Acetamide is treated separately with the following reagents. Which one of these would give methyl amine ?

Options

A. `NaOH//Br_(2)`
B. Sodalime
C. Hot conc. `H_(2)SO_(4)`
D. `PCl_(5)`

Answer

A

Explanation

Key Idea The reagent which can convert `-CONH_(2)` group into `-NH_(2)` group is used for this reaction.
Among the given reagents only `NaOH//Br_(2)` converts `-CONH_(2)` group to `-NH_(2)` group, thus it is used for converting acetamide to methyl amine. This reaction is called Hoffmann bromamide reaction, in which primary amides on treatment with `Br_(2)//NaOH` form primary amines.
`underset("Acetamide")(CH_(3)CONH_(2))+NaOH+Br_(2) to underset("Methyl amine")(CH_(3)NH_(2))+NaBr+Na_(2)CO_(3)+H_(2)O`