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#1 2008-12-05 15:35:23


E1 and E2 elimination reactions of 3-bromo-2,2-dimethylbutane

Yahoo Answers chemistry wrote:

(a) When 3-bromo-2,2-dimethylbutane is heated with dilute solution of sodium ethoxide in ethanol or with ethanol alone, reaction follows first-order kinetics; along with substitution, there also occurs elimination to yield alkenes I and II (illustrated below). Propose a likely mechanism for the reaction which these alkenes are formed under these conditions.

(b) When the same halide is allowed to react with a concentrated solution of sodium ethoxide in ethanol, reaction follows second-order kinetics; again elimination accompanies substitution, this time to yield, not alkenes I and II, but alkene III. Propose a likely mechanism or mechanisms for the elimination taking place under these conditions.

(c) What substitution product or products would you expect in each case? Draw and name these substitution product or products.

The first reaction is an E1 elimination reaction. In the absence of sodium ethoxide, some of the substitution product results. In the presence of sodium ethoxide, the amount of elimination will increase. I don't know a way to predict this effect and I like to quote Roush on this that the product of E1 reactions are difficult to control or predict.

The carbocation can rearrange to a tertiary carbocation to lead to products I and II. These will be accompanied by the corresponding substitution products and another elimination product. It is my sense that literature data for these rearrangements gives a mixture of rearranged and unrearranged products. While rearrangements occur, their extend is rarely as great as indicated in many organic chemistry textbooks.

The second reaction will be an E2 elimination as indicated below. The reaction conditions present will mainly give this elimination reaction. Success of substitution reactions declines as the base strength increases, Brown, Foote, and Iverson suggest pKa ~11 being the cut off where greater will give mainly elimination. So, this favors elimination. The second factor is the quaternary carbon next to the bromide. This will also retard substitution and therefore favor elimination.


There are similar examples to this in A Guide to Organic Chemistry Mechanisms.


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