT IOSCiI Quantum Mechanical Atom The orbital quantum numb-r for thc ciectron hydrogcn atom What the smallest possibl- valuc for tha tota cncrgy of this clcctron? Emin ~0 3778 03778 No (-03778) Enter Help...
5 answers
Multiply. See Examples 1 through 4.$$5(-3)$$
Multiply. See Examples 1 through 4. $$ 5(-3) $$...
5 answers
Points) Let A and B be the following matrices.A = [9- 41,B =~8Perform the following operations:A'B =B . A =
Points) Let A and B be the following matrices. A = [9 - 41, B = ~8 Perform the following operations: A'B = B . A =...
5 answers
Simplify each expression.$$6 a-2 b-3 a+9 b$$
Simplify each expression. $$ 6 a-2 b-3 a+9 b $$...
5 answers
Application of sodium borohydride is recommended in thetreatment of foxing, which is a term used broadly to describe thestaining and discoloration of old books and papers. While foxingcan have many causes, one cause is from the oxidation of cellulosein the paper. Below is the oxidation reaction of cellulose to oneof its oxidative products. Explain why sodium borohydride would bea good treatment for the foxing of paper caused by oxidation ofcellulose.
Application of sodium borohydride is recommended in the treatment of foxing, which is a term used broadly to describe the staining and discoloration of old books and papers. While foxing can have many causes, one cause is from the oxidation of cellulose in the paper. Below is the oxidation reaction ...
5 answers
Find P(2 < X < 5) using the cumulative function given by: (0, X<0 F(x) = 1-e % X >0e {~e %4-e$e-1_e {
Find P(2 < X < 5) using the cumulative function given by: (0, X<0 F(x) = 1-e % X >0 e {~e % 4-e$ e-1_e {...
5 answers
How do you determine if the effect of cytochalasin D isreversible? Describe a simple experiment to determine this?
How do you determine if the effect of cytochalasin D is reversible? Describe a simple experiment to determine this?...
5 answers
Suppose the amount of certain radioactive substance in sample decays from 7.30 mg to 5.70 mg over period of 68. years_ Calculate the half Iife of the substance:Round your answerto significant digits.vcars
Suppose the amount of certain radioactive substance in sample decays from 7.30 mg to 5.70 mg over period of 68. years_ Calculate the half Iife of the substance: Round your answerto significant digits. vcars...
5 answers
1 8MiWhch Select one What can 341 1 They are They anet They are radioactive They produce are 1 6 mabliee The compound contains negamve about 1 CHEMCA W isotopes unde 9o2s shints in NMR with nitrogen signals spin this neaats broad peaks quantum concept doesnt 1 numbers apply Ehe of one such nucleophlletthe H?
1 8 Mi Whch Select one What can 341 1 They are They anet They are radioactive They produce are 1 6 mabliee The compound contains negamve about 1 CHEMCA W isotopes unde 9o2s shints in NMR with nitrogen signals spin this neaats broad peaks quantum concept doesnt 1 numbers apply Ehe of one such nucleo...
5 answers
Solve ' 0/5 POINTS "Lonenba (Entacxy PREVIOUS Your anawers ANSWERS comm LARPCALC1O L 3.026. integer constant. 1/2 Submissions Used 1 1 SJLON AW{upidp ASK YOU
Solve ' 0/5 POINTS "Lonenba (Entacxy PREVIOUS Your anawers ANSWERS comm LARPCALC1O L 3.026. integer constant. 1/2 Submissions Used 1 1 SJLON AW {upidp ASK YOU...
5 answers
Determine the domain and range of the function using the graph below: 15 44 13 12 H H05Domain:<r <Range:<y <
Determine the domain and range of the function using the graph below: 15 44 13 12 H H0 5 Domain: <r < Range: <y <...
5 answers
Use the following three identities to evaluate sin 1Ox cos 3x dxsin SX cOS tx2 [sin (s+ t)x + sin (s - t)x]sin SX sin tx2 [cos (s +t)x - cos (s - t)x] COS SX COS tx 2 [cos (s + t)x + cos (s - t)x]sin 1Ox cos 3x dx
Use the following three identities to evaluate sin 1Ox cos 3x dx sin SX cOS tx 2 [sin (s+ t)x + sin (s - t)x] sin SX sin tx 2 [cos (s +t)x - cos (s - t)x] COS SX COS tx 2 [cos (s + t)x + cos (s - t)x] sin 1Ox cos 3x dx...
5 answers
You are driving car at a speed of 38 mileshhr in Canada_ (a) What is its speed in meter per second? (b) Is it exceeding the 55 kmh speed limit?
You are driving car at a speed of 38 mileshhr in Canada_ (a) What is its speed in meter per second? (b) Is it exceeding the 55 kmh speed limit?...

-- 0.018819--