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The selectivity of chlorine radical is 1:4.5 for 1?, 28 , and 32 hydrogens, respectively- Assuming that only monochlorides are products in the radical chain chlorin...

Question

The selectivity of chlorine radical is 1:4.5 for 1?, 28 , and 32 hydrogens, respectively- Assuming that only monochlorides are products in the radical chain chlorination of 2 dimethybutane_ what would be the expected ratio of the two isomeric alkyl chlorides formed in that reaction? ( [5 points_Considering the reaction between 2-methylpropane and Iodine.Propouse all the possible products, and mechanism for each product: ( [0 points) Using the values from appendix 3_ Calculate the AH? for each re

The selectivity of chlorine radical is 1:4.5 for 1?, 28 , and 32 hydrogens, respectively- Assuming that only monochlorides are products in the radical chain chlorination of 2 dimethybutane_ what would be the expected ratio of the two isomeric alkyl chlorides formed in that reaction? ( [5 points_ Considering the reaction between 2-methylpropane and Iodine. Propouse all the possible products, and mechanism for each product: ( [0 points) Using the values from appendix 3_ Calculate the AH? for each reaction, and for the propagation steps. (5 points) Reasoning from Hammond $ postulate, predict the regioselectivity of radical iodination (5 points)



Answers

Radical chlorination of $CH_3CH_3$ forms two minor products $\textbf{X}$ and $\textbf{Y}$ of molecular formula $C_2H_4Cl_2$ .
a. Identify the structures of $\textbf{X}$ and $\textbf{Y}$ from the following $^1H$ NMR data:
Compound $\textbf{X}$: singlet at 3.7 ppm
Compound $\textbf{Y}$: doublet at 2.1 and quartet at 5.9 ppm
b. Draw a stepwise mechanism that shows how each product is formed from $CH_3CH_3$.

You are now with your problem. 79 79. We know usually in the halogen halogen group. The mall. The halogen stand about the other halogen can. Aussie died as I s I on. Okay, so chlorine when request Rome I iron so Lauren can Osei Dy Roman! Roman! Roman! I am become Roman, and we'll deal to Laura. High on this is hospitalizing h oxidation reaction. And when we see this, Lauren from free energy. Right Free element zero. Now, Lauren minus one. Here. Oxidizing agent. They see Warren so reducing, Agent. Never mind. Remind from minus one become zero The oxidation tech in Greece. They And now you've got this one. You see this? Mhm. So chlorine. And this one. But so we have e sale.

This question provides a mechanism for which were asked to determine the overall reaction. The overall rate law and the intermediates. The first step is fast. The second step is slow, which is the rate determining or rate limiting step. There are two intermediates intermediates are created and then later consumed Those being cl and CCL three. When we sum up these reactions, the Cl and the CCL three's cancel and the overall reaction of CL two plus C H C L three goes to hcl plus CCL four. The rate law will be determined from the molecular charity of the slow step. Giving us K prime is equal to a slow step being this one concentration of cl raised to the first power because the coefficient is one and then C H C L three raised to the first power because it's coefficient is one. But this is not a satisfactory rate law because CL is not part of the overall reaction. So we need to get rid of that intermediate by recognizing the first step is fast and reversible, meaning it has achieved in equilibrium. So the rate in the forward direction equals the rate in the reverse direction, right in the forward direction, expressed this way, right in the reverse direction, expressed this way with a super scripted too. Because in the reverse direction Cl has a coefficient of two. This then gives us an expression for CL, which we can plug in here to get rid of the cl and rate will be equal to a ratio of rate constants multiplied by C. L concentration, raised to the one half power, the square root multiplied by the ch cl three concentration. All of those rate constants can be combined into a single rate constant and this will be our rate law for the overall reaction.

