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Answer questions 9-14 using the following reaction: eq HCIKumtl #' % t09. What is the 1,2-product of the given reaction? 10. What is the 1,4-product of the giv...

Question

Answer questions 9-14 using the following reaction: eq HCIKumtl #' % t09. What is the 1,2-product of the given reaction? 10. What is the 1,4-product of the given reaction? E 11. What is the kinetic product of the given reaction? 12. What is the thermodynamic product of the given reaction? € 13. What product is favored if the reaction is conducted at low temperatures? 14. What product is favored if the reaction is conducted at high temperatures? € 15_ Which of the following will undergo

Answer questions 9-14 using the following reaction: eq HCI Kumtl #' % t0 9. What is the 1,2-product of the given reaction? 10. What is the 1,4-product of the given reaction? E 11. What is the kinetic product of the given reaction? 12. What is the thermodynamic product of the given reaction? € 13. What product is favored if the reaction is conducted at low temperatures? 14. What product is favored if the reaction is conducted at high temperatures? € 15_ Which of the following will undergo a Diels-Alder reaction with . dienophile? Choose all that apply:



Answers

Give the major products obtained from the reaction of one equivalent of HCl with the following: a. 2,3-dimethyl-1,3-pentadiene b. 2,4-dimethyl-1,3-pentadiene For each reaction, indicate the kinetic and thermodynamic products.

All right, when we look at these answer choices, the definition of a catalyst is that it lowers the activation energy of a reaction which overall causes it to speed up. We look at the rest of the choices, okay, we can narrow it down because catalyst does not affect any of the bond energies in any way. It only effect affects that activation energy. It also does not change the reaction of the overall energy of the reaction, which is the difference, or the distance between the energy of the reactions in the products. So our best choice is that a catalyst lowers the activation energy and therefore speeds up the reaction.

Here were asked for the major products obtained from the reaction of each of these compounds with one of Corbyn of HCL, and they were also asked to indicate the kinetic and thermodynamic products. The first one is to three dimethyl 13 pence a dying which I've drawn on screen here. And so we're gonna number this or we're gonna number are dying one through four. Just those four carbons that participate in the dying and we're gonna number. It's what we get the most stable carbon cat I in, uh, couple Catalans overall. So if I go from this side, my car broke a tie. Ins are always on carbons two and four. So if I go from this side over here, I would have tertiary on two and a secondary on four. If I went from the other direction, I'd have a tertiary onto in a primary on four. So I'm gonna go from this direction, so I get tertiary and secondary instead of tertiary in primary. This will be one too 34 on, and then the first step is to have that 12 double bond. Grab our hydrogen off of the chlorine and give us A car broke out. I am so that carbon cut. I would look like this, um, be hydrogen here. Went on to carbon one is that we can have the call carbon to be the carbon cat I And since that will be a Lilic, um, so that is our first intermediate. And then there's this also has a resonance contributor. If we move that double bond right there and that residents contributor looks like this, it's no never bonds here, and our car broke out I and is up here and then we'll get a product from each of these just by attacking the car. Broke a tie ins with the C L minus the Korean nuclear file that was left over from the HCL. So from this top one right here will have the chlorine on this carbon up here and are double bond stays over here. This is our 12 product, and then from the bottom, we will have double bond right here and a chlorine over here. And this is our 14 So a zipper, kinetic and thermodynamic, the one to product is always the kinetic product. Who's, um now, keep in mind that the Connecticut promoting them. It could be the same products. But the kinetic product is always 12 because of the proximity effects soapy for proximity effect. And that just means that because when this hydrogen is pulled off the chlorine, the Korean then is right next to this carbon cat are in a position to. So it's really easy for just going attack just because of the proximity. So that will be the Connecticut product. And then the thermodynamic product is going to be the most stable product on. And we're talking about thes reactions that usually means the most stable out keen. And so here the most Abel Kane is usually going to be the AL Keen. That is some of substituted, Um, and so here, that is going to be the 14 So this will be our thermodynamic product because the Al Keen is more stable. Um, and then let's part want our part A and then for part B, we're going to the same thing. We're gonna number our, um, carbons to get the most stable carbon cattle. And so if I go from the left here, I would have tertiary on number two in Turkey. Area number four. I went from the right. I would have secondary unto in primary and four, so I definitely want both of those tertiary res. So we're gonna go this direction and then do the same process so my one to bond will grow up the hydrogen and kick off that chlorine and we'll get a carbo cat ion that looks like this for that one and then not will have a residence contributor. That looks like this for that over here. Um, and if you'll notice these, they're actually the same thing. Just flipped over. So if you were to have your hand on a table with your palm facing up and you flipped it over so your problem is facing down, you would get, uh, the other version of this actually, the same the same structure. So, um, here we're actually on only going to get one product. It's what that is the 12 and the 14 product. It will be exactly the same. Um, and so that products. Oops. Well, look like this. I'm just gonna draw it on this side of the glory will be here, and the double bond will be over here, so that is our products. That is obviously going to be both the kinetic and thermodynamic products, since it is the 12 and the one for product. So this is Medic and thermo, since it's 12 and 14 at the same time.

