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Question 1What is the geometric requirement for an E2 reaction in substituted cyclohexane?Anti periplanarHydro periplanarSyn periplanartrans diequatorial...

Question

Question 1What is the geometric requirement for an E2 reaction in substituted cyclohexane?Anti periplanarHydro periplanarSyn periplanartrans diequatorial

Question 1 What is the geometric requirement for an E2 reaction in substituted cyclohexane? Anti periplanar Hydro periplanar Syn periplanar trans diequatorial



Answers

Although anti periplanar geometry is preferred for E2 reactions, it isn't absolutely necessary. The deuterated bromo compound shown here reacts with strong base to yield an undeuterated alkene. Clearly, a syn elimination has occurred. Make a molecular model of the reactant, and explain the result. (EQUATION CAN'T COPY)

So here we have the falling reaction. So we have a duty aerated bro compound which is being reacted with the base in order to form A and under iterated al cane. So this occurs. We're told this occurs via any to elimination reaction, and clearly we see that is occurring with the sin elimination. And this is kind of counter to what we normally think. We think that he to elimination reactions will always occur in anti para planner geometry. However, this is not always true. So we kind of want to try to understand why, why this is occurring. So, uh, the question asked us to give a model of our sin para planner geometry so I can go ahead and do that real quick for us. Right? So this is easily done. We just can rotate the carbon carbon bond here. This particular carbon carbon bonds. Sorry, this particular carbon carbon bond here, so we can just rotate that a little bit inward and that's going to give us the following configuration here. Right? So what we see here is that we have sin pair plainer geometry and we have this in order to optimize the orbital overlap in the transition state leading to the AL Keen product. So clearly in our model here we see that the deuterium bromide in the two carbon atoms that will constitute the double bond they all lie in the same plane and therefore Senate sent elimination will occur as opposed to anti elimination, because the beta hydrogen and the bromide cannot lie in the same plane. It's so therefore, citizen sit elimination will look her as opposed to anti elimination. So we can write a little bit about that hair. And we can just say that uh, oops. Sorry. That's an elimination will occur because our deuterium, our bromide and our two carbons I can, uh, why within be same plane?

So we have the following organic molecule. We have won our to our one to dive aroma one to die final ethane and we want to draw a new map projection. We also want to show what would happen if we performed in 82 elimination on this organic compounds. Um, we just want a double check to make sure that we have the right, uh, configuration. So it's going to look at the left carbon here and a sign our priorities so quickly going here we know that our berm Oh, he's gonna be our priority one. We know that this carbons gonna be already too, because I was a connecting roaming to it. Ah, the finals gonna be party three and our hajj is gonna be proud of for all right, so that looks like it's going Ah, counter clockwise, which would appear s, however, says we want the hydrogen going away from us, and right now we haven't going towards us. We would say it's the opposite. So it's actually our configuration. If we do the same thing for the right carbon, uh, we would assign our highest party group here, which is the roomy legal that one, Uh, this carbon view to because of the connecting bro Mean the fennel would be three and hydrogen with before here it looks like it's also going counterclockwise, but since again we have the hydrogen going towards us. But we wanted to go away from us. It's actually are as well. So now that we have, ah established that it is indeed the right configuration, how we draw on human projection of this. So let's say we haven't I hear and we're looking down the organic molecule in that direction, looking down that bond so the resulting Newman projection would look something like this. So we'll go ahead and draw the ah, front carbon donated by, ah period here. And we know that the bones were 120 degrees from each other so well, look, something like this looks like the final group is pointing upwards. The drumming group, disappointing to left and behind your group, is pointing to the right. Now we want to draw the carbon in the back and it's resulting groups as well, or it's connecting groups. So you do note that with a circle and the falling lines, so it appears that the remain is on our left. Hydrogen is on the right. And this Ah, the group is going down. All right, so we want to perform a two reaction or any to elimination reaction. An important thing to know is when we're doing it to elimination. Reaction is that we want Thea reacting hydrogen and Al Kyohei Light, which is in this case roaming to be an anti pere plainer geometry from each other. So we want them to be 180 degrees from each other, in which case we see that this particular bro mean circle and this particular hydrogen circled in red are 180 degrees from each other. And this particular bro mean in this particular Hodgins, circled in red are also 180 years from each other. But since we know that there is a lot of symmetry here, we could essentially say that these are essentially the same. Ah, same groups. It would undergo the same e to reaction, elimination, reaction. So we're just gonna focus on the green green bromate and green hydrogen groups for now. So in order to a more easily visualize the ante para planner geometry, we go ahead and rotate these bonds. So would say we're focusing on the front carbon here, and we want to rotate the front cover and by 120 degrees. So, in other words, we want Well, im sorry we're looking at Thea. We're looking at the ray we're looking at this one circled on bread. So the bro mean And the hydrant circled in red. Since those are in the front, let's say we wanted to rotate 100 that 120 degrees. So this particular bro mean? What I mean by that is this particular bro me would go here to this position of the federal group. This group would go down. Disposition of hydrogen group in this hydrogen would go to this position of the Roaming group. Now, I could do the same thing for the back carbon as well. I want to rotate everything by 120 degrees so I could perform the same thing for the back. I won't draw that out, but I'm gonna go and draw the resulting human projection of what that would look like. So now for our front carbon, we now have our roaming group on the top here our fellow group on the bottom right and our hydrogen on the bottom left. And now for our backs, Carbon. The groups coming out of that since we also rotated that 120 grease which I didn't illustrated here. But if you were to do that on your own, you would see you would have something like this. We have the final group on the top left here. You would have the roaming group on the top right here, and you would have the hydrogen on the bottom here. Now we could easily see that those that bro me and a hydrogen that was circling red. They exhibit ante para planner geometry and they are good to undergo any to elimination now so we can go ahead and draw. Hurry to elimination, which is a one step reaction in which a base deep resonates this hydrogen. And when the Basie permeates the hydrogen, the electrons go get shared between the carbon here, the front carbon and the back carbon and this Roman leaves. And as a result, we know that we for Manal Keen and when we foreman all cane, we get ah, plainer geometry. So the resulting Newman projection will look something like this where this fennel and brew mean are now plainer. And this hydrogen and fennel are now plainer. And when are we? Ah, the bro mean. And the hydrogen that were circling, read or no longer in the equation are no longer in the Newman confirmation. Human projection. Rather Sorry. All right, so now how we would draw this in a bond line structure is something like this. Notice how the fennel in, bro mean are on the same side. And how the hydrogen fennel are on the same side here and now the question asks us, Teoh determined steri chemistry are this Al came, And in order to do that, we signed priority. Um, so we compare high carbon one here which having a little one here and in this carbon, it's gonna be a little too. But we're looking at carbon number one here, and we see that the final group has a higher party. So priority Group goes to the final. And I was looking at carpet number two and we see that bro Ming has priority over the final group because so that's carbon. The, uh, connecting carbon to the Al Kane is a carbon. So the brewing has higher already group here, and we see that the private groups are on the same side as each other. So we would say that is Zeus, Amun or Z, and therefore the resulting Al Kane is in the Z exhibit C stare a chemistry.

