Two stereoisomers of a bromodecalin are shown. Although the di...

Question

Two stereoisomers of a bromodecalin are shown. Although the difference between these stereoisomers may seem trivial, one isomer undergoes elimination with KOH much faster than the other. Predict the products of these eliminations, and explain the large difference in the ease of elimination.

Two stereoisomers of bromodecalin, labeled for fast and slow elimination reactions with KOH.

Accepted Answer

All right. Hello everyone. So this question is asking us to explain why one of the stereo isomers below undergoes heat induced elimination with potassium hydroxide more rapidly than the other. What are the products of the eliminations? Note that the hydrogens are drawn wedged and dashed as a guide? OK. So here we have structures of the two stereo isomers in question. And it's worth mentioning that the stereo isomer that has the fast elimination is the one where chlorine is depicted on a wedge, excuse me, on a dash and therefore pointing downwards. So before we can understand the role that this plays in the elimination reaction, we have to distinguish between which elimination reaction is taking place here if that's E one or E two. So let's recall a few differences between these two mechanisms or these two reactions in general, recall that an E one reaction is a bi molecular uni molecular reaction that's characterized by the formation of a carbo cion intermediate. Now to favor an E one reaction, we generally look for a neutral or relatively weak base for E one reactions. In addition to this a leaving group that is on a carbon atom that is highly substituted is more favorable for E one. It is also a multi step reaction that is not stereo specific. Now, on the other hand, we can talk about the E two reaction that is bi molecular and only has a single step mechanism. So for the E two reaction, the base should generally be strong or in other words, should generally contain a negative charge in addition to this. However, the E two reaction unlike E one is stereo specific and it is going to require anti coplanar stereo chemistry. This means that the leaving group and the hydrogen on an adjacent carbon must be anti coplanar to each other. Now, in this particular case, our leaving group that's chlorine is attached to a secondary carbon which can favor E one or E two. However, our base is potassium hydroxide with hydroxide having a negative charge because of this, it's considered a strong base. So this reaction is going to proceed via E two for that reason. So because we know this is going to precede via E two, we have to understand the orientation of our leaving group and the hydrogen of the adjacent carbon that gets removed to facilitate the formation of a pie bond in the product. And to understand this better, I'm going to go ahead and draw cheer confirmations for both stereo isomers. Keep in mind, however, that I am going to only draw a chair confirmation of the cyclohexane ring on the very right side of both structures. This is because this is going to be the ring that participates in the actual elimination. So I'm going to abbreviate the rest of the ring here as are. Uh So for this question, I'm going to go ahead and abbreviate the remaining rings as R because I want to focus on the ring that's going to participate in the actual elimination. So for the purposes of drawing a cheer confirmation chlorine in the first ceo isomer here is depicted on a wedge. This means that in my chair confirmation it's going to be pointing upwards and in the equatorial position. Now adjacent to this, I have two hydrogen atoms that are going to be pointing upwards in the axial or vertical positions. Lets compare this now to our second stereo isomer in the second stereo isomer, chlorine is depicted on a dash and therefore is pointing downwards. So on the bottom, right carbon of the chair confirmation, I'm going to draw chlorine on the axial position. It's going to point downwards now and the hydrogens on either adjacent carbon are once again going to be pointing upwards in the axial position. So now let's go back to the concept of anti coplanar stereo chemistry. E two reactions generally require an antic coplanar orientation so that orbitals can overlap and allow for the formation of that P bond. So in the case of this specific problem where we do have a cyclohexane ring, the hydrogen of the adjacent carbon, as well as the living group, chlorine should be transaxial. And that is precisely the the configuration that we see when considering the stereo isomer in which chlorine is pointing downwards. Because notice how here chlorine and either hydrogen are trans to each other because they point on opposite sides of the ring and both of those are axial. So they are transaxial which is ideal for elimination. And as for the product, in this case, two products, the reason there are going to be two products is if you imagine the formation of a pi bond by removing a proton from either carbon atom adjacent to the one containing chlorine. So let's go ahead and showcase that and there you have it. So as for the explanation, I am going to go ahead and briefly pause the recording so I can write out the final answer. But actually, before I do that, I just want to clarify that. In both cases, the hydrogen atom on the adjacent carbon that was removed can be referred to as a beta hydrogen. So because the carbon atom containing chlorine had two unique beta hydrogens on either side of it, this explained the formation of two akines after the elimination. All right. So now I have resumed the recording. Here are the two elimination products. And our explanation reads as follows. One of the stereo isomers has its beta hydrogen and chlorine in transaxial confirmation which is optimum four E two enabling rabid elimination. So with that being said, thank you so very much for watching and I hope you found this helpful.

Two stereoisomers of a bromodecalin are shown. Although the difference between these stereoisomers may seem trivial, one isomer undergoes elimination with KOH much faster than the other. Predict the products of these eliminations, and explain the large difference in the ease of elimination.

Key Concepts

Stereoisomerism

Elimination Reactions

Zaitsev's Rule

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