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2) Starting with a compound with 6 (or less carbons) propose a synthesis for the assigned compounds: Provide all reagents and show intermediates (3 points)4) Provid...

Question

2) Starting with a compound with 6 (or less carbons) propose a synthesis for the assigned compounds: Provide all reagents and show intermediates (3 points)4) Provide all the Resonance structures of the cyclopentadienyl anion using arrows show the flow of electrons (delocalization) in each resonance structure: (2 points)NOz

2) Starting with a compound with 6 (or less carbons) propose a synthesis for the assigned compounds: Provide all reagents and show intermediates (3 points) 4) Provide all the Resonance structures of the cyclopentadienyl anion using arrows show the flow of electrons (delocalization) in each resonance structure: (2 points) NOz



Answers

Draw all possible resonance structures for each of these compounds. Determine the formal charge on each atom
(a) $\mathrm{O}_{3}$
(b) $\mathrm{SO}_{2}$
(c) $\mathrm{NO}_{2}-$
(d) $\mathrm{NO}_{3}-$in each of the resonance structures:

Hey, guys. So in this question were asked to find resonance structures for each of these compounds, given the curved arrows that were given in the textbook. So in part A, we have this compound and we have this curved arrow right here denoting that electrons from this triple wand are moving to this nitrogen here. So remember, when we have a curved arrow with two ah arrowheads right here, that means a pair of electrons are moving as opposed to one. So when we re draw this structure, we have our ch three remains the same R C is going to be here, and instead of three bonds, we're gonna have a double bond here instead, since the electrons are now moving to the nitrogen, so we'll have two pairs of loan of two lone pairs of electrons. Now that we have this, we just need Teoh. Make sure we have our for more formal charges. So we're going to have a negative charge on the ah nitrogen. Since nitrogen has five valence electrons, we're going to subtract the number of bonds we have. So it's too, and we're also going to subtract the number of ah lone pairs of electrons we have. So we have four electrons, so that's five minus six. So that's a negative one. And then on this carbon we have four valence electrons and then we're going to subtract the number of bonds. So that's three bonds. So that's four months. Three. So that's a positive one. So remember when we do residence structures, we want to make sure that the net charge of our formal charges is the same for both structures. So when we add these together we get a net of zero, which is the same as what we started with moving on to part B. We have this compound right here. So we see that the one pair of electrons is moving to this carbon or to this bond right here to fill the ah octet of the middle carbon right here. So let's redraw that. So instead we're going to have one pair of electrons on our oxygen, a double bond on this carbon, and then the rest of our compound remains the same. So now that this carbon has a full octet, it has a formal charge of zero, and now the positive charge seemingly moves over to oxygen, But that's just due to the movement of electrons. So oxygen has a ah go violence of six. So we have six minus 123 bonds, plus two lone pair to a lone electron. So that's six minus five. So that's a positive one. So that's why are positive charges now on oxygen? So that that is this resonance structure. And for part C, let's do this last structure. So it's a pretty similar structure to the last one, but instead we have no hydrogen on this oxygen. It's just three lone pairs of electrons bonded to carbon, and we're moving. All of our electrons were moving the extra lone pair of electrons from the oxygen to this bond. And since carbon can't have five bonds removing one of the bonds from this double bond to this carbon so let's draw this out. So one of the lone pairs from this oxygen is moving to this bond right here to form a double wand and the double one that was here now forms Ah, a lone pair right here. So the, um, Onley formal charge that we need to worry about right now Is this carbon right here? So we have a valence of four and then we have to subtract to the number of bonds we have. So that's three. And then the number of lone Electron. So that's two. So four minus, um, or excuse me. 123 we'll four minus five is negative one. So that's where we get this negative formal charge on this carbon from to These are all three of our residents structures, and this concludes.

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.

Ch. three and 3. So what we have is carbon, hydrogen, hydrogen, hydrogen and then we have a nitrogen double bond nitrogen double bond nitrogen. So the fast nitrogen is neutral, the second is positive and the final nitrogen is negative to these charges. Councils give us a neutral species. Alternatively, we can dump this electron density. What we get is the following C h H. H and and triple bond. And so we still have a positive charge on the middle nitrogen. However, the negative charges now on the fast nitrogen because what we've done is done this electron density here and then moved this electron density. But what we've done is still maintained neutral species, as you can see these charges in both of the resonant structures do in fact, cancel out

There are three possible resonant structures from ethel aside, let's first determine the number of valence electrons available for these three resonant structures. There's one carbon, 3 nitrogen and three hydrogen. Carbon has four valence electrons, hydrogen has one but there are three of them, nitrogen has five and there are three of them Giving us a total of 22 valence electrons. If we bond the three hydrogen to carbon and we bond the three nitrogen is together bonded to carbon, we've used up 2468 10, 12 of the valence electrons. So we have 10 left over If we put two here Another to hear that's four, six, eight, 10. So now we've used up all 22. This nitrogen has eight valence electrons, 88 and eight. So everything has the required eight valence electrons except for hydrogen, which only need to. And now we'll determine formal charge Carbon brought in four valence electrons. It has four valence electrons surrounding it. Each of these pairs of electrons that are bonding are shared, so they're split in half. So each corresponds to just one valence electron. So 4 -40, Nitrogen brought in five valence electrons. It has 1, 2, 3, 4, 5 surrounding it. 5 -5 is zero. This nitrogen brought in five. It has 1,234, 4 -5 is negative one or the other way around. Sorry? Um 5 -4 is positive. One brought in five. It only has four. Then we go to the next one that has 1, 3456. Mhm. So it brought in five. It has 65 minus six is negative one. Its formal charges minus one For the next one. We can move this double bond over to here and then move this lone pair over to here. And we get this LewiS structure here, nitrogen brought in five. It has 1234 So five minus four is plus one here, nitrogen brought in five. Still it has 1234 so it's plus one here. Also, this nitrogen brought in five, it has 1234567 So 5 -7 is -2. And then this carbon is the same at for a peace brought in for it has for now, so it's zero. And now for the last one we'll draw the same based structures we had And then we had 10 electrons remaining. If we put two here into here, That gives us six left 12 34 56 This nitrogen has eight. This one has eight. This one has eight, carbon has eight. So this is the third resonant structure. Carbon brought in four valence electrons. It still has 1, 2, 3, 4, 4 -4 is zero. Nitrogen always brings in five. It has 123456 so five minus six gives us negative one, it has a negative one formal charge. This nitrogen, Also bringing in five, has 1, five -4 is plus one for its formal charge, and then this nitrogen has 12345 It brought in five, so five minus five is zero on its formal charge.


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