Question

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?

Answer 1

Both halides will undergo dehydrobromination to form styrene.

Explanation:

1-Bromo-2-phenylethane

`underbrace("C"_6"H"_5"-C"color(red)("H")_2"CH"_2"Br")_color(red)("1-bromo-2-phenylethane") + "OH"^"-" stackrelcolor(blue)("Alc.", Δcolor(white)(mm))(→) underbrace("C"_6"H"_5"-CH=CH"_2)_color(red)("styrene") + "H"_2"O" + "Br"^"-"`

1-Bromo-1-phenylethane

`underbrace("C"_6"H"_5"-CHBrC"color(red)("H")_3)_color(red)("1-bromo-1-phenylethane") + "OH"^"-" stackrelcolor(blue)("Alc.", Δcolor(white)(mm))(→) underbrace("C"_6"H"_5"-CH=CH"_2)_color(red)("styrene") + "H"_2"O" + "Br"^"-"`

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.