This question asked us to draw the major 12 and 14 addition products for each reaction and then indicate the kinetic and thermodynamic products. So for part A, we're going to start off by numbering are dying. Remember that we number dying's only the four carbons that are participating in the double bonds. And we never them to give us the most stable carpet kinds. So our car broke. Headlines are always on carbons two and four. So here, if we start the top to make this carbon one carbon to would be a tertiary carbo cada line and then four would be a secondary. If we went the other direction to would be secondary and four would be secondary. So we're gonna go from the top to get that tertiary. So this will be one too three for, um And so the first step is to grab the hydrogen with the 12 double bund kick that chlorine off. This gives us two different intermediates because of residents of the first one. Just looks like this. That's our number two carbo cat ion. And there's also a residence form of this. We just move this over here that gives us a 14 So now this is a pier and our car broke. Iodine is down here. So then all we have to do to get to the actual products is attack. These carbon combines with the C L minus. That's left over. And so we get products that look like this. So are 12 looks like this and our one desire want to and are 14 looks like this, um, and then as faras, kinetic and thermodynamic. The kinetic product is always the one to product because of the proximity effects. And then the thermodynamic product is whichever one has the more stable AL Keen. So they could both. Like the same products could be both the kinetic and promoting a product. If that was the most stable l keen in this case, it's not. In this case, the 14 is the more stable Al cane because it's more substituted. This will be the thermal product because it has be more stable. L king for part B, we're gonna do the same thing. So if we started numbering over here, we would have secondary or started tertiary for number two and secondary for number four. If we went the other, I would a secondary for number two and secondary for number four. So we're gonna go first way, So this will be 1234 So, again, the first step is to grab hydrogen and kick off our quarry in to put the nuclear fire layer. And then our first, um, intermediate looks like this. So our car broke iodine. Is that to show you one right there? Um and then we can get another intermediate. That looks like this. The residents form here, where now we have our car broke out. I am carbon four like this. And then we'll attack both these with the Korean to get our products. So for the first one, we have this product. This is our 12 And for the second one, we'll have this product. This is our 14 and then a Sephora's Connecticut thermodynamic. Like I said before, the 12 is always the kinetic because of the proximity effects. And then the thermodynamic is which everyone has a more stable Al Keen. So more substituted Al Keen. And in this case, that is the one for So this is the thermal product. And that is because of the AL keen stability. This one is more substituted

Okay, so the reaction here is three a. Which is a gas because B. Which is a solid gives us to see which is a chris and D. Which is Aquarius, I'm not gonna worry about are solid. And let's let's work with everything in polarity here at 1.1, 3 more And 2- seven more. And for a, I'm told that I have 2.48 moles And that's going to be in our volume of 1.87 L. That's gonna be 1.33 moller. And that's true at equilibrium. So we'll go ahead and write RK for this. It's going to be the concentration of C squared time's D. All over the concentration of a cube. Okay, We're gonna ignore because it's a solid. So we'll plug in our concentrations at equilibrium. We need them all to be in polarity. So we're going to get a K. c. here equal to 1.23. So that's going to be answered. A.


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