Okay. This problem is describing a situation in which we have isolated the are I summer of two core Penton and we're going to react it with seal to on HV, which are the regents needed for radical coronation to make 24 die for plantain. So before we start, let's just ensure that this is the r I summer. So we have a hydrogen in the back that's in the bank. My Korean is Priority one. My, this group is number two, and this group is priority number three that is going in a clockwise direction, which corresponds to the R I summer. So that is correct. As for the reaction, if we erect Seal two and HB with this compound and we should get a situation in which my chlorine in the front is unaffected, my ah, right summer is unaffected and we're just adding a Korean to the fourth position. The reason why it's unaffected is because the this chlorine, the one that position for, is not affecting carbon number two. So there's no changing sterile chemistry. As for the stair chemistry of the Korean at position over four, that is up for grabs because it could be in the front like this. Pour it could be in the back. But no matter what, my chlorine at position number two remains unchanged. It does not. It is not affected by radical coronation by carbon number four because it's uncover number two there unassociated with each other. So you should be. My two answers my two different products. So the question is also asking. And it what? Whether they're president, different ratios. And the answer to that is probably no, because we have a radical coronation in car number four, there is no difference in whether it was on a secondary treasure primary carbon cut on. It was just what whether it was on in the front or in the back. So as far as which one to be president more. I would say that this one would probably be present in a little bit more than this one, but not by much. So it might be approximately 51 49 but more so 50 50. So the reason why I would say this one might be president little bit more is because if this is in the front, this might be taking up more space in the front, making it harder for a radical coronation to be happening in the front as well. So it would occur in the back just because there's more space. Okay, so as for whether these are optically active, that is just a fancy name to say. Are they Cairo? So let's analyze thes. Okay, I'm gonna race to use. Okay, So are these Cairo or not? Let's analyze this one first. So Kyra ality is basically the assumption that a carbon has four different multi four different atoms or molecules attached to that central carbon. So let's analyze this carbon right here. We have 1234 different groups attached to that central carbon. Okay? And we also have four different groups attached to this central carbon. OK, But as as for whether or not their cargo, let's say that we were to marry this image so that we would end up with these Koreans in the back, whether I have the corns in the back or have them in the front as long as they're both in the front or both in the back there considered identical. So this would be even though they have caramel parts, they are identical whether we mirror image them or not. So these would be me. So compounds so not optically active. As for this one, it has Carol parts because we have the four different atoms slash molecules attached to my central carbon and the same for this site. But if we were to mirror image this, I would have a corn in the back and corn in the front. These are not the same. So these this molecule will be considered Cairo and this one would be considered me so or not Carol. So this one is optically active.

Here for the duration C to watch for. Plus, that beer is to see to edge effect br solution for the equation. See, to watch for plus at Be a has it goes toe C two h five the no if move on equal to 165 reactor Mhm because to have toe Yeah, here we know that X c two h five be, uh, he called tau zero point 517 17 and x at be Yeah ah, according toe zero 0.173 No x c two h four Is it going toe? 0.31? No assuming that assuming dead mall friction yeah. Off C two at four in feed equal to x and mall Friction off h b r in feed equal toe one minus x and F one equal toe 16 five mall per second. No, but as force you to watch for mhm balance for C two watch four hit x f one according toe. Zero 0.5 and seven after right plus 0.31 after after putting the value Mhm. Yeah. Where have fun equal to 165 heavy. Find the value of X which is equal toe zero point 8 to 7, have to diverted by 165 no for HBR, one minus x f. One a call to Zito 0.69 after and break it zero point 173 have to kill us zero 0.517 after no putting the value of X, which has which we have find this value and the creation and the value of f the find the value off have toe This is equal toe 108 0.77 more right Bye. Second, no to find x X equal toe Zito 0.82 So one F two divided by 165 which will be called to zero point 8 to 7. Divided by 165 He called to 0.5 Yeah, 45 This will be the value f x No more friction off as to edge foreign feed. No more friction off. C two h four in feed according toe X equal toe 0.5 45 Now the mole fraction off HBR Indeed. Mole fraction off at BR and feed equal to one minus x equal toe 0.4 five five from stick geometry off the reaction we require the C two h four on HBR in one day show one. So here C two h four is in access amount and HBR is in the limiting amount limiting reactant so mhm for the second part person days access reactant equal toe 0.5, Force five minus zero point 455 divided by a zero point 455 multiply pinto 100 and mhm. We found that person Tae's e access reactant equal toe 19 0.78 percent soc toe edge four present in the 19.78 percent exists in feed storm. Now for the third part, which is fractional conversion. Yeah, so see toe edge four in feed equal to yeah 0.45 Sorry. Uniformed five for five in 2165 equal to 89.9 to 5 moles per second and edge br in. Freed he called go 0.4 55 and two 165 equal to 75 0.0 75 moles per second. No fractional conversion off the limiting reactant HPR equal toe. Mhm reflection conversion off. Limiting reactant H p r yeah. will be cool too. 75 0.75 minus 18 0.8 17 Diverted play 75 0.7 five. Yeah. Which will be equal to 0.7 49 three on, um, some will be 0.75 So fraction conversant equal to 0.75 or 75%. So fractional. Yeah. Conversion equal toe 0.75 or so in 25 percent. Thank you.


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