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Question: The reaction \({{C}}{{{H}}_{{3}}}{{CON}}{{{H}}_{{2}}}\xrightarrow[{{{NaOBr}}}]{}{{gives}}\) A.\(C...

The reaction CH3CONH2→NaOBrgives{{C}}{{{H}}_{{3}}}{{CON}}{{{H}}_{{2}}}\xrightarrow[{{{NaOBr}}}]{}{{gives}}
A.CH3BrC{H_3}Br
B.CH4C{H_4}
C.CH3OBrC{H_3}OBr
D.CH3NH2C{H_3}N{H_2}

Explanation

Solution

NaOBrNaOBr converts amide group to amine group. CH3CONH2C{H_3}CON{H_2} is an amide. In the Hoffmann bromamide reaction, an amide is treated with bromine in presence of sodium hydroxide solution. It is a degradation reaction.

Complete step by step answer:
We know that CH3CONH2C{H_3}CON{H_2} is a primary amide.
Bromine treated with an aqueous or ethanolic solution of sodium hydroxide gives NaOBrNaOBr.
The reaction which involves the treatment of amide with bromine in an aqueous or ethanolic solution of sodium hydroxide is known as the Hoffmann bromamide reaction. So this is a Hoffmann bromamide reaction. In this type of reaction, amide degraded to give primary amine
Here NaOBrNaOBr converts the amide to an amine.
In Hofmann degradation reaction, primary amine is obtained having one carbon less than the starting amide.
The reaction is shown below:
CH3CONH2+2NaOBr→CH3NH2+2Br+Na2CO3C{H_3}CON{H_2} + 2NaOBr \to C{H_3}N{H_2} + 2Br + N{a_2}C{O_3}
Thus the product for the reaction given in the question is a primary amine with one carbon atom less than the starting amide.
One carbon atom is less in the product when compared to the reactant, therefore this reaction is called a degradation reaction.
The mechanism for Hoffmann degradation reaction is as follows:
The hydroxide ion of the strong base reacts with the amide. Here water is the leaving group. Amide gets deprotonated and forms its anion.
The anion of amide then attacks the bromine molecule. It breaks the bromine-bromine bond. Thus N-Bromamide and a bromine anion are formed.
Base again attacks the N-bromamide leading to its deprotonation and thus water is formed along with the anion of the bromamide.
Then a rearrangement takes place in the bromamide anion where the R group gets bonded to the nitrogen instead of the carbonyl carbon. The bromide anion leaves the compound and thus an isocyanate is formed.
Water is added to the isocyanate and it gives carbamic acid.
In the last step, carbamic acid loses carbon dioxide and then protonation takes place leading to give primary amine as the product.
Thus the answer is an option (D) CH3NH2C{H_3}N{H_2}.

Note:
Hofmann degradation reaction is used to prepare primary amine from primary amide. Isocyanate is the intermediate formed during the reaction. Hoffmann degradation reaction cannot be used for secondary and tertiary amides to produce primary amine. Hoffmann's reaction is also used to convert aryl amide to arylamine in presence of bromine and a strong hydroxide base.