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Choose the resonance hybrid for HOCOz9 Hio9 K[Dm Hyo8 :MH Hefoi...

Question

Choose the resonance hybrid for HOCOz9 Hio9 K[Dm Hyo8 :MH Hefoi

Choose the resonance hybrid for HOCOz 9 Hio 9 K [Dm Hyo 8 : MH Hefoi



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Draw a second resonance structure for each ion. Then, draw the resonance hybrid.

The questions on the topic the basis of chemical bonding. And this question is asked that what is the residence habit and how does it differ from the resonant structures drawn for molecules. So basically they start writing the definition of as an exhibit first, prisoners have bird. It is the it is basically defined as the average of two or more two or more. There is an instructor. Okay that the money good house positions. So just to understand how to actually represent it, you can see that suppose carbonate as the structure as one of the resurrecting phone also got less economical structure in the other form that government can exist in it look like this. Yeah. Another phone inhabitable one and think last left. What will you do just this? So all of these are all less canonical structures are presenting structures but none of them are the have everyone. How did they, how did it look like they have read of this would look somewhere like this where we have double bond first we have single months everywhere because that's the minimum every atom had. And the program negative charges dispersed equally among all the bones. And now we have double one partial development here, partial government here in Barcelona one here. And this is what resonance infrastructure look like you do, which they say that yes resonance habits structures are different from enormous touches. Reason is highly structure are different from um, kind of legal structure, canonical structure. These uh, normal resulting structures that we have also called as canonical structure. These are the two answers. Let's so this is the answer to the first part. This is the answer for this again.

Hey, guys. So in this question were given some resonance structures and we're ask defying which residents structure would be the major contributor to the residents hybrid for each system. So let's go ahead and get started. So in our first, um, part part A, we have this compound right here, and our major residents contributor would be this one on the right here. This is because this, um, structure has the positive charge. This coward book had I own it parted of charges secondary or it's actually tertiary. Excuse me. Tertiary meaning that this carbon has three attachments through ah, alcohol groups attached to it making a tertiary as compared to this carbon cat ion which is only a primary since this carbon only has one attachment because this is tertiary, this car big hat I own have, ah, more stability due to like inductive effects and ah, things like that. So this would be the major structure for Part B. We have a similar thought process. This first carbo cat ion is only primary and this car book have I on is secondary since this has more alco groups attached to it, since it's a secondary carbo cat eye on this will be our major structure because of the inductive effects part C, Our major structure, when looking at this would be this one to the right. This is because we have complete OC tents and more bonds. And due to this, this compound is going to be more staple than this structure here in part D. We look at these two structures and right away we can say this one on the left is going to be our major structure. This because there's no charge separation, whereas this one on the right house charge separation where, Although the oxygen can handle a negative charge, this oxygen has a positive charge. And but since it has a high electro negativity, it can't handle a positive charge. Now, for our last one or second to last one, we'll use the same thought process as touch A as a car book hat. I on Look at each of these. We see that the ones on our ends have only one alcohol group attached to them, so they're going to be primary. But this one in the middle is secondary, since it has to alcohol groups attached to this, um, carbon so because of that, this middle structure is going to be our major structure. Now for our last one, we see right away like the one previously with the oxygen's. This one's going to be our major structure, since there is no charge separation, unlike this one where there's positive charges on the nitrogen and negative charges on the nitrogen, so that's basically all of our residents.

This question asked us which residents contributor in each pair makes the greater contribution to the residents hybrid. So basically, which contributors? More stable is what they're asking here. And so for part A. I believe the book, at least the PF that I have of the book misprinted this molecule. They had be positive charge on this carbon, but then this one was missing a bond. Son had a positive had four bonds, so I'm pretty sure this is what it's supposed to be. So if we take this molecule as I've drawn it, Um, And compared to this one here, this carbo Katelyn is a secondary a Lilic Carbo Calyon. And this one is a primary Ilic. And so a secondary A Lilic is more stable than a primary, Elik and the stability of the cargo Calyon does trump the stability of the double bond in this case. So this one is going to be the greater resonance contributor here Report be here. The difference between here is where we have the night of charge show on first molecule. We haven't honest carbon, and second, when we have it on the oxygen, carbon does not like taking any charges. All oxygen is much more willing to take both a positive and a negative charge than a carbon atom. So this one is going to be the more stable contributor here. Also, another thing to note on this molecule is that here we have this aromatic ring intact. And here we have broken that residents no residence, but like broken the aromatic ring here, there is still six pie electrons, but this is a much more stable version of that ring. So this one is the greater residents contributor. Um, for part C here again, we're gonna compare the carbon cat I in. So this is a secondary a Lilic, and this is a tertiary Elik. Um, and again, tertiary is better than secondaries, so this one will contribute more. And lastly, here we have, um, a Benzie ring groups with, um, a, uh, purple chain attached to it. So this one, we have this secondary. Oops. Secondary. Benzel Like karma. Cat eyes not cooperating today. Um, And over here we have a secondary a Lilic carbo Calyon. Um so here again, we this time we really have broken the aromatic city of this ring because we've taken one of the pairs of electrons out of it. So here we've made the ring less stable, Would meet it, um, non aromatic. Now, um, while here, the the, um, disease intact and the carbon kind is actually more stable here. So this one is going to be the greater contributor for party.

This is the answer to Chapter one. Problem number 15 from the Smith Organic Chemistry textbook on this problem gives us nitrous acid are Pardon me? Um, nitrous acid? Yes, and asks us Ah huh. To draw resident structure for this molecule and then to label, which is the major contributor of the minor contributor and to draw the residents hybrid. So the first thing we're asked to do is draw resident structure for this. Um, this is fairly straightforward. So we have a double bond bond. We have lone pairs adjacent to it, so we can just move alone pair and the double bond. So one of the lone pairs on this oxygen can add in here to make in your double bond. In one of these two bonds can break in. The electrons can revert to that oxygen. Ah, and so what this is going to give us is this resident structure, so we'll have this oxygen double bound to the nitrogen. And now the nitrogen only single rebound to the second oxygen here. Um, this oxygen will now have three lone pairs. And so, of course, it will have a negative charge. This oxygen will only have one lone pair, and so it will have a positive charge. Um, And so given these two resident structures, it's fairly straightforward and simple to say that this one is the major contributor of the original structure we were given on this structure that we've drawn is the minor contributor. And we should know that, right, Because the structure that we've drawn has two charges in it. So even though it has a net charge of zero, Um, and even though both of these charges are on pretty electro negative Adams, so they should be somewhat stable. The difference between a charged and uncharged resident structure and stability is still significant. So the more stable structure is gonna be the one without charges. And since it's more stable, it's going to be the major contributor s. So the last thing that we're asked to do here eyes to draw the residents hybrid. So that's gonna I look like this. So the connectivity is, of course, exactly the same. Um, and then what we're gonna have, uh, we can go ahead and draw, uh, one lone pair on each of these molecules or on each of these atoms. Rather. Okay, Um And so then we just need Thio account for, um, the charges in our in our minor contributor. So even though it is it, a minor contributor does contribute. And so what we're actually gonna have, um, is sort of partial double bonds between each oxygen and the nitrogen. So we indicate that with this thes dashed lines on and then we need to account for the fact that there's a partial charge on each of these Oxygen's as well. So we used the delta symbol, um, and so Delta Plus for the part for positive Delta minus for the partial. Negative. Ah. And so this is our hybrid structure right here. Um, and of course, the hybrid exists more closely to the major contributor over. The minor contributor does play a part as well on. So that's why we have to show it like this. Ah, and that's the answer to Chapter one problem number 15


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