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For the structures shown below, state the number of pi electrons present in the molecule.HHHHH-H-BHHNumber of pi electrons Number of pi electrons Is the molecule ar...

Question

For the structures shown below, state the number of pi electrons present in the molecule.HHHHH-H-BHHNumber of pi electrons Number of pi electrons Is the molecule aromatic according to the Huckel criteria?If the molecule were planar; would it be antiaromatic?

For the structures shown below, state the number of pi electrons present in the molecule. H H H H H- H- B H H Number of pi electrons Number of pi electrons Is the molecule aromatic according to the Huckel criteria? If the molecule were planar; would it be antiaromatic?



Answers

Complete a Lewis structure for the compound shown below, then answer the following questions. How many carbon atoms are $s p^{2}$ hybridized? How many $\mathrm{C}-$ N bonds are formed by the overlap of an $s p^{3}$ hybridized carbon with an $s p^{3}$ hybridized nitrogen? How many lone pairs of electrons are in the Lewis structure of your molecule? How many $\pi$ bonds are present?

This is the answer to Chapter 17. Problem number 31 fromthe Smith Organic chemistry textbook. This problem asks us how many pie electrons are contained in each molecule. Ah, and so, um, we need thio. Basically. Just count the pie bonds and remember that each pi bond eyes made of two electrons. And so, um, for a we have five pi bonds on so five pi bonds is gonna be 10 electrons for B. Ah, we have, um, three pie bonds. It's gonna be six electrons, but then we have an unparalled electron as well. So be is gonna be seven electrons, uh, for C. We have, um, 12345 pi bonds. So that's gonna be 10 electrons offer D. We have 357 pi bonds for a total of 14 pie electrons. Ah, and then any? We have six pie bonds for a total of 12 pie electrons. Okay, so that's it. Um, the only riel outlier here is is part B where we have that un paired electron that we have to account for. Otherwise, we can just town the number of double bonds and each of these and multiply by two on then for being. We re do the same thing and just add one for that un parent electron that we have on DSO. That's the answer to Chapter 17 problem number 31.

The problem says consider the levees is structured shown below. So here is the levees structure. Mhm. Right. And it is asked that does the levees structure depict a neutral molecule or an iron. If it is an iron, what is the charge on an iron? Right and be a disaster. What is the hybridization of? Is exhibited by each of the carbon articles? Right. And are there multiple equivalent resonance structure for the species? Right. And 3rd, 4th 1 and the last one. It is asked how many electrons are in the pie system of the species? So great. I don't know. So for the first one, yeah, this level is structured depicts D I am first one. Mhm It is not New Jersey. It is but it is the iron. Right. So how is it possible there you see here is the charge that is right again. This is the one who and this is edge. This is a literacy, this is it. This is everybody. See little edge. And this isn't. Thank you. So who has basically two lone pairs. Right, but it has one more loan Paris present. That means there is a negative side here. Right. Mm. So this is basically it makes it makes I am judged iron charge If he was to my last one. Right. So, formal ties for little charge is given by what a formal charge on oxygen intensive the economy number of balance electrons minus Yeah, okay, I n electrons and minus bonding electrons fight. Yeah. So this becomes a number of valence electrons is nothing but this is six. Right? It's the number of valence electrons is 6-. I only electrons is six again and minus bonding electrons is nothing. But this is too. So this is to buy two. This will give us six minus 7. 8 to- Her. So, this is the final formal chat. Okay, right. So, now for this 2nd 1, we can see the hybridization. Right, So hybridization let us right, You know, this is carbon 1, 2 and this is today. So for carbon you can see see when is sp two hybridized? Why? Because double bond is present a Yeah, Right. Similarly, it will have C2. Okay, We'll have sp two and again C3 will have also as the two. Thanks. Mhm. So, no. What is us net next next? It is asked that how many electrons are thirdly disaster are there multiple equivalent prison and structures for this trip for these species? Yes, Yes, it is intense. So, it is asked. Yes. So we were right. There are multiple equivalent resonance structures for the specialist because there is symmetry senior. So, it would be like this also. We can put this double bond with this also. Right, so it is right. How did this explain? How? So does it see how there is a structure is possible resonance structure is possible in this way. C double born. Oh, this is edge. Is it C And this is all minus. And this transfers here. This a good time. Right. And this you can see yeah, this is taken by this, you can see. And another way the reason infrastructure could people like this suitable wonder See my list. And this is seen Deborah Bondo because his age, this is edge in this message. Okay, so there are two structures. Okay, so now, lastly it is asked how many electrons are in the pie system? Multi species. So how many electrons are there? So it is possible for the pie system? We will see here. So in the pie system there are four electrons by because this is one, this is two. We went and joined. There's nothing but sharing of electrons. So this will be 1234. So there could be whole electrons. Right? So answer is forward electrons. I hope you understood the concept. Thank you for watching.