Okay, So the first part of this question asks what structural requirement is necessary for an AL keen to have assist in a trance I summer. So let's review some vocab first and Al Keen, remember, is a C double bonded see and a cysts. Ice immer is when to functional Groups exist in the same planes. Assist being same. So these two functional groups exist in the same plane as each other. This is our plane. So a trans by summer is when two functional groups exist in opposite planes or the plane across from one another. No notice. When I say functional groups, we aren't counting hydrogen as a functional group. We have to look at something other than hydrogen. It could be a method acumen, Ethel. It could be anything just a functional group. And notice how in the trans ice summer, these functional groups lie in opposite planes. Now, one feature that allows thes two I simmers to exist is this double bond. This double bond is stiff and cannot allow these two carbons to rotate. It doesn't work that way. So by keeping this stiff double bond that we see in both of these, I simmers. It keeps thes functional groups locked in their plane. They can't move, which is why they're considered ice Immers. So the structural requirement that is necessary front Al Qaeda have a cyst in a trance. I summer is one. There needs to be a double bond, and those functional groups seem to be on either side of that double bond and to the functional groups. So functional groups. I need to be on either side of the carbon of the double bond so we can't have two functional groups. For example, we can't have two functional groups coming off of the same carbon because there is no Sister Trans. They have to be on either side. So this isn't it. So now, moving on to the second part of the question Can these cysts in trans I simmers exist for an al cane? So let's remember that an Al cane is a C single bond C. Now, this single bond allows thes carbons toe actually rotate something that a double bond is not allowed for. So if we were to draw out something that looks like ASIS I summer with our functional groups appearing to be in the same plane. All right, so this one kind of looks like assists. Actually, what happens is that this double this single bond allows this entire carbon to rotate or this entire carbon to rotate. And what we end up with is something that may look like a trans ice immer but really is just the same. These two are equal, even though the functional groups appear to be in the same plane here and appear to be in different planes here, it's solely due to the rotation of this single bond. So al Canes can not make CIS trans ice MERS in the last part of the questions asks, Can you make us to send a trans? I simmer for a now Fine. Well, it's like it. Or all kinds are all kinds, or C, with a triple bond to another seat. This triple bond is much like a double bond and does not allow for much rotation. But with a triple bond, it causes these two groups two people to be plainer, So the geometry around a triple bond is is a plainer. So these two elements right here in here, these two functional groups there on the same plane So while that might be considered a quote unquote cysts because they're in the same plane, there is absolutely no possibility to make a trans ice summer because of the triple bond. So to answer that question, you you cannot have assist in a trance I simmer existing in an al kind. The Onley, the only type of hydrocarbon sis Trans I simmer, Nice MERS. They only exist in Al Keene's.

Now hear the cushion is that Draw the NT very planner geometry for the Italy action. Off seats three Joyce yet see it? Studio beast. Then draw the products that reserves after elimination off a trivia. Now, the first question is, what is this? Ended very blender geology. And the very plainer geometry is when hydrogen and hello, John Adams are oriented on opposite sides off the molecule. But in the same place they go opposite side of the molecule Hydra Zen here, hydrogen must represent. And here it should be hemorrhaging. Oppose your side of the molecule. But all hide risen this to carbon and X should be in the scene. This is Auntie Penny Planner geology. So we have to draw this structure here. First of all are for and be Doug Urban. You're doing so. Our folk are bun is here, which is containing hydrogen. Hello, Jin is here the cumin. So you're renting beer here and anti bury Brendan geology should have hydrogen here. After that, all are in the simply then the demanding molecule one should be able to bring when hydrogen here one should be below the plane when hydrogen must be here. Similarly here. Well, about the green, me tango and well below deeply me take. And this is dependent. Jumpy ended. Very Brenner. And after that, you have drone draw. The piece from Berit is coming. Base must be here. And it is for me the bond here. This phone is shifted here. And this we're is leaving his beard mints. So this is the junkie. And but it the product which is formed is ch see? Double born should appear here on scene. Dean, I'd risen. Addresses must come here and tired group you when you take on service to you and after it which is good inch fear edge less via spoons. So this is the product formed by the two elimination off the beach.


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