cis-1-Bromo-4-tert-butylcyclohexane and trans

Question

cis-1-Bromo-4-tert-butylcyclohexane and trans-1-bromo-4-tert-butylcyclohexane both react with sodium ethoxide in ethanol to form 4-tert-butylcyclohexene. Explain why the cis isomer reacts much more rapidly than the trans isomer.

Accepted Answer

Hi, everyone. Welcome back. Our next problem says the elimination reactions of the cis one iota for isopropyl cyclohexane and trans one IOTA for Isopropyl psych vaccine with sodium oxide in methanol form four isopropyl cyclo one in explain the reason why the CIS version is more reactive than the trans version. Choice. A, the isopropyl substituent is in a more stable axial position. B the isotop substituent is in a more stable equatorial position. C the isotop substituent is located three carbons away or d the isopro substituent is in a less stable equatorial position. Well, without even needing to do any real figuring when we look at our multiple choice options. Two of these are incorrect right off because they cannot be the correct answer. When we think about cyclohexane rings and the equatorial and a axial positions, the equatorial position for a substituent is always going to be more stable because it has less torsional strain than the axial position. So choice A cannot be our answer because it says that the Tor isotop substituent is in a more stable axial position, but that's just a contradiction. The axial position is not more stable. So this will not be the reason. And that's the same with choice D which is the isotop substituent is in a less stable equatorial position again, because the equatorial position is not less stable. So choice D cannot be our answer. Then when we look at choice C it says the isopropyl substituent is located three carbons away. But that would not be a difference between the cis and trans isomers. They would have the same connectivity the isopropyl substituent as we see in the name is on carbon four in both isomers. So this is the same for both isomers and therefore cannot be the reason why there's a difference in reactivity. So choice C cannot be our answer either. So we have justified process of elimination. Choice B must be our answer. The isotop substituent is in a more stable equatorial position. Well, that satisfies sort of letter in the law. When it comes to answering this problem, we have our answer. But really the goal here is to understand why. So let's think about these structures a little bit. We have two isomers of a cyclohexane ring. So I'm going to draw two chair configurations. So we can compare the two draw two little chairs a bit weird looking. But now we need to think about where our substituents will be with relation to the ring. Will they be above the ring below the ring? I'm just going to designate carbon one as being an upper right corner here So I'm going to have iodine on carbon one. So let's start by thinking about if iodine were in the axial position. So carbon one that means pointing straight up. Well, now I need to think about what type of elimination reaction I have, I have a strong negatively charged base because I have sodium metox side as my base. So strong base and I have a secondary. He lied as my reactant. So that means I'm expecting to have an E two reaction. And the E two reaction means that I'm going to have this concerted removal of a beta hydrogen as the leading group will use. But I'm going to remove a beta hydrogen that is anti coop cleaner to my leaving group. So when I decide where I'm going to put my iodine, I have to make sure there's a beta hydrogen in the anti coplanar spot. So let's draw our hydrogens in and look at a Newman projection here. So if my iodine is going straight up in the axial position and at the equatorial position, let's go down and there's a hydrogen there, my alkene is going to form. When I look at the name of my product that's given uh it's a four isopropyl cyclo hex one E. So my double bond will be from carbon one to carbon two. I'm gonna make carbon two down from carbon one. So in that case, the hydro there's two hydrogens here, one will be going straight down in the axial position and one will be coming up in the equatorial position. So to draw new and projection, I'm going to start with carbon two in the front, I'm looking down along the bond back towards carbon one. So for carbon two, I have pointing straight down my axial hydrogen and then upwards to the left, I'll put r because that's the carbons in my ring, upwards to the right is my equatorial hydrogen. Then for carbon two or carbon one, excuse me, I have carbon two in the front with iodine in the axial position, I have iodine going straight up. I have my equatorial hydrogen going down and to the right and then my ring going down to the left. So yes, with iodine in the axial position, I have this hydrogen beta hydrogen that is anti cop laner to my iodine. So we'll leave iodine where we put it in the axial position. So I will draw that on my other structure as well. So I've got iodine placed on carbon one and then I know I have an isopropyl group on carbon four. So in the cis isomer, let's go to carbon four. So that's the opposite side of the ring. Well, if iodine's an axial position, it's pointing upward, then the cyst position would have isopropyl group also pointing upward, which at this point is the equatorial position. And then the trans version if iodine is pointing upward, isopro must be downward. And so the isopropyl group will be pointing straight down in the axial position. So there we have it, we see that since we have this E two reaction, our cis isomer will have the isopropyl group in this more stable equatorial position, making it more reactive since it will form the more stable alkene product. So choice B the isopropyl group is in a more stable equatorial position is our answer to why the cis isomer is more reactive than the trans iser. See you in the next video.

cis-1-Bromo-4-tert-butylcyclohexane and trans-1-bromo-4-tert-butylcyclohexane both react with sodium ethoxide in ethanol to form 4-tert-butylcyclohexene. Explain why the cis isomer reacts much more rapidly than the trans isomer.

Key Concepts

Steric Hindrance

E2 Elimination Mechanism

Conformational Analysis

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