Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: lithiumuranium on June 21, 2012, 07:36:13 PM
-
Why is this the major product? For steric reasons?
-
Yes. What you have there is the kinetic product (as opposed to the thermodynamic product) because the base it very hindered.
-
You've had your question answered, however I would suggest you prove this to yourself by constructing models for the cyclohexane and tert-butoxide. Try abstracting anything but an exocyclic hydrogen and get a feel for the sterics involved. Even professors and industrial Ph.D.s use model kits; it's very important for students to use them.
-
You've had your question answered, however I would suggest you prove this to yourself by constructing models for the cyclohexane and tert-butoxide. Try abstracting anything but an exocyclic hydrogen and get a feel for the sterics involved. Even professors and industrial Ph.D.s use model kits; it's very important for students to use them.
Yes. What you have there is the kinetic product (as opposed to the thermodynamic product) because the base it very hindered.
I am not wanting to become involved in another discussion that I pick up the flag for with my skepticism, but I think if one wanted a more suited answer, I'd consider other ideas. I grant that I too have used the hindered argument. It is easier to explain and students are generally happy to accept it. However, this reaction does not work as well with the bromide or iodide and it presumably works even better with a fluoride. t-Butoxide is used in many elimination reactions and certainly with cyclohexanes so it is difficult to think it is steric hindrance. I have made models in it was on that basis that I found it difficult to understand why this should be so. Perhaps I was not seeing the interactions others may see.
It is my belief that this is a simple pKa result. Methyl groups are more acidic than methylenes or methines. If bond breaking or stretching is not advanced, then neighboring CH participation is reduced and acidity determines the product. That is a virtual description of a Hoffmann elimination, one that gives the least substituted (read from the most acidic hydrogen). What is different in a Hoffmann elimination, an R3N(+) does not result in as significant level of bond breakage and gives the product from the deprotonation of the most acidic hydrogen.
The Zaitsev is often with a bromide or iodide in ethanol with ethoxide. These conditions are SN1-like conditions. I presume the bond stretches more, CH participation comes from the most substituted carbons, and elimination gives the most substituted alkene.
In each instance the kinetic product is produced, but the actual steps involve subtle changes that affect the proton acidity by neighboring or intermolecular group effects.