JEE Main · 2006 · 30 Apr 2006 · Offline · Haloalkanes and Haloarenes

This is an E2 elimination by alcoholic KOH, which demands that the departing bromide and the $\beta$-hydrogen lie anti-periplanar. In trans-2-phenyl-1-bromocyclopentane the hydrogen on the phenyl-bearing carbon is on the same face as the bromine, not anti to it, so elimination cannot run toward that carbon and the conjugated 1-phenylcyclopentene is not formed. The only correctly placed $\beta$-hydrogen lies on the carbon away from the phenyl group, and its loss gives 3-phenylcyclopentene as the major product.