Each question in this section has four suggested answers of which ONE OR MORE answers will be correct. The intermediate(s) involved in the reaction of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CONH}_{2}\) with bromine and aqueous \(\mathrm{KOH}\) is/are (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CONBr}_{2}\) (b) \(\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CONBr}\right]-\mathrm{K}^{+}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NCO}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NC}\)

The given reaction is: $$\mathrm{C}_{6} \mathrm{H}_{5}\mathrm{CONH}_{2} + \mathrm{Br}_{2} + \mathrm{KOH} \longrightarrow \text{product(s)}$$

The reaction of benzamide with bromine and aqueous potassium hydroxide undergoes the Hofmann Bromamide Reaction mechanism. This reaction is used to convert amides to primary amines with one carbon atom less.

According to the Hofmann Bromamide Reaction mechanism, the reaction proceeds in the following steps: 1. Bromination of benzamide: $$\mathrm{C}_{6} \mathrm{H}_{5}\mathrm{CONH}_{2} + \mathrm{Br}_{2} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5}\mathrm{CONBr}_{2}$$ 2. Reaction with \(\mathrm{KOH}\): $$\mathrm{C}_{6} \mathrm{H}_{5}\mathrm{CONBr}_{2} + 4\, \mathrm{KOH} \rightarrow \left[ \mathrm{C}_{6} \mathrm{H}_{5}\mathrm{CONBr} \right]^{-}\mathrm{K}^{+} + \mathrm{KBr} + 3\, \mathrm{H}_{2}\mathrm{O}$$ 3. Migration of \(\mathrm{R}\) (-phenyl) group and formation of isocyanate: $$\left[ \mathrm{C}_{6} \mathrm{H}_{5}\mathrm{CONBr} \right]^{-}\mathrm{K}^{+} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5}\mathrm{NCO} + \mathrm{Br}^{-}\mathrm{K}^{+}$$ 4. Hydrolysis of isocyanate: $$\mathrm{C}_{6} \mathrm{H}_{5}\mathrm{NCO} + \mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5}\mathrm{NH}_{2} + \mathrm{CO}_{2}$$ After analyzing the reaction mechanism, we identify the following intermediates directly from the reaction steps: (a) \(\mathrm{C}_{6} \mathrm{H}_{5}\mathrm{CONBr}_{2}\), (b) \(\left[\mathrm{C}_{6} \mathrm{H}_{5}\mathrm{CONBr}\right]^{-}\mathrm{K}^{+}\), and (c) \(\mathrm{C}_{6} \mathrm{H}_{5}\mathrm{NCO}\). The last compound, (d) \(\mathrm{C}_{6} \mathrm{H}_{5}\mathrm{NC}\), does not appear during the mechanism. Based on the reaction steps, one or more correct answers will be: (a), (b), and (c).