In the specific bimolecular E2 elimination of 2-bromobutane with aggressive alcoholic KOH, the therm — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

The E2 mechanism is rigorously stereospecific, demanding an exact anti-periplanar arrangement of the $\beta$ -hydrogen and the leaving group. For 2-bromobutane, rotation around the C2-C3 single bond allows for two distinct anti-periplanar conformations. In one conformation, the two bulky methyl groups are positioned *gauche* ($60^\circ$ apart) to each other, creating massive steric strain. In the lower-energy, more stable conformation, the two methyl groups are positioned *anti* ($180^\circ$ apart). When concerted elimination dynamically occurs from this sterically relaxed transition state, it geometrically locks the methyl groups onto opposite sides of the newly formed double bond, perfectly yielding *trans*-2-butene.