A) \[RCON{{H}_{2}}\]
B) \[RN{{H}_{2}}\]
C) \[B{{r}_{2}},OH\]
D) \[{{H}_{2}}S{{O}_{4}}\]
Correct Answer: D
Solution :
Primary amide reacts with bromine in the presence of sodium hydroxide (or other base) to give primary amine. This reaction is called Hofmann bromamide reaction. \[\underbrace{R-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-N{{H}_{2}}+B{{r}_{2}}+4NaOH}_{\text{reactant}\,\text{+}\,\text{reagent}}\xrightarrow{{{H}_{2}}O}\] \[\underbrace{\underset{\begin{smallmatrix} \text{primary} \\ \text{amine} \end{smallmatrix}}{\mathop{RN{{H}_{2}}}}\,+2NaBr+N{{a}_{2}}C{{O}_{3}}+2{{H}_{2}}}_{\text{product}}\] Hence, in this reaction \[{{H}_{2}}S{{O}_{4}}\]is not a reactant, reagent or product.
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