A) \[RCON{{H}_{2}}\]
B) \[RN{{H}_{2}}\]
C) \[B{{r}_{2}},O{{H}^{-}}\]
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. \[\underbrace{\underset{amide}{\mathop{R-\overset{\begin{smallmatrix} O \\ |\,| \end{smallmatrix}}{\mathop{C}}\,-N{{H}_{2}}}}\,+B{{r}_{2}}+4NaOH}_{reactant+reagent}\xrightarrow[{}]{{{H}_{2}}O}\] \[\underbrace{\underset{\begin{smallmatrix} \Pr imary \\ a\min e \end{smallmatrix}}{\mathop{RN{{H}_{2}}}}\,+2NaBr+N{{a}_{2}}C{{O}_{3}}+2{{H}_{2}}}_{product}\] Hence, in this reaction\[{{H}_{2}}S{{O}_{4}}\]is not a reactant, reagent or product.
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