Major products $A$ and $B$ formed in the following reaction sequence are: 2-methylcyclohexan-1-ol $\stackrel{\mathrm{PBr}{}_3}{\to }$ $A$ (major) $\underset{\mathrm{\Delta }}{\overset{\text{alc. KOH}}{\to }}$ $B$ (major)

Solution & Explanation

$\mathrm{PBr}{}_3$ replaces the $-\mathrm{OH}$ by $-\mathrm{Br}$ with retention of position, giving 1-bromo-2-methylcyclohexane ($A$). Alcoholic KOH causes $\beta$-elimination (dehydrohalogenation); by Saytzeff's rule the more substituted alkene is the major product. Eliminating toward the methyl-bearing carbon gives the tetrasubstituted double bond of 1-methylcyclohexene ($B$).