Elimination reaction of 1-bromo-2-phenylcyclopentane In the following reaction, treating the trans-1-bromo-2-phenylcyclopentane with ethanolic KOH gives an alkene. But in this case, the double bond is on the left side (less substituted) instead of being the more substituted one. 
  


But why does this happen? Zaitsev's rule favours the highly substituted alkene. Even if a carbocation is formed, a hydride shift would have landed it on the carbon with the phenyl group, as the carbocation would be stabilised by resonance. Does the trans configuration forbid such reaction?
There are three common elimination mechanisms: $\mathrm{E1, E2}$ and $\mathrm{E1_{cb}}$. $\mathrm{E1}$ and $\mathrm{E1_{cb}}$ both imply that an ionic intermediate is formed first; $\mathrm{E1}$ implies the same carbocationic intermediate as in an $\mathrm{S_N1}$ reaction. The $\mathrm{cb}$ in $\mathrm{E1_{cb}}$ means conjugate base, i.e. the first step is deprotonation of the compound to give an carbanionic intermediate.

$\mathrm{E1}$ reactions require good leaving groups and a moderately stabilised carbocation; this is not the case in your reactant.

$\mathrm{E1_{cb}}$ requires a very strong base and an at least mildly stabilised anion; this is not the case with your reagent and reactant.

Thus, the mechanism must be according to $\mathrm{E2}$, where the abstraction of a proton and the displacement of the leaving group happen at the same time. $\mathrm{E2}$ mechanisms require a $180^\circ$ anti-periplanar configuration of the $\ce{Br-C-C-H}$ bonds for easier attack and favour those double bonds where the hydrogen is less stericly hindered. Additionally, your reactant is trans-configured, thus meaning that there is no available proton on the higher-substituted carbon that can participate in elimination. Altogether, this leads to $\mathrm{E2}$ eliminations generally following Hofmann’s rule rather than Zaitsev’s rule. Hofmann initially observed what later became the Hofmann rule in the Hofmann elimination, which follows an $\mathrm{E2}$ mechanism.

As the ethanolic $\ce{KOH}$ is not strong enough to isomerise the double bond, the Hofmann product will be observed.