Question

Can someone explain to me for E1 mechanism, why OH is the one attacking the hydrogen instead of the conjugate (H2O)? One of the examples that my professor gave me was KOtBu/tBuOH.This one the congugate (tBuOH) is the one attacking the hydrogen instead of the base (KOtBu). I'm confused about if I'm suppose to use the conguate to attack the hydrogen or the base. Please help! He gave me a list of base that will be on the test (NaOH/H2O, KOtBu/tBuOH, NaOEt/ EtOH, KNH2/ NH3, LDA, and KCPh3).

Answer 1

No need to get confused here. You just have to think in terms of the base having enough strength to abstract the proton (fast step) α- to the intermediate carbocation generated in the slow step (rate determining step) in the E1 reaction (two step process).

For e.g., let's take the reaction of 2-bromopropane with NaOH/H₂O. Here, in the first step, a stable 2^o carbocation generates in presence of NaOH/H₂O by the elimination of NaBr, which is a slow and rate-determining step.

Note: In this step, NaOH is a strong base, so it dissociates into Na⁺ and OH^- ions; the leaving group/ion (Br^-) combines with Na+ and forms NaBr.

Note: H₂O is not such a strong base to abstract proton in presence of a strong base such as OH^-.

In the given list of bases, such as NaOH/H ₂O, KO^tBu/^tBuOH, NaOEt/ EtOH, KNH₂/ NH₃, LDA, and KCPh₃., NaOH, KO^tBu, NaOEt, KNH₂, LDA, and KCPh₃ are strong enough bases to abstract proton α- to the generated carbocation.