= Solution
\b[Both reactions are <E2 eliminations> requiring an axial <chloride> and an axial beta <hydrogen> in an <antiperiplanar conformation> relative to the C–Cl bond.] The reacting C–H bond supplies the <alkene> $\pi$ bond as the C–Cl bond breaks; <hydroxide> accepts the <proton>.
For the trans-1,2 substrate, the low-energy chair has methyl and <chlorine> both equatorial. <ring inversion> is needed to place <chlorine> axial, but this also places methyl axial. The population of the reactive chair is consequently small, explaining the slow reaction. At the methyl-bearing neighboring <carbon> the axial position is occupied by methyl, not <hydrogen>. Only the other beta <carbon> has the needed anti axial H, so elimination gives \b[3-methylcyclohexene] rather than the more substituted <alkene>.
For the cis substrate, a favorable chair has methyl equatorial and <chlorine> axial. It can react without paying the axial-methyl penalty, explaining the faster rate. Both adjacent <carbons> now supply an anti axial <hydrogen>. Removing the H on the methyl-bearing <carbon> gives \b[1-methylcyclohexene, the major product]; removing the other gives \b[3-methylcyclohexene, the minor product]. The former has the more substituted <double bond>.
Thus <alkene> stability selects between accessible pathways in the cis case, but it cannot create the missing <antiperiplanar conformation> H in the trans case.
\Image[/past-exam-of-the-chemistry-b-course-of-the-university-of-cambridge/2021/paper-1-b3-openings.png]
{title=Original stereochemical maps for trans-diaxial elimination and the epoxide-opening products}
Back to article page
