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.

d. trans-1-chloro-3-methylcyclohexane + high concentration of CH₃O⁻

Accepted Answer

Hey, everyone and welcome back 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 mars. The reaction that we are analyzing is a reaction between trans one bromo repre bill cycle vaccine and a concentrated solution of meth oxide. Now, we are going to draw Cycle Hexen which is a six member dring Position one. We have our bromo substitution. Let's suppose that it's pointing up and you will soon understand why Position three, we are going to draw our proposal substitution pointing down. That's because we have a trans arrangement. So we can either have BMO pointing up and propel pointing down or to make sure that they are pointing in opposite directions. We can also have BROMO at position one pointing down And pro pill at position three pointing up. So we have two possible structures for the trans arrangement and we need to think about the elimination products for each. Now, before we think of the elimination, let's keep in mind that meth oxide is a strong base. We have a negative charge on oxygen and it's a small base, it's not hysterically hindered. If it's a strong base, then we are interested in an E two mechanism. And if it's a small base, the major product will be the state of product or a more substitute at Al King. If we look at the two structures, and keep in mind that our leaving group is our BROMO substitution, we are going to identify our beta positions. So there are two high regions that we are going to expand. And simply, we may notice that we can get two products for the first structure when we eliminate those two beta hydrants, because both of them will be equally substituted al Kings. And similarly, we will get two products for the second structure because there will be two equally substituted Elkins formed, Meaning all four will be predominant products in our reaction. For simplicity, we are going to draw a chair confirmation of each structure. Now let's begin with the first one. We're going to draw our promos obsession pointing out just as it is given in the structure. So let's add promos obsession pointing up than if we go counterclockwise at position number three, bromo is at position number one. Then at position number three, we are going to add an equatorial propio group Ch two Ch 2 Ch three. And the reason is because they're in trans arrangement, if we go counterclockwise, that's up down up. So essentially there's an axial position at carbon three pointing up And if we assign an axial position to the propeller group, there actually be says we want to have trans. So instead we're signing an equatorial position. Now, the two hydrants that we're interested in are the two high Jin's in axial positions adjacent to booming because we want to have an anti co planer arrangement for our elimination. So we have expanded to hydrants that we want to eliminate. And now we are just going to draw the nand humor of that chair because if we think about the two structures that we have drawn previously there, they are in an tumors. So the easiest way is just to draw the mirror image of the given structure. So if we draw its mirror image, we are going to assign our proposed group. Now let's draw sees to first, then we have another siege to and C string in that position. Number one, we're going to have a BROMO substitution as well as to axial hydrogen atoms. And now we already think about the elimination, we will get for different products. Now, meth oxide, first of all can eliminate the first proton in the first structure adjacent to broom. In in a need to mechanism. There's an attack formation of a pi bon followed by the loss of the leaving group. So that's how we get the first structure. And to draw the first structure, we are just going to redraw the chair and we may notice how we are essentially forming a pi bond between carbon number one and carbon number six. So that's how we get our first al king. And now we are just going to replicate that process. What if we remove the other hygiene from the first structure? We are going to use our base attack. The proton formed the pi bon and remove the leaving group. This gives us a structure with a pi bond at a different position, right. So we're going to change the pi bond to a position between carbon number one and carbon number two. Wonderful. So we got our structures. Now, we are ready to think about the in anti armor of one bromo three propose cycle hexane or basically the second structure that we have drawn previously. So now, once again, we have two possibilities with oxide can attack the first hydrogen, we are going to form a pi bond for by the loss of the leaving group. And we can essentially remove romaine from the structure, remove the implicit hydrogen and the one that has been d protein ated, right. So we are forming our 1st 5 bond and now we are ready to think about the formation of the yellow structure. So if we remove the other proton, meth oxide comes in, remove the proton, we are forming a pi bond for by the loss of the leaving group. And this essentially gives us a pi bond at a different position. So simply we're going to replace The pipe bomb from the previous structure with another one. Okay. And we noticed that we got all of our four structures for different Elkins. Now, it's very important to understand that because they're equally substituted, equal again, each of our products, each Elkin is we have di substituted Calkins because they're equally substituted, they are all predominant products. So there are no major or minor products and we can label them as our final answers. We have equally yield of an anti MERS. And based on our products, we can state that the correct answer to the multiple choice question is a Thank you for watching.

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.
d. trans-1-chloro-3-methylcyclohexane + high concentration of CH₃O⁻

Key Concepts

Elimination Reactions

Stereochemistry

Regioselectivity and Stereoselectivity

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