For each of the following reactions, draw the major eliminatio...

Question

For each of the following reactions, draw the major elimination product; if the product can exist as stereoisomers, indicate which stereoisomer is obtained in greater yield.

c. trans-1-chloro-2-methylcyclohexane + high concentration of CH₃O⁻

Accepted Answer

Welcome back, everyone. In today's video, we are going to draw the major elimination products of the following reaction show their stereo chemistry and indicate the major and minor stereo is MERS. Now we're looking at the reaction between trans one bromo to propose cycle hexen and a concentrated solution of meth oxide. Let me simply droll cyclo hexane Position one has a bromo substitution. We're going to draw it pointing up on a wedge and because it's a trans structure at the adjacent position, carbon number two, we will make sure that the propel group is pointing down, right, because one of the groups should be pointing up, the other one should be pointing down in a trans arrangement. Now, for the other structure, we can have another possibility, we can actually assign our BROMO substitution pointing down on a dash. And then the propose suspicion will be pointing up on a wedge that's really important to keep in mind Because trans arrangement implies that there are two possibilities. And you may notice that the two structures are a pair of an anti MERS. Now to predict the products, the best idea is to actually draw the chair confirmation of each structure. Let's begin with the first one. So we're going to add a chair and we're going to assign our substitutions now because the t adjacent positions are occupied in trans arrangement, BROMO, substitute print and propel substitution, they will be both axle. So for the first chair, we can add an Excel bromine and the adjacent position, we will add an Excel Propel Group Ch two Ch 2 Ch three. That's indeed trans because one of them is pointing down the other one is pointing up. Now, we want to find our beta hydrogen. First of all, there's one beta hydrogen ad position to monitor the same carbon as the propeller group. And it's in equatorial position because the axle position is already occupied. Now, on the other hand, Position six or basically the carbon that is to the left of blooming That would have two hydrogen atoms, one of them will be axial, the other one will be equatorial. The reason why it's important is just because we want to make sure that in an elimination, our hae jin is anti per planer to blooming. Specifically, this will be an E two reaction in which meth oxide would take the proton, we would form a pi bon followed by the loss of the leaving group. That's because meth oxide this is strong base. So we got our first share of confirmation and we will draw it's an anti armor. So the best way to do this is just to draw a mirror image of it. And if we draw the mirror image of that chair, we will essentially add our propio group ch two CH two CH three, our bromo substitution, both of them are in axial positions and have similarly to what we had before. We will have one equatorial hydrogen. And the other one will be AXL. The only hygiene that we can remove is the axle one because it's anti per planer to our booming. The other one is in equatorial position. So we cannot remove it. And that means we only have one possibility if we think about the formation of the first product myth oxide comes in attacks. The proton, we form a pi bond called by the loss of the leaving group. And similarly, actually let's fix that right. It's not the equatorial position that we are attacking. So we are just going to change our arrow and we will clearly indicate that we are attacking the axial hygiene. There are two of them forming our pipe on four by the loss of the leaving group. And now in the second structure, that's the same thing. Meth oxide tags the axial hye Jin, we're forming a pi bond followed by the loss of the leaving group And we are getting two products. Now, if you think about the first one, you may simply remove that hydrogen in the excel position and you may also get rid of your implicit hydrogen which is an equatorial position. And now we get rid of grooming and we are simply forming a pi bond. We may also remove high to Shan in the equatorial position bonded to the same carbon as the proposed group. And that's how we get our first product. For the second possibility. We are going to replicate the process. We're going to notice that there's the loss of grooming. We're removing our hygiene in the excel position. We may hide the implicit hydrogen from the carbon that is bonded to the proposed group. And we are now simply adding our pi bon. We got our products. We may also notice that they are mirror images of each other. So they are basically an anti MERS. We can state that the mixture is raise emmick Sends the two products are theoretical formed in equal amounts, they're equally substituted, right. So that means both of them are our major products. But to be accurate, there are no major minor products. They are both formed in equal quantities. So we can state that they're just two products formed. And based on the products that we got, we may conclude that the correct answer to the multiple choice question is D Thank you for watching. And I will see in the next video.

For each of the following reactions, draw the major elimination product; if the product can exist as stereoisomers, indicate which stereoisomer is obtained in greater yield.
c. trans-1-chloro-2-methylcyclohexane + high concentration of CH₃O⁻

Key Concepts

Elimination Reactions

Stereochemistry

Regioselectivity and Stereoselectivity

Watch next