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Radical chlorination of pentane is a poor way to prepare 1 -chloropentane, but radical chlotination of neopentane, $left(mathrm{CH}_{3}ight)_{4} mathrm{C}$, is a go...

Question

Radical chlorination of pentane is a poor way to prepare 1 -chloropentane, but radical chlotination of neopentane, $left(mathrm{CH}_{3}ight)_{4} mathrm{C}$, is a good way to prepare neopentyl chloride, $left(mathrm{CH}_{3}ight)_{3} mathrm{CCH}_{2} mathrm{Cl} .$ Explain.

Radical chlorination of pentane is a poor way to prepare 1 -chloropentane, but radical chlotination of neopentane, $left(mathrm{CH}_{3} ight)_{4} mathrm{C}$, is a good way to prepare neopentyl chloride, $left(mathrm{CH}_{3} ight)_{3} mathrm{CCH}_{2} mathrm{Cl} .$ Explain.



Answers

Like alkenes, conjugated dienes can be prepared by elimination reactions. Draw a stepwise mechanism for the acid-catalyzed dehydration of 3-methyl-2-buten-1-ol [(CH$_3$)$_2$C = CHCH$_2$OH] to isoprene [CH$_2$ = C(CH$_3$)CH = CH$_2$] .

This is the answer to Chapter 10. Problem number 55 Fromthe smith Organic chemistry. Textbook on this problem asks what's three Al Keane's excluding stereo ice mers can be used to prepare three claro three metal heck, sane by the addition of HCL. Um And so, um, what we can notice right off the bat here is that if we draw this structure without the HC, well, without the chlorine in it s so if we just draw like this, um, we have, ah, tertiary carbon there. That carbon number three, where the metal group attach is is a tertiary carbon. And so remember that the addition of HCL to an Al Keane is going to follow my Markov niqabs rule. So the chlorine is going to go to the more substituted carbon, which is always gonna be that tertiary carbon carbon number three. So, basically, all we have to do eyes redraw this backbone three times, so draw three metal hack seen three times. Um and then we just need to draw the three possible positions for this double bond. So if we put the double bond there, um, Or if we put the double bond here or if we put the double bond, here are the result is always gonna be the same because this reaction is going to follow. Mark Avnet Cops roll. So that carbon number three, that tertiary carbon is always gonna be the one that gets the chlorine on. And then whatever the other carbon is, that's part of the double Bond will always be the one that gets the hydrogen. So any of these three Al Keane's plus hcl will give us our target product. Three claro three. Nothing, Maxine. Um, okay. And so that's the answer to Chapter 10. Problem number 55.

This is the answer to Chapter 15. Problem number 39 Fromthe Smith Organic chemistry textbook. Ah, and this problem asks us which Al Kane is needed to make each alcohol. Hell, I'd by radical halogen nation. Okay. And so, in order to solve this, we need to remember that radical halogen ation replaces a carbon hydrogen bond with a carbon halogen bond. Ah, and so to determine what Al cane each of these alcohol headlights came from. We just need to replace the halogen with a hydrogen s. So it's ah, it's very straightforward. Um, we just need to replace each of these challenges with a high 100 Jin. So for a, uh, we just have cycle plantain for B. We're just gonna have to metal butane. Ah, for C um, we will just have purple hack scene, uh, and then for D. Um, we will just have ah, 11 dime, Ethel. Propane. So there we go. Eso In each case, we just replaced the halogen with a hydrogen to get to our starting al canes on its Ah, that's really all there is to this problem. It's pretty straightforward on that's the answer to Chapter 15. Problem number 39

