Jan 4, 2010
I'm having trouble with question number 5 from the 9th passage. It refers to the reaction of 2-bromopropane with EtO-/EtOH as having 25% of the product being the substitution product and 75% being the elimination alkene product. Now, the answer's explanation says that the elimination product is derived from an E2 reaction because a strong base is used in the reaction. I understand that, but isn't E2 preferred using aprotic solvents (EtOH is definitely a protic solvent)? Ultimately, I am not entirely hung up on what the right answer is, but is it a stead-fast rule to say that an elimination reaction proceeds via the E2 mechanism when there is a strong base used in the reaction even if the solvent is protic and can H-bond? And, thus, the E1 mechanism only proceeds when the base is weaker?

If you guys have any other tips for Sn1/Sn2/E1/E2 reactions, I'd really appreciate it. I get the basics, but when I have to carefully decide which reaction mechanism is occurring when it isn't extremely obvious (e.g. a methyl group or tertiary Carbon), I just get frustrated.



Call me Bond...PiBond
7+ Year Member
Aug 12, 2009
Resident [Any Field]
E2's favored conditions are polar aprotic solvents but they will still eliminate even if the conditions are not the preferred ones, given that the base is still strong (ie KOtBu, LDA, OH, alkoxides). So for the MCAT, if you see a strong base (and usually heat) consider E2 immediately.

And yes, E1 is favored with weaker bases.

For Sn2, consider steric hindrance of the electrophile. Less steric hindrance promotes the reaction.
For Sn1, we are forming a carbocation intermediate so we want a more substituted electrophile for a more stable intermediate (and weak nucleophile).