Elimination Reaction

[HowAboutDAT]

hey guys, i posted a reaction that I have a question on, the question is in the document. thanks


View attachment Rxn.docx

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[BiomajorPreDent]

Someone correct me if I am wrong, but I believe it is because when your doing an elimination reaction where there is an aromatic ring, the elimination will create a double bond which is conjugated with the ring, as that is most stable

Im pretty sure this rule applies if you are also doing an elimination reaction where there already is a double bond in the reactant, it wants to create a stable conjugated dbl bond

 

[HowAboutDAT]

it's not an aromatic ring, just good old cyclohexane. Im thinking that large bases produce the hofmann product (least substituted alkene) while smaller bases, like KOH, NaOH produce the zaitsev product

 

[nze82]

hey guys, i posted a reaction that I have a question on, the question is in the document. thanks


View attachment 12866

Look at your base. It's a BULKY base. Bulky bases favor the Hoffman Product (Least substituted double bond) over the Saytzeff Product (Most substituted double bond). Why?
In Elimination reactions, the base is responsible for removing a proton as the halide group leaves. The base ALWAYS removes the MOST ACCESSIBLE PROTON. Now, in your example, the most accessible proton is the one associated with the methyl (-CH3) group. Since the base is bulky, it can't approach the ring due to steric hindrance. So, it won't be able to remove the proton associated with the ring!

Hope this helps!