Question

27. An alkyl halide by the formation of its Grignard reagent and heating with water yields propane. What is the original alkyl halide?

Answer 1

The original alkyl halide is: 1-methyl ethyl bromide

Chemical reaction: $\underset{1-\mathrm{methylethylmagnesium}\mathrm{bromide}}{{\left(\mathrm{CH}3\right)}_2\mathrm{CH}-\mathrm{MgBr}}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}}\to \underset{\mathrm{Propane}}{{\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_3}+\mathrm{Mg}\left(\mathrm{OH}\right)\mathrm{Br}$

$\underset{1-\mathrm{methylethyl}\mathrm{bromide}}{{\left(\mathrm{CH}3\right)}_2-\mathrm{CH}-\mathrm{Br}}+\underset{\mathrm{Magnesium}}{\mathrm{Mg}}\to \underset{1-\mathrm{methyl}\mathrm{ethylmagnesium}\mathrm{bromide}}{{\left(\mathrm{CH}3\right)}_2-\mathrm{CH}-\mathrm{MgBr}}$