[orgoclear]

(Me)3-C-CH2-C(Br)-(Me)2 ---> E1 ??

I thought that after removal of Br- the carbocation is tertiary. So the H+ will be eliminated to give more substituted alkene.

But the answer was given the less subsituted alkene.

As no reagent is mentioned (whether the base causing elimination) is hindered or small.. shouldn't the more substituted alkene be the major product

If the reagent was hindered then of course Hoffmann product would have formed. But only it is asked to find E1 product.

[orgoclear]

anyone??

[orgopete]

I did spend some time searching on this problem (web), but I didn't find anything to report. Further, no conditions are given and as I had been researching more generally on SN1/E1 reactions, it can be a lot more complicated that what is given in introductory chemistry books.

The question of whether a reaction might be E1 or E2 is commonly asked. It is very difficult to answer that question as this can be very subjective. If 2-bromo-2,4,4-trimethylpentane were solvolyzed in a polar solvent, a mixture of SN1 and E1 products would reasonably be produced. If a base were added, that would ostensibly increase the amount of elimination. If the concentration of base were continually increased, the reaction mechanism would change from E1 to E2. Based upon examples I have seen, I surmise that researchers would be unable to predict the concentrations in which [E1] = [E2]. However, we could make a reasonable estimate that in the absence of base, an E1 reaction should occur and increase in proportion as the temperature is increased. Similarly, if a high concentration of base were used, especially if a polar solvent were not used, then an E2 mechanism should prevail. This question asks for speculation of products in which no conditions are given.

Re: Hoffmann v Zaytsev
This is what sparked my interest. I have a premise that Zaytsev products come from E1 types of reactions while Hoffmann products are from E2 types of reaction. Let me go further in this. I argue that if there are no concerted reactions, then we should analyze by which reaction rate explains the formation of a product. As I have looked at many examples, if a reaction goes by an E1 process, then elimination of a proton from the most electron rich carbon (most substituted) will predominate. If the rate or extent of bond breakage is limiting, or as it becomes limiting, then hydrogen acidity will predominate. This can be an E1cb or an E2 type of elimination. For 2-bromo-2,4,4-trimethylpentane, the hydrogens of a methyl group are more acidic and could lead to increased amount of Hoffmann product. Generally, a poorer leaving group is required in order to decrease bond breakage and the E1 reaction.  This problem appears designed to ask whether a t-butyl group will interfere with the deprotonation process. Can one increase the amount of Hoffmann product by interfering with the proton removal step?

Re: concerted v stepwise processes
As I recall, Arigoni argued that there is no such thing as a concerted reaction. I don't remember if he used this example. If you can divide time into infinitely small units, then you would find that it is impossible for three billiard balls to collide at the same time. One ball will always arrive before the other. If reactions are molecular collisions, then electron movements might be considered similarly. Some electron movements should occur faster (or sooner) than others.

I argue that a reaction need not be an E1 reaction in order to lead to a Zaytsev product. For example, in the elimination of 2-bromobutane in ethoxide/ethanol, as bond lengthening occurs, then it will attract the electrons from the more substituted carbon. This process can occur without breaking the C-Br bond and the intermediacy of a carbocation intermediate. Similarly, in the elimination of 2-fluorobutane, because little bond lengthening will have occurred, then electron donation from the longest CH bonds will not drive the reaction. Rather, deprotonation of the most acidic hydrogens will now be the faster process and drive the reaction to the Hoffmann product.

[researcher]

I was looking it up on Allinger and I don't understand why the answer is the less substituted alkene, if it is less stable.

[Marvinthefish]

Hey, perhaps it's to do with the fact that there are 6 protons in the "Me₂" group, but only 2 protons (obviously) in the -CH₂- group? I.e. 6 protons available to give the terminal alkene, only 2 give the internal alkene?

I'm pretty sure this isn't the full solution though! Also, maybe if a base is involved in removing the proton, it's going to go for the easily accessible protons, rather than the -CH₂- ones which do look slightly hindered.