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Compant, would like to provide ientity cards with Jetter and mumbers for its empojtes- Hox EL diffennt ideutity cuuds Iacl' no lulter and TMUACC Mend more than...

Question

Compant, would like to provide ientity cards with Jetter and mumbers for its empojtes- Hox EL diffennt ideutity cuuds Iacl' no lulter and TMUACC Mend more than once:TPloc) mR+x Pz4) 2C*wo G3R 7o

compant, would like to provide ientity cards with Jetter and mumbers for its empojtes- Hox EL diffennt ideutity cuuds Iacl' no lulter and TMUACC Mend more than once: TPlo c) mR+x Pz 4) 2C*wo G 3R 7o



Answers

Ilcrc, whire precipirare $\{\mathbb{R})$ obraincd on rrcarment wirh aqucous solurion of $\mathrm{BaCl}_{2}$ is of (a) $\mathrm{AgC}$. (b) $\mathrm{CaCl}_{2}$ (c) $\mathrm{BaSO}_{4}$ (d) $\mathrm{PbC}^{\prime} \mathrm{I}_{2}$

In this problem, I can write DDX in it see http CH two C 00 H. This compound in pageants of Cl two A red P. That to be dad reaction. It will give the compound edge, see http C h C L C. O AT and this compound in perchance of alcoholic kO edge. The hydro chlorination happened here and the compound formed the CH two double 1 C h c o egg. Therefore, according to the option of indeed correct option, the is correct answer. I hope you understand the solution.

Chinese solution when directed with stool and avoided give some right sector own additional access this compound. So could you inform general which hard place here's the correct terms.

This is the answer to Chapter 30. Problem number 32 from the Smith Organic Chemistry textbook. In this problem asks us to draw a mechanism for this cycle ization reaction between foreigner sealed I phosphate and I So Penton Neil Penton he'll die phosphate. Um And then it asks us to comment on some unusual features of this reaction. So to start drawing foreign steel die phosphate here, So the first thing that's gonna happen, Um, And remember, we've seen this before in other problems in this chapter. So the dye phosphate group, and far too sealed by phosphate, is willing to leave. So the electrons will go to the oxygen, um, and will form carbo cat eye on. Right. So So after the dye phosphate leaves, we're here where we have that carbo cat eye on. Um And so at that point, that's when the ice Oh, Penton, he'll die phosphate eyes going to come in and react. So here is our S o pente, Neil die phosphate. And, uh, the electrons in this double bond will attack the corporal cat eye on that re formed in the first step. And so that, uh, will give us this intermediate. So give me a moment to draw it out. Oh, and actually, I just noticed that I made a mistake in the previous intermediate. So this double bond should not be here. It should instead be here. So I apologize for that. Uh, and so back to drawing this intermediate. So here is the product of, uh, this addition. Let me just particularly note that nitrogen there on. Of course, we have a carbo cat eye on here now, um, and so it's important to note that nitrogen there because that's the nitrogen that a base in solution. And we're just gonna represent base with a capital B. They're so base is gonna come, um, and grab that proton. Grab that hydrogen. Um, and the electrons will revert to that bond to make it a double bond. Um, and so product here, eyes. Now, um, this So basically all that we've done is, uh, if you look at our starting for in a sealed by phosphate, we've basically just extended it by one unit. Right? So it looks like we just just tacked on another repeating sub unit. Another another one of these, um, and in effect, that's that's what's happened so far. So, um, what we need to do now is redraw this in such a way that it's easy to visualize the cycle ization that's going to occur here. So in order to do that, um, you can start at one end here. Just just bear with me. This made take me a moment to draw. Oops. Okay. And then here's our ropp or die phosphate. Okay, so the obviously the point of redrawing it like this, uh oops. And I dropped a metal group. There we go. Okay, so the point of redrawing it like this is, um just because it makes it much easier to show this cycle ization that's going to occur. So once again, the O. P. P. Group is very willing to leave, so we can just show that bond breaking the electrons going to that oxygen there. Um and so what will happen next? Um, is that you know what? I'm not even gonna redraw this entire thing. So here's what I'll do. So there's one terminus, then the other terminus is going to look like this now, so we'll do this, and we can use this sort of squiggle curvy line connecting these just to show that there's a whole ring going on over here and that cuts down on time having to redraw the ring, making mistakes and drawing rang. And then right. So nothing, nothing from this portion of the molecule here has changed it all and is really affected by the rest of this reaction at all. So we might as well just leave it out, and I will redraw it for the final product. Ah, but until then, there's there's not really any point. Um, and so we now have a carbo cat eye on here. And very similarly to the first reaction in this mechanism, this double bond is going to attack this carbo cat eye on. And so the product that is going to be this Ah, and again, I'm going to draw this simplified version that is ugly, right? So I'm actually now, of course, that closed the ring so and again, we can just draw squiggle there, um, and so will draw, actually, and draw it this way. Uh, Proton there, hydrogen there. Um And then, of course, we will have a carbo cat eye on here. Um, and so, uh, what will happen now again, Exactly. Has happened in. The first reaction in this sequence is that, um base here will show up. Snag this hydrogen. These electrons will revert to their to reform the double bond. Um, you know, unfortunately, it looks like I'm gonna have to draw, uh, my product on a new page. So? So there we go. There's the final product. Um, flecks of bean. Right. So, uh, so the other part of this question that were asked is what is unusual about this cycle? Ization reaction and what's unusual about this cycle? Ization reaction. Is this right here where I'm gonna show you in red. So a corporate cat ion has formed here in the mechanism that I've drawn. Um, and that's a secondary carbo cata. Uh, whereas ah, corporal cat I own could form uh, elsewhere. Our team make a, uh, tertiary carbo chi on. So if if instead of of that corbeau cat eye on we formed Ah, Corbeau cat eye on, uh right. So if instead of that, if instead of that one we formed the carbo cat eye on here, um so this the ring wouldn't close here. It would close here instead, Um, and that would be a tertiary carbo cat eye on. And, of course, tertiary carbo cat ions are more stable, but it would result in a different ring to it would result in a 14 member grain instead of 15 members. Um, so, uh, the the reason for for this the reason that this is able to happen is that, um you know, in reality, uh, the electrons here are sort of bouncing back and forth. So each of these positions has somewhat of a cargo cat eye on by the difference being when the carbo cat eye on is at this position, the position it's actually in it allows us to close the string down and make this ring. Um, which for whatever reason, is preferential. Ah, and so that's what's unusual is that we would expect this to to form a tertiary carbo cat eye on, or at least to rearrange to a tertiary carbo cat. Eye on over. That's not what we see. Um, and of course, the result. A CZ, I said, Is this molecule flecks of being? Um and so that's the answer to Chapter 30. Problem number 32

As we know that so too can act both heads acts both as occident as well as reluctant. It has temporary bleaching acts and also it has employ Laurie bleaching accent, also bleaching action also, therefore, according to the option obscenity, each correct answer.


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