Reaction of benzamide, KOH and bromine yields :

Options

A. Benzene
B. Bromobezene
C. Aniline
D. Acetanilide

Explanation

To solve the question regarding the reaction of benzamide with KOH and bromine, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**: The reactants are benzamide (C6H5CONH2), KOH (potassium hydroxide), and bromine (Br2).

2. **Deprotonation of Benzamide**: The KOH acts as a strong base and deprotonates the amide nitrogen in benzamide. This results in the formation of a negatively charged amide ion (C6H5C(O)NH).

\[ \text{C}_6\text{H}_5\text{C(O)NH}_2 + \text{KOH} \rightarrow \text{C}_6\text{H}_5\text{C(O)N}^- + \text{H}_2\text{O} \]

3. **Formation of N-Bromoamide**: The negatively charged amide ion can now react with bromine. The bromine molecule (Br2) will react with the amide ion, resulting in the substitution of a bromine atom for the hydrogen atom on the nitrogen. This forms N-bromoamide.

\[ \text{C}_6\text{H}_5\text{C(O)N}^- + \text{Br}_2 \rightarrow \text{C}_6\text{H}_5\text{C(O)NBr} + \text{Br}^- \]

4. **Formation of an Imine**: The N-bromoamide can undergo further reaction with KOH. The hydroxide ion (OH^-) can abstract a proton from the nitrogen, generating a double bond between carbon and nitrogen, resulting in the formation of an imine.

\[ \text{C}_6\text{H}_5\text{C(O)NBr} + \text{KOH} \rightarrow \text{C}_6\text{H}_5\text{C=NBr} + \text{H}_2\text{O} \]

5. **Hydrolysis of the Imine**: The imine can be hydrolyzed in the presence of water (from KOH) to yield benzamide again, but with the release of carbon dioxide (CO2) and the formation of a primary amine.

\[ \text{C}_6\text{H}_5\text{C=NBr} + \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_5\text{C(O)NH}_2 + \text{CO}_2 + \text{Br}^- \]

6. **Final Product**: The final product of the reaction is benzamide (C6H5C(O)NH2) along with the release of carbon dioxide and bromide ion.

### Conclusion:
The reaction of benzamide with KOH and bromine yields benzamide again, along with the release of CO2 and Br^-.

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