This is the answer to Chapter 15. Problem number 79 Fromthe Smith Organic chemistry textbook. Ah, and this problem asks us to look at the radical chlorination of ethane. Um, and we're told that it forms two minor products X and why we're told that their molecular formula is C two h four C L, too. We're giving a little bit of any more data about each one now and asked to derive structures. And then lastly, we're asked to draw Draw a stepwise mechanism. Pardon me? That shows how each product is formed. Okay. Ah. And so to begin, whenever we were given a formula and asked to come up with the structure, I think that the best way to start is to calculate a hydrogen deficiency index or the degrees of on saturation. Ah, And so for this one, it's gonna be two times two plus two minus four minus two. So that's two times the number of carbons plus to minus. The number of hydrogen is minus the number of Halla Jin's on. And so ah, in this case, we end up with zero. And so there's no one saturation here, which makes sense. It's only a two carbon molecules, so we can't really have a ring. Um, and it's unlikely that we would have a double bond. Uh, okay, so, um, we're not given integrations for these. NMR signals were just given chemical shift and splitting pattern. Um, and so, uh, right away. What I noticed for X is that there's only a single signal. Uh, and so the fact that there's only a single signal, uh, means that the hydrogen is in this molecule have to be equivalent. Um, and the only way that can really be true, uh, is if this molecule has, as an axis of symmetry right through the middle. And so it's it's gonna look like this one chlorine on each of the carbons. Um, and that's why we only get a single signal a single it. Because all of the hydrogen is in this molecule are equivalent us. Isn't looking at why? Um, there's really only one other possibility that we can make since this formula is, uh, so small. Only two carbons. Um, And so to get to signals, the hydrogen is on each of the carbons. We're gonna have to be non equivalent. And the only way that we can really draw that is like this. And so in this case, um, our doublet a 2.1 p p. M's call that, uh, signal one. So that's gonna be these higher regions then our other signal single, too. Eyes going to be this single hydrogen split into a quartet by its three neighboring protons. Okay, s. So there's our structures for X y. So were also asked to draw a mechanism that accounts for the formation of these two products. So to start, we will have our chlorine or chlorine gas sealed, too. Um, when that eyes treated with, um, let's say heat. So when it's when it's heated, um, this breaks into to chlorine radicals. So there we go. Um, and actually, let me ah, move this over. Okay, because I do want to make sure to indicate that this is the initiation step of our reaction. So I for initiation. Um Okay, so that's the initiation step. So we now have, um, an active radical species in this reaction. So next we're gonna have umm our propagation steps and the propagation steps here Going to look like this. Um So Okay, we have one of our chlorine radicals that we formed in the initiation step. And it is going to, uh, come along and grab a hydrogen. So that's gonna look like this, uh, the electrons of this bond cleave home elliptically eso After this first propagation step, we have this. Um, yeah, I'm actually I'll just erase this, Um, because I do want to fit the second part of this step on this line us to just understand that that HCL is also produced here. But we'll just look at the active species. So we now have ah radical on one of our carbons here. Um, and so that's actually going to react with another equivalent of chlorine gas. So C l, too. It's always a good practice to draw your electrons in, and so this is going to look like this. So the CLC Oh, Bond cleaves home elliptically. And, uh, this is going to regenerate a chlorine radical. Okay, um, and so that's Ah, that's like our common propagation step. So that step is common to both, uh, x and why? Um however, uh, here is where they diverged a little bit. So if we are going to draw the formation of why first. So why is gonna look like this? So we have this molecule that we just made. It looks like this. Ah, and we have our chlorine radical that we just generated or regenerated. Uh, and so they're going to react. So we now have a radical on this carbon, which has already been chlorinated once, um, and, uh, just like in the second part of the propagation, step above that can never react with another equivalent of chlorine gas. So like that, and that's gonna get us to our product. Why, Okay, And again that that does regenerate the chlorine radical to keep this reaction going. So this is part of the propagation step. Um, I'm just distinguishing the propagation steps that make why, from the propagation steps that make X so X is gonna be largely similar. The only difference is that for X, we're going to take ah, hydrogen from the other carbon. And so that's exactly what I'm drawing here. Um, pretty much identical to the steps that make why Just happening at the other carbon. So the carbon hydrogen bond cleaves, um, the carbon now has the unpaid electron on it, and that will react with another equivalent of chlorine gas. So again, the radicals here Now, this will react Justus we've seen before immediately above it. Ah, and so that is gonna give us our product. Ex as well as regenerating the chlorine radical. Oops. I said one thing, and I do something else. All right, So ch to see l see age too C l s. Oh, there's our product, X. Um, And then again, a zay said this will also regenerate the chlorine radical. So, um, the termination step I'm gonna put up to the top right here. Um, just to save having to start a new page for this very short step s o the termination step, remember? Can be, um, any two radicals coming together? Ah, and quenching one another so that after this step, there is no more reactive, radical species. And so I'm just gonna use two of these chlorine sze, So to chlorine radicals can come together. And when they do, uh, we get Seo to back. Um, and there's now no active radical to continue propagating this reaction. So that's gonna be our termination step. Ah, and this is the the stepwise mechanism that shows how x and wire formed. Um, And then again, uh, the structures of X and wire here. But just to review this first page, this is how we derived them from any more data. Ah, and that's the answer to Chapter 15. Problem number 79.

This is a Chapter twelve problem number sixty four of the Smith Organic Chemistry textbook. Ah, And in this problem, we're being asked to essentially, I summarize sis to beauty mean to trance tube, you teen. Unfortunately, it can't be completed in a single step on DSO. We're asked to devise a small synthesis and the way to solve this problem is too a begin by bro emanating the al cane which gets you toothy Die, bro mo product on the die Brahma product is ideal here because you can then a treat the dye Brahma product with two equivalence of sodium a meid and the two equivalence of sodium a mite will do to elimination reactions to take you all the way to the al kind on. Then the al kind can be reduced with the dissolving metal reduction so sodium in liquid ammonia. Ah, and that will give you the trans Al Keen. Uh and so it's ah, pretty straightforward problem solvable in those three steps


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