The hydrocarbon styrene `(C_(6)H_(5)CH=CH_(2))` can be prepared by the dehydrohalogenation of either 1-Bromo-2-phenylethane or 1-bromo-1-phenylethane using alcoholic KOH. Which alkyl halide will take part in the reaction?

The hydrocarbon styrene `(C_(6)H_(5)CH=CH_(2))` can be prepared by the

1.	The hydrocarbon styrene `(C_(6)H_(5)CH=CH_(2))` can be prepared by the dehydrohalogenation of either 1-Bromo-2-phenylethane or 1-bromo-1-phenylethane using alcoholic KOH. Which alkyl halide will take part in the reaction?
Answer» 1-Bromo-1-pehnylethane is a secondary alkyl halide while the other is of primary nature. The alkene is expected to be formed by `E_(2)` elimination reaction. In this reaction, secondary alkyl halide is expected to give the desired alkene. `underset("1-Bromo-1-phenylethane")(H_(5)C_(6)-underset(Br)underset(\|)(C)H-CH_(3))+KOH(alc)tounderset("Styrene")(C_(6)H_(5)-CH=CH_(2)+KBr+H_(2)O`

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The hydrocarbon styrene `(C_(6)H_(5)CH=CH_(2))` can be prepared by the dehydrohalogenation of either 1-Bromo-2-phenylethane or 1-bromo-1-phenylethane using alcoholic KOH. Which alkyl halide will take part in the reaction?

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