Solving problems. 12 off chapter 15. How many electrons that does each of the four nitrogen Adams imputing contribute to the aromatic per system? So first of all, let's trow pure in structure. Then we have defy Remember ring. Wow. Okay, so let's number the nitrogen as one. Nitrogen to nitrogen three and nitrogen for so nitrogen. One tune and three They contribute each with one Elektrim because, um, there in an sp two espirito hybridization and saw only one electron contribute to the pi toe the pie system so we can draw them, um, in this way and they each contribute with one a letter. One one one electron nitrogen number four instead, with this one is one of these when nitrogen number four instead, us both electrons off the lone pair contributing toe the pie systems or nitrogen number four. He contributes with a total of two electrons toe the by system. So now if we have one electron for each nitrogen 1 23 These are, um, three electrons and from nitrogen number four. We have two electrodes because both electrodes off the lone pair, um, contribute our part of the romantic system. So we have a here toe electrician's and then we air to count the electrons from the car booze. So it will be 34 567 eight and will be nine and 10 with the electrons from this nitrogen and so then electricity total make pureeing.

All right, guys, We're going to be doing a thing from sixty for chap nine in chemistry. Central Science Sin was to write a single Louis structure for s o three and turn the hybridization of the soul for Adam. So we have our Louis structure here. There we can brew only think of Ah, hybridization In this case, if you have if you have to Ah, to electron density surrounding you. You're S P if you have three. I'm sorry if you have as if you have to lecture on the seas surrounding you, you're S P you're S P. If you have three electron density surrounding you, you have your SB two. And if you have for electron dances running, you have your s P three. So we have one, two, three, one, two and three even though we have a double bourbon here and that still technically counts as one electron density. So since we have three electron density is flying around did and we don't have any any loan parents. That means that we're going tohave oven s p to hybridization. Now they want our their equipment loose. Fractiousness mature. Well, yes, They're our resident structures So the deal bond over here That can go onto here or rather let me. Sure, sure. The resident structure. No, don't mind this. This was just used to calculate the number of electrons for getting our our Louis structure are the If you have twenty four Elektron as if you have twenty four bounce like machines have twenty four electrons in your Louis structure. Now they're two things that can happen. We can We can do this where the life signs from here Come overmix Dubon here while the electron here go Well, the Dubon breaks and reform electrons go received this oxygen or we can have it where this oxygen forms of double bond. And this oxygen here, Yes, as in this tub on brakes informs the oxygen's so so these There are two different structures Mess of round one that we have the blonde on the right oxygen and on the red B. So in this case, we have a bond on elect oxygen and sing a bundle on the top in the reduction. So these air all resonant structures of one another. So since we have restaurant structures where the where you have a double on on on each of these oxygen's in a specific resident structure. That means that you are going to have the localized pie bonding. You are goingto have a partial double bond character on each of these bees, as you are not a part in a bond on each of these eso eso bonds. And that's going to be this Ben of angry. Don't need to draw the Louis structure. We can draw about showing the electrons. So we're gonna have a partial double bond characters. You're here and here.


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