Question

# $\text{RCON}{{\text{H}}_{2}}+\text{xNaOH}+\text{B}{{\text{r}}_{2}}\to \text{R}-\text{N}{{\text{H}}_{2}}+2\text{NaBr}+\text{N}{{\text{a}}_{2}}\text{C}{{\text{O}}_{3}}+{{\text{H}}_{2}}\text{O}$. Number of moles of $\text{NaOH}$ used in the above Hoffmann bromamide reaction is:A. 3B. 4C. 5D. 6

Hint: Hoffmann bromamide reaction converts amides to amines using a base and bromine. This rearrangement is an organic reaction of a primary amide to primary amine of one lesser carbon atom. Either count the number of sodium atoms on both sides and balance or write the reaction of the process.

Complete step by step answer:
There are two methods to solve this question, let us see them one by one;
Method (1): Writing the reaction: As the name suggests, Hoffman bromamide degradation produces an amine one less carbon atom using bromine molecule and amide by reacting with aqueous solution of sodium hydroxide. One carbon atom is out with sodium forming sodium carbonate $\left( \text{N}{{\text{a}}_{2}}\text{C}{{\text{O}}_{3}} \right)$. Both aliphatic and aromatic amines can be produced by this reaction. The reaction involved to form amines from amides is

$\text{RCON}{{\text{H}}_{2}}+B {{\text{r}}_{2}}+4\text{NaOH}\to \text{RN}{{\text{H}}_{2}}+\text{N}{{\text{a}}_{2}}\text{C}{{\text{O}}_{3}}+2\text{NaBr}+2{{\text{H}}_{2}}\text{O}$.

Method (2): Counting the number of sodium atoms and balancing: The number of sodium atoms in the left hand side are ‘x’. The number of sodium atoms in right hand side are 4 (2 atoms from sodium bromide or $\text{NaBr}$ and 2 atoms from sodium carbonate or $\text{N}{{\text{a}}_{2}}\text{C}{{\text{O}}_{3}}$). According to the law of conservation of mass, the number of atoms of an element on both sides of the reaction should be the same. So, the value of ‘x’ will be 4.
The correct option is ‘b’. Number of moles of $\text{NaOH}$ used in the above Hoffmann bromamide reaction is 4.
So, the correct answer is “Option B”.