An halide ${\mathrm{C}}_5{\mathrm{H}}_{11}\mathrm{Br}$  on treatments with Alc. KOH given 2–Pentene only the halide will be

Two possible halides which on elimination of HX can form pent-2-ene are:
CH₃CH₂CH₂CHXCH_3(I) , CH₃CH₂CHXCH₂CH_3(II)
The (II) structure being symmetrical gives only pent-2-ene one on elimination of HX
CH₃CH₂CHXCH₂CH₃→CH₃CH=CHCH₂CH₃

                                                              Pent-2-ene
Thus, the halide is 3-halopentane.