Question

How will you bring the conversion of Methyl Bromide to Acetone?

Answer 1

Step 1: Formation of Grignard reagent
Methyl bromide reacts with Magnesium and dry ether, to form Methyl magnesium bromide(Grignard reagent).
$\underset{\mathrm{Methyl}\mathrm{bromide}}{\mathrm{C}{\mathrm{H}}_3\mathrm{Br}}\underset{\mathrm{dryether}}{\overset{\mathrm{Mg}}{\to }}\underset{\mathrm{Methyl}\mathrm{magnesium}\mathrm{bromide}}{{\mathrm{CH}}_3\mathrm{MgBr}}$

Step 2: Addition of cyanide ions
When Methyl nitrile reacts with Methyl magnesium bromide, it forms an adduct.
$\underset{\mathrm{Methyl}\mathrm{magnesium}\mathrm{bromide}}{\mathrm{C}{\mathrm{H}}_3\mathrm{MgBr}}\stackrel{\mathrm{C}{\mathrm{H}}_3\mathrm{CN}}{\to }\mathrm{C}\underset{\mathrm{Adduct}}{{\mathrm{H}}_3-\mathrm{C}\left(\mathrm{C}{\mathrm{H}}_3\right)\mathrm{N}-\mathrm{MgBr}}$

Step 3: Formation of acetone
When the adduct undergoes hydrolysis, acetone is formed.
$\underset{\mathrm{Adduct}}{\mathrm{C}{\mathrm{H}}_3-\mathrm{C}\left(\mathrm{C}{\mathrm{H}}_3\right)\mathrm{N}-\mathrm{MgBr}}\stackrel{{\mathrm{H}}_3{\mathrm{O}}^{+}}{\to }\mathrm{C}\underset{\mathrm{Acetone}}{{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_3}+{\mathrm{H}}_2\mathrm{N}-\mathrm{MgBr}$