What is (are) the elimination product(s) of the following reaction?

Question

Quizplus · Accepted Answer

The reaction involves the dehydration of an alcohol, which typically results in the elimination of a water molecule (H2O) and the formation of an alkene. In this case, since the alcohol is a tertiary alcohol, it is likely that the preferred elimination mechanism is E1.

Product I: Elimination of the OH group on the tertiary carbon would form a tertiary carbocation. This carbocation is highly stabilized due to the presence of three alkyl groups. The loss of a proton from any of these groups would result in the formation of the same product, 2-methylpropene.

Product II: Elimination of the OH group on the primary carbon would form a primary carbocation. This carbocation is less stable than the tertiary carbocation and would require the loss of a proton from a nearby group. However, even if this were to occur, the resulting alkene would be less substituted and therefore less stable than the product formed in I. Additionally, the E2 mechanism is less likely to occur since elimination of the proton would result in a 1,2-hydride shift, which is unlikely due to the steric hindrance of the nearby tert-butyl group. Therefore, no significant formation of the alkene shown in II is expected.

Product III: Elimination of the OH group on the secondary carbon would form a secondary carbocation. This carbocation is less stable than the tert-butyl carbocation but more stable than the primary carbocation. However, in this case, there are two possible protons that could be eliminated, resulting in the formation of two different alkenes: 2-methylprop-1-ene and 2-methylprop-2-ene. Both of these alkenes are less stable than the product formed in I, but more stable than the one formed in II. Therefore, the formation of both products II and III is expected.

What Is (Are) the Elimination Product(s) of the Following Reaction