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Use curved arrow notation and draw a resonance structure for each of the following: points each) OHCHz...

Question

Use curved arrow notation and draw a resonance structure for each of the following: points each) OHCHz

Use curved arrow notation and draw a resonance structure for each of the following: points each) OH CHz



Answers

Using the curved arrows as a guide to placing the electrons, write a resonance structure for each of the compounds shown. The resonance structure should include formal charges where appropriate.

Hey, guys in this question were given three compounds and wrasse defying residents structures for each of them, given the curved lines using and using them as guides. So, in our first problem, in part A, we have a benzene ring with two metal attachments, and we have curved arrows more moving in the thes directions. Remember, when we have curved arrows like this with the full arrowhead, that means a full pair of electrons is moving from that bond to the next. So let me go ahead and draw my ah benzene ring benzene ring first, and then we have our two metal attachments. So when we move this pair of electrons over to this bond right here, this is going to form a double bond here. And when we do that, this carbon can't support five bond. Since we'll have a double bond here, one bond here and a double one here. So that's why this double bond on my right side is going to move to this bottom right part of the hex or the benzene ring. Now, when we get to this carbon, this carbon is going to have five bonds on it. Because of this double bond and this double bond. So this Oh, that. Yeah, this double one, this double bond. And there's going to be a hydrogen here as well. So that's five bonds. Carbon can't hold five bonds. So that's why this double one then moves to this bond right here. And when we look at this, we don't have to put any formal charges or anything because none of the, um, octet it's of the, um, carbons making up the benzene ring were changed in any way because they're basically the bonds are basically in the same places. So this is our complete residents structure for this part a moving on to part B. We have this, uh, Alka off keen and with ah, look what looks to be what was an alcohol attachment. And this Ah, oxygen is donating a pair of electrons to this bond right here. And this is going to go ahead and push this pair of electrons from this bond over to this bond which will complete this carbons octet. So let's go ahead and draw this out first. So when one pair of electron moves over to this carbon, we have a bond here. This carbon would have five bonds because of the double bond here and the making of a double bond here, as well as the hydrogen here. That's why this double bond must move over to this carbon. Now that we have that looking at all of our carbons and all of our other Adams, none of the formal charges have changed or the formal charges are all now zero. So when we have resident structures, we want to make sure that our net formal charge is the same as it was in the beginning. In the beginning, initially, we have a net formal charge of positive one here and negative one here. So that's a net of zero. And now for a residence structure. Here we have a net of zero because all of our four more charters are now zero. So this is there now complete resident structure. And now, for part C, we have this pen maintain ring and with some attachments to that, So we have a lot of electrons moving, so lets go ahead first, draw out our structure. We have the paint in ring carbon here, hydrogen here, oxygen on and oxygen. So starting off with this oxygen right here. This oxygen is donating a pair of electrons to this bond right here. So let's go ahead and do that. This carbon would have five Barnes, if this double one doesn't move, so it has to move to this pond right here in the painting ring. And now this carbon right here would have to support five bonds, which it can't to it. It moves the double wand electrons to the oxygen since its very electra negative, and it can handle the negative charge. So this now has a negative formal charge, and we have a negative negative one formal charge to start off with. So this structure is now are a complete residents structure.

He has curved arrows to draw resident structures, showing how you convert the first compound into the second. So the way to start this problem is to physically look at what's changed. Well, in this common Hey, this compound, we have a positive charge and carbon in this compound. The positive charge is on the third carbon, the pie bond. The double bond has also physically moved from between the two central carbons to the terminal one. So when we're drawing resident structures, we want to be pushing electrons never charges, so we don't push a positive charge. We on Leigh push either loan pairs of electrons or pie bonds. The only loan pairs of electrons are pi bonds we have within this compound is right here. We have a single pie bond between these two carbons in order to get this pie bond here or in this case, between these two carbons, what we can do is pushed the electron density from between this carbon two. Between these two, whenever we're pushing a lone pear or a pie bond, it must be a double headed arrow. So two heads to this arrow. What that does is it satisfies the positive charge on carbon. This carbon now has four bonds. It no longer has a formal plus one charge. But now this carbon is lacking a bond. It only has three bonds. If you were to split each bond in half and calculate the formal charge, he would see it plus one. It also doesn't have a full architect. But that's how you would draw the arrow formalism from the first compound to the second. They're just resident structures. Yeah, down here, we can look at the difference. Nothing changes with this thes two oxygen through the same in the product on ly changes. Here we have a pie bond between oxygen and carbon that disappears, and this oxygen gains another loan pair of electrons. So if we're lacking a pie bond here in the product and adding a lone parent this oxygen, we must be moving this pie bond up on to oxygen because you have to have the same amount of electrons in your starting resident structure as in your product. So what we can do is we can draw an arrow from the pie bond up on toe oxygen, so this pie bond is becoming one of the loan pairs on oxygen. When that happens and I apologize, I forgot to dry it. You also form a positive formal positive part on the central carbon because it's now is lacking electron density. It has a formal plus one charge the oxygen Gaines, a formal negative charge.

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

Let's start this problem by drawing the resident structure of Thie car box a late shown here. Well, we have a negative charge on this oxygen, so to distribute that charge, what we can do is we can push electron density from a lone parent oxygen into form a new pie bond. And if you do that, you can't have five pounds to carbon, so you have to break one. And pylons are easiest to break, and you can push that pie bond of onto the other oxygen. That gives you the negative charge on the opposite oxygen. So the overall residents hybrid is going to be some mixture of these two compounds, and they're going to be an equal contribution. The way we can draw that is to draw essentially a bond and a half between every carbon that has a double bond in our two resonance forms. So this bond overall it's half double bond have single bond. So that's what we're trying to show with this resident structure or the overall hybrid structure. And then, of course, you also just want to note that there is an overall negative charge in this species. Let's do the same process again to the second compound. So we have a positive charge on nitrogen. So we're gonna have to figure out how to draw a resident structure. We need to satisfy the positive charge of nitrogen. To do so, we can push electron density from the pie bond and give it to nitrogen as a lone pair of electrons. If you do that, that gives carbon a formal positive charge, because now it only has three single bonds. So if we're going to draw the overall resonance hybrid, we're gonna have some mixture between a double bond and a single bond between the carbon in the nitrogen in the way we can draw. That is just by drawing kind of a dashed line. So not quite a single bond. Not quite a double. I'm somewhere in between. And there's overall a positive charge in the species. The next one we have a carbon. I am to satisfy the negative negative charge here. The on ly a direction we can push is up towards the carbon eel. You can't push this one pair of electrons down because then you break the octet rule. It isn't a valid resident structure to push this long pair in the downward direction. So we have to push up towards the car. Venial. If you do that, this carbon can't have five funds. But luckily there is a bond we can break, which is a pylon, and we can push that up on to oxygen. This is a valid resident structure. Thie Overall resonance hybrid is therefore going to be a single slash double bond in this case somewhere between a single and a double and somewhere between a several double and a single between the carbon in the oxygen with an overall negative charge for the final structure here, oxygen has a positive charge. In order to take care of that, the only thing we can do is push up a pie bond from between the oxygen and carbon and give itto oxygen so that it no longer has a formal for us. One charge that would look something like this giving oxygen all of its electrons now. But now carbon has the full positive charge. If we're to overlap these two resident structures, you'LL see that we have a double bond here, a single bond here, So we're going to be somewhere in the middle in our product somewhere between a double and a single body. We still have one lone pair. That doesn't change between the two forms, and we have an overall positive charge in the resident's hybrid.


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