The hydrocarbon styrene (C6CH5CH=-CH2) can be prepared by the dehydrohalogenation of either 1bromo-2 phenyl Ethane or 1 bromo 1-phenyl Ethane using alcoholic KOH. which alkyl halide will take part in the reaction?
Both halides will undergo dehydrobromination to form styrene.
Both compounds have an α-hydrogen (marked in red), so both can undergo elimination reactions.
However, the first compound will probably react more rapidly because the benzylic hydrogen is more acidic.