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Predict the major product in each of the following reactions and show if it originated: from an SN1, Si2, E1,or E2 reactionCH;NaOCH4)Cl CH;CH;B)Cl CH;CH;OHC)ACH; ~C...

Question

Predict the major product in each of the following reactions and show if it originated: from an SN1, Si2, E1,or E2 reactionCH;NaOCH4)Cl CH;CH;B)Cl CH;CH;OHC)ACH; ~CI Eha

Predict the major product in each of the following reactions and show if it originated: from an SN1, Si2, E1,or E2 reaction CH; NaOCH 4) Cl CH; CH; B) Cl CH; CH;OH C) ACH; ~CI Eha



Answers

Predict the product or products of each of the following reactions: (a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{HCOOH} \longrightarrow$ (b) $\mathrm{H}-\mathrm{C}=\mathrm{C}-\mathrm{CH}_{3}+\mathrm{H}_{2} \longrightarrow$ (c)

In the given reaction. Roaming replaces the hydroxide and we form this major product. Mr. Reaction between our alcohol and potassium hydrogen is removed and potassium takes its place in this reaction between our alcohol and our reactant, that is our product on our last reaction between these two carbon molecules will not happen and we're left with no reaction.

This problem assets to draw the products of each of these reactions. So starting with a, um, we have HBR with a pie bond. What's gonna happen is the h. Woods, uh, be attacked by one of those pipelines and we'll kick the br off. Some will have to. H is on one of the carpets, right? And we always put the h on the carbon that is less substituted. Look, the other one can become a carbo cattle because they want a more stable cargo Catterton and then are roaming will come in and attack that carbo cat ion. So here we will get each to see of our BR and then the ch three. So that's if we have one equivalent of HBR. If we have access HBR or two equivalents HBR, then we can do that. So both pi bonds and the exact same thing will happen. But now we'll have three. Hydrogen is on this end, carbon and to grow means in the middle because again, we're still gonna have that more stable carbon can I? And on the internal carbon with br two, we have a little bit different of a reaction. So what happens here is a draw. We are, uh, one of the pie bonds will come and attack one of the be ours, And the BR will also come in attack one of the carbons and will kick off the other one. So we'll get a transition state that looks like this have that be, are here attached to bold carbons, and it will be positive. And then our other bro mean which is now negative, will come in and attack one of these carbons from the opposite side and will push this off those only on one. So then we'll have See you. Three is a double bond. The bro means have to be opposite of each other because they add anti because of the period transition state, there's no room for it to come in intact when the sides would have to come in from the opposite side. If we've accessed HBR, um, we'll end up with an Al Cain now, um, and we'll just put another bombing on both of those carbons. So now there will be two here, and to hear now, if we have, um, a an internal l kind with extra in access of HPR. Um it doesn't matter which carbon the BR goes on because they're equal. They're both secondary, right? But they will both go on the same carbon. So say we we put the 1st 1 on. We put it on this one. Oops. If we put it on this one now, when we go to former Carbo Cat I and, um you know, just like we did up here. If we when we go to form that carbon, can I on this molecule? Now, that's after one equivalent of HPR. This will be the more stable Carbon Catomine. Because that pro mean on there, it's more substituted. So now the second grooming will always go on the same carbon that the 1st 1 went on. So our products will look like this with two bro means on the same carbon. Um, so that is a symmetrical internal l kind so this side and this side of the same So it doesn't matter which carbon we pick. We're going to the same product either way. But if we have a non symmetrical l kind like F, then it does matter and we get to different possibilities. So if the first bro mean goes on this carbon than they both will go in that carbon. It's a little look like this, but if the 1st 1 goes on this carbon, then it'll look like this where they will both be on that third carbon. So there's two products for

Predict the major product off additional. HBR drew fallen Arkin's and the direction proceeds through a crab, dying for him on a tertiary carbon So this couple will receive berman being reaction proceeds. Flu comeback Italian form a nursery carbon so that carbon received woman See, the reaction can proceed for either secondary complicate time here or secondary capital time there. But this convict ABAC Italian. At this couple, I got into the tertiary one but shifting H minus from the next carbon. It generates tertiary crumpacker time at this carbon and this carbon will receive berman.

In this problem, the reaction will happen. Something like this. Just look at it carefully. Siesta key. See people born see Ch two Cht. So this compound in pageants of 03 or CCL four as well, add, add to all will give the final product ID CS three or edge less CS three CH 2 c o H. So according to the option option B H, correct answer about this problem.


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