Question

Which of the following reactions is appropriate for converting acetamide into methenamine.
A. Gabriel’s Phthalimide synthesis
B. Carbylamine reaction
C. Hoffmann bromamide reaction
D. Stephens reaction

Answer 1

Hint: The acetamide can be converted into methenamine by reacting to the acetamide with bromine and sodium hydroxide. The chemical formula for acetamide and methenamine is $C{H}_{3}CON{H}_2$ and $C{H}_{3}N{H}_2$.

Complete step by step answer:
During the conversion of acetamide into methenamine, the amide functional group is converted into an amine functional group.
Gabriel’s Phthalimide synthesis is used for the conversion of primary alkyl halide into a primary amine. So, option (A) is incorrect.
The Carbylamine reaction is used for the conversion of the amine functional group into the isocyanide functional group. So, option (B) is incorrect.
The amide reacts with bromine and sodium hydroxide to give amine. The reaction is known as the Hoffmann bromamide reaction.
The reaction of acetamide with bromine and sodium hydroxide is as follows:
$\underset{\text{acetamide}}{\text{C}{\text{H}}_3\text{CON}{\text{H}}_{\text{2}}}\text{ + }{\text{Br}}_2\text{ + }\text{4NaOH}\to \underset{\text{methylamine}}{\text{C}{\text{H}}_{\text{3}}\text{N}{\text{H}}_{\text{2}}}\text{ + }\text{2}\text{NaBr}\text{ + }\text{N}{\text{a}}_{\text{2}}\text{C}{\text{O}}_{\text{3}}\text{ + }\text{2}{\text{H}}_{\text{2}}\text{O}$
Initially, the base sodium hydroxide abstract proton forms acetamide to generate an anion of amide.
The anion of amide now attacks on bromine, so an N-bromoamide generates.
Again the base attacks on N-bromoamide so an anion of bromoamide forms.
Then the methyl group attached with the carbonyl group shifts to the nitrogen atom and bromide ion leaves forming an isocyanate structure.
Nucleophilic attack of water on isocyanate takes place which loses the carbon dioxide and forms a structure in which nitrogen has a negative charge and one hydrogen and one methyl group.
Then the protonation of this nitrogen takes place which gives methylamine.
The Hoffmann bromamide reaction is used for the conversion of amides into amines. So, option (C) is correct.
The Stephens reaction is used for the conversion of cyanide into an aldehyde. So, option (D) is incorrect.

Therefore, option (C) Hoffmann bromamide reaction is correct.

Note: The reagent used in Gabriel’s Phthalimide synthesis potassium Phthalimide. The reagent used in the Carbylamine reaction is chloroform and a base. The Carbylamine reaction is also known as Hofmann isocyanide synthesis. The reagent used in the Hoffmann bromamide reaction is base and bromine and the reagent used in Stephens’s reaction tin (II) chloride and hydrochloric acid.