question_answer

The hydrocarbon styrene \[({{C}_{6}}{{H}_{5}}CH=C{{H}_{2}})\] 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-phenylethane 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. \[\underset{\text{1-Bromo-1-phenylethane}}{\mathop{{{H}_{5}}{{C}_{6}}-\underset{\begin{smallmatrix}  | \\  Br \end{smallmatrix}}{\mathop{C}}\,H-C{{H}_{3}}}}\,+KOH\,\,\text{(alc}\text{.)}\xrightarrow{\,}\]  \[\underset{\text{Styrene}}{\mathop{{{C}_{6}}{{H}_{5}}-CH=C{{H}_{2}}}}\,+KBr+{{H}_{2}}O\]