Cis vs trans elimination reaction products

[Snarkalark]

To preface: I haven't taken ochem yet, so I'm not sure if this is a topic covered by the MCAT as I have not yet started to study. Anyway, in my gen chem class we're going over nucleophilic substitution and elimination reactions and there's a question on the study guide that I don't recall covering in class.

It's basically showing a molecule with 2 benzene rings on either side of 2 single bonded carbons. In addition to the rings, attached to the left carbon is a Cl and a H and attached to the right carbon are 2 Hs. The products shown are the -cis and -trans versions of what you would get after an elimination reaction. The question posed is how you would yield a majority of the -trans version.

I know it has to do with choosing either the E1 or E2 pathway, but I don't really understand why. I assume the answer is going by E2 because I think E1 would result in racemerization and production of both, but I don't understand why E2 would necessarily yield the trans vs the cis product. Can anyone help me out with this?

---

Members don't see this ad.

 

[ksmi117]

Now, I'm not positive with this answer, but this is how my logic worked.

The trans-, cis- thing doesn't really depend on the mechanism, as the bonds are free to rotate however they so please before the reaction.

The reason they are trans- is that they are two very large substituents. Having them on the same side of the molecule would lead to steric hindrance and such. Therefore, having them trans- to each other would create the most stable product.

 

[ronaldo23]

It yields a trans product, because for E2 the two leaving groups need to be coplanar...aka anti-periplanar transition state. I think.

 

[chemtopper]

Here the mechanism is E1 because all bezylic halides undergo E1 mechanism through a carbocation formation.Bezylic carbocation is more stable due to delocalization of charge

ksmi117 is right that now it is free to adopt more stable trans form which has minimum steric hindrance.

 

[sv3]

for E1: most substituted product is favoured through a carbocation
for E2: Works best if the proton to be removed is anti to the leaving group. The stereochemistry of the alkene is determined by the anti arrangement of the B-proton and leaving group.

For E2 reactions, I generally will rotate bonds (mentally) to set up the anti arrangement and then eliminate. I don't think you necessarily get a trans product just because it's an anti elimination though. Chemtopper's point above seems to be the key (along with ksmi117's) that the planar geometry allows the molecule to reduce steric hinderence in E1 - at least that's my interpretation which coudl be wrong though.

Can someone confirm this or call me out for being wrong? I had a difficult time understanding this part as well while first learning it. Similarily, I used to think anti and syn addition automatically mean trans and cis formations respectively, but i've seen some examples where this isn't the case. Either my books have some errors or there isn't a steadfast rule linking these two similar concepts.

Thanks
steve