On a recent exam I was asked to identify the mechanism when 1-bromo-1-methylcyclohexane was reacted with KOt-Bu in HOt-bu, no other conditions were given (temp, concentration, etc).
I inferred the lack of conditions to imply equal amounts of base and solvent at room temperature; I said E2; this was wrong. My professor explained that because the base was so bulky, the substrate was both tertiary and bulky, and the solvent was protic, a C+ would form before the base could "get" over to the substrate and this would proceed through an E1 mechanism.
I am having a difficult time understanding this explanation because it is my understanding that the conjugate acid of a strong base is often used as the solvent, that strong bulky bases will proceed through E2 (Hoffman) mechanism given a 1°, 2°, or 3° substrate, with 3° being the best (unless the substrate is a benzene or allylic which prefer E1).
I have found many examples using the same substrate and base, but without a solvent listed; they are all E2. I have found several references stating that sterics do not effect basicity. I have also found many examples of bulky 3° alkyl bromides, that are not 1-bromo-1-methylcyclohexane, reacting with KOt-Bu in HOt-Bu which proceed via E2.
Can anyone explain this and/or outline some guidelines for differentiating between E1 and E2 given similar scenarios?
I am unsure as to why this is marked as “assumed answered”, It is not.
Any responses would be greatly appreciated!
Hi, Emily - I have to agree with you on this one. The large base would promote the E2 pathway to give the less substituted alkene (i.e. deprotonating on the methyl group). The corresponding alcohols are frequently used as solvents with the alkoxide salts, so I'm not sure why your instructor is hung up on that. For what it's worth, I have taught organic chem for seven years.
In the presence of a base, an E2 mechanism occurs, where the base attacks a beta hydrogen and removes it and the leaving group to form a pi bond. The bulky base is still E2, it just tends to favor the formation of the less substituted alkene. Often called an anti-Zaitsev or Hoffmann elimination.