[davidenarb]

Hi,

I would like to understand why E2 favors tertiary substrates (rather than a primary substrates). I have a reason for that, but I would like to check if it is a logical reason.

E2 favors tertiary substrates because in E2 mechanism of tertiary substrates, the proton can easily (more likely) be abstracted   rather than a primary substrates.

Thanks by confirming by yes or no. (if no, please give an alternative reason)

[orgopete]

Maybe, sort of. It is true, but probably not for the reason stated. If you look at SN1/E1 reaction rates, tertiary halides react the fastest. That tells us the rate limiting step is cleavage of the C-X bond. I argue the SN1/E1 reaction also tell you something about the nature of that bond. Since carbon atoms are weak electron donors, adding neighboring atoms weakens the C-X bond. That can have two effects. The tertiary carbon is more resistant to attack and solvolysis is faster.

From the perspective of CH acidity, one might expect the hydrogen to which the halogen is attached to be the most acidic. How then can elimination occur? I reason that lengthening the C-X bond will pull electrons from the neighboring atoms in an hyperconjugation effect. This will increase the acidity of the beta hydrogens, especially antiperiplanar hydrogens.

Two different types of E2 elimination reaction occur, Zaitsev and Hoffman. The Zaitsev are SN1-like. They give the most stable alkene. These reactions are best from an alkyl bromide or iodide in EtOH/EtONa. A chloride or better fluoride can give a Hoffman product with KOtBu/tBuOH (a less polar medium). A Hoffman product can be reasoned as elimination from the most acidic hydrogen, for example a CH3.

E2 eliminations occur from abstraction of the most easily removed hydrogen, but that hydrogen is greatly dependent on the reaction conditions.