A chemist initiates the dehydration/elimination of 3,3-dimethyl-2-butanol using highly acidic condit — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Protonation and loss of water from 3,3-dimethyl-2-butanol [$CH_3-C(CH_3)_2-CH(OH)-CH_3$] directly yields a secondary ($2^\circ$) carbocation: $CH_3-C(CH_3)_2-C^+H-CH_3$. Due to the extreme internal steric bulk and the thermodynamic drive to form a more stable tertiary ($3^\circ$) carbocation, a 1,2-methyl shift rapidly occurs, forming the rearranged $CH_3-C^+(CH_3)-CH(CH_3)_2$. Elimination of a proton from this new $3^\circ$ carbocation can occur from two distinct adjacent positions: 1) Removing a proton from the internal $CH$ group yields the highly substituted tetramethylethene (2,3-dimethyl-2-butene), which is the major Saytzeff product. 2) Removing a proton from one of the equivalent terminal $CH_3$ groups yields 2,3-dimethyl-1-butene. 3) Before rearrangement occurs, the initial $2^\circ$ carbocation can rapidly lose a terminal proton to form 3,3-dimethyl-1-butene as a minor product. Total distinct alkene structural isomers = 3.