It should be no surprise that E 1 occurs mainly for secondary and tertiary carbons, since the mechanism proceeds through a carbocationic intermediate. For primary carbons, formation of the carbocation does not occur, but elimination still occurs to give an alkene product. This time, the mechanism is different. An E 2 elimination proceeds in a concerted manner (similar to S N 2). The reaction is called a dehydration . The deprotonation, formation of the \(π\) bond, and release of H 2 O all occur in the same step. This step is rate-determining and involves two molecules, so it is second order. Since these curly arrows must occur simultaneously, there are specific restrictions as to what conformations will provide product. In order for correct orbital alignment to occur, the \(β\)-proton and H 2 O must be antiperiplanar , allowing for alignment of the \(σ_{C-H}\) and \(σ^{*}_{C-O}\). Protonation of the oxygen makes the \(σ^{*}_{C-O}\) even lower in energy. While E 2 reactions are certainly possible with alcohols, the more synthetically useful way to make alkenes is by dehydrodehalogenation . Contrary to E 1 /E 2 with alcohols (acidic conditions, heat), this reaction occurs under basic conditions at lower temperature. All of the same rules of stereoelectronic control apply for E 2 : Deprotonation of a \(β\)-proton must occur antiperiplanar to the halide. \(σ_{C-H}\) and \(σ^{*}_{C-X}\) must overlap properly. Zaitsev’s rule is followed The reaction is stereospecific (stereochemistry of the product is determined by the mechanism) and stereoselective (two possible products, more stable one is favored). So, what determines how fast this reaction goes? Well, if the rate-determining step involves both the base AND the alkyl halide, then making each of these more reactive will increase the rate so that: rate = k [base] [alkyl halide]. The factors effecting rate are: leaving group ability – again, the pK a H of the leaving group must be < 4. structure of alkyl halide – primary and secondary only strength of the base – stronger bases increase the rate The most common strong bases are LDA (lithium diisopropylamide, pK a H = 42), NaNH 2 (sodium amide, pK a H = 38), NaH (sodium hydride, pK a H = 36), KOtBu (potassium tert-butoxide, pK a H = 18), NaOMe (sodium methoxide, pK a H = 15). Bulky bases will favor the Hoffman product in both E 1 and E 2 because during the deprotonation, the formation of the more highly substituted alkene is more sterically encumbered. One consequence of these factors can be seen in the elimination of cyclohexyl halides. We know that substituted cyclohexanes exist in two different chair conformations, but in only one of them is there proper orbital overlap to give the alkene. This might not necessarily be the most stable conformation , but it is indeed the reactive conformation . A general rule is that the leaving group must be in an axial position in order for an E 2 reaction to occur. This has an important effect on the product observed, as shown in the example below: