Question

Acetamide is treated with the following reagents separately. Which one of these would yield methylamine?
(a) NaOH-$B{{r}_{2}}$
(b) Soda lime
(c) Hot conc. ${{H}_{2}}S{{O}_{4}}$
(d) $PC{{l}_{5}}$

Answer 1

Hint: Acetamide is an organic compound, it is the simplest amide which is derived from acetic acid. Preparation of amine is usually done by the catalytic hydrogenation of metal.

Complete step by step answer:
The preparation of amines by the reduction of the nitro compound is usually done by the catalytic hydrogenation of metal and acid, it can be used to carry out the reduction and the product formed will be acidic.
Hoffmann bromamide degradation is the reaction in which amines only primary amines are prepared. The amine formed will have one carbon atom less than the amide.
In this reaction there is use of an alkali as a strong base which attacks the amide which leads to the deprotonation and generation of an anion. The conversion of a primary amide to a primary amine with one less carbon is accomplished by heating the amide with a mixture of halogen, a strong base and hydrogen.
The mechanism involved is:

When Acetamide is treated with bromine and NaOH, methyl amine is produced as potassium bromide, sodium carbonate and water molecule is produced as a byproduct.
$C{{H}_{3}}CON{{H}_{2}}+B{{r}_{2}}+4NaOH\to C{{H}_{3}}N{{H}_{2}}+2KBr+N{{a}_{2}}C{{O}_{3}}+2{{H}_{2}}O$

Note: Anion reacts with $B{{r}_{2}}$ forms bromoamide. Deprotonation of bromoamide occurs leads to the formation of bromoamide anion which undergoes rearrangement and forms isocyanate. Isocyanate forms carbamic acid when a water molecule is added. Methylamine has a strong fish-like odor. Many commercial available compounds are formed using methylamine. Whereas Acetamide is used to reduce the pressure in the eyes. It is a white crystalline solid in